#$&* course PHY 202 3/23 about 3:30 pm 009. `query 9
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Given Solution: The maximum possible efficiency is (T_h - T_c) / T_h, where T_h and T_c are the absolute max and min operating temperatures. T_h is (580 + 273)K = 853 K and T_c is (380 + 273) K = 653 K, so the maximum theoretical efficiency is max efficiency = (T_h - T_c) / T_h = (853 K - 653 K) / (853 K) = .23, approx. This means that the work done by this engine will be not greater than about 23% of the thermal energy that goes into it. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: Openstax: Prin only: A gas-cooled nuclear reactor operates between hot and cold reservoir temperatures of 700ºC and 27.0ºC . (a) What is the maximum efficiency of a heat engine operating between these temperatures? (b) Find the ratio of this efficiency to the Carnot efficiency of a standard nuclear reactor (found in Example 15.4). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Th = 700 +273 = 973 K Tc = 27+ 273 = 300 K Eff = (973 K - 300 K)/973 K = 0.692 or 69.2%. The nuclear power reactor from Example 15.4 has an efficiency of 47.6% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 engine’s efficiency? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Qin = Qout + W = 8.50 kJ + 10.0 kJ = 18.5 kJ Eff. = W/Qin = 10.0 kJ / 18.5 kJ = 0.541 or 54.1% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The energy to do the work and the energy transferred to the environment must be put into the engine. So this engine requires an input of 10 kJ + 8.5 kJ = 18.5 kJ of heat. It does 10 kJ of work, so its efficiency, which is the ratio of work done to energy input, is e = (work done) / (energy input) = 10 kJ / 18.5 kJ = .53, or 53%, roughly. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: Openstax: (a) What is the work output of a cyclical heat engine having a 22.0% efficiency and 6.00×10^9 J of heat transfer into the engine? (b) How much heat transfer occurs to the environment? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Eff. = W/Qin 0.22 = W/6.00*10^9 J W = 1.32*10^9 J Qout = Qin - W = 6.00*10^9 J - 1.32*10^9 J = 4.68*10^9 J confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: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: Eff1 = 0.36 Eff2 = 0.3932 W1 = eff. * Qin = 0.36 * 2.5*10^14 J = 9.0*10^13 J W2 = 0.3932 * 2.5*10^14 J = 9.83*10^13 J ‘dW1-2 = 8.3*10^12 J Qout1 = 2.5*10^14 J - W1 = 1.6*10^14 J Qout2 = 2.5*10^14 J - W2 = 1.5*10^14 J ‘dQout1-2 = 1.0*10^13 J confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Previously the energy produced was .36 * 2.5 * 10^14 J = 9 * 10^13 J. After the improvement the efficiency is about 39.3% so the energy produces is .393 * 2.5 * 10^14 J = 9.8 * 10^13 J, roughly. The difference is about 8 * 10^12 J. This heat now goes into the electrical energy produced by the plant, and the heat transferred to the environment decreases by the same amount. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: Openstax: A coal-fired electrical power station has an efficiency of 38%. The temperature of the steam leaving the boiler is 550ºC . What percentage of the maximum efficiency does this station obtain? (Assume the temperature of the environment is 20ºC .) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Th = 550 +273 = 823 K Tc = 20 + 273 = 293 K Eff = (823-293)/823 = 0.64 %eff. = 0.38 / 0.64 = 0.59 or 59% of max efficiency confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ********************************************* Question: query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Eff = ((550+273)-Tc)/(550+273) = 0.64 Tc = 592.56 K or approx. 593 K or 320 C Th = Tc / (1 - eff.) = 593 / (1 - 0.35) = 912.308 or approx. 912 K or 639 C 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:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!