course PHY 202
.................................................??s?Z???????]??assignment #008
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11:10:55 Query video experiment 9 give the heights which water was raised, the potential energy increase of the water for each height, the thermal energy required to heat the air in the bottle at each height, and the efficiency of the system at each height
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RESPONSE -->
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11:11:06 ** Data as submitted by one student: Bottle Engine Data At heights of 20, 53, 80, 100 and 120 cm the amounts of water raised were about 165 ml, 140 ml, 100 ml, 50 ml and 30 ml, indicating equal amounts of gas expansion. The precise values give the following results: Column headings are: 1. Height of beaker (cm) 2. Work done by raising water to beaker (J) 3. Pressure in bottle (Pa) 4. Temperature in bottle (K) 5. Thermal energy transfer (J) 6. Efficiency (%) 1 2 3 4 5 6 20 .35316 102962 285.4 7.79 4.53 53 .701906 106199 294.4 20.76 3.38 80 .82404 108848 301.8 31.43 2.62 100 .4905 110810 307.2 39.22 1.25 120 .35316 112772 312.6 47 .75 Details of the analysis would include the following: Figuring the weight of water raised and multiplying it by the height to which it is raised gives the work done (2d column). For example at 80 cm height the 100 ml of water raised has a mass of of 100 grams or .1 kg, so the weight is .1 kg * 9.8 m/s^2 = .98 Newton. Raising .98 Newton a distance of .8 meters results in work .98 N * .8 m = .8 Joules, approx.. The more precise value is given in the table. The bottle contains about .1 mole of air and specific heat capacity at const volume is about 5/2 R = 5/2 * 8.31 J/(mol Kelvin) = 20 J / (mol Kelvin), approx. Starting at 280 K and increasing, for example, to 301.8 K therefore requires about (20 J / (mol Kelvin) ) * .1 mol * 21.8 Kelvin = 43 Joules. You should modify this for the correct initial temperature and the correct number of moles. Then expanding the gas requires energy P * `dV; at the 80 cm height the gas expands by 140 ml = .14 liters and pressure is about 109 kPa, which indicates a thermal energy transfer of P `dv = .14 * 10^-3 m^3 * 109,000 N/m^2 = 15 Joules, approx. The total thermal energy transfer would therefore be about 43 J + 15 J = 58 J. Using a more precise value for the number of moles in the volume of air in the container might lead to the results in the table for thermal energy transfer (column 5). You should check this out for yourself and see if you agree with the values obtained here. Efficiency is work done / thermal energy added. Corrected results should probably show that efficiency is greater at 53 cm height than at 20 cm height. **
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RESPONSE --> i think I missed something???
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11:17:45 query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%?
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RESPONSE --> 550C + 273C = 823K Tc = 826K * (1=0.3) = 578K eff = (Th-Tc)/Th Th= Tc / (1-eff) Th = 578K / ( 1-0.35) = 889K
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11:18:10 ** Carnot efficiency is eff = (Th - Tc) / Th. Solving this for Tc we multiply both sides by Th to get eff * Th = Th - Tc so that Tc = Th - eff * Th = Th ( 1 - eff). We note that all temperatures must be absolute so we need to work with the Kelvin scale (adding 273 C to the Celsius temperature to get the Kelvin temperature) If Th = 550 C = 823 K and efficiency is 30% then we have Tc =823 K * ( 1 - .28) = 592 K. Now we want Carnot efficiency to be 35% for this Tc. We solve eff = (Th - Tc) / Th for Th: Tc we multiply both sides by Th to get eff * Th = Th - Tc so that eff * Th - Th = -Tc and Tc = Th - eff * Th or Tc = Th ( 1 - eff) and Th = Tc / (1 - eff). If Tc = 576 K and eff = .35 we get Th = 592 K / ( 1 - .35 ) = 592 C / .6 = 912 K, approx. This is (912 - 273) C = 639 C. **
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RESPONSE --> right and then you would subtract 273C to get the degrees back to celcius
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11:18:16 univ phy problem 20.44 (18.40 10th edition) ocean thermal energy conversion 6 C to 27 C At 210 kW, what is the rate of extraction of thermal energy from the warm water and the rate of absorption by the cold water?
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RESPONSE --> ok
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11:18:18 ** work done / thermal energy required = .07 so thermal energy required = work done / .07. Translating directly to power, thermal energy must be extracted at rate 210 kW / .07 = 30,000 kW. The cold water absorbs what's left after the 210 kW go into work, or 29,790 kW. Each liter supplies 4186 J for every degree, or about 80 kJ for the 19 deg net temp change. Needing 30,000 kJ/sec this requires about 400 liters / sec, or well over a million liters / hour. Comment from student: To be honest, I was suprised the efficiency was so low. Efficiency is low but the energy is cheap and environmental impact in the deep ocean can be negligible so the process can be economical, if a bit ugly. **
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RESPONSE --> ok
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