#$&* course Phy 122 11:45 am June 19 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Given Solution: The maximum possible efficiency is (T_h - T_c) / T_h, where T_h and T_c are the absolute max and min operating temperatures. T_h is (580 + 273)K = 853 K and T_c is (380 + 273) K = 653 K, so the maximum theoretical efficiency is max efficiency = (T_h - T_c) / T_h = (853 K - 653 K) / (853 K) = .23, approx. This means that the work done by this engine will be not greater than about 23% of the thermal energy that goes into it. Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: Openstax: Prin only: A gas-cooled nuclear reactor operates between hot and cold reservoir temperatures of 700ēC and 27.0ēC . (a) What is the maximum efficiency of a heat engine operating between these temperatures? (b) Find the ratio of this efficiency to the Carnot efficiency of a standard nuclear reactor (found in Example 15.4). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 700 + 273.15 = 973.15 K 27 + 273.15 = 300.15 K (TH - TC) / TH = (973 K - 310 K) / (973 K) = 65% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok Your Self-Critique Rating: ok ********************************************* 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 engines efficiency? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: input of 10 kJ + 8.5 kJ = 18.5 kJ of heat 10 kJ of work e = (work done) / (energy input) = 10 kJ / 18.5 kJ = 54 % confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The energy to do the work and the energy transferred to the environment must be put into the engine. So this engine requires an input of 10 kJ + 8.5 kJ = 18.5 kJ of heat. It does 10 kJ of work, so its efficiency, which is the ratio of work done to energy input, is e = (work done) / (energy input) = 10 kJ / 18.5 kJ = .53, or 53%, roughly. Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: Openstax: (a) What is the work output of a cyclical heat engine having a 22.0% efficiency and 6.00×10^9 J of heat transfer into the engine? (b) How much heat transfer occurs to the environment? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: dW = .22 * 6 * 10 ^9 J = 1.3 * 10 ^9 J The rest of the energy input would go to the environment confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok ********************************************* 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: .36 * 2.5 * 10 ^14 J = 9 * 10 ^13 J .393 * 2.5 * 10 ^14 J = 9.8 * 10 ^13 J difference is about 8 * 10^12 J confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Previously the energy produced was .36 * 2.5 * 10^14 J = 9 * 10^13 J. After the improvement the efficiency is about 39.3% so the energy produces is .393 * 2.5 * 10^14 J = 9.8 * 10^13 J, roughly. The difference is about 8 * 10^12 J. This heat now goes into the electrical energy produced by the plant, and the heat transferred to the environment decreases by the same amount. Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: Openstax: A coal-fired electrical power station has an efficiency of 38%. The temperature of the steam leaving the boiler is 550ēC . What percentage of the maximum efficiency does this station obtain? (Assume the temperature of the environment is 20ēC .) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 550 + 273.15= 823.15 K 20 + 273.15 = 293.15 K (Th - Tc) / (Th) = (823.15 K - 293.15 K) / (823.15 K) = 64% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok Your Self-Critique Rating: ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!