We also have 3 4 to be isentropic so t 3 t 4 p 3 p 4

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We also have 3 4 to be isentropic, so T 3 /T 4 = ( P 3 /P 4 ) ( k 1) /k . But P 2 = P 3 and P 1 = P 4 . So T 2 T 1 = T 3 T 4 , (10.83) T 4 T 1 = T 3 T 2 . (10.84) So η = 1 T 1 T 2 , (10.85) = 1 1 parenleftBig P 2 P 1 parenrightBig k - 1 k . (10.86) A plot of η versus the pressure ratio P 2 /P 1 for k = 7 / 5 is plotted in Fig. 10.14. As the pressure ratio, P 2 /P 1 rises, the thermal efficiency increases for the Brayton cycle. It still is much less that unity for P 2 /P 1 = 20. To approach unity, very high pressure ratios are needed; η = 0 . 9 requires P 2 /P 1 3200. Note in terms of temperature, the efficiency looks like that for a Carnot cycle, but it is not. The highest temperature in the Brayton cycle is T 3 , so the equivalent Carnot efficiency would be 1 T 1 /T 3 . Example 10.4 Consider a CPIG air standard Brayton cycle with fixed inlet conditions P 1 and T 1 . We also fix the maximum temperature as the metallurgical limit of the turbine blades, T max . Find the pressure ratio which maximizes the net work. Then find the pressure ratio which maximizes the thermal efficiency. We have T 2 = T 1 parenleftbigg P 2 P 1 parenrightbigg k 1 k , T 3 = T max , T 4 = T 3 parenleftbigg P 4 P 3 parenrightbigg k 1 k . (10.87) CC BY-NC-ND. 2011, J. M. Powers.
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10.2. BRAYTON 297 0 5 10 15 20 0.2 0.4 0.6 0.8 1.0 P 2 /P 1 η k = 7/5 Figure 10.14: Thermal efficiency versus pressure ratio for air standard Brayton cycle, k = 7 / 5. We also have P 4 = P 1 and P 2 = P 3 . So T 4 = T max parenleftbigg P 1 P 2 parenrightbigg k 1 k = T max parenleftbigg P 2 P 1 parenrightbigg 1 k k . (10.88) Let us let the modified pressure ratio θ be defined such that θ parenleftbigg P 2 P 1 parenrightbigg k 1 k . (10.89) Really θ is the temperature ratio, T 2 /T 1 . When the pressure ratio goes up, the temperature ratio goes up. Now the net work is w net = ( h 3 h 4 ) ( h 2 h 1 ) , (10.90) = c P ( T 3 T 4 T 2 + T 1 ) , (10.91) = c P ( T max T max θ 1 T 1 θ + T 1 ) , (10.92) = c P T 1 parenleftbigg T max T 1 T max T 1 θ 1 θ + 1 parenrightbigg . (10.93) To find the maximum w net we take dw net /dθ and set to zero: dw net = c P T 1 parenleftbigg T max T 1 θ 2 1 parenrightbigg , (10.94) 0 = c P T 1 parenleftbigg T max T 1 θ 2 1 parenrightbigg , (10.95) θ = ± radicalbigg T max T 1 . (10.96) CC BY-NC-ND. 2011, J. M. Powers.
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298 CHAPTER 10. CYCLES We take the positive root, since a negative pressure ratio does not make sense: θ = radicalbigg T max T 1 . (10.97) The second derivative tells us whether our critical point is a maximum or a minimum. d 2 w net 2 = 2 c P T max θ 3 . (10.98) When θ > 0, d 2 w net /dθ 2 < 0, so we have found a maximum of w net . The maximum value is w net = c P parenleftBigg T max T max parenleftbigg T max T 1 parenrightbigg 1 / 2 T 1 parenleftbigg T max T 1 parenrightbigg 1 / 2 + T 1 parenrightBigg , (10.99) = c P T 1 parenleftBigg T max T 1 T max T 1 parenleftbigg T max T 1 parenrightbigg 1 / 2 parenleftbigg T max T 1 parenrightbigg 1 / 2 parenrightBigg , (10.100) = c P T 1 parenleftBigg T max T 1 2 parenleftbigg T max T 1 parenrightbigg 1 / 2 parenrightBigg . (10.101) Note w net = 0 when θ = 1 and when θ = T max /T 1 .
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  • Spring '10
  • Powers
  • Thermodynamics, Heat engine, Carnot cycle, Gas turbine, Thermodynamic cycles, J. M. Powers

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