course Phy 122 Qܴӎx~Зassignment #018
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23:19:32 prin phy and gen phy problem 15.19 What is the maximum efficiency of a heat engine operating between temperatures of 380 C and 580 C?
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RESPONSE --> According to the Summary of Topics for Introductory Problem Set 5, the maximum efficiency of a heat engine can be defined as (TH - TC)/TH, where TH is the maximum operating temperature and TC is the minimum operating temperature. This relationship is based on the efficiency formula: (Qin - Qout) / Qin. The temperatures must first be in units of Kelvin. TH = 580 C + 273 = 853 K TC = 380 C + 273 = 653 K Max Efficiency = (853K - 653K) / 853K = 0.23 = 23%
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23:20:00 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.
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RESPONSE --> I understand this concept and I feel that I answered the problem adequately.
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23:20:11 query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%?
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RESPONSE --> This problem was not assigned to me.
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23:21:13 ** 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 --> I have read the solution and understand the process. This problem was almost identical to problem #20 in Chapter 15.
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23:21:24 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 --> I am not a university physics student
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23:22:22 ** 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 --> I have read the solution for this problem.
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