Now when is the thermal efficiency maximum consider ?

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Now when is the thermal efficiency maximum? Consider η = 1 θ 1 , (10.102) = 1 θ 2 . (10.103) So we must have θ → ∞ in order to have η reach a maximum. But we are limited to θ T max /T 1 . So the efficiency at our highest allowable θ is η = 1 1 T max T 1 = 1 T 1 T max . (10.104) But at the value of peak efficiency, the net work is approaching zero! So while this is highly efficient, it is not highly useful! Lastly, what is the efficiency at the point where we maximize work? η = 1 radicalbigg T 1 T max . (10.105) A plot of scaled net work, w net /c P /T 1 versus modified pressure ratio, ( P 2 /P 1 ) ( k 1) /k is given for T max /T 1 = 10 in Fig. 10.15. For this case the θ which maximizes w net is θ = 10 = 3 . 162. At that value of θ , we find w net /c P /T 1 = 4 . 675 . Example 10.5 Consider the Brayton power cycle for a space craft sketched in Fig. 10.16. The working fluid is argon, which is well modeled as a CPIG over a wide range of T and P . We take the pressure in the heating process to be isobaric, P 2 = P 3 = 140 kPa , and the pressure in the cooling process to be CC BY-NC-ND. 2011, J. M. Powers.
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10.2. BRAYTON 299 0 2 4 6 8 10 0 1 2 3 4 5 w net /(c P T 1 ) θ = (P 2 /P 1 ) ((k-1)/k) 4.675 3.162 Figure 10.15: Scaled net work versus modified pressure ratio for Brayton cycle with T max /T 1 = 10. isobaric, P 4 = P 1 = 35 kPa . We are given that T 1 = 280 K , T 3 = 1100 K . The compressor and turbine both have component efficiencies of η t = η c = 0 . 8. We are to find the net work, the thermal efficiency, and a plot of the process on a T s diagram. For argon, we have R = 0 . 20813 kJ kg K , c P = 0 . 5203 kJ kg K , k = 5 3 1 . 667 . (10.106) Note that c P = kR/ ( k 1). Let us start at state 1. We first assume an isentropic compressor. We will quickly relax this to account for the compressor efficiency. But for an isentropic compressor, we have for the CPIG parenleftbigg P 2 P 1 parenrightbigg k 1 k = T 2 s T 1 . (10.107) Here T 2 s is the temperature that would be realized if the process were isentropic. We find T 2 s = T 1 parenleftbigg P 2 P 1 parenrightbigg k 1 k = (280 K ) parenleftbigg 140 kPa 35 kPa parenrightbigg 5 / 3 1 5 / 3 = 487 . 5 K. (10.108) Now η c = w s /w , so w = w s η c = h 2 s h 1 η c = c P ( T 2 s T 1 ) η c = parenleftBig 0 . 5203 kJ kg K parenrightBig (487 . 5 K 280 K ) 0 . 8 = 135 . 0 kJ kg . (10.109) Now w = h 2 h 1 = c P ( T 2 T 1 ), so T 2 = T 1 + w c P = (280 K ) + 135 kJ kg 0 . 5203 kJ kg K = 539 . 5 K. (10.110) Notice that T 2 > T 2 s . The inefficiency (like friction) is manifested in more work being required to achieve the final pressure than that which would have been required had the process been ideal. In the heater, we have q H = h 3 h 2 = c P ( T 3 T 2 ) = parenleftbigg 0 . 5203 kJ kg K parenrightbigg ((1100 K ) (539 . 5 K )) = 291 . 6 kJ kg . (10.111) CC BY-NC-ND. 2011, J. M. Powers.
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300 CHAPTER 10. CYCLES q H heat source compressor turbine heater cooler low temperature reservoir 1 2 3 4 q L P 1 = 35 kPa T 1 = 280 K P 4 = 35 kPa T 3 = 1100 K P 3 = 140 kPa P 2 = 140 kPa η c = 0.8 η t = 0.8 W Figure 10.16: Schematic of Brayton power cycle for spacecraft.
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  • Spring '10
  • Powers
  • Thermodynamics, Heat engine, Carnot cycle, Gas turbine, Thermodynamic cycles, J. M. Powers

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