MTL-501

MATL-0501

Materials Engineering

Physics for Teachers

Ralph Zee

Bill Baird

Spring 1999

Lesson 1. Introduction, basic energy information.

Basic energy concepts:

 Dr. Baird gives the slide presentation.

 The energy problem of today is that there is an unlimited demand and a limited supply of energy. The rise of modern civilization has been paralleled by a steady increase in the world’s use of energy. The more energy we have at our command, the better we can meet our desired needs of food, clothing, shelter, warmth, light, transportation, and communication.

The sun has provided almost all the energy we use today. Light and heat come directly from the sun; food owes its energy content to photosynthesis; water power exists because solar heat evaporates water and later falls as rain, wind power comes from convection in the atmosphere due to unequal heating of the earth’s surface by the sun. The fossil fuels coal, oil, and natural gas were formed from plants and animals that lived and stored energy derived from sunlight millions of years ago. Only nuclear energy and heat from inside the earth cannot be traced to the sun.

The expected breakdown of energy sources for 1996 are:

Oil 40%

Coal 26%

Natural gas 21%

Nuclear 8%

Hydro 5%

Together, these energy sources should produce around 10²⁰ Joules of energy. At this rate, it is easy to see how our natural sources of energy will soon be gone. Oil and natural gas will be first exhausted, and coal in another two to three centuries. The advantages and disadvantages of future energy sources will have to be compared.

 Coal: Over one half of the electrical power generated in the United States is obtained from coal-fired steam turbine generating plants. The coal gasification process is when coal is subjected to intense heat and the gases that are released from combusting coal are captured. It is like wood burning in a fireplace - jets of flames come out, heat drives volatile gases out of the fuel that is decomposing in the center of the log or piece of coal. Modern boilers can transfer 80-90% of the heat of combustion of coal into steam. The efficiency of large generators in converting mechanical power to electrical is very high - about 99%. The down side is that there is an increase in concern about the environmental impact of coal burning plants. Coal burning contributes to the greenhouse effect and global warming. Besides air pollution, mining leaves land unfit for use. Coal plants actually expose surroundings to more radioactivity than normally operating nuclear plants because of radon gases.

 Nuclear Power: In a nuclear power plant the heat to generate steam is supplied by a nuclear reaction rather than the chemical reaction of burning coal. The steam turbines in nuclear power plants have the same theoretical efficiency limit as those in coal fired plants. It is not practical to run nuclear reactors at temperatures and pressures as high as coal boilers so efficiency is lower.

In 1990, 19% of the United States electrical power came from nuclear power plants. High construction costs, questions of public safety, and the problem of disposal of radioactive waste have slowed development of additional nuclear power.

Nuclear power comes from the fission of the Uranium isotope. The task is to disrupt the heavy nucleus without using more energy than we get out. In 1938 it was discovered that absorbing a neutron causes a nucleus of the Uranium isotope ₉₂U²³⁵ to undergo fission. ₉₂U²³⁶ is so unstable that it explodes almost at once into two fission fragments. Because stable light nuclei have proportionally fewer neutrons than heavy nuclei, the fragments are unbalanced when they are formed and release one or two neutrons each. Usually the fragments are still unstable and undergo radioactive decay. This is the highly radioactive product of fission. In the nuclear reactor exactly one neutron per fission causes another fission and energy if given off at a constant rate. A nuclear reactor is a very efficient energy source. About 1MW is produced by the fission of 1 gram of ₉₂U²³⁵ per day. To get the same output from coal, 2.6 tons must be burned.

The energy liberated in nuclear power plants is heat and it is extracted by circulating a coolant liquid or gas. The hot coolant can then be used as the heat source of a steam turbine which can power an electric generator.

 Solar Energy: Bright sunlight can deliver over 1kW of power to each m² it falls on. A tennis court receives the solar energy equivalent to that in a gallon of gasoline every 10 minutes. The power per unit surface area in the sun’s radiation (before it passes the earth’s atmosphere) is about 1.4 kW/m². A maximum of 1.0 kW/m² reaches the surface of the earth on a clear day and the average over a 24 hour period is about 0.2 kW/m².

One way to use the sun’s energy involved a series of mirrors that track the sun and concentrate its radiation on a boiler whose steam drives a turbine and thence an electric generator. Photovoltaic cells convert light energy directly into electricity. No heat is involved so the cells avoid thermodynamic limitations. Capturing energy by photovoltaic cells has a high capital cost and the price of energy they produce is about 25 cents/kW-h. This price compared to 5 cents for coal powered steam turbine plants is not a deal. But there are many advantages to solar power. Some of these are lack of noise, lack of moving parts, minimum maintenance requirements, and freedom from pollution.

 Wind Power: An energy source that is fundamentally solar in power is wind. Energy can be gathered and converted by "forests" of windmills. A large modern windmill with rotor blades 100 meters in diameter might produce 3MW of energy when the wind speed is 12 m/s. About 17,000 wind turbines (85% of those in the United States) are located in California. Windmills are cheaper per Watt than solar cells but still more expensive than conventional power stations.

 Biomass Energy: Biomass is another mechanism for using solar energy. Plants absorb solar energy and store it; the energy can be used by burning the plant matter or by using fermentation and distillation to make alcohols. Ethanol made from fermented corn is at present crude oil price levels. These plant carbohydrates can be used for fuel for vehicles. This causes little pollution but the fuel needed for harvesting may exceed the amount of energy recovered. The land needed to run one car on alcohol can provide food for 8 to 16 people.

Wood has always been a cheap source of energy. The problem is that we are using it up faster than we can replace it and wood smoke is more polluting than coal. Plus, wood is too valuable for use in construction and for paper to simply burn it. An advantage of biomass as energy is that the materials take as much CO₂ out of the atmosphere when they grow as released when they burn.

 Water: Anyone that has been to the ocean and seen crashing waves knows that there is much power behind them. There is a great tidal range and enormous energy is evolved. To capture this energy, waves run up a sloping funnel-like channel to a reservoir above. Water from the reservoir then powers a turbine generator on its way back to the ocean. Power of falling water is responsible for 60,000 MW of electricity in the United States.

Another indirect use of solar energy would use the temperature gradient in the ocean. Water temperature can be a difference of 15^oC per few hundred meters could run a heat engine between the two temperatures. The warm water is used as a heat source for vaporizing a fluid that would drive a turbine. The thermodynamic efficiency of this method is very low, but there is a vast energy supply.

 There is always a supply of water, wind, and sunshine, and plants can always be planted. Therefore, these energy sources can be called "renewable". A disadvantage of all renewable energy installations is that they take a lot of space. A medium size city needs 1000 MW of power. Less than 150 acres is enough for a 1000 MW nuclear plant, where 5000 acres might be needed for solar collectors and over 10,000 acres for windmills. Converting crops into fuel might require 200 square miles of farmland to yield 1000 MW per year.

 Review of basic scientific units and energy. Name the fundamental units. How do we express very large and very small numbers? Can you give some examples of very large and very small numbers? Compare and contrast power and energy. Review pico (p), nano (n), micro (µ), milli (m), kilo (k), mega (M) and giga (G) and so on.

 Can you name the different types (or forms) of energy in the universe?

 Heat energy is given in units of calories (cal). Food energy is also measured in Calories, but 1 Cal of fat is equivalent to 1 kcal of heat or mechanical work (bad for people trying to lose weight.

Mechanical energy is given in unit of Joule (J). One J is 1 kg m²/s². 1 cal=4.186 J.

Electrical energy is the product of V, I and t. Electrical power is VI.

Energy in the atomic scale is usually given in eV unit (electron volt). One electron volt equals 1.6x10^-19 J.

Mass and energy are interchangeable via the famous Einstein relationship of E=mc².

One electron weighs 9.1x10^-31 kg. Destroying this one electron produces 8.19x10^-14J which equals 511,000 eV or 0.511 MeV.

How is kW-h related to other energy units?

See Appendix A for more details in conversion units.

 First law of thermodynamics: you can never win. The first law is a generalization of the law of conservation of energy that includes possible changes in internal energy. Mathematically it is expressed as: )U=Q-W where U is the internal energy, Q is the heat input and W is the work done.

 Heat content of materials. Discuss with table on gasoline, oil, coal, world reserve and world resources etc.

 Conduct simple experiments to show different types: electrochemical cell (wet) to dry cell (cutup), heat from heat lamp, heat from water and temperature measurement methods.

 Think about energy and our human body. Aren't you glad that we are inefficient to build fat to mechanical work?

 Example problem of energy usage.

I am the king of a new country (King Ralph) that has a population of 120,000 people. My country is very nuclear oriented and it generates 10¹⁸J of energy per year using nuclear, 10¹⁷ J using coal, 10¹⁷J from oil and 2x10¹⁷J from solar.

(a) What fraction of each type of energy do my people use?

(b) If my country pays $100,000,000 per year energy, what is the rate of energy cost?

 Assignment:

1. It is expected that U.S. will consume 10²⁰ J of energy in 1995. This energy will be supplied by oil (50%), coal (30%) and natural gas (20%). Calculate the amount of these three resources that we need for the year (in units of barrels of oil, tons of coal and cubic feet of gas).

Lesson 2: Basic energy concept, energy slave problem and calorie

 Ancient people did not have electricity and internal combustion engines. These are products of the late nineteenth century. So how did ancient people stay cool in the summer, get juice from fruit, and move themselves around? Easy. They used servants or slaves. Humans can be employed to do work if engines and motors and electric light filaments are not available. But the good life through servants is quite different from having easy access to electrical switches and ignition keys. For one thing, humans have only one-tenth the power of a good horse. And just one car has the power of over 100 horses. So we use the equivalent of many humans every day. The average American uses energy equivalent to more slaves than the grandest Egyptian Pharaoh could have dreamed of. Let's calculate how many...

Lesson 3: Continuation of calorie work

 The human body requires energy for fundamental processes like maintaining constant temperature and achieving movement. This energy is derived from the food we eat (and stored reserves of fat). Different types of food contain different amounts of energy per unit of volume or weight. For example, water contains no calories at all; a banana split contains over 1,000 calories. If you consume more calories than your body burns, you accumulate more body fat. Let's explore this question with a laboratory exercise in which you will calculate the number of calories your body needs during a 24-hour day (based on your gender, metabolism and exercise regimen) and compare this with the calories you take in. The results should tell you something about your potential to gain or lose weight through body fat.

Lesson 4: Energy conversion illustrations

 Electrical energy can be used to heat water. No surprise there, eh? We all know how to plug in the coffee pot. But have you ever wondered how this process works? Let's find out by examining the power of a small immersion coil to heat a cup of water. The energy absorbed by water when we raise its temperature is measured in units called calories. A calorie is defined as the amount of heat energy needed to raise the temperature of one gram of water one degree Centigrade. {One thousand of these is a kilocalorie -- the Calorie used to measure energy in food.} We'll measure the power of the immersion coil in watts, defined as the product of voltage and current. Electrical energy is measured in watt-hours, which is what Alabama Power charges us for. But, enough! Let's see what happens when we heat a cup of water with electrical energy. Let's have tea...

 All energy are convertible from one form to the other and the second law of thermodynamics says that you cannot break even. You are going to lose useful energy when converting from one form of energy to the other. The second law is frequently referred to as increase in entropy (or disorder or chaos)

 In general, efficiency is defined as: Efficiency = (Output/Input) x 100%

Efficiency of energy conversion in a heat engine where there is a heat source and a cold sink can be expressed as: Efficiency = (T_high - T_low)/T_high < 1 or 100%

The best efficiency one can expect from most engines is about 30% (usually 10%).

What are some of the consequences of the lack of efficiency?

How can one develop higher efficiency systems?

 The conversion process is usually between electrical energy, thermal energy, light (photo or electromagnetic) and mechanical work. These basic conversion concepts and measurement of their efficiency is demonstrated in the following units.

 Suppose we are interested in how much electricity is required to heat up a cup of water to compare this with the energy accounted for in calories when the water temperature rises. We've all heard that energy cannot be created or destroyed. So what if some energy gets "lost" in conversion from electrical to thermal (heat). Like, a lot of energy gets lost! We can examine the process of energy conversion by comparing the energy output in a process with the energy input to that process. The ratio (converted to percent) is called efficiency. Let's make tea again...

 Sample problem:

I have a microwave oven that uses 110V and draws 5 amps of current. I find that it is 80% efficient in heating water. How long do I have to set my time if I want to bring 50 grams of water from 30^oC to 99^oC?

 Homework assignment:

Remember seeing those water wheels that turn by water from a reservoir. I want to use such a wheel to produce electricity. I find that its efficiency is 5%. The height of the waterfall is 100 m. If I want to generate 1,000 W of electricity, how many grams of water do I need to pass through the wheel in one hour?

Lesson 5: Energy conversion illustration continues

 Use electricity to heat water into steam. The electrical energy is measured by monitoring the electrical current (ampere) with time. The product of voltage, current and time yield energy (Volt-Amp-S=J and V-A=Watt). Part of this electrical energy is converted into thermal energy by boiling water to form steam. The beaker will be carefully weighed to determine the amount of steam formed during the demonstration. The thermal energy equals the water loss times the latent heat of vaporization. The efficiency of converting electrical energy into steam is quite high.

First conversion: Electrical to thermal

Voltage of heater =

Current of heater =

Power of heater =

Duration of boiling =

Electrical energy consumed =

Weight of water+pressure cooker before boiling =

Weight of water+pressure cooker after boiling =

Amount of steam produced =

Heat of vaporization of water =

Energy in vaporization =

Efficiency of boiling =

 Introduce electromagnetic spectrum (optical, x-ray, IR and then microwave, see attached sheet). Determine the efficiency of microwave oven heating by again monitoring the electrical energy consumption of the microwave and the heating of water. Converting electrical energy into microwave energy which then convert to heat (mass of water times temperature rise times specific heat). Notice the unit of conversion from electrical (J) to thermal (calorie).

Power to the microwave oven =

Duration of heating =

Electrical energy consumed =

Mass of water used =

Temperature rise =

Specific heat of water =

Energy absorbed by water =

Efficiency of microwave =

 Can we extend this experiment with microwave to account for both heating of water and boiling?

Power to the microwave oven =

Duration of heating (need to heat longer to induce boiling) =

Electrical energy consumed =

Mass of water before heating =

Initial temperature of water =

Mass of water after boiling =

Specific heat of water =

Latent heat of boiling for water =

Energy absorbed by to heat to boiling point =

Energy used for boiling =

Total energy transferred to water for boiling =

Efficiency of microwave =

 The concept of thermoelectricity will be illustrated using modular to convert heat into electricity. (if time permits)

 See Appendix B for more activities in this area.

 Example problem: I have a solar panel on a spacecraft near the sun and the power density is 10 kW/m² and the area of the panel (I am a stubborn American who still uses stupid English units) is 3 ft x 3 ft. The voltage of the equipment is 12 V. (a) What is the maximum current it can draw? (b) What is the minimum resistance (or impedance) of the load? (c) How much water can the astronaut boil in 1 day if his boiler has an efficiency of 50%?

Solution: 3ft x 3ft = 0.84 m². Power is 10 kW/m² x 0.84 m² = 8.4 kW = 8,400 W

Power is Voltage x Current = 12 V x Current = 8,400 giving a current of 700 A

Minimum resistance = Voltage/Current = 17.1 mS.

In one day the total energy available is 8,400Wx(3600x24)s =7.26x10⁸J, 50% efficiency yields 3.63x10⁸J = 8.67x10⁷ cal.

The latent heat of boiling of water is 540 cal/gm. It can therefore boil 1.61x10⁵gm of water or 161 kg.

 Assignment:

Imagine that you are a romantic in France and you bought an air conditioner in Paris (it uses 220 V) and the output power of the compressor is rated at 1 horsepower and the efficiency of the compressor is 40%. But you are a very scientifically minded romantic and you decided to measure these values yourself. You find that the output horsepower is correct but the efficiency is only 20%. (a) Calculate the current draw of the air conditioner. Hint: first calculate the two horsepower values needed for the two efficiencies. Then calculate the current needed for those two horsepowers. (b) Speculate on the danger you are in. (remember that there are current ratings for different conductors).

Lesson 6: Energy conversion illustration continues

We have thus far learned about mainly conversion of electrical energy to thermal energy. Now we will look at conversion of mechanical energy into mechanical energy. As Tim Taylor would say: "Erh??"

We will illustrate conversion of kinetic energy (in the form of running water) into potential energy (by lifting a weight). Water from a faucet can be used to drive a turbine which in turn winds up a string and lifts a weight through a known height. In this case we are basically using the kinetic energy of the running water to drive a turbine to increase the potential energy of the weight. To solve the problem, we need to calculate how much kinetic energy is used to produce the potential energy. Hopefully if everything works, the potential energy gain (output of this stupid engine) is smaller than the kinetic energy of the running water. Or else, we will have an engine with efficiency larger than 100% (make sure we patent this then and have a laugh).

(A) First, let’s determine the kinetic energy of the running water.

This energy (KE) = (½)mV².

This energy must be a function of:

(1) how long the water runs (this is the mass term),

(2) the exit velocity at the outlet, and distance to the turbine (contribute to the velocity term).

(2) Let’s do the velocity first.

To determine the exit velocity, we open the faucet to the desired position and point the nozzle upward. How high the water goes is related to the exit velocity by:

mgh (height reached) = (½)mv²

or: v (outlet) = /(2gh) where g is the gravitational acceleration constant (32ft/s²or 9.8m/s²)

The final velocity we are interested in (V) also depends on how far is the faucet to the turbine:

V = v + gt

or V² = v² + 2g)x (where )x is the distance)

(1) Now determine the mass. This mass is the mass of the running water that is required to lift the weight. So we simply time how long it takes to lift the weight and measure how much water is needed.

Mass of water:

Total kinetic energy consumed: (½)mV² =

(B) Now determine the potential energy gained. Potential energy gained = mgh

Mass of weight =

Gravitational acceleration constant =

Distance lifted =

Potential energy gained =

(C) Efficiency =

Homework: I live in the Pacific Northwest and my home is powered by a hydroelectric generator. 50,000 kg of water passes through the generator every second. The water enters the generator at a velocity of 1 m/s and exit the generator at 0.4 m/s. The efficiency of my generator is 40%. What is the electrical output power of my generator? (Kinetic energy is 0.5mv²)

Lesson 7: Use of electricity (concept of resistance, power and energy)

 Review last lesson.

 Develop the ohm's law which relates electrical current, potential and resistance. Use a resistance and a simple ohmmeter. Then illustrate the measurement based on ohm's law using a power supply and an ampere meter.

V=IR or I=V/R or R=V/I

 In the simplest microscope model of conduction in a metal, each atom in the crystal lattice gives up one or more of its outer electrons. These electrons are then free to move through the crystal lattice, colliding with stationary positive ions. If there is no electric field, the electrons move in straight lines between collisions, the direction of their velocities random, and on average they never get anywhere. But if an electric field is present, the paths curve slightly because of the acceleration caused by electric field forces (see the description in the handout).

 Discuss the contribution to the resistance of a long wire (diameter, length and the material).

 Introduce and contrast the idea of electrical conductivity, electrical resistivity, electrical conductance and electrical resistance.

D (resistivity, unit of S-m) (This is independent of shape, intrinsic property)

F (conductivity, unit of 1/Sm) (this is independent of shape also, intrinsic)

R (resistance, unit of S) = V/I = D (L/A) (depends on shape, extrinsic)

C (conductance, unit of mho) = 1/R (depends on shape, extrinsic)

 Illustrate the concept of resistivity and resistance through the use of different wires (composition and size). Use copper and glass to illustrate the effect of conductivity. Measure conductivity of different materials in class.

 Show the relationship between resistivity and resistance (geometry).

 Copper is usually used for wiring in houses. Compare the resistance (R) and conductance (1/R) of copper and steel, aluminum, glass, and wood. Glass has a very high resistance and therefore is used as an insulator.

Resistivities at Room Temperature

Metal	D (S m)	Substance	D (S m)
Silver	1.47x10^-8	Pure Carbon	3.5x10^-5
Copper	1.72x10^-8	Amber	5x10¹⁴
Gold	2.44x10^-8	Glass	10¹⁰ - 10¹⁴
Aluminum	2.63x10^-8	Lucite	>10¹³
Steel	20x10^-8	Mica	10¹¹ - 10¹⁵
Lead	22x10^-8	Quartz	75x10¹⁶
Mercury	95x10^-8	Sulfur	10¹⁵
		Teflon	>10¹³
		Wood	10⁸ - 10¹¹

 Any change in physical condition may give a body a new resistance different from its previous. Impurities, change in temperature, changes in hardness, and mechanical strains effect it. The relation between resistance and temperature is shown in the following formula:

R = R_o(1 + aT)

where R_o = resistance at 0^oC

R = resistance at T^oC

a = constant for the material.

T = temperature in ^oC

Temperature dependence of resistivity? In an ideal crystal lattice with no atoms out of place, a correct quantum-mechanical analysis would let the free electrons move through the lattice with no collisions at all. But the atoms vibrate about their equilibrium positions. As temperature increases the amplitude of these vibrations increase, and collisions become more frequent. Therefore resistivity of a metal increases with temperature.

What does this mean for wires carrying current in your home and why do we need circuit breaker?

 Illustrate the temperature dependence of resistivity using a light bulb power to different power levels.

 Example Problems

(1) Why do wires carrying a current get warm?

Electrons gain energy between collisions through the work done on them by the electric field. During collisions they transfer some of this energy to the atoms of the material of the conductor. This leads to an increase in internal energy and temperature of the material.

(2) In household wiring, a copper wire commonly known as 12-gauge is often used. Its diameter is 2.05 mm. Find the resistance of a 30.0 m length of this wire.

Resistivity of copper is 1.72x10^-8 S-m

Resistance R = D x L/A = 1.72x10^-8 S-m x 30m / (B 1.025²mm²)

R =1.72x10^-8 S-m x 30m / (B 0.001025²m²)

R= 0.156 S

(3) If I have a heater wire which has a resistance of 100 S at 300^oC and a resistance of 120 S at 800^oC. What is its resistance at 2000^oC?

R = R_o(1 + aT) where T is temperature in C.

Substituting the two conditions into the equation, we have

100S = R_o(1+300a) or 100S = R_o+300aR_o [1]

120S = R_o(1+800a) or 120S = R_o+800aR_o [2]

Subtract equation [1] from [2] we have: 20S = 500aR_o

so aR_o=0.04 and it has a unit of S/^oC [3]

Substituting aR_o=0.04S/^oC into equation [1] we have

100S = R_o+300x0.04 or R_o=88S [4]

Finally substituting aR_o=0.04 and R_o=88S into the resistance equation:

R = R_o(1 + aT) = R_o + aR_oT = 88S + 0.04S/^oCx2000^oC

We have R=168S

 Assignment

(1) What length of copper wire 0.600 mm in diameter has a resistance of 100 S?

What length of glass wire 0.600 mm in diameter has a resistance of 100 S?

(2) I have an electrical wire with an a value (in the temperature dependent equation) of 2x10^-4/^oC and its resistance at room temperature (25^oC) is 200S. What is its resistance at 2000^oC?

Lesson 8: Continuation of electricity (The electric pickles)

What would happen if you took an ordinary dill pickle, stuck a nail in each end, connected each nail to an electrical source, and turned on the power? The result is the ELECTRIC PICKLE , a potential new light source! Here is a technical report that will serve as a model for your class project in this course. It was found on the world-wide web, and modified for use in this class.

The electric pickle

Characterization of Organic Illumination Systems

Bill Hamburgen, Jeff Mogul, Brian Reid, Alan Eustace, Richard Swan, Mary Jo Doherty, Joel Bartlett

Digital Equipment Corporation Western Research Laboratory

April 1, 1989

Abstract

Recent anecdotal reports of novel principles of illumination have stressed qualitative aspects. This note presents a quantitative study of an organic illumination system, characterizing the temperature and current-flow properties of the system as functions of time and device parameters. Theoretical and practical implications of these measurements are discussed.

Introduction, History of Incandescent Illumination Devices, Theory of Organic Illumination Devices, Experimental Media, Experimental Method and Setup, Measurements, Analysis, Gastronomical effects, Practical considerations, Further Work, Acknowledgments, References

1. Introduction

There has been a great deal of interest of late in triboluminescence and electroluminescence in organic materials. Triboluminescence in wintergreen Life Savers ™ + ® has been investigated by many over the years [8], while electroluminescence in organic thin films is an active area of current research both here and abroad [10].

In early December 1988, our attention was called to work by Bill Bidermann on electroluminescence in pickles [23]. It was reported that inserting iron electrodes into a dill pickle and energizing with modest alternating currents caused the pickle to glow. Subsequent reports reached us in January 1989 regarding corroborating experiments [15, 21]. We decided to investigate the phenomenon with the aim of improving our understanding of the underlying mechanisms and examining the potential for commercial applications.

2. History of Incandescent Illumination Devices

Our experiments indicate that pickles are a form of incandescent lamp. In this light, it is useful to consider the historical development of such devices.

Sir Humphrey Davy first demonstrated in 1802 [3] that platinum strips heated in the open air with electricity emit light. Frederick de Moleyns was granted a patent for an incandescent bulb in 1841 [3]. He used charcoal between platinum wires. Sir Joseph Wilson Swan (no relation to one of the authors) produced the first lamp with carbon filaments in evacuated glass bulbs [3]. This exciting invention was brazenly copied a year later by a minor American inventor and industrialist [3].

3. Theory of Organic Illumination Devices

While the exact mechanisms are unclear, our observations lead us to propose the following model of light generation. Upon initial application of power, the pickle conducts strongly. This is not surprising since the pickle is thoroughly impregnated with a highly ionic sodium chloride salt solution. Resistive losses cause the pickle to heat. One would expect the heating to be the greatest in the vicinity of the electrodes where the current flux is highest.

When the temperature at the surface of the electrode reaches about 100 degrees C, boiling occurs. The water vapor generated locally blankets the electrode. This vapor is non-ionic and not conductive, and if sufficiently thick, current can no longer flow from that point on the electrode surface. Of course as soon as the local current flux ceases, the heating at that point ceases as well, at the area begins to cool. When the area has cooled sufficiently, the vapor blanket collapses and conduction and heating resume. At some point during the transition to or from the conducting condition, an arc is supported and light is produced. It appears that a quasi-equilibrium state is reached providing a relatively steady light source. The composition of the plasma in the arc is not known but may contain hydrogen (from decomposition of water and the sample's organic constituents), carbon (from the sample), and various atmospheric gases. A spectroscopic observation of the arc, perhaps through a fiber optic probe, would help elucidate the composition.

4. Experimental Media

We performed experiments on five different experimental media. As previous work had focussed exclusively on pickles, we acquired three different varieties of pickles. Because we were attempting to characterize the properties of the various media, we purchased high-quality samples from Draeger's market, at somewhat higher than prevailing costs. Commercial exploitation of this phenomenon would of course require bulk purchases to obtain reduced prices, with the attendant quality control issues.

Two of the pickle varieties, the "Kosher" and the "Dill", were substantial specimens, measuring approximately 1.5" diameter. The ``Kosher'' pickle was 5" long, while the "Dill" pickle was 5.5" long. The third variety of pickle, chosen for experimentation with miniaturization of this technology, was a "Cornichon" pickle, measuring about .5" in diameter and 1.5" long. The two larger pickles were slightly below ambient temperature at the start of the experiment; the smaller pickle had reached thermal equilibrium. All pickles used were whole and undamaged.

Standard preparations for some of these media are presented in Appendix I; we do not know if the pickles we used in fact followed the standard preparations.

We also tried to elicit electroluminescence in two non-pickle media, a segment of "Mandarin Orange" (raw) and a piece of stir-fried "Bok Choy." Although a pickle is technically a "fruit" ["fruit: 1 d) a product of fertilization in a plant with its modified envelopes or appendages; specifically: the ripened ovary of a seed plant and its contents." [22] ] it is not commonly thought of as such; an orange is indisputably a fruit. Also, although both pickles and oranges are acid media, they differ in that pickles are ionic ("salty") whereas oranges are non-ionic ("sweet"). Bok Choy, on the other hand, is clearly not a fruit, and is only slightly ionic; in fact, it has rather less taste than the other media. Our Mandarin Orange sample was about 1" long; the Bok Choy sample was about 2" long, .75" wide, and not very thick.

5. Experimental Method and Setup

Our apparatus consisted of a fused AC line cord, the ends terminated with bare .1631"/.1571" diameter CDA10100 copper electrodes. To facilitate insertion, the ends of the electrodes were ground to a conical shape with a 60 degree included angle. The line cord was powered through a YEW Model 2509 Digital Wattmeter, which was used to monitor the current during the experiment. To protect the wattmeter, a 10 Amp instrument fuse was used; this proved adequate for the five trials described herein, though it did blow later during some informal testing. Small currents were noted prior to energizing some of the test samples. The source of these currents was not determined, but as the magnitude was typically small compared to that measured while energized, it is not felt that this was an important source of error. The two electrodes were placed in the test sample in either an axial or parallel configuration, depending on the shape of the test sample. The electrode orientation, penetration, and separation were recorded in each instance. The voltage was checked at the beginning of the series of experiments and found to be approximately 113 VAC RMS.

Temperature in the vicinity of the electrode was sensed with a .030" diameter stainless steel sheathed grounded type K thermocouple probe. To ease penetrating the tough skin of the test sample, the thermocouple probe was bonded into a 21 gage syringe needle with the tip of the probe projecting approximately .080". The thermocouple probe was positioned with the tip in contact with the end of the #1 electrode. It was found that as the experiment progressed, heating caused deformations in the test sample that could shift both electrode and thermocouple locations. Since the temperature gradients could be expected to be large in the vicinity of the electrode, our temperature measurements should be considered as more a qualitative than a quantitative indication of conditions near the arc.

Table 1 shows the electrode orientation, separation, and penetration depths for each of the five experimental media that we used. Figure 1 shows a schematic depiction of the experimental setup (axial orientation).Dimensions in inches.

Table 1: Electrode positions

Electrode Electrode Electrode #1 Electrode #2

Subject orientation separation penetration penetration

Bok Choy parallel 0.5 0.5 0.5

Mandarin Orange parallel 0.38 0.38 0.38

Cornichon axial 0.5 0.5 0.5

Kosher Pickle axial 2.5 1.0 1.5

Dill Pickle axial 2.5 1.5 1.5

6. Measurements

In each of the five cases, the temperature and current were recorded at the start of the experiment, just prior to energizing the sample. Additional data was taken at elapsed times of 5, 10, 20, 40, and 80 seconds. Temperature and current data are presented in tables 3 and 4, respectively. We also present graphs of both temperature and current versus time, in figures 2, 3, 4, 5, and 6.

Observations were also made of the approximate time of arc initiation and quenching. These observations are summarized in table 2. An existing light photograph was taken of one of the samples while its light output was at a peak.

Table 2: Qualitative observations

Subject Arc size Arc started Comments

Bok Choy Irregular At about 20 sec

Mandarin Orange Never really lit up

Cornichon Small At about 20 sec

Kosher Pickle Large At about 15 sec Arced on top side of electrode, sank as it burned. Pickle expert reports "it still tastes a lot like a pickle" afterwards.

Dill Pickle Really good At about 10 sec Arced on top side of electrode, sank as it burned. Photograph taken. Arcing decreased at about 35 seconds, increased again at about 55 seconds.

Table 3: Subject Temperature vs. Time

Subject 0 sec 5 sec 10 sec 20 sec 40 sec 80 sec

Bok Choy 19.2 19 159 167 147 127

Mandarin Orange 21.8 25 43.4 96 99 97

Cornichon 14 110 103 101 87 72

Kosher Pickle 13 110 122 104 100 97

Dill Pickle 14 128 126 87 53 45

Table 4: Subject Current vs. Time

Current (Amperes)

Subject 0 sec 5 sec 10 sec 20 sec 40 sec 80 sec

Bok Choy .16 .63 .90 .99 .30 .16

Mandarin Orange .4 .23 .4 .37 .29 .27

Cornichon .13 .35 .36 .52 .13 .13

Kosher Pickle .12 4.24 4.83 3.99 1.99 1.08

Dill Pickle .1 5.9 5.6 .87 .37 .7

7. Analysis

It appears that liquid phase ionic conduction is necessary for heating leading to arc initiation. The Mandarin Orange, which was the only sample not prepared with a brine solution, was also the only sample that did not light up.

Informal experiments beyond those reported in section 6 support the importance of the sample's "salty" aspect. Domestic sweet pickles (Del Monte) were found to produce a very disappointing arc in comparison with the saltier Dill and Kosher pickles. (They also tended to fall off the electrodes.)

In all cases arcing commenced only after vapor evolution was observed from the vicinity of the electrode surface. The initiation of the arc also corresponded closely with the measured temperature reaching the boiling point of water. In all cases with strong arcing, the current decreased markedly with the onset of arcing. This supports the hypothesis of the vapor blanketing mechanism as the initiating effect.

We also found that arcing pickles smell bad. While this is not a serious impediment to laboratory investigation, it may prove to severely limit opportunities for exploiting this technology commercially.

8. Gastronomical effects

We also investigated the effect of electrical simulation upon the gastronomical qualities of pickles. Subsequent to electrostimulation, the "Kosher" pickle was dissected using standard laboratory techniques to extract a .2 inch slice, roughly equidistant between the two electrodes. Careful examination revealed no tissue scarring, but a small loss of moisture content. Surprisingly, this slice did not exhibit the egregiously noisome odor noted in section 7. Further testing indicated that the taste was neither enhanced nor diminished, but remained "very much like a pickle." Our conclusion is that the culinary potential of electrical stimulation is limited.

9. Practical considerations

The primary advantage of pickles as light bulbs is that they can be eaten, either before or after providing illumination. Thus they are to be preferred for long sea voyages. Pickles are also organically grown and so do not contribute to pollution. However, whereas incandescent lamps can be manufactured by a single machine at a rate of 20 or 30 per minute [4], proper pickling takes several weeks and requires careful control [5]. Thus the challenge to economically exploit the rediscovery of pickle light sources comes down to developing techniques for the massive growing of cucumbers and efficient vast vats for pickling. This may be an excellent industry for the Developing world.

10. Further Work

In the area of commercial viability, it has been noted that the very small color range of organic illumination systems is a serious drawback [10]. Certainly concentrating entirely on pickles will limit us mainly to green hues. Ripe (black) olives, which are also prepared in a brine solution, may provide an extension to the available color spectrum, and also result in a more compact fixture than pickles. Issues involving the translucence of olives and the smaller fittings required will have to be explored.

Our investigation leaves a number of unanswered questions; the most fundamental relate to the critical electrode/sample interface. Use of direct current would show whether a stable arc can be supported in cases where the voltage does not periodically drop to zero, and help determine if the arc forms preferentially at the positive or negative electrode. Spectroscopic observation could determine which plasma species contribute the light, and a photosensor and appropriate recorder could aid in elucidating arc dynamics. Novel experimental electrode designs might further promote understanding of the phenomenon by permitting better control of geometry and more direct observation of the arc. High speed photography may help answer the question concerning which phase produces the light; transition to or from the conducting state. It is hoped that other investigators will pick up the torch (with due attention to avoiding electrocution) and carry on this important work.

11. Acknowledgments

We wish to acknowledge the contributions of our laboratory assistants, Carol Peters and Bob Alverson, without whom our experiments could not have taken place.

12. References

[1] Australia Bureau of Census and Statistics. Manufacturing industries: jam, fruit, vegetables, pickles, sauces, condiments, etc.

[2] James A. Beard. Olive oil pickles. Beard on Food. Alfred A. Knopf, New York, 1974, Chapter 5.

[3] ----. Incandescent Lamp. The New Encyclopaedia Britannica, 15th Edition, Volume 6. Encyclopaedia Britannica, Inc., Chicago, 1986, pages 277.

[4] ----. Development of Lighting Technology. The New Encyclopaedia Britannica, 15th Edition, Volume 23. Encyclopaedia Britannica, Inc., Chicago, 1986, pages 29.

[5] ----. Food Processing. The New Encyclopaedia Britannica, 15th Edition, Volume 19. Encyclopaedia Britannica, Inc., Chicago, 1986, pages 360.

[6] Robert J. Courtine (editor). Larousse Gastronomique, 2@+[e] Edition. Librairie Larousse, Paris, 1984.

[7] Lloyd Lawrence Gallardo. An evaluation of United States Department of Labor policy regarding wages paid Mexican Nationals: Michigan pickles: a case study. PhD thesis, University of California, Berkeley, 1962.

[8] Simpson L. Garfinkel. Athena Report, Part III. Technology Review 92(2):MIT7-MIT10, February/March, 1989.

[9] Arlo Guthrie. Motorcycle (Significance of the Pickle) Song. Arloco/ASCAP. 1967.

[10] ----. NEC Develops Organic Thin Film EL Element, translated from: Nikkan Kogyu Shimbum, 11/16/88. Japan Review (53):10-11, January, 1989.

[11] Bill Kalsow and Violetta Cavalli-Sforza. Pickles: An Implementation of Persistent Data Structures. Research Report 29, Digital Equipment Corporation Systems Research Center, November, 1988.

[12] Jeanne Lesem. The pleasures of preserving and pickling. Alfred A. Knopf, New York, 1975.

[13] Thomas Lupton. The way to get wealth, or, An easier way to make wine: also to make cyder, mead, rum the true art of distilling of brandy to make pickles and divers sorts of vinegars ... to help a bad memory ... the duty of all sorts of servant maids ... to make China varnish ... a help to discourse ... G. Conyers, London, 1702.

[14] Pandya. Indian Chutneys, Raitas, Pickles & Preserves. Inner Traditions International Ltd., 1986.

[15] Carol Peters. Private communication. 1989.

[16] Beatrix Potter. Ginger & Pickles. Frederick Warne, London, 1909.

[17] Cherry Pyron and Clarissa Silitch. Pickles, Relishes, Chutneys, Sauces, Catsups, Mincemeats, Beverages, and Syrups. Yankee Books, Dublin, New Hampshire, 1978.

[18] ----. Reinventing the Pickle: Mutant bacteria put crunch back into koshers. The San Francisco Chronicle :P-4, February 19, 1989. Originally published in The New York Times.

[19] Tobias George Smollett. The adventures of Peregrine Pickle, in which is included Memoirs of a lady of quality. Limited Editions Club, Oxford University Press, Oxford, 1936.

[20] Susan Terris. Pickle. Four Winds Press, New York, 1973.

[21] Charles P. Thacker. Private communication. 1989.

[22] Webster's New Collegiate Dictionary G. & C. Merriam Company, Springfield, MA, 1979.

[23] Richard T. Witek. Private communication. 1988.

Lesson 9: Energy storage (battery)

 Review last lesson.

 Today we will discuss the most common form of portable energy source: battery.

 Illustration of the main shortcoming of battery using electric vehicle

See attachment for GM IMPACT 3's specifications.

See attachment for battery components for future commercial electric car (GM IMPACT 3).

What are the attractiveness of using electric cars?

What are the problems with their usage?

The capacity of the batteries is rated as Ampere-hour (A-h) and Volt. The two together will make Watt-hour (W-h).

How do I make use of these two numbers (A-h and V) to calculate what is the total energy stored in the battery?

Compare the various parameters on the GM specification sheet for IMPACT 3 to see if they make sense (Is the specifications self consistent?).

 Demonstration of storage capacity of battery:

Use two AA alkaline batteries for a small car and monitor their voltage and current output over time until they die.

What we will have then are two curves: voltage versus time, and current versus time.

Take the product of the two curves will give power versus time curve.

Then the area under the curve (product of power times time) will give the energy stored in the battery.

Determine the weight of the batteries and then extrapolate to find what size battery do we need for the energy requirement of the vehicle.

 Example problems:

(1) I just purchased an electric vehicle (cost me $200,000). It has a 30 HP motor and has a battery pack with an energy storage capacity of 20 kW-h. My maximum speed is 70 mph and I can maintain that speed when I operate the motor at full power.

(a) Under this maximum power condition, what is my range?

To determine the range, we first need to calculate the time

Energy = Power x time 20 kW-h = 30 HP x Time

or 20,000 W-h = (30 x 750W) x Time

Time = 0.89 hour

So the range is: Time x Speed = 0.89 h x 70 mph = 62.3 miles.

(b) If my motor is operating at 200 V, what is the ampere-hour rating of my battery pack?

20 kW-h = 200 V x (Ampere-Hour Rating)

20,000 W-h = 200 V x (Ampere-Hour Rating)

Ampere-Hour Rating = 100 A-h

Extra information on energy density in batteries

According to the test conducted on two alkaline AA batteries on an electric model car, the two batteries generated approximately 2V at 0.38A for 2.5 hours. This gives a combined energy of Volt x Amp x Time of 1.9W-h or approximately 1W-h per battery. Each battery weighs 20 grams.

So the energy density of the alkaline AA battery is 1W-h/20g or 0.05 W-h/g.

How does this compare with lead acid car batteries?

I have determined the weight of a full size lead acid car battery today, it is 35 lbs or about 15 kg. It is of course 12V and it has an ampere-hour rating of about 50 A-h. So what is its energy density? Energy in the battery is Volt x Amp x Time = 12V x 50 A-h = 600 W-h. The energy density of a lead acid car battery is then 600W-h/15kg = 0.04 W-h/g.

How does this compare with the car batteries used in the Sol of Auburn?

Earlier: Batteries: 136kg (300lb), 96V, capacity at 52 A-hr.

So: The energy density in Sol of Auburn batteries is 96V x 52A-h/136kg = 0.037 W-h/g.

 Homework Assignment

(1) I am an engineering working for Chrysler to develop electric cars driven solely by batteries. In my experiments on the test track, I found that my range is 100 miles on the highway if the car is driven at a constant speed of 60 mph whereas the range dropped to 80 miles if the speed is at 70 mph. I further found that at 60 mph, my electric motor was operating at a power level of 15 kW.

(a) How much time can my car run at 60 mph? How much time can my car run at 70 mph?

(b) What is the power level that my electric motor operates at 70 mph?

Lesson 10: Solar energy

 Review of last lesson.

 Solar energy is a viable means to supply power to homes and automobile.

 What are the advantages of solar energy?

 What are the disadvantages and problems of solar energy?

 Solar cells are also commonly known as solar photovoltaic. Sun energy (photons especially in the visible range) excites electrons in a semiconductor P-N junction. This excitation generates an electrical current. This is quite similar to how the light meter in your camera works. The P-N junction is sensitive to light.

 Solar powered cars use sun energy to run an electric motor. The excess energy from bright sunlight also charges up storage batteries. In the off sun time (either at night or cloudy period), the batteries will be used to power the motor or other devices.

 Most solar devices are experimental at this time. For automobile, the one at Auburn is called the Sol of Auburn. It has the following specifications.

Dimensions: 6m (20') long, 2m (6.6') wide and 1.25m (4') high.

Motor: 6kW(8HP) DC, 600rpm, 96V, 92% efficient at operating level, 94% peak, 12kg

Solar array: Peak power at 1400W at a solar power density of 1kW/m², 96V.

Batteries: 136kg (300lb), 96V, capacity at 52 A-hr

Chassis: Tubular aluminum frame, 1.5" diameter and 0.065" wall, carbon fiber, kevlar composites and carbon body.

Brakes: motorcycle hydraulic disk brakes.

Wheels: Two 36 spokes and Two 48 spokes, 51 cm wheels.

Tires: 51x4cm (20"x1.75") tires at 85psi.

Maximum speed: 57 mph.

Weight of car (without driver): 800 lbs.

Driver: Under 176 lbs, under 6'.

What is the efficiency of the solar cell used in this car?

 Can you relate the problems of solar car discussed above with the Sol of Auburn?

 Efficiency of solar conversion:

Solar cell to convert light into electricity directly. The idea of converting light energy into electricity will also be demonstrated. In this case, an electrical light bulb (100W clear GE) will be used. This is a two step conversion process (really stupid idea if you think about it). The first step is to apply 120VAC to the light bulb and convert electrical energy into light energy and then the light energy into electrical energy using photovoltaic (commonly called solar cell). (Why is this stupid? But a good demonstration of stupidity nevertheless)

First step: We will measure the electrical power into the light bulb by monitoring the voltage and current from the line to the light bulb. Then we will use a light meter to measure the amount of light (photo) energy it produces.

Voltage of light bulb =

Current of light bulb =

Electrical power consumption of light bulb =

Light meter reading = (1ft-candle=1lumen/ft², 1lumen=1/680 lightwatt)

Distance of bulb to light meter (also where the solar cell will be) =

Area of sphere at where the meter is = 4Br²=

Total light power from the light bulb =

Efficiency of light bulb for visible light production =

Second step: We will use the light from the light bulb to produce electricity using the solar cell. This will be done by measuring the power absorbed by the solar cell and the power generated by the solar cell.

Total light power from the light bulb =

Area of solar cell =

Area of entire sphere =

Power absorbed by the solar cell =

Load resistor =

Load voltage =

Power to load produced by the solar cell =

Efficiency of solar cell =

Overall efficiency =

Repeat the same experiment but at reduced levels of power to the light bulb.

Voltage of light bulb =

Current of light bulb =

Electrical power consumption of light bulb =

Light meter reading = (1ft-candle=1lumen/ft², 1lumen=1/680 lightwatt)

Distance of bulb to light meter (also where the solar cell will be) =

Area of sphere at where the meter is = 4Br²=

Total light power from the light bulb =

Efficiency of light bulb for visible light production =

Total light power from the light bulb =

Area of solar cell =

Area of entire sphere =

Power absorbed by the solar cell =

Load resistor =

Load voltage =

Power to load produced by the solar cell =

Efficiency of solar cell =

Overall efficiency =

Plot your results appropriately.

How do you conduct a test to see if the solar cell derives its power from the optical (visible), IR and/or UV portion of the electromagnetic spectrum?

 Example problem

I have invented a new improved solar cell which has very high efficiencies (which makes me a millionaire or possibly a billionaire, but I still teach at Auburn just for fun). In my experiments, I found that when the solar power input was 1 kW, my solar cell output was 300 W. What is the efficiency? Answer: 30%

I also obtained the following data:

Input Power (kW) Output Power (kW)

2 0.60

3 0.90

4 1.15

5 1.30

6 1.55

7 1.75

8 1.95

9 2.10

10 2.20

Plot the conversion graph and discuss what is happening.

 Assigned Problem

I want to use solar cell to produce all the electricity I need for my home. I want to be completely independent of the Alabama Power Company. I find that my power needs during the day time hour of 9 am to 6 pm is 5,000 W and at night from 6 pm to 9 am the following morning is 2,000 W. My plan is to use sufficient solar cells in the daytime to provide the power needs and also to charge up a set of batteries for use after the sun has set. I am using 12V car batteries. Each battery is capable of storing 50 Amp-hour of charge when fully charged (that is at each full charge, each battery has enough stored electrical energy to produce 50 Amp for 1 hour or 1 Amp for 50 hour or any product of current-time of 50 Amp-hour). It just so happen that I live in an area where the sun always shines from 9 am to 6 pm with a power density of 1 kW/m² and the solar energy is not available from 6 pm to 9 am. My solar cell efficiency is 9%.

(a) What is the minimum area of solar cell do I need to satisfy my daytime needs and to charge up my batteries?

(b) How many batteries do I need?

Lesson 11: Solar energy continues

Lesson 12: Thermal energy loss (heat conduction)

 Review previous lesson.

 In the kitchen you use an aluminum pot for good heat transfer from stove to your food, but your refrigerator is insulated with styrofoam to prevent conduction of heat. How do we describe the difference between the two materials?

If you hold one end of a copper rod and place the other end in a flame, the end you are holding gets hotter and hotter, even though it is not in direct contact with the flame. Heat reaches the cooler end by conduction through the material. On the atomic level, the atoms in the hotter regions have more kinetic energy, on the average, than their cooler neighbors, and so it goes through the material. The atoms themselves do not move from one region to another but their energy does.

In metals, electron motion provides another mechanism for heat transfer. Most metals are good conductors of electricity because some electrons can leave their parent atoms and wander through the crystal lattice. These "free" electrons can carry energy from the hotter to the cooler regions of the metal. Good thermal conductors such as silver, copper, gold, and aluminum are also good electrical conductors.

Heat transfer occurs only between regions at different temperatures and the direction of heat flow is always from higher to lower temperature. Develop the basic formula for heat current. Some thermal conductivities (k) are listed in the charts below.

 Introduce the basic equation for thermal conductivity and emphasize the intrinsic and extrinsic factors, heat flow, thermal conductivity and temperature gradient. (review units)

Q(heat flow rate) = k(thermal conductivity) x )T x A (area) / L (length)

Thermal Conductivities of Some Common Metals

Metal	Thermal Conductivity, k (in unit of W/m-K)
Aluminum	205.0
Brass	109.0
Copper	385.0
Lead	34.7
Mercury	8.3
Silver	406.0
Steel	50.2

Thermal Conductivities of Other Materials

Materials, k (in unit of W/m-K)	Materials, k (in unit of W/m-K)
Brick (insulating), 0.15	Nylon, 0.3
Brick (red), 0.6	Polystyrene, 0.033
Charcoal, 0.055	Rubber (92% dense), 0.16
Concrete, 0.8	Rubber (sponge), 0.05
Cork, 0.04	Styrofoam, 0.01
Felt, 0.05	Dry sand, 0.33
Fiberglass, 0.04	Oak wood, 0.16
Glass, 0.8	Pine wood, 0.26
Ice, 1.6	Air, 0.024
Sheet rock, 0.1

 Use fire (high temperature), liquid nitrogen (low temperature), copper and quartz to illustrate the effect of conductivity. We will NOT be able to measure thermal conductivity in class (it is not easy to do quickly). WHY is it difficult? Need propane torch, LN, copper bar, quartz or glass bar and thermometers.

 Use the wall panel and thermometers at different locations and a heat lamp to illustrate thermal conductivity of different materials (good and poor insulation). Use wall with thermal insulation to illustrate the construction of home to minimize heat transfer. Use thermometers to monitor temperature at different locations and illustrate good versus poor insulation. Need simulated walls and temperature measurement device.

 Use a single pane and double pane window to illustrate that glass is a relatively good thermal conductor (compared to styrofoam), although it is a good insulator relative to copper. Illustrate that a good conductor would result in a small temperature gradient (difference).

 Use the parameters in references to illustrate the energy loss from the wall and windows of a typical home and calculate cost.

 Example using conduction through a picnic cooler: A styrofoam box used to keep drinks cold at a picnic has total wall area (including the lid) of 1.2 m² and a wall thickness of 1.6 cm. It is filled with ice, water, and drinks at 0^oC. What is the rate of heat flow into the cooler if the temperature of the outside wall is 28^oC?

 Assignment

Heat conduction through a wall with a window: During the winter, heat flows out of a house through the walls. If it is 0^oC outside and 21^oC inside and I have a wall with a total area of 25 m² with a window of 1 m². The window is made of a single sheet of normal glass 0.2" thick. The rest of the wall is made of a 0.5" thick sheet rock over a 3" thick fiberglass insulation and then a 0.5" thick styrofoam and finally a 2" thick insulating brick.

(a) What is the heat flow through the window? (your answer should have a unit of Watt)

(b) What is the heat flow through the rest of the wall? (should also have the same unit of Watt) (Hint: Use the thermal conductivity table, take reciprocal to get thermal resistivity and determine the thermal resistance values for the components. Sum them up and take reciprocal again to get thermal conductance and then finally apply the heat flow equation)

(c) If the average difference in temperature between the inside and the outside of the wall is 21^oC and the heating cost is 8 cents/kW-h, how much does it cost for the lost heat this wall for 1 month?

Lesson 13: Thermal energy loss (conduction and radiation)

 Review last lesson in heat conduction.

 Illustrate the poor thermal insulation property (or high conductivity) of glass using a heat lamp and thermocouple. Notice almost the same temperature across the glass. Glass is a poor conductor amongst metals but a relative good conductor amongst insulators. Emphasize that a good insulator will result in a large temperature difference across it and visa versa.

 R-value: In the winter, it is warm inside a building and cold outside. The inside surface of a wall or roof assembly is warm and the outside surface is cold. This concept can be explained by the concept of thermal resistance or conductivity using the R value.

The effectiveness of a building assembly in resisting the conduction of heat is expressed in terms of thermal resistance (R). R is expressed in either English units of ft²hr^oF/Btu or metric units of ^oC-m²/W. The higher the R value, the higher the insulating value. Metals have very low R values, concrete and masonry materials only have slightly better values. Wood has a substantially higher thermal resistance, but not nearly as high as that of an equal thickness of any of the common insulating materials. Thermal insulation is any material added to a building assembly for the purpose of slowing conduction of heat through the assembly.

Common Construction Materials and Their R-values

Materials	R-values (in unit of ft²hr^oF/Btu)
½" gypsumboard	0.45
½" plywood	0.62
Common brick (per inch thick)	0.2
½" insulated board sheathing	1.32
Carpet with fibrous pad	2.08
3/8" built-up roof	0.33
2" insulation	7.00
3.5" insulation	11.00
6" insulation	19.00

 Example problem: Heat conduction through a wall with a window: During the winter, heat flows out of a house through the walls. If it is 0^oC outside and 21^oC inside and I have a wall with a total area of 25 m² with a window of 1 m². The window is made of a single sheet of normal glass 0.3" thick. The rest of the wall is made of a 0.5" thick sheet rock over a 5" thick fiberglass insulation and finally a 2" thick normal red brick.

(a) What is the heat flow through the window? (your answer should have a unit of Watt)

Q_window = k_windowx A_window x )T/ L_window

= 0.8 W/m-K x 1 m² x 21 K / 0.3" = 2,200 W or 2.2 kW

(b) What is the heat flow through the rest of the wall? (should have the same unit of W)

We know from the wall experiment that the heat flow through the various parts must be the same and we set them equal to Q. However the temperature drop across each part is different but the sum must be 20 degrees. Also the areas are the same (24 m²). So we can set up the following equations.

Q = k_{sheet rock}x A x )T_{sheet rock} / L_{sheet rock} [1]

_{Q = k_fiberglass x A x )T_fiberglass / L_fiberglass [2]
_{Q = k_brick x A x )T_brick / L_brick[3]}}

)T_{sheet rock} + )T_fiberglass +)T_brick = 21 K [4]

substituting the values into these equations:

Q = 0.1 W/m-K x 24 m² x )T_{sheet rock} / 0.013 m = 189 )T_{sheet rock}

Q = 0.04 W/m-K x 24 m² x )T_fiberglass / 0.13 m = 7.4 )T_fiberglass

Q = 0.6 W/m-K x 24 m² x )T_brick/ 0.051 m = 283 )T_brick

)T_{sheet rock} + )T_fiberglass +)T_brick = 21 K

Q = 189 )T_{sheet rock} [5]

Q = 7.4 )T_fiberglass[6]

Q = 283 )T_brick [7]

)T_{sheet rock} + )T_fiberglass +)T_brick = 21 K [8]

Four equations four unknowns. Perfect. We need to solve for Q.

Rewriting the equation [8]: )T_{sheet rock} = 21 - )T_fiberglass -)T_brick [9]

Combining [6] and [5]: 7.4 )T_fiberglass = 189 )T_{sheet rock}

or )T_fiberglass = 25.5 )T_{sheet rock} [10]

Combining [7] and [5]: 283 )T_brick = 189 )T_{sheet rock}

or )T_brick= 0.67 )T_{sheet rock}[11]

Substituting [10] and [11] into [8] and solve for )T_{sheet rock}

)T_{sheet rock} = 21 - 25.5 )T_{sheet rock}- 0.67 )T_{sheet rock}

giving: 27.17 )T_{sheet rock} = 21

or )T_{sheet rock} = 0.77 K

Finally substitute this into [5] gives: Q=146 W

Now let be sure that we did all the math correctly (not necessary but a good check):

Substituting Q=146W into equations [5], [6] and [7] yields:

)T_{sheet rock} = 0.77 K

)T_fiberglass= 19.7 K

)T_brick = 0.52 K

Giving )T_{sheet rock} + )T_fiberglass +)T_brick = 20.99 K (close enough to 21 K)

Can you solve the same problem using R-values?

Total power loss through the window and wall =2,200W+146W=2346W=2.346kW

Energy for 1 month=Power x time=2.346kW x 30 days x 24 hrs/day = 1.7x10³ kW-hr

Cost = 1.7x10³ kW-hr x $0.08/kW-hr = $135

 Assignment for thermal conduction

Heat conduction through a wall with a window: During the winter, heat flows out of a house through the walls. If it is 0^oC outside and 21^oC inside and I have a wall with a total area of 20 m² with a window of 1 m². The window is made of a single sheet of normal glass 0.2" thick. The rest of the wall is made a 3" thick fiberglass insulation and a 2" thick insulating brick.

(a) What is the heat flow through the window? (your answer should have a unit of Watt)

(b) What is the heat flow through the rest of the wall?

Thermal energy loss (radiation)

 Radiative heat transfer is a second way to transfer heat from one place to another. It does not require a medium.

 Introduce the electromagnetic radiation spectrum and the various aspects and applications of radiation. Use the chart on page 155 of notes to illustrate different electromagnetic regions. Notice radio/TV, microwave, IR (infrared, heat), optical (visible),UV (ultraviolet), x-ray, gamma ray.

 Demonstrate the microwave region using a microwave oven. Determine the efficiency of microwave oven heating by again monitoring the electrical energy consumption of the microwave and the heating of water. Converting electrical energy into microwave energy which then convert to heat (mass of water times temperature rise times specific heat).

Power to the microwave oven (voltage times current) =

Duration of heating =

Electrical energy consumed =

Mass of water used =

Initial temperature =

Final temperature =

Temperature rise =

Specific heat of water =

Energy absorbed by water =

Efficiency of microwave =

Lesson 14: Thermal energy loss (radiation and convection)

 Review previous lesson on radiation.

 Demonstrate using a prism to split light into different colors in the visible region.

 Demonstrate using a heat lamp and a normal lamp the infrared (or heat) portion of the spectrum.

 Demonstrate the transmission of visible light through different glasses using a light meter and two sources of light (normal and heat lamp).

 Demonstrate the transmission of infrared radiation (heat) through different glasses using a light meter and two sources of light (normal and heat lamp) by feel and thermocouple.

Comparison of transmission through different glasses

Light meter reading without glass:

Temperature reading without glass:

	Visible Light (visual and light meter)	IR (heat) (feel and thermocouple)
Normal glass
E glass (energy efficient)
Stain glass
Dark double glass

 Describe the basic concept of energy (or heat) transfer by radiation. Develop the basic equation for radiative heat transfer, the Stefan's law.

 Introduce the idea of surface emissivity for heat absorption through surfaces with different colors. Use the T⁴ equation for radiation to describe heat transfer by radiation. A material with high emissivity is efficient in both absorbing radiation energy as well as emitting it. There a good absorber is also a good emitter.

 Place a black sheet and a white sheet near a heat lamp and feel the difference by hand and by measurement.

 Relate the importance of selecting thermally efficient external color to energy conservation. Significant amount of heat in the summer in Alabama houses is generated through radiation absorption of the roof.

Emissivity of Some Common Materials

Materials

Emissivity Values (no unit, dimensionless)

Carbon

0.85-0.95

Aluminum

0.11

Brass (oxidized)

0.61

Brass (unoxidized)

0.03

Copper (oxidized)

0.60

Copper (unoxidized)

0.02

Black gloss paint

0.90

Gold (polished)

0.02

Fire brick

0.75

 Solar Energy: Bright sunlight can deliver over 1kW of power to each square meter it falls on. A tennis court receives a solar energy equivalent to that in a gallon of gasoline every 10 minutes. The power per unit surface area in the sun’s radiation (before it passes the earth’s atmosphere) is about 1.4 kW/m². A maximum of 1.0 kW/m² reaches the surface of the earth on a clear day and the time average over a 24 hour period is about 0.2 kW/m².

One way to use the sun’s energy involves a series of mirrors that track the sun and concentrate its radiation on a boiler whose steam drives a turbine and thence an electric generator. Photovoltaic cells convert light energy directly into electricity, no heat is involved so the cells avoid thermodynamic limitations. Capturing energy by photovoltaic cells has a high capital cost and the price of energy they produce is about five time that of coal powered steam turbine plants is not a deal, but there are many advantages to solar power. Some of these are lack of noise, lack of moving parts, minimum maintenance requirements, and freedom of pollution.

 Concentrate on the use of IR camera and compare different materials for different transmission between IR and visible.

Demonstrations using Infrared Camera

Demonstration 1 requires a high sensitivity setting on the IR.

1. Video tape with palm print and also heat transfer through thin metal sheet using hand.

Demonstrations 2 to 8 require a lower sensitivity setting on the IR

2. Heat lamp on full and direct view from IR.

3. Compare heat lamp intensity with normal 100W light bulb with setting for heat lamp.

4. Insert normal glass and observe high heat transmission.

5. Repeat and compare with E-glass.

6. Repeat and compare with stain glass.

7. Repeat and compare with double stain glass.

8. Green house effect by observing the low transmission of heat through plexiglass.

Demonstrations 9 to 11 require a higher sensitivity setting on the IR.

9. Increase sensitivity of IR to observe the transmission of IR through the Silicon and compare with metal plate.

10. Examine the low heat production of fluorescent tube.

11. Finally look at human faces.

 Example problem: You work part time as a contractor building houses (to supplement your teaching income). Your client ask you what differences are there (if any) between using a light color roof and a dark color roof in the south where summer is very hot (we all know that). The homeowner wants to use a very dark shingles with an emissivity of 0.9. The roof has a 6" thick fiberglass insulation and the attic temperature is always maintained at 80^oF (26.7^oC=300K) and thermal conduction through the roof to the attic and radiative loss are the main sources of heat transfer. Determine the equilibrium temperature in the day time for the shingles if the solar power is 1 kW/m².

Energy absorption rate = 1kW/m² x 0.9 = 900 W/m².

Heat loss through conduction = (0.04W/m-K) / (6x0.0254m) )T = 0.26 (T- 300) W/m²

Heat loss through radiation = ,FT⁴ = 0.9 x 5.67x10^-8W/m² T⁴ = 5.6x10^-8 T⁴ W/m²

Balancing heat absorption and loss yields:

900 = 0.26T - 78 + 5.6x10^-8 T⁴

5.6x10^-8 T⁴ + 0.26T - 978 = 0 (NOW WHAT DO WE DO????)

You can either go and consult a mathematician to solve the equation or make some approximations. If we are lucky, we can get pretty close to the correct answer.

Assume that the 5.6x10^-8 T⁴ term is small compared to the other two terms (this is true for small T). Then the solution is simply T=3761 K. This is pretty high temperature and violate the assumption of small T. Actually if we substitute this solution into the original full equation, we have: 1.12x10⁷ + 978 - 978. This certainly is not equal to zero.

Other assumption is that 0.26T is small, then we have as an approximation the equation: 5.6x10^-8 T⁴ - 978 = 0 or T= 364 K (much more reasonable assumption). To check, we can substitute this solution back into the original equation and we have:

978 + 94.5 - 978 which indicates that our answer is about 10% off. Quite acceptable in this approximation.

 Assignment

After teaching high school science for 20 years, you decided to change career and became a space thermal scientist for NASA. You were charged with the design of a solar panel in space (the only heat transfer is radiation, no heat conduction at all). Your solar panel received 1 kW/m² of solar energy and its emissivity (also its absorptivity) is 0.6 (or 60%). Determine the equilibrium temperature of the panel. Repeat the calculation for an emissivity of 0.3 and see what you find.

 Assignment

Using the information provided on Bill Baird's home electric and gas bills for 1994 (given on next page), answer the following questions.

1) Determine and then graph the monthly electric consumption in kW-hr for the year.

a) Write a one-sentence explanation that you believe explains the graph you drew.

b) What is the average cost of electricity in $ per kW-hr for residential customers of Alabama Power Co.? If the use fell to zero, would the monthly bill be $0.00?

c) What is the annual cost of electricity in this house of 2400 square feet?

2) Determine and then graph the monthly gas consumption in 100's of cu. ft..

a) Write a one-sentence explanation that you believe explains the graph you drew.

b) What is the average cost of gas in $ per 100 cu.ft. for residential customers of Alabama Gas Co.? If the use fell to zero, would the monthly bill be $0.00?

c) What is the annual cost of gas in this house which heats its water with gas?

3) Does this house heat its living space with gas or electricity?

4) Does this house cool its living space with gas or electricity?

Alabama Power Co. Electric bills for Bill Baird's house, Auburn, AL-1994

Month Previous Current Energy Used Charge

(kw-hr) (kw-hr) (kw-hr)

Jan. 0802 1376 $44.59

Feb. 1376 1745 31.15

March 1745 2139 32.79

April 2139 2467 28.45

May 2467 2768 26.67

June 2768 3242 37.99

July 3242 4456 86.47

Aug. 4456 5746 91.30

Sept. 5746 6890 81.76

Oct. 6890 7537 49.25

Nov. 7537 7902 30.80

Dec. 7902 8315 33.94

Jan. '95 8315 8766 36.43

Alabama Gas Co. bills for Bill Baird's house in Auburn, AL - 1994

Month Previous Current Energy Used Charge

(x100 cu.ft.) (x100 cu.ft.) (x100 cu.ft.)

Jan. 5608 5703 $69.49

Feb. 5703 5854 97.75

March 5854 5950 71.87

April 5950 6006 49.95

May 6006 6036 31.19

June 6036 6051 19.44

July 6051 6063 16.57

Aug. 6063 6075 16.33

Sept. 6075 6085 14.89

Oct. 6085 6095 14.89

Nov. 6095 6112 20.97

Dec. 6112 6138 29.27

Jan. '95 6138 6186 42.66

Lesson 15: Thermal energy loss (convection)

 Review thermal conduction and radiation

 Introduce thermal loss through convection and contrast the three methods of heat loss in terms of conditions and requirements.

 Convection is defined as the conveying of heat through a liquid or gas by motion of its parts. How does this happen and what is the mechanism?

 Introduce thermal radiation and use the vacuum bell jar to illustrate that energy from the sun is mainly transmitted via radiation.

 The basic mechanism is a combination of conduction and fluid motion. Convection occurs whenever a surface is in contact with a fluid at a temperature that is different from its own. Consider a hot vertical wall in contact with a cold fluid. As time passes, the fluid in intimate contact with the wall is heated by conduction, causing the fluid to become less dense. Due to the difference in density, a buoyancy force results, causing the lighter fluid to rise and to be replaced by the cooler fluid, with this process repeated continually. In the above process, the motion of the fluid is set up by natural forces, so the process is called "natural" or "free" convection. However, if the motion of the fluid is set up by external forces, then the process is called "forced" convection.

 Example: The flow of cigarette smoke in a still room is free convection whereas the flow of smoke in a room with a fan running is forced convection.

 Newton's law of cooling: heat flux is proportional to the temperature difference. Heat flux (q') is defined as the amount of energy flow through a unit area per unit time.

q' (heat flux, in unit of W/m²) = h )T where h is the convection heat transfer coefficient

Newton's law can also be written as heat flow (q) which is defined as the amount of energy through a specified area (A) per unit time.

q (heat flow, in unit of W) = hA)T (where A is the area)

 The parameter h describes the ability of the system to transfer heat via convection. Higher h values correspond to better heat transfer characteristic. The exact value of h depends strongly on the condition of the system as shown in the table.

Typical h Values for Different Systems

Conditions	h (W/m²-K)
Air (free)	5 - 15
Air or superheated steam forced	13 - 300
Oil, forced	50 - 1,700
Water, forced	300 - 12,000
Water, boiling	3,000 - 55,000
Steam, condensing	5,500 - 100,000

 Use the vacuum bell jar with heater and top and bottom thermocouples to compare radiative heat transfer and convection.

 Also use a bell inside the bell jar to show that transmission of sound also require a medium. ONLY electromagnetic radiation can be transmitted through a vacuum.

 Green House

The proper function of a green house relies on both radiation and convection. The idea of the green house is to trap the sun's energy during the cold months of winter. Remember that the sun's energy consists of radiation over a wide range of wavelength or frequency. All electromagnetic radiation travels at the same speed, the speed of light. The main interest here is optical (or visible) and IR (infrared or heat). During the winter months the earth's axis is tilted at an angle that makes the northern hemisphere receive less visible light from the sun. This means that fewer photons of light strike objects like rocks and soil, and less infrared radiation is emitted from these objects. Thus, less heat energy is radiated to the earth's atmosphere where it could be re-radiated back to the earth to maintain warm air temperature.

Back to the green house. We want to build a structure which will convert the visible light into heat and then trap the heat so it does NOT escape. So we use glass or plastic which is transparent to visible light so it lets all the visible energy through. Then the soil and flower pots and plants (which have dark colors and therefore high emissivity and absorption) convert the visible light into IR (heat energy). Now the glass of the green house does NOT let the heat out. Therefore the proper glass for the green house should have a LOW transparency to IR radiation (longer wavelengths). We need to select glasses or plastics which have a high transmission for visible light but low transmission for IR.

 Demonstrate the green house effect using models, thermocouples and IR camera.

 What about the Ozone layer and the green house gases?

 Example problems:

Part 1: When I was young and energetic many many many years ago, I worked at a window design company. I invented the double pane window concept and filed a patent for it. Since then I have become a millionaire with royalty coming in like a tidal wave. In this design, I have two panes of window separated by a small gap. The h-value (convection heat transfer coefficient) was 5 W/m²-K. The window measures 3'x6' and the outside temperature is 35^oC and the inside temperature is maintained at 25^oC. How much heat comes in through the window every hour?

Solution: Area of window = 3'x6' = 18 sq.ft. = 1.67 m²

Heat flow = h x area x )T = 5 W/m²-K x 1.67 m² x 10 K = 83.6 W = 83.6 J/s

Heat flow per hour = 83.6 J/s x 3600 s =301,000 J or 301 kJ.

Part 2: When I was a little more mature and working for myself, I discovered another great phenomenon. I found that by inserting a thin plastic sheet in between the double pane glass, I could break up the convection current and reduce the h-value by 40% (from 5 W/m²-K to 3 W/m²-K. It was then that I became a billionaire. How much heat comes in through this new design every hour with the same dimensions as before?

Solution: Simply multiply the answer to the previous question by 0.6 (or 60%) and the answer is 180 kJ.

 Assignment

(1) You just built a house with double pane windows. In one of the windows you measured the heat flux to be 100 W/m² when the difference in the outside and inside temperature is 25^oC. What is the h-value of this window?

(2) My attic consists of a floor which is 100 ft. by 50 ft. The roof ridge runs down the center of the symmetrical attic, and is 4 ft above the floor. I would like to keep the attic space from getting so hot that it melts the plastic bags that cover my hanging clothes, and adds lots of money to my air conditioning bill. I figure that 120 deg. Fahrenheit is the maximum temperature I am willing to tolerate during the summer months. To keep from exceeding this maximum temperature I must change the air in my attic completely at least once every 3 minutes.

Help me decide the most economical way to keep my attic cool. My two choices are:

(a) Install a power fan that draws 5 amps at 120 volts and costs $199.95 including installation. The fan moves 10,000 cubic feet per minute. It will cost me $0.08 per kilowatt-hour for the electricity, which I will use for 10 hours of daylight every summer day - 3 months. To allow 10,000 cubic feet per minute into my attic I must pay an additional $1100.00 to cut soffit vents into the horizonal surfaces under the edges of my roof and install gable vents.

(b) Use a "passive" system one that requires no attic fan. This involves cutting the same number of soffit vents into the horizonal surfaces under my roof edges, but instead of gable vents, I will cut out a 2-inch wide slot in the very top ridge of my roof to allow the heated air inside the attic to escape out the top of my roof. This system allows 7,000 cubic feet of air to move through my attic every minute, taking inside attic air with it. Total cost for cutting all these holes and installing vent covers is $2,000.00. Over five summers, which of these two options is most economical and how much more economical? Show all calculations that lead up to your answer.

Lesson 16: Mechanical construction (tensile and creep)

 Review last lesson

 Introduce the idea that when a force is applied to a material (stressed), it will change its shape or deform (strain) accordingly.

 Stress (F) is defined as force divided by the cross sectional area or F=F/A_o. What is the unit of stress?

 Strain (,) is defined as fractional change in shape ,=)L/L_o. What is the unit of strain?

 Young's modulus or Modulus of Elasticity (E) is the ratio of stress and strain in the elastic range (shape recovers after stress is removed), E=F/,. What is the unit of E?

 Materials that are soft, rubbery or weak possess low Young's moduli (such as rubber and lead).

 Materials that are hard and strong possess high Young's moduli (such as steel and machine tools).

Table of Modulus of Elasticity

Materials	Modulus of Elasticity (10⁶ psi)
^Lead	^2.0
^Magnesium	^6.5
^Aluminum	^10.0
^Silver	^10.3
^Gold	^11.3
^Copper	^18.1
^Nickel	^29.9
^Iron	^30.0
^Molybdenum	^43.4
^Tungsten	^58.5

^{ Demonstrate a simple tensile test using a rubber band and a metallic wire.
Materials examined =
Cross sectional area of the sample =
Original Length of the sample (L_o^{) =}
Gravitational constant =}

^Load (gm)	^{Load Force (N)}	^{Stress (N/m2)}	Length, L (mm)	Increase in L )L (mm)	Strain )L/L_o

 Plot the stress strain curve and discuss. This graph should have stress as the vertical axis and strain as the horizontal axis. Determine the modulus of elasticity.

 Repeat similar illustration using bending.

 Example problems

1. I have a gold bar which is 0.5" in diameter and 2" long. If the final bar length is 2.1 inches, what is the weight that I have applied?

Solution: Area = A_o = Br² = B(0.25")² = 0.2 in²

Strain = , = )L/L_o = (2.1-2)/2 = 0.05

Modulus = E = 11.3x10⁶ psi

Stress = F = E, = 11.3x10⁶ psi x 0.05 = 565,000 psi

F = F/A_o therefore F=FA_o = 565,000 psi x 0.2in² = 113,000 lbs.

2. I have two wires tied together end to end to form one long wire. One wire is a 0.05" diameter 3 inch long aluminum and the other is a 0.07" diameter 4 inch long lead. If I applied a load of 5,000 lbs, what is the final length of the composite wire?

Solution: Since the wires are attached end to end (in a series), both wires must experience (or carry) the same force (but different strains).

Area of aluminum wire = A_Al = 2x10^-3 in² and

Area of lead wire = A_Pb = 3.8x10^-3 in²

Stress on the Al wire = F_Al = F_Al/A_Al= 5,000lb/2x10^-3in² = 2.5x10⁶ psi

Stress on the Pb wire = F_Pb = F_Pb/A_Pb= 5,000lb/3.8x10^-3in² = 1.32x10⁶ psi

Modulus of aluminum is 10x10⁶psi and that of lead is 2x10⁶ psi

Strain of the Al wire = ,_Al = F_Al/E_Al = 2.5x10⁶psi/10x10⁶psi = 0.25

Strain of the Pb wire = ,_Pb = F_Pb/E_Pb = 1.32x10⁶psi/2x10⁶psi = 0.66

)L_Al= L_o,_Al x ,_Al= 3" x 0.25 =0.75". Therefore the final Al wire is 3.75" long.

)L_Pb= L_o,_Pb x ,_Pb= 4" x 0.66 =2.64". Therefore the final Pb wire is 6.64" long .

Total final length = 10.4 inches.

3. Now if I tied two bars together, this time not end to end but side by side. Both bars have a length of 3". One is made of aluminum and the other lead and both are 0.4" in diameter. If I applied a force of 1,000,000 lbs, what is the final length of the assembly?

Solution: Since the bars are attached in parallel, the load of 1,000,000 lbs is shared between them.

Area of bar = B(0.2")² = 0.13 in²

Force on Al + Force on Pb = F_Al + F_Pb = total force = 1,000,000 lb (force is split between the two bars. The question is how is the force split?

The strain of the two bars must be the same (denote as ,)

For aluminum: , = F_Al/E_Al = (F_Al/A_Al)/E_Al and

For lead: ,= F_Pb/E_Pb = (F_Pb/A_Pb)/E_Pb

Equating the two, we have (F_Al/A_Al)/E_Al = (F_Pb/A_Pb)/E_Pb

or F_Al/ A_AlE_Al= F_Pb/ A_PbE_Pb (1)

A_Al =A_Pb = 0.13in² the same value, the two areas in equation 1 cancel.

Modulus of aluminum is 10x10⁶psi and that of lead is 2x10⁶ psi

Equation (1) becomes: F_Al / E_Al = F_Pb / E_Pbor

F_Al/ 10x10⁶ = F_Pb / 2x10⁶ or

F_Al= 5 F_Pb

So total force = 1,000,000 lb = F_Al + F_Pb = 5 F_Pb+ F_Pb= 6 F_Pb

Giving F_Pb= 166,000 lb.

We are almost there. So the force on the lead bar is 166,000 lb.

Stress on the lead bar is = 166,000lb/0.13in² = 1,277,000 psi

What is its strain? ,=F/E = 1,277,000/2x10⁶ = 0.64

So )L=,xL_o = 0.64 x 3 inches = 1.9 inches

Final length = 1.9 + 3 = 4.9 inches.

 Assigment problems

1. If I tied two wires together end to end to form one long wire. One wire is a 0.05" diameter 3 inch long aluminum. The other wire is an unknown material with a diameter of 0.06" and its length is 4" long. If I applied a load of 10,000 lbs and the total final length is 9.5" long. What is the modulus of elasticity of my unknown material?

2. Now if I tied two bars together, this time not end to end but side by side. Both bars have a length of 4" long. One is made of gold and the other tungsten. The gold bar is 0.6" in diameter (I am rich) and the tungsten is 0.3" in diameter. If the final length is 4.1", what is the load or force that is applied? (For parallel bars, the load is shared between the two bars. The bars have the same strain)

Extra problems for practice (will not be collected)

1 Energy conversion: I have a natural gas burner which I use to heat a special oil mixture. This oil mixture has a specific heat of 3 cal/gm^oC and I find that it takes 10 cubic feet of natural gas to heat 10 kg of this oil mixture from 25^oC to 100^oC. What is the efficiency of this burner? (1 BTU =252 cal)

2 Energy conversion: I live in the Pacific Northwest and my home is powered by a hydroelectric generator. 50,000 kg of water passes through the generator every second. The water enters the generator at a velocity of 1 m/s and exit the generator at 0.4 m/s. The efficiency of my generator is 40%. What is the electrical output power of my generator? (Kinetic energy is 0.5mv²)

3 Thermal Conduction: You were born and raised in Alabama so you really like your home to be nice and warm. But unfortunately you married a person who is an Eskimo and enjoys cold surroundings. You built a home and decided that the two of you cannot possibly sleep in the same room. So you have two rooms separated by a thin wall. You are too cheap to have the wall insulated so it is just a sheet of 0.5" thick sheetrock. Obviously you have a high power heater and your spouse does. The wall separating the two rooms measures 8 ft by 20 ft. You keep your room at 90^oF and your spouse's at 50^oF. What power heater do you need?

4 Radiation heat transfer: I have a system in space that is generating a lot of waste power that I need to get rid of by radiation into the deep cold space. The power that I need to get rid of is 1,000 Watts and my radiator has a surface area of 1mx2m (=2m²) and it has an emissivity of 0.99. What is the equilibrium temperature of my radiator?

5. Radiation heat transfer: I have a piece of radiator material with an unknown emissivity value. However, I do know that at 700 K, its radiation loss rate is 10,890 W/m². (a) What is its emissivity? (b) What is its radiative power density at 900 K? (c) However much energy can a panel with an area measuring 2mx3m dissipate in one day at 900 K?

6. Thermal Conduction: Sheets of brass and steel, each of thickness 1 cm, are placed in contact. The outer surface of the brass is exposed to 100 deg. Celcius, and the inner surface of the steel is kept at 0 deg. Celcius. What is the temperature of the common interface? [See p. 13 for thermal conductivities.]

Solution:

1 From energy conversion table: 1 cu.ft. of natural gas produces 1032 BTU

10 cubic feet produces 10,320 BTU

To raise the temperature of the mixture = mass x temperature increase x specific heat

= 10,000 gram x 75^oC x 3 cal/gm^oC = 2,250,000 cal = 2.25x10⁶ cal = 8,929 BTU

Therefore efficiency = 8,929BTU / 10,320BTU =0.87 or 87% which is quite good.

2 In one second, the energy decrease of the water passing through the generator is

0.5 (50,000 kg) (1m/s)² - 0.5 (50,000 kg) (0.4m/s)² = 21,000 J

So the power of water going through the generator is simply 21,000 J/s = 21,000 Watts

Efficiency is 40%. So the electrical output is 21,000W x 0.4 = 8,400 Watts or 8.4 kW.

3 Amount of heat flow rate through the wall = k )T A / L

= (0.1 W/m-K) (40^oF) (8ftx20ft) / (0.5")

= (0.1 W/m-K) (22K) (14.86 m²) / (0.0127 m)

= 2,574 Watts

4 Power going to the radiator is 1,000 W

Radiation power loss = ,FT⁴ times Area

= 0.99 x 5.67x10^-8W/m²K⁴ T⁴ x 2m²= 1.12x10^-7 T⁴ W

The two must equal each other: 1,000 W = 1.12x10^-7 T⁴ W

giving T⁴=1000 K⁴/1.12x10^-7 = 8.9 x 10⁹ K⁴ giving T=307 K

5. Heat loss through radiation = ,FT⁴ = 0.9 x 5.67x10^-8W/m²K⁴ T⁴

(a) At 700 K, we have: 10,890 W/m²= , 5.67x10^-8 W/m²K⁴ (700K)⁴

10,890 W/m² = 13,613 , (W/m²)

, =0.80

(b) At 900 K, its radiation loss rate is ,FT⁴ = 0.80 x 5.67x10^-8W/m²K⁴ x (900K)⁴

=29,760 W/m².

= 0.8 x 5.67x10^-8 W/m²K⁴ x (900K)⁴ x 6 m² x 24 hr x 3600s/hr

= 1.54x10¹⁰ J

6. Q_brass = k_brass x Area x (100^oC-T) = 109W/m-K x Area x (100^oC-T)

Q_steel = k_steel x Area x (T-0^oC) = 50.2W/m-K x Area x (T-0^oC)

Set Q_brass = Q_steel and the two areas cancel out.

109 (100^oC-T) = 50.2 (T-0^oC) NOW ALL WE HAVE TO DO IS SOLVE FOR T

10900 - 109T = 50.2T

10900 = 159.2T or T = 68.5^oC

Electrical resistance of two segment of wires connected together

Electrical resistance of square cross section

Energy density comparison of two different sizes

Conversion of cal and Cal and J...

Designing a portable energy source using water and a nozzle - how does this compare with battery

MTL 501 Materials Engineering Physics for Teacher

Spring 1998 Examination One

Constant and conversion factors you may need

1 cal = 4.18 J, 1 HP = 746 Watts

1 inch=2.54 cm, Gravitation constant = g = 9.81 m/s²,

1 HP = 746 Watts, 1 BTU = 1054 J = 252 calories

1 m = 3.28 ft, 1^oF = 9/5 ^oC exactly, 0^oC = 273 K

Box your answer at the right hand side at the bottom of the calculation (include proper units) before moving to the next problem.

Make sure you show your work for partial credit.

1. [15pts] I have a natural gas burner which I use to heat a special oil mixture. This oil mixture has a specific heat of 3 cal/gm^oC and I find that it takes 10 cubic feet of natural gas to heat 10 kg of this oil mixture from 25^oC to 100^oC. What is the efficiency of this burner?

(1 cu.ft. Of natural gas contains 258 kcal).

2. [20pts] I am doing a remodeling job at home and I find a long wire with two segments connected end to end to form a long wire (evidently the previous builder did not have enough length of one wire). The total resistance of the long wire is 0.700 ohm. One part the long wire is made of copper with a diameter of 1.50 mm and a length of 100 feet. The other segment is made of aluminum with a rectangular cross section measuring 1.0mm x 2.0mm. How long is the total length of the wire (aluminum plus copper)? (Room temperature resistivity of copper is 1.72x10^-8Sm and that of aluminum is 2.63x10^-8Sm)

3. [20pts] There is an unknown electrical conducting cable which has a temperature dependent coefficient of resistance (a_o) of 3.00x10^-3 /^oC. Its resistance at 200^oC is 10.0 S. What is its resistance at -50^oC?

4. [25pts] I have a hydroelectric generator using a waterfall to produce electricity. Water enters the generator at a speed of 1.00 m/s and exits the generator at 0.25 m/s. The generator is 35% efficient and the output power of the generator is 100,000 Watts.

[10pts] (A) How high is the waterfall if water starts at the top with zero velocity?

[15pts] (B) How many kilogram of water goes through the generator in one day?

5. [20pts] I am a solar cell designer and made a cell with an efficiency of 25%. The solar power on it is 1,000 W/m².

[15pts] (A) I find that I can use electricity from the cell to heat 15,000 grams of water from 25^oC to 99^oC in one hour. What is the surface area of the panel? (Specific heat of water is 1 cal/gm/^oC)

[10pts] (B) What is the new surface area of the panel if in one hour energy from it is able to heat 15,000 grams of water from 25^oC to 100^oC and then vaporize half of the water? (Latent heat of vaporization of water is 540 cal/gm)

Name:

MTL 501 Materials Engineering Physics for Teacher

Spring 1998

Final Examination

Constant and conversion factors you may need

1 cal = 4.18 J, 1 HP = 746 Watts F= 5.67x10^-8W/m²-K⁴ Stefan's constant

1 inch=2.54 cm, Gravitation constant = g = 9.81 m/s²,

1 HP = 746 Watts, 1 BTU = 1054 J = 252 calories

1 m = 3.28 ft, 1^oF = 9/5 ^oC exactly, 0^oC = 273 K

Box your answer at the right hand side at the bottom of the calculation (include proper units) before moving to the next problem.

Make sure you show your work for partial credit.

Use 3 significant figures throughout the exam.

1. [20pts] I am doing a remodeling job at home which requires electrical wiring. I am using a copper wire with a diameter of 2.00 mm and 250 feet long. The electrical resistivity of copper is 1.72x10^-8 S-m. However I find that this wire is too small (in cross sectional area of course) for what I need. I need the resistance to be less than 0.0400 S. How many such copper wires do I have to connect in parallel to satisfy my requirement?

2. [20pts] Given the equation for resistance as a function of resistivity, length and cross-sectional area of a conductor is

R = DL / A

[6pts] (A) Write a plain English sentence describing the relationship among resistance, length, and cross-sectional area of a conductor. Do not use words like "divided by" or "over" or "numerator" or denominator" in your sentence. In other words, we want a scientific description of this relationship.

[6pts] (B) Which of the terms in the above equation are/is intrinsic properties and which are/is extrinsic properties?

[8pts] (C) Calculate the cost of a LONG copper wire with a 0.500 inch diameter to be used to carry current from Brown’s Ferry Nuclear Power Plant to Auburn (a distance of 215 miles) if the density of copper is 555 pounds per cubic foot and the price of copper is $1.60 per pound. There are 5,280 feet in one mile and the resistivity of copper is 1.72x10^-8 S-m.

3. [30pts] I have a piece of radiator material with an unknown emissivity value. However, I do know that at 700 K, its radiation loss rate is 10,900 W/m².

[10pts] (A) What is its emissivity?

[10pts ] (B) What is its radiation loss rate at 900 K?

[10pts] (C) However much energy can be dissipated from one side of a panel with an area measuring 2.00m by 3.00m in one day at 900 K?

4. [20pts] The R-value of a certain thickness of fiberglass is found to be 3.40 m²K/W.

[5pts] (A) What is the R-value in units of ft²F^ohr/Btu. Use dimensional analysis and show all your work.

[10pts] (B) If the k-value of fiberglass is 0.0470 W/m-K, what is the thickness of this fiberglass in inches. Show all your work.

[5pts] (C) Is the k-value an intrinsic or extrinsic property of fiberglass? Is the R-value an intrinsic or extrinsic property of fiberglass?

5. [30pts] I am a farmer who lives in a very remote area where I have no city water and no electricity from Alabama Power Company. I have a well for all my water supply. I use a motor to operate a well pump and I use a solar photovoltaic panel to generate the electricity for the motor. The pump is rated at 2.00 HP (needs 2.00 HP of electricity from the solar panel to operate) and it is 90.0% efficient in pumping water. My solar cells are 10.0% efficient and provide sufficient power to operate the pump when the power density of the sun is exactly at 500 W/m².

[10pts] (A) What is the size of my solar panel?

[10pts] (B) If my pump runs at 120V, how much current does it need?

[10pts] (C) As I mentioned, my pump is 90.0% efficient in pumping water and my well is 100 m deep, how many kilograms of water can it pump up in exactly 1 hour? (You may like to know that potential energy of a substance is related to mgh, mass times gravitational constant times height)

6. [30pts] I am an engineer working for Toyota to develop electric cars driven solely by batteries. I have twenty 12.0V batteries in the car and each battery has a 100A-h rating. In my experiments on the test track, I found that my range is 120 miles on the highway if the car is driven at a constant speed of 60.0 mph.

[6pts] (A) How much energy is stored in my twenty batteries? (I will accept any energy unit)

[6pts] (B) At 60.0 mph, how long will the batteries last?

[6pts] (C) What is the power (in unit of kW) drawn by my motor when I run my car at 60.0 mph?

[6pts] (D) However, the range dropped to 100 miles if the speed is at 70.0 mph. What is the power (in unit of kW) drawn by my motor when I run my car at 70.0 mph?

[6pts] (E) If my motor runs at 20.0 kW at 80.0 mph, what is the range at 80.0 mph?

I am very energy conscious and I build my house with very well insulated walls. My walls have an insulating brick (2" thick) on the outside with a 6" thick fiberglass insulation and then a 1" thick sheetrock.

(a) What is the effective R-value of the wall?

(b) The winter temperature outside is 0^oC and I want to maintain my inside temperature at 23^oC. The total wall area is 50,000 sq. ft. What is the minimum power rating of my heater (in units of Watts) if the only heat loss is through my walls?

3. In a home energy system design, it is necessary to remove 100,000W of waste power using a 10m x 2m radiator panel. I find that at when the power going to the panel is 100,000W, the panel is at a temperature of 648 K. What will the panel temperature be if the power to the panel is 200,000W?

4. You are very energy conscious and you use energy efficient double pane windows in your home. One of your windows measure 2m by 2m. It has two panes of energy efficient glass (5mm thick each) with a thermal conductivity of 0.2 W/m-K. The two panes are separated by a 10mm gap giving a convection heat transfer coefficient h of

5 W/m²-K. If the amount of heat flow through this particular window measures 100W, what is the temperature difference between the outside and the inside?

I have a piece of radiator material with an unknown emissivity value. However, I do know that at 700 K, its radiation loss rate is 10,890 W/m². (a) What is its emissivity? (b) What is its radiative power density at 900 K? (c) However much energy can a panel with an area measuring 2mx3m dissipate in one day at 900 K?

5. I am a hydroelectric power generator designer living in Idaho. In one of my projects, the water flow rate in my hydroelectric generator is 100,000 kg/second. The water enters the generator at a velocity of 10 m/s and my generator is 50% efficient with a output rating of 500 kW (kilowatt). What is the exit velocity of my water? (Kinetic energy is 0.5mv²)

MTL 501 Materials Engineering Physics for Teacher

Spring 1995 Examination One

Constant and conversion factors you may need

1 inch = 2.54 cm

F= 5.67x10^-8W/m²-K⁴ Stefan's constant

Box your answer at the right hand side at the bottom of the calculation (include proper units) before moving to the next problem.

Make sure you show your work for partial credit.

1. I am doing a remodeling job at home which requires electrical wiring. I am using a copper wire with a diameter of 1mm and 100 feet long. Unfortunately, the job requires 150 feet of wire and the only other wire I have is 1.4mm diameter aluminum wire. If I connect the two types of wires end to end to form the long wire that I need, what is the resistance? (Room temperature resistivity of copper is 1.72x10^-8Sm and that of aluminum is 2.63x10^-8Sm)

2. There is an unknown electrical conducting cable which has an electrical resistance of 10S at 100^oC and 11S at 200^oC, what is its temperature dependent coefficient of resistance (a_o)?

5 W/m²-K. If the amount of heat flow through this particular window measures 100W, what is the temperature difference between the outside and the inside?

6. The table below gives the (Alabama Power) power consumption rate at three-hour intervals for Dr. Baird's house in the summer for one day (midnight to midnight). Alabama Power has a program to encourage power usage at night where the rate is $0.08/kW-hr from 6 am to 9 pm everyday and $0.04/kW-hr from 9 pm to 6 am. How much does Alabama Power charge Dr. Baird for the 24-hour period?

Time Period Average Power Energy Cost

Consumption (kW) Consumption (kW-hr)

Midnight - 3am 1

3 am - 6 am 1

6 am - 9 am 3

9 am - noon 2

noon - 3 pm 2

3 pm - 6 pm 2

6 pm - 9 pm 5

9 pm - midnight 3

Strain (,) is defined as fractional change in shape ,=)L/L_o. What is the unit of strain?

 Young's modulus or Modulus of Elasticity (E) is the ratio of stress and strain in the elastic range (shape is recovered after stress is removed), E=F/,. What is the unit of E?

 Materials that are soft, rubbery or weak possess low Young's moduli (such as rubber and lead).

 Materials that are hard and strong possess high Young's moduli (such as steel and machine tools).

Table of Modulus of Elasticity

Materials	Modulus of Elasticity (10⁶ psi)
^Lead	^2.0
^Magnesium	^6.5
^Aluminum	^10.0
^Silver	^10.3
^Gold	^11.3
^Copper	^18.1
^Nickel	^29.9
^Iron	^30.0
^Molybdenum	^43.4
^Tungsten	^58.5

^Load (gm)	^{Load Force (N)}	^{Stress (N/m2)}	Length, L (mm)	Increase in L )L (mm)	Strain )L/L_o

 Plot the stress strain curve and discuss. This graph should have stress as the vertical axis and strain as the horizontal axis. Determine the modulus of elasticity.

 Example problems

1. I have a gold bar which is 0.5" in diameter and 2" long. If the final bar length is 2.1 inches, what is the weight that I have applied?

Solution: Area = A_o = Br² = B(0.25")² = 0.2 in²

Strain = , = )L/L_o = (2.1-2)/2 = 0.05

Modulus = E = 11.3x10⁶ psi

Stress = F = E, = 11.3x10⁶ psi x 0.05 = 565,000 psi

F = F/A_o therefore F=FA_o = 565,000 psi x 0.2in² = 113,000 lbs.

Solution: Since the wires are attached end to end (in a series), both wires must experience (or carry) the same force (but different strains).

Area of aluminum wire = A_Al = 2x10^-3 in² and

Area of lead wire = A_Pb = 3.8x10^-3 in²

Stress on the Al wire = F_Al = F_Al/A_Al= 5,000lb/2x10^-3in² = 2.5x10⁶ psi

Stress on the Pb wire = F_Pb = F_Pb/A_Pb= 5,000lb/3.8x10^-3in² = 1.32x10⁶ psi

Modulus of aluminum is 10x10⁶psi and that of lead is 2x10⁶ psi

Strain of the Al wire = ,_Al = F_Al/E_Al = 2.5x10⁶psi/10x10⁶psi = 0.25

Strain of the Pb wire = ,_Pb = F_Pb/E_Pb = 1.32x10⁶psi/2x10⁶psi = 0.66

)L_Al= L_o,_Al x ,_Al= 3" x 0.25 =0.75". Therefore the final Al wire is 3.75" long.

)L_Pb= L_o,_Pb x ,_Pb= 4" x 0.66 =2.64". Therefore the final Pb wire is 6.64" long .

Total final length = 10.4 inches.

Solution: Since the bars are attached in parallel, the load of 1,000,000 lbs is shared between them.

Area of bar = B(0.2")² = 0.13 in²

Force on Al + Force on Pb = F_Al + F_Pb = total force = 1,000,000 lb (force is split between the two bars. The question is how is the force split?

The strain of the two bars must be the same (denote as ,)

For aluminum: , = F_Al/E_Al = (F_Al/A_Al)/E_Al and

For lead: ,= F_Pb/E_Pb = (F_Pb/A_Pb)/E_Pb

Equating the two, we have (F_Al/A_Al)/E_Al = (F_Pb/A_Pb)/E_Pb

or F_Al/ A_AlE_Al= F_Pb/ A_PbE_Pb (1)

A_Al =A_Pb = 0.13in² the same value, the two areas in equation 1 cancel.

Modulus of aluminum is 10x10⁶psi and that of lead is 2x10⁶ psi

Equation (1) becomes: F_Al / E_Al = F_Pb / E_Pbor

F_Al/ 10x10⁶ = F_Pb / 2x10⁶ or

F_Al= 5 F_Pb

So total force = 1,000,000 lb = F_Al + F_Pb = 5 F_Pb+ F_Pb= 6 F_Pb

Giving F_Pb= 166,000 lb.

We are almost there. So the force on the lead bar is 166,000 lb.

Stress on the lead bar is = 166,000lb/0.13in² = 1,277,000 psi

What is its strain? ,=F/E = 1,277,000/2x10⁶ = 0.64

So )L=,xL_o = 0.64 x 3 inches = 1.9 inches

Final length = 1.9 + 3 = 4.9 inches.

 Assigment problems (Due Thursday May 4)

MTL 501 Materials Engineering Physics for Teacher

Spring 1997

Examination One

Constant, conversion factors and formulae you may need

1 inch=2.54 cm, Gravitation constant=9.81 m/s²,

1 HP (horsepower) = 746 Watts, 1 BTU = 1054 J = 252 calories

1 m = 3.28 ft, 1 J = 1 kg-m²/s²exactly, 1 W = 1 J/s exactly, 1 cal = 4.18 J

1^oF = 9/5 ^oC exactly, 0^oC = 273 K

P = I²R = V²/R, E=VIt, V=IR,

E=½mv², E=mgh,

R=Dl/A, R=R_o(1+a_oT)

Thermal energy = m c )T

Efficiency = output power / input power Or output energy / input energy

Box your answer at the right hand side at the bottom of the calculation (include proper units) before moving to the next problem.

Make sure you show your work for partial credit.

[20 pts] 1. I have a natural gas burner which I use to heat a special oil mixture. This oil mixture has a specific heat of 3.00 cal/gm^oC and I find that it takes 10.0 cubic feet of natural gas to heat 10.0 kg of this oil mixture from 25^oC to 100^oC. What is the efficiency (in unit of %) of this burner?

[30 pts] 2. I am an Alabama Power Company electrical engineer. My boss asks me to design a cable made of copper (room temperature resistivity of copper is 1.72x10^-8Sm ) to carry electricity from the power station to his private cottage. The length of the cable is 10.0 km (10.0 kilometer) and its resistance must not exceed 1.00 S for the entire length.

(A) If the cross section of the cable is a square, what is the size of the side of the cable?

(B) Is this a maximum or minimum value (in other words, can the size be larger or smaller than this value)?

(D) Under the condition of passing 1000 amps through the cable, how much power is generated in this cable?

(E) How much electrical energy would the cable consume under the 1000 amps condition in one day?

[20 pts] 3. I am building a house and I have decided to use aluminum as my wiring material. The resistivity of aluminum at exactly 0^oC is 2.63x10^-8Sm. I have a segment of the wire which is rectangular in cross section (measuring 1.00 mm by 2.00 mm) and it is 100.0 m long. I find that when I pass 30.0 amps of electrical current through, the temperature increases to 100^oC. If the resistance temperature coefficient (a_o) is 5.00 x10^-3/^oC, what is the resistance of the wire when it is carrying 30.0 amps?

[30pts] 4. I am a hydroelectric power generator designer living in the state of Oregon. In one of my projects, the water flow rate in my hydroelectric generator is 100,000 kg/second. The water enters the generator at a velocity of 10.0 m/s and the water exits the generator at a velocity of 5.00 m/s. The output power of my generator is 1,000 kW (kilowatt).

(A) What is the efficiency of my generator?

(B) If I work extremely hard and get the efficiency to 100%, what is the output power of my generator? (Assume all other conditions remain unchanged).

1. I am remodeling my home and find that one section of my electrical wiring has two different wires tied end to end. Evidently the previous owner ran out of proper wires and was too cheap to buy more. The total resistance of the long wire is 1S. One section of it is made of copper and it is 1 mm in diameter and 100 feet long. The second section is made of aluminum and but it has a rectangular cross section measuring 1mm x 2mm. (and that of aluminum is 2.63x10^-8Sm)

(a) What is the length of the aluminum section?

(b) If under operating condition, the voltage drop across the first section (copper) is 1 V, what is the current throught the wire?

(c) If under operating condition, the voltage drop across the first section (copper) is 1 V, what is the voltage drop across the aluminum section?

(d) Under the condiction of (b) and (c) above, what is the heat generated in the total wire?

3. I recently bought an electric car (powered solely by battery) from Ford. It uses a single battery at 100 V and the battery is rated at 50 A-h. According to Ford, the range of the vehicle is 200 miles if it is driven at a constant speed of 30 mph (miles per hour).

(a) How much energy is stored in the battery?

(b) What is the power of the motor (in unit of Watts) when it is driven at 30 mph?

(c) Unfortunately I have a slow leak (or drain) in the electrical circuit. The current leak (or drain) is exactly 1 Ampere at all times. If I fully charge my battery and then unplug and park it somewhere, how long does it take to completely drain my battery?

4. I have been driving for 20 years. I tabulated the cost every year I have spent on gasoline. I found that the annual cost followed the formula

Cost ($) = 140 x (Year)^1/2

(a) Make a table that shows the annual cost from my first year of driving (Year 1) to the present (Year 20).

(b) Plot the annual cost (y-axis) versus Year (x-axis).

Time Period Average Power Energy Cost

Consumption (kW) Consumption (kW-hr)

Midnight - 3am 1

3 am - 6 am 1

6 am - 9 am 3

9 am - noon 2

noon - 3 pm 2

3 pm - 6 pm 2

6 pm - 9 pm 5

9 pm - midnight 3

MTL 501 Materials Engineering Physics for Teacher

Spring 1995

Final Examination

Constant and conversion factors you may need

1 inch=2.54 cm, Gravitation constant=9.8 m/s²,

1 HP (horsepower) = 746 Watts, 1 BTU = 1054 J = 252 calories

1 m = 3.28 ft, 1 J = 1 kg-m²/s², 1 W = 1 J/s, 1^oF = 9/5 ^oC, 0^oC = 273 K

Box your answer at the bottom of the calculation (include proper units).

Make sure you show your work for partial credit.

Electrical Resistivities

Metal	D (S m)	Substance	D (S m)
Silver	1.47x10^-8	Pure Carbon	3.5x10^-5
Copper	1.72x10^-8	Amber	5x10¹⁴
Gold	2.44x10^-8	Glass	10¹⁰ - 10¹⁴
Aluminum	2.63x10^-8	Lucite	>10¹³
Steel	20x10^-8	Mica	10¹¹ - 10¹⁵
Lead	22x10^-8	Quartz	75x10¹⁶
Mercury	95x10^-8	Sulfer	10¹⁵
		Teflon	>10¹³
		Wood	10⁸ - 10¹¹

Common Construction Materials and Their R-values

Materials	R-values (in unit of ft²hr^oF/Btu)
Sheetrock (per 1/2" thick)	0.45
1/2" plywood	0.62
Insulating brick (per inch thick)	1.2
1/2" insulated board sheathing	1.32
Carpet with fibrous pad	2.08
3/8" built-up roof	0.33
2" insulation (fiberglass)	7.00
3.5" insulation (fiberglass)	11.00
6" insulation (fiberglass)	19.00

1. [35pts] I have a thermoelectric converter. It delivers 10V when one side is at 100^oC and the other at 0^oC.

(a) What is the maximum voltage this device can generate when the hot side is at 100^oC?

(b) What is its Seeback coefficient (S)?

6. [40pts] I want to make a bar that can support a weight of 4,000 lbs. The length of the bar is 100 ft long when relaxed (no load). I want the length of the bar to be 101 ft when loaded to 4,000 lbs. I find that I do not have a single bar strong enough to do the job in my garage and I have to tie 10 identical small bars (each 100 ft long) together side by side to have the strength needed. The bars that I have are 0.1" in diameter.

(a) What is the Young's (or elastic) modulus of my bars?

(b) I find that a single small bar that I have can hold the weight without breaking, but will stretch beyond my specification of 101 feet. If I use just one bar of this material, what will be its length when loaded to 4,000 lbs?

Week 7: Lesson 1 (May 9 Tuesday, Meet at Haley Center):

Auto (mechanical components of the drive train)

 Begin new topic: Energy for Automobiles

 How many ways can energy conversion and sources be used for transportation, especially to make wheels go round in an automobile (limitations and requirements).

 Slides show.

 When thinking about energy, one of our main concerns is transportation. Gasoline is now the main fuel for cars, but other sources can also be used. Ethanol which is a renewable product from plant materials, agricultural wastes, and garbage burns clean. Gasohol is another idea which is better for the environment. It has 10% alcohol added to gasoline. Electricity is a form of energy is used but to a very limited extend. It costs more and does less than conventional fuel. A flywheel is used in electric vehicles to store kinetic energy. Electric cars require little maintenance and have less bad effect on the environment. On the down side, electric cars are sluggish and the battery has to be recharged about every 50 miles. Solar energy is also a form of power for the solar car.

 An alternative to different sources of energy for cars is to make them more fuel efficient. To get more miles per gallon, smaller lightweight cars would be beneficial. Shaving about 10% off of the weight of a car by making it smaller or replacing heavy materials with lighter substitutes reduces fuel consumption by about 7%. Lighter materials lead to a problem of safety. Safe, fuel efficient cars should be made of lightweight but strong materials. Less than 10% (more like 6%) of the energy contained in a gallon of gas is used to propel the vehicle forward and the rest is lost as waste heat or to overcome engine or transmission friction or run the air conditioner. With the use of better design or lubricants, energy loss can be reduced.

 Because of the rising cost of petroleum, lowered sales, and export limitations it is hard to cut the cost of fuel by mass production. There is a weight reduction demand for automobile manufacturers to cut the weight 10-15kg/m² per project area. Simple weight reduction tends to adversely affect safety and increase noise and vibration. The use of plastic is also an idea for use on cars. Its light weight and strength make it a good candidate. In the future it may even be used on the body and the wheels.

 Different components of the internal combustion engine:

- The whole idea of an internal combustion engine is to controllably ignite the gasoline to move pistons up and down to propel your automobile or whatever.

- Fuel is injected into the ignition chambers via the carburetor/fuel injector using a fuel pump. The carburetor (in old cars) is to control the fuel and air mixture into the engine. New cars use fuel injectors. The fuel is pumped into the engine combustion chambers by a fuel pump. Most fuel pumps are mechanical (some European cars have electric pump).

- Valves on top of the combustion chamber (usually 2 values for each cylinder or piston) control the injection of fuel and exhaust of combustion gases.

- The number of cylinders varies from 4 to 6 to 8 (also 5 now in some European cars).

- The displacement of an engine is basically the amount of volume that the pistons move in one complete cycle. Measured in cubic inches (cu.in.) or cubic centimeter (cc). A large engine would have eight cylinders (V8) with a displacement of 400 cu.in. The Jaguar XJS has (V12).

- The ignition of the fuel is accomplished by a high voltage applied to the spark plugs from a high voltage coil using either points/solid state ignition (electronic ignition).

- Pressure from the combustion (controlled explosion) pushes the piston down. The up-down motion of the piston rotates a crankshaft.

- The crankshaft delivers power to the wheels via the transmission.

- Automatic transmission shifts automatically and power is delivered using fluid coupling. Manual transmission needs the operator to shift. Power is delivered through the clutch.

- Cars after 1975 has catalytic converters which has Pt (platinum) in it. The converter is extremely hot and burns the unburnt gasoline so as to minimize emission.

- The muffler system does not control emission but just reduce noise.

- Motor oil is pumped to the moving parts of the engine to reduce fiction and wear to prolong engine life and efficiency.

- Coolant (antifreeze) is pumped through the engine to prevent overheating, especially in the hot summer.

 What are the main materials for the engine? Cast iron steel for the engine block, aluminum for the heads, possibly light weight composites for the piston and other components.

 Example problem. (From p.140 of the course packet) According to the World Almanac, the mileage traveled by over 107 million passenger cars registered in the US in 1975 was estimated at 1.03x10¹²miles. The US population in 1997 was 213 million souls. The energy content of one gallon of regular gasoline is 1.25x10⁵Btu. The average car in 1975 obtained 14 miles/gallon of gas consumed. The energy equivalent of one human being working full time at full output is 193 Btu/day. Calculate the average human energy equivalent to our per capita consumption of gasoline for use in passenger cars.

Miles per person per day =

Gallons per person per day =

Energy (Btu) per person per day =

Human equivalents per person =

 Assigned problems. Due on May 11th, Thursday.

According to the chart on page 220 of the course packet, the relative energy cost per passenger-mile of various modes of urban transportation are as follows:

Mode of Btu per Total % of US transportation

Transportation Passenger Miles demand in 1980

Airplane 7,150 9.3

Automobile 5,400 88.4

Train 2,620 0.7

Bus 1,700 1.2

Walking (3 mph) 524 less than 0.1

Bicycle (8 mph) 310 less than 0.1

(1) The thermal efficiency of a typical automobile engine is 29%, meaning that for every 100 Btu of energy input to the engine 71 Btu is lost (out the exhaust, radiator, etc.) as waste heat. The mechanical efficiency of the engine is typically around 71%, meaning that the moving parts lose about 29% of the input energy due to friction and other energy conversion "costs". The rolling efficiency of automobile drive trains is around 30%, meaning that after the carburetor and combustion chamber (thermal conversion), the piston and cam shaft (mechanical conversion), the transmission and tires lose 70% of the energy they receive. (a) What is the overall efficiency of the car in converting gasoline into motion? (b) If a gallon of gas represents about 125,000 Btu of energy, how much energy is wasted from each 10 gallon tank-full we use? (c) There are 125 million registered automobiles in the U.S. Each is driven about 10,000 miles per year at a fuel efficiency of about 12 miles per gallon. Calculate the Btu's of heat energy "wasted" per year in the U.S. through automobile use. Could this help explain global warming?

(2) Assume that in the same year the above figures were valid that the U.S. public traveled a total of 1.2 X 10⁹ passenger miles. Assume that the Btu content of regular gasoline for automobiles is 1.25 X 10⁵ Btu/gal (a) How many gallons of gasoline were used in 1980 for automobiles in the U.S.? (b) What was the average fuel efficiency in miles per gallon for this fleet? Does this figure seem realistic to you?

Name:

MTL 501 Materials Engineering Physics for Teacher

Spring 1997

Final Examination

Constant, conversion factors and formulae you may need

1 inch=2.54 cm, Gravitation constant=9.81 m/s²,

F= 5.67x10^-8W/m²-K⁴ Stefan's constant

1 HP (horsepower) = 746 Watts, 1 Btu = 1054 J = 252 calories

1 m = 3.28 ft, 1 J = 1 kg-m²/s²exactly, 1 W = 1 J/s exactly, 1 cal = 4.18 J

1^oF = 9/5 ^oC exactly, 0^oC = 273 K

P = I²R = V²/R, E=VIt, V=IR,

E=½mv², E=mgh,

R=Dl/A, R=R_o(1+a_oT)

Thermal energy = m c )T

Efficiency = output power / input power Or output energy / input energy

Heat transfer power = kA)T/L

Radiative power = ,FT⁴

Box your answer at the bottom of the calculation (include proper units).

Make sure you show your work. You may receive partial credit.

Thermal Conductivities of Other Materials

Materials, k (in unit of W/m-K)	Materials, k (in unit of W/m-K)
Brick (insulating), 0.15	Nylon, 0.3
Brick (red), 0.6	Polystyrene, 0.033
Charcoal, 0.055	Rubber (92% dense), 0.16
Concrete, 0.8	Rubber (sponge), 0.05
Cork, 0.04	Styrofoam, 0.01
Felt, 0.05	Dry sand, 0.33
Fiberglass, 0.04	Oak wood, 0.16
Glass, 0.8	Pine wood, 0.26
Ice, 1.6	Air, 0.024
Sheet rock, 0.1

1. [20pts] You are very energy conscious and you use energy efficient double pane windows in your home. One of your windows measures 2.0m by 2.0m. It has two panes of energy efficient glass (5.0mm thick each) with a thermal conductivity of 0.20 W/m-K. The two panes are separated by a 10mm gap of air. If the amount of heat flow through this particular window measures 100W, what is the temperature difference between the outside and the inside?

2. [30pts] I have a piece of radiator material with an unknown emissivity value. However, I do know that at 700 K, its radiation loss rate is 10,890 W/m².

(A) What is its emissivity?

(B) What is its radiation loss rate at 900 K?

(C) However much energy can a panel with an area measuring 2m by 3m dissipate in one day at 900 K? In this case, you only have one surface radiating.

3. [20pts] The R-value of a certain thickness of fiberglass is found to be 3.4 m²K/W.

(A) What is the R-value in units of ft²F^ohr/Btu. Use dimensional analysis and show all your work.

(B) If the k-value of fiberglass is 0.047 W/m-K, what is the thickness of this fiberglass in inches. Show all your work.

(D) Is the R-value an intrinsic or extrinsic property of fiberglass? Why?

4. [30pts] I am a farmer who lives in a very remote area where I have no city water and no electricity from Alabama Power Company. I have a well for all my water supply. I use a motor to operate a well pump and I use a solar photovoltaic panel to generate the electricity for the motor. The pump is rated at 2.0 HP (needs 2.0 HP of electricity from the solar panel to operate) and it is 100% efficient. My solar cells are 10% efficient and provide sufficient power to operate the pump when the power density of the sun is at 500 W/m².

(A) What is the size of my solar panel?

(B) If my pump runs at 120V, how much current does it need?

(C) My pump is 100% efficient in pumping water and my well is 100 m deep, how many kilograms of water can it pump up in 1 hour? (You may like to know that potential energy of a substance is related to mgh, mass times gravitational constant times height)

5. [30pts] I am an engineer working for Toyota to develop electric cars driven solely by batteries. In my experiments on the test track, I found that my range is 120 miles on the highway if the car is driven at a constant speed of 60 mph whereas the range dropped to 100 miles if the speed is at 70 mph. I have twenty 12V batteries in the car and each battery has a 100A-h rating.

(A) What is the power (in unit of kW) drawn by my motor when I run my car at 60 mph?

(B) What is the power (in unit of kW) drawn by my motor when I run my car at 70 mph?

(D) If my motor runs at 20 kW at 80 mph, what is the range at 80 mph?

6. [20pts] I am doing a remodeling job at home which requires electrical wiring. I am using a copper wire with a diameter of 2.0 mm and 250 feet long. Unfortunately, the job requires 400 feet of wire and the only other wire I have is aluminum wire with a rectangular cross section of 1.0mm x 2.0mm. If I connect the two types of wires end to end to form the long wire that I need, what is the total resistance of the 250 feet wire? (Hint: don’t forget when two resistors are in series, one after the other, the two resistances add.)

(a) What is the effective R-value of the wall?

5. [25pts] I want to make a bar that can support a weight of 4,000 lbs. The length of the bar is 100 ft long when relaxed (no load). I want the length of the bar to be 101 ft when loaded to 4,000 lbs. I find that I do not have a single bar strong enough to do the job in my garage and I have to tie 10 identical small bars (each 100 ft long) together side by side to have the strength needed. The bars that I have are 0.1" in diameter.

(a) What is the Young's (or elastic) modulus of my bars?

I recently bought an electric car (powered solely by battery) from Chrysler. It has a single battery at 100 V and it is rated at 50 A-h. According to Chrysler, the range of the vehicle is 120 miles if it is driven at a constant speed of 30 mph (miles per hour).

(a) How much energy is stored in the battery?

(b) What is the power of the motor (in unit of Watts) when it is driven at 30 mph?

(d) The slow electrical leak (or drain of 1 Ampere mentioned in part c above) occurs when the car is turned off as well as during operation. What is the new range (in miles) of my electric car when driven at 30 mph?

Week 8: Lesson 1 (May 16 Tuesday, Meet at Haley Center):

Auto (Solar Power)

 Solar energy is a viable means to supply power to cars.

 What are the advantages of solar powered cars?

 What are the disadvantages and problems of solar powered cars?

 Solar cells are also commonly known as solar photovoltaics. Sun energy (photons especially in the visible range) excites electrons in a semiconductor P-N junction. This excitation generates an electrical current. This is quite similar to how the light meter in your camera works. The P-N junction is sensitive to light.

 All solar cars are experimental at this time. The one at Auburn is called the Sol of Auburn. It has the following specifications.

Dimensions: 6m (20') long, 2m (6.6') wide and 1.25m (4') high.

Motor: 6kW(8HP) DC, 600rpm, 96V, 92% efficient at operating level, 94% peak, 12kg

Solar array: Peak power at 1400W at a solar power density of 1kW/m², 96V.

Batteries: 136kg (300lb), 96V, capacity at 52 A-hr

Chassis: Tubular aluminum frame, 1.5" diameter and 0.065" wall, carbon fiber, kevlar composites and carbon body.

Brakes: motorcycle hydraulic disk brakes.

Wheels: Two 36 spokes and Two 48 spokes, 51 cm wheels.

Tires: 51x4cm (20"x1.75") tires at 85psi.

Maximum speed: 57 mph.

Weight of car (without driver): 800 lbs.

Driver: Under 176 lbs, under 6'.

 Can you relate the problems of solar car discussed above with the Sol of Auburn?

 Efficiency of solar conversion:

Voltage of light bulb =

Current of light bulb =

Electrical power consumption of light bulb =

Light meter reading = (1ft-candle=1lumen/ft², 1lumen=1/680 lightwatt)

Distance of bulb to light meter (also where the solar cell will be) =

Area of sphere at where the meter is = 4Br²=

Total light power from the light bulb =

Efficiency of light bulb for visible light production =

Total light power from the light bulb =

Area of solar cell =

Area of entire sphere =

Power absorbed by the solar cell =

Load resistor =

Load voltage =

Power to load produced by the solar cell =

Efficiency of solar cell =

Overall efficiency =

How do you conduct a test to see if the solar cell derives its power from the optical (visible), IR and/or UV portion of the electromagnetic spectrum?

Week 4: Lesson 1: HOME, electricity use (electrical consumption and heating)

 Review previous lesson.

 Introduce the concept of electrical energy and power through the voltage, current, resistance and time. Show that kW-h and energy and kW is power. Relate all units of energy and power by showing the relationship between calories, joules, watts, horsepower and so on.

 Use a volt meter and a current meter (either manual record or by computer) the energy consumption rate of different appliances

 Demonstrate the dependence of electrical resistance as a function of temperature using different metallic wires and a light bulb.



What is the profit margin of the power company?

A coal burning power plant

A coal fired steam turbine power plant has a mechanical power output of 500 MW and a thermal efficiency of 35.0%. a) At what rate must heat be supplied by burning coal? b) If the heat of combustion of coal is 2.5x10⁴ J/g, what mass of coal is burned per second? c) At what rate is heat discarded by the system?

Solar heating

A well-insulated house of moderate size in a temperate climate requires a maximum heat input rate of 20.0 kW. If this heat is to be supplied by a solar collector with an average (night and day) energy input of 200 W/m² and a collection efficiency of 60.0%, what area of solar collector is required?

Answer : 167 W/m^{2 (check)}

Week 4: Lesson 2: HOME. construction (mechanical structure for insulation)

- Show the construction needed for structure and to house insulation.

- Use a small wood beam and with a car jack to illustrate the loading of the floor. and new construction concepts for high strength to weight ratio.

Week 5: Lesson 1: Midterm

Week 5: Lesson 2: Review of exam and discuss class project

Week 6: Lesson 1: AUTO, forms of energy sources for cars.

Power for Automobiles

When thinking about energy, one of our main concerns is transportation. Gasoline is the main fuel for cars, but some of the sources mentioned above can also be used. Ethanol which is a renewable product from plant materials, agricultural wastes, and garbage burns clean. Gasahol is another idea which is better for the environment. It has 10% alcohol added to gasoline. Electricity is a form of energy that is used, but it costs more and does less than conventional fuel. A flywheel is used in electric vehicles to store kinetic energy. Input or output can be electrical, hydraulic, mechanical or a combination. These cars require little maintenance and have no bad effect on the environment. On the down side, electric cars are sluggish and the battery has to be recharged about every 50 miles. Solar energy is also a form of power for the solar car.

An alternative to different sources of energy for cars is to make them more fuel efficient. To get more miles per gallon, smaller lightweight cars would be beneficial. Shaving about 10% off of the weight of a car by making it smaller or replacing heavy materials with lighter substitutes reduces fuel consumption by about 7%. Lighter materials leads to a problem of safety. Safe, fuel efficient cars should be made of lightweight but strong materials. Less than 15% of the energy contained in a gallon of gas is used to propel the vehicle forward and the rest is lost as waste heat or to overcome engine or transmission friction or run the air conditioner. With the use of better design or lubricants, energy loss can be reduced.

Two requirements for the body of a car are high strength and corrosion resistance. The material with the best corrosion resistance is Zn coated steels. Two sides of the steel are hot dipped. This galvanized steel is more corrosion resistant than undipped steel. This material is hard to form, assemble, and finish so it is not used on doors and fenders. Zinc coated metal is used for exterior panels. Aluminum is used for hoods of cars because it has good corrosion resistance properties.

Because of the rising cost of petroleum, lowered sales, and export limitations it is hard to cut the cost of fuel by mass production. There is a weight reduction demand for automobile manufacturers to cut the weight 10-15kg/m² per project area. Simple weight reduction tends to adversely affect safety and increase noise and vibration. The use of plastic is also an idea for use on cars. Its lightweight and strength make it a good candidate. In the future it may even be used on the body and the wheels.

- Discuss the different forms of energy sources for cars and their limitations and requirements. Include gasoline, electric, solar, steam and even nuclear.

- Show the different forms of cars and their attributes and problems.

- Use energy chart.

- Use a battery (AA or D) to run a car to determine the energy density (in terms of both size and weight) and to show the limitation of electric car) battery. Show using know energy requirement to determine the size of the battery for cars. Cost.

Week 6: Lesson 2: AUTO, comparison of energy capacity.

- Review last lesson and continue.

- Use different solar cells and different sizes and a light bulb of know power to determine the efficiency of these cells. Calculation will enforce the idea of geometry and common sense thinking. Show the size of the panel needed for solar car and cost.

Week 7: Lesson 1: AUTO, main components of a car and its energy relationship

- Use a model to show the different components of a car and illustrate the different requirements and materials selection.

Week 7: Lesson 2: AUTO, compare the impact energy absorption of different materials

- Introduce the idea structural integrity of a car in an accident.

- How much energy is involved and where does the energy go?

- Use a rolling mill (in Wilmore Lab) to show the generation of heat during deformation.

- Use an impact tester (made of a drop ball or a pendulum) to compare the energy absorption capability of glass, metal and rubber. Measure energy absorption and their usage in various components of cars.

- Use a real windshield to illustrate safety glass versus normal glass and the concept of tempered glass.

Week 8: Lesson 1: AUTO, compare the corrosion and temperature response of different materials

- Discuss the concept of oxidation and the different response of different materials in terms of their atomic structures: metallic bond, covalent bond and ionic bond.

- Soak different materials (glass, iron and plastic) in jar of water under three different environments (normal, boiled and frozen) to illustrate the requirements for rust.

- Heat the various materials to show the temperature response.

- (See if we can show negative thermal expansion of rubber)

Week 8: Lesson 2: AUTO, recycling of automobile parts

- Discuss recycling and its role in the environment and the economy.

AUTO, glass and its usage (energy impact) and safety and tempered glass

Home - Thermal

Lesson One: Introduction of thermal conductivity and demonstration using wall insulation.

1. Using the inquisitive approach, develop the thermal conductivity equation. Discuss in detail the physics behind the equation.

2. Include both material intrinsic parameter (thermal conductivity) and extrinsic parameter (size and geometry).

3. Provide table of conductivity.

6. Show thermal insulation in series is additive.

7. Define R-value.

8. Homework and examples: Calculate using given parameters (and possibly parameters) determined in class), heat transfer during winter and summer in a typical home and energy cost. (need size, energy rate and type...). Determine if the wall is made of copper, how thick it must be to have the same insulation property.

Home - Thermal

Lesson Two: Thermal conductivity of windows and concept of surface emissivity (using different colors).

1. Review learning from last class.

2. Introduce the idea of radiation heat transfer using solar energy.

7. Homework and examples: Calculate the effects of different color shingles for an average home in terms of summer cooling and winter heating. Determine how much energy is needed for each window in a home.

Home - Thermal (Optional)

Lesson Three: Heat content of materials (heat capacity and energy content).

1. Review

2. Define heat content and equate heat and energy (equivalence of joules and calories, two types of calories).

3. Burn different types of fuel to measure their energy content (introduce the idea of energy density). These include hardwood, softwood, propane, kerosine. Determine how much is need to heat up a beaker of water to a certain temperature. Keep in mind the exhaust gas. Need a beaker of water, temperature measurement device and different sources of energy as stated above. Also need to scale to weigh materials

Home - Electrical and Energy

Lesson One: Introduction of electrical conductivity equation and demonstration of electrical heat.

1. Using the inquisitive approach, develop the electrical conductivity equation. Discuss in detail the physics behind the equation. Define electrical conductivity and resistivity.

2. Include both material intrinsic parameter (electrical conductivity) and extrinsic parameter (size and geometry).

3. Provide table of conductivity.

5. Introduce the idea of wire gage and use different gage wires to illustrate electrical heat and fire (also use a neck to show this effect). Need two copper construction wires with very different gages and high current power supply.

Home - Electrical and Energy

Lesson Two: Electrical power generation. Comparison of different forms of energy production. Introduce the idea of chemical and mechanical energy (battery, direct conversion and generator).

1. Review

2. Introduce the three concepts of electricity production: battery (chemical, direct conversion and generator). Dissect a chemical battery and illustrate the chemical inside it. Use thermoelectric modular to illustrate the idea of direct conversion and a bicycle generator to show how Alabama Power Company generates electricity. Need a battery, thermoelectric modular and a bicycle generator.

3. Lead to the development of the fundamental equations for electrical power and energy:

Power = Voltage x Current = Current² x Resistance = Voltage² / Resistance

Energy = Power x Time = Voltage x Current x Time

Show the current monitoring device from the Power Company

4. Show the relationship between power and energy (Power = Energy / time). Use a car battery, an energizer alkaline battery, a thermoelectric device and wall socket to illustrate the idea of power, energy and cost. Hook up a light (or a motor) to each of the source and see how long it last. Use a voltmeter and an amp meter to monitor the power and the energy consumption. From the data, determine the total amount of energy contained in each of the devices and the cost and the convenience. Need different power sources, ampere meter, voltmeter, light or motor and recording device.

5. Use a refrigerator and an fan to illustrate the power rating and introduce the idea of energy efficiency. Need a refrigerator and fan and measurement devices.

6. Homework and example: Use the basic cost of oil etc and the cost of electricity to see what the profit margin of the power company is. Determine how many ampere service does a normal home need.

Home - Structural

Lesson One: Introduce the different materials used for structural component of a house. Conduct three point bending test of wood.

1. Introduce lumber, shingles, brick, plywood, sheetrock, siding, nail, concrete and their functions and properties. Need these example materials.

2. Conduct three point bending test on a piece of wood (scale down) until failure. Use this to introduce the three point bending equation. Need weight or car jack, wood piece, clamps and measurement device.

3. Illustrate using the three point bending test on why we use different wood geometries and sizes for different part of the house.

4. Use a rectangular and a triangular geometry to illustrate the use of cross member to strengthen a structure. (make a square with and without cross member). Need lumber and nail and a hammer. Lead to the discussion of the roof structures.

5. Homework and example: Given the maximum weight of an occupant and the span of the structure, calculate the minimum wood size required for a floor joist.

Home - Structural

Lesson Two: Use different geometries to illustrate scaling law and different designs (such as I-beam construction) to illustrate strength to weight ratio.

1. Review.

2. Use the three point bending test to illustrate the concept of scaling law. Vary the span, and size of the beam to determine their effects on bending and on the three point bending equation. Show the scaling law from these data. Need the equipment for three point bending and different wood beams.

3. Use plotting extensively for above.

4. Introduce the idea of advanced design such as I-beam and pressed wood. (Purchase some I-beam from the lumber yard for illustration). Show the strength to weight ratio gain by such a design using wood beam with added top and bottom reinforcements (our own I-beam). Need basic three point bending test rig and I-beam. Can also use the aluminum I-beam in machine shop. Weight and compare strength to weight ratio.

Solution: Weight of product = 4.002603/6.02x10²³ gm = 6.64884x10^-24 gm

Weight of reactants = 4x1.0078252/6.02x10²³+2x9.1x10^-28=6.69833x10^-24 gm

Weight loss/reaction = 4.9493x10^-26gm = 4.9493x10^-29 kg

Energy release/reaction = 4.9493x10^-29 kg x (3x10⁸ m/s)² =4.45x10^-12 J

Number of hydrogen atoms in the sun = 2x10³³ gm x 6.02x10²³ H/mole

= 1.2x10⁵⁷

Number of reaction = 1.2x10⁵⁷/4 = 3x10⁵⁶

Total energy of the sun = 1.34x10⁴⁵ J

How much energy would 1 gram of natural uranium generate and what is that equal to in terms of oil, gas and coal?

Natural uranium contains 0.72% of ₉₂U²³⁵ (that can fiss...). Each ₉₂U²³⁵ atom will react with a neutron to form ₅₄Xe¹³⁶ and ₃₈Sr⁹⁷ plus three neutrons. The reaction is given as:

_{0n¹ + ₉₂U²³⁵ = ₅₄Xe¹³⁶ + ₃₈Sr⁹⁷ + 3₀n¹
The precise atomic weight of a neutron is 1.0086652, ₉₂U²³⁵ is 235.04394, ₅₄Xe¹³⁶ is 135.90722 and ₃₈Sr⁹⁷is 96.92939. (Remember atomic weight is defined as gram mass of the element per mole, or 6.02x10²³ atoms)
Total weight before fission for each atomic reaction is:}

(1.0086653+235.04394)/6.02x10²³ =3.921139x10^-22 gm

Total weight after fission for each atomic reaction is:

(3x1.0086653+135.90722+96.92939)/6.02x10²³ = 3.917983x10^-22 gm

Weight Loss = 3.155513x10^-25 gm

Energy release/reaction: E=mc² = 3.155513x10^-28 kg x (3x10⁸m/s)² = 2.84x10^-11 J

Number of U²³⁵ in 1 gm=6.02x10²³/235x0.0072 = 2.56x10²¹ x 0.0072 =1.84x10¹⁹ atoms

Total energy release by 1 gram of uranium = 5.23x10⁸J = 523 Megajoules = 523 MJ.

2. Determine the amount of energy that our sun will produce during its lifetime. The sun's mass is 2x10³⁰ kg. It is made of hydrogen and the energy it produces is by the reaction of combining these hydrogen atoms to form more complex atoms. For the simplicity of calculation, you may assume that the reaction to be combining four hydrogen atoms with two electrons to form a helium.

4₁H¹ + 2_-1e⁰ = ₂He⁴

The precise atomic weight of hydrogen is 1.0078252 and that of helium is 4.002603. The weight of an electron is 9.1x10^-28 g.

(b) Solar cell to convert light into electricity directly. The idea of converting light energy into electricity will also be demonstrated. In this case, an electrical light bulb (100W clear GE) will be used. Again this is a two step conversion process (really stupid idea if you think about it). The first step is to apply 120VAC to the light bulb and convert electrical energy into light energy and then the light energy into electrical energy using photovoltaic (commonly called solar cell). (Why is this stupid? But a good demonstration of stupidity nevertheless)

Voltage of light bulb =

Current of light bulb =

Electrical power consumption of light bulb =

Light meter reading = (1ft-candle=1lumen/ft², 1lumen=1/680 lightwatt)

Distance of bulb to light meter (also where the solar cell will be) =

Area of sphere at where the meter is = 4Br²=

Total light power from the light bulb =

Efficiency of light bulb for visible light production =

Total light power from the light bulb =

Area of solar cell =

Area of entire sphere =

Power absorbed by the solar cell =

Load resistor =

Load voltage =

Power to load produced by the solar cell =

Efficiency of solar cell =

Overall efficiency =

Materials	Emissivity Values (no unit, dimensionless)
Carbon	0.85-0.95
Aluminum	0.11
Brass (oxidized)	0.61
Brass (unoxidized)	0.03
Copper (oxidized)	0.60
Copper (unoxidized)	0.02
Black gloss paint	0.90
Gold (polished)	0.02
Fire brick	0.75