course PHY 202 ??e????????????assignment #018
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21:25:17 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 --> First, when dealing with efficiency, I must always convert Celsius temps to Kelvin. 380 C = 673 K, 580 C= 607 K. Then, I must use the formula, 1-TL/TH, so 1-607/673=1-.90=.10 .10= efficiency of the system./
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21:34:42 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 --> oops, I calculated the right Kelvin temps in my notes, but I think I typed them into my solution wrong,(it's getting late and all of the numbers are starting to look the same :) so I see my simple error there. Also, the formula I used to get the max possible efficiency was right, I just transposed the high and low temp.
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21:57:42 query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%?
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RESPONSE --> E=1-(TL/TH) where e=efficiency, TL=Lower operating temp, and TH=higher operating temp. So, is the temp is .28 and source is 550 C; Take 550 C and add 273 to convert it 823 Kelvins. Thus: .28=1-(TL/823K), solving for TL gives us 597 Kelvins. Using 597 as TL in the same formula with .35 as the efficiency gives us 918.K.
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21:58:09 ** 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 --> remember to convert answer back into Celsius.
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????????k??K?? assignment #018 018. `query 8 Physics II 07-11-2007