Asst 8

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16:42:10 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 --> FIND: Maximum efficiency of a heat engine operating between temps of 380 to 580 C Difference between temps 200 C. 380 c + 273 = 653 K 580 c + 273 K= 853 K max eff = (853 - 653 ) /853 K = .234

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16:42:34 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 --> Okay once I reached that point I just use it to express the percentage of thermal energy that goes into the system. Okay.

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16:47:02 query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%?

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RESPONSE --> Same equation (T_h - T_c) / T_h We know 550 c = 823 K 823 K * (1-.28) = 592 (T_c) We already know efficiency T_h = 592 K / (1-.35) T_h=910 K

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16:47:22 ** 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 --> Take 910 K back down to C = 637.7 C

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16:47:26 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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16:47:29 ** 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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Very good. Let me know if you have questions.