#$&* course PHY 202 2/12/15 at 12AM 009. `query 9
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Given Solution: The maximum possible efficiency is (T_h - T_c) / T_h, where T_h and T_c are the absolute max and min operating temperatures. T_h is (580 + 273)K = 853 K and T_c is (380 + 273) K = 653 K, so the maximum theoretical efficiency is max efficiency = (T_h - T_c) / T_h = (853 K - 653 K) / (853 K) = .23, approx. This means that the work done by this engine will be not greater than about 23% of the thermal energy that goes into it. Your Self-Critique: OK Your Self-Critique Rating: ********************************************* Question: Openstax: Prin only: A gas-cooled nuclear reactor operates between hot and cold reservoir temperatures of 700ºC and 27.0ºC . (a) What is the maximum efficiency of a heat engine operating between these temperatures? (b) Find the ratio of this efficiency to the Carnot efficiency of a standard nuclear reactor (found in Example 15.4). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The maximum possible Carnot efficiency at two temperatures is based on the absolute temperatures of the hot and cold reservoirs. For this problem we get e_max = (T_H - T_C) / T_H = (973 K - 310 K) / (973 K), or around 65%. Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: Openstax: A certain heat engine does 10.0 kJ of work and 8.50 kJ of heat transfer occurs to the environment in a cyclical process. (a) What was the heat transfer into this engine? (b) What was the engine’s efficiency? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: If output is 10,000J and 8,500J is simultaneously lost, then input is equivalent to their sum, or 18,500J. e = W/QH e = 10,000J / (10,000J + 8,500J) e = 0.54 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The energy to do the work and the energy transferred to the environment must be put into the engine. So this engine requires an input of 10 kJ + 8.5 kJ = 18.5 kJ of heat. It does 10 kJ of work, so its efficiency, which is the ratio of work done to energy input, is e = (work done) / (energy input) = 10 kJ / 18.5 kJ = .53, or 53%, roughly. Your Self-Critique: OK Your Self-Critique Rating: ********************************************* Question: Openstax: (a) What is the work output of a cyclical heat engine having a 22.0% efficiency and 6.00×10^9 J of heat transfer into the engine? (b) How much heat transfer occurs to the environment? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: e = W/QH 0.22 = W/(6.0*10^9 J) W = 1.32*10^9 J confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The work done is 22% of the energy input, or `dW = .22 * 6 * 10^9 J = 1.3 * 10^9 J. The rest of the energy input goes to the environment. Your Self-Critique: OK Your Self-Critique Rating: ********************************************* Question: Openstax: Assume that the turbines at a coal-powered power plant were upgraded, resulting in an improvement in efficiency of 3.32%. Assume that prior to the upgrade the power station had an efficiency of 36% and that the heat transfer into the engine in one day is still the same at 2.50×10^14 J . (a) How much more electrical energy is produced due to the upgrade? (b) How much less heat transfer occurs to the environment due to the upgrade? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: e = W/QH 0.36 = W/(2.5*10^14 J) W = 9*10^13 J 3.32% of 36 is 1.1952, so efficiency will increase from 36% to approximately 37.2% 0.372 = W/(2.5*10^14J) W = 9.3*10^13 J Thus, the increase in efficiency leads to an increase in energy output of 3.0*10^12 J. Had the producer wanted the same energy output with the newer efficiency to have been produced at the old efficiency, 0.36 = (9.3*10^13 J)/QH QH = 2.58*10^14 J output Waste = Output - Input = 2.58*10^14J - 9.3*10^13J = 1.653*10^14 J released into environment Thus, the increased efficiency has saved approximately 1.7*10^14 J of energy. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Previously the energy produced was .36 * 2.5 * 10^14 J = 9 * 10^13 J. After the improvement the efficiency is about 39.3% so the energy produces is .393 * 2.5 * 10^14 J = 9.8 * 10^13 J, roughly. The difference is about 8 * 10^12 J. This heat now goes into the electrical energy produced by the plant, and the heat transferred to the environment decreases by the same amount. Your Self-Critique: I misunderstood the original problem, thinking that the efficiency was improved by 3.32% of that efficiency, not that it was a simple summation. This makes sense. The rest of my calculations would have been correct had I had the correct newer efficiency. Was my re-calculation of the energy saved had it been produced under the old efficiency unnecessary? 0.3932 = W/(2.5*10^14J) W = 9.83*10^13 J Thus, the increase in efficiency leads to an increase in energy output of 8.3*10^12 J. Your Self-Critique Rating:2 ********************************************* Question: Openstax: A coal-fired electrical power station has an efficiency of 38%. The temperature of the steam leaving the boiler is 550ºC . What percentage of the maximum efficiency does this station obtain? (Assume the temperature of the environment is 20ºC .) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: e = 1 - QL/QH 0.38 = 1 - (550+273K)/QH 0.62 = 823/QH QH = 1327.42 K = 1054.42C I’m not sure I understand the actual question the problem is posing and where to go from here. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The maximum efficiency of an engine running between 550 C and 20 C is e_max = (T_h - T_c) / (T_h) = (823 K - 293 K) / (823 K) = 63%, roughly. Operating at 38%, the station is at 38% / (63%) = 60%, roughly, of the maximum possible efficiency. Your Self-Critique: Hmm, I would have never thought that this is what the problem was asking. I knew the 20C would come into play somewhere, but I had no idea that the point was to find the efficiency of an engine running between 550C and 20C. Your explanation makes sense though. Your Self-Critique Rating:3 ********************************************* Question: query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 0.28 = 1 - QL/(550+273K) 0.72 = QL/823K QL = 592.56 K 0.35 = 1 - 592.56/QH QH = 911.6K = 638.6C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Carnot efficiency is eff = (Th - Tc) / Th. Solving this for Tc we multiply both sides by Th to get eff * Th = Th - Tc so that Tc = Th - eff * Th = Th ( 1 - eff). We note that all temperatures must be absolute so we need to work with the Kelvin scale (adding 273 C to the Celsius temperature to get the Kelvin temperature) If Th = 550 C = 823 K and efficiency is 30% then we have Tc =823 K * ( 1 - .28) = 592 K. Now we want Carnot efficiency to be 35% for this Tc. We solve eff = (Th - Tc) / Th for Th: Tc we multiply both sides by Th to get eff * Th = Th - Tc so that eff * Th - Th = -Tc and Tc = Th - eff * Th or Tc = Th ( 1 - eff) and Th = Tc / (1 - eff). If Tc = 576 K and eff = .35 we get Th = 592 K / ( 1 - .35 ) = 592 C / .6 = 912 K, approx. This is (912 - 273) C = 639 C. ** Your Self-Critique: OK Your Self-Critique Rating: ********************************************* Question: univ phy problem 20.45 11th edition 20.44 (18.40 10th edition) ocean thermal energy conversion 6 C to 27 C At 210 kW, what is the rate of extraction of thermal energy from the warm water and the rate of absorption by the cold water? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 = 3,000 kW. The cold water absorbs what's left after the 210 kW go into work, or 2,790 kW. Each liter supplies 4186 J for every degree, or about 17 kJ for the 4 degree net temp change of the water entering and exiting the system. Needing 3,000 kJ/sec this requires about 180 liters / sec, or about 600 000 liters / hour (also expressible as about 600 cubic meters per hour). Comment from student: To be honest, I was surprised 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. ** Your Self-Critique: Your Self-Critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ________________________________________ #$&*