Now consider an ideal turbine t 4 s t 3 parenleftbigg

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Now consider an ideal turbine: T 4 s T 3 = parenleftbigg P 4 P 3 parenrightbigg k 1 k , (10.112) T 4 s = T 3 parenleftbigg P 4 P 3 parenrightbigg k 1 k , (10.113) = (1100 K ) parenleftbigg 35 kPa 140 kPa parenrightbigg 5 / 3 1 5 / 3 , (10.114) = 631 . 7 K. (10.115) CC BY-NC-ND. 2011, J. M. Powers.
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10.2. BRAYTON 301 But for the real turbine, η t = w w s , (10.116) w = η t w s , (10.117) = η t ( h 3 h 4 s ) , (10.118) = η t c P ( T 3 T 4 s ) , (10.119) = (0 . 8) parenleftbigg 0 . 5203 kJ kg K parenrightbigg ((1100 K ) (631 . 7 K )) = 194 . 9 kJ kg . (10.120) Thus, since w = h 3 h 4 = c P ( T 3 T 4 ), we get T 4 = T 3 w c P = (1100 K ) 194 . 9 kJ kg 0 . 5203 kJ kg K = 725 . 4 K. (10.121) Note that T 4 is higher than would be for an isentropic process. This indicates that we did not get all the possible work out of the turbine. Now for the cooler, q L = h 4 h 1 = c P ( T 4 T 1 ) = parenleftbigg 0 . 5203 kJ kg K parenrightbigg ((725 . 4 K ) (280 K )) = 231 . 7 kJ kg . (10.122) We are now in a position to calculate the thermal efficiency for the cycle. η = w net q H , (10.123) = w turbine w compressor q H , (10.124) = c P (( T 3 T 4 ) ( T 2 T 1 )) c P ( T 3 T 2 ) , (10.125) = ( T 3 T 4 ) ( T 2 T 1 ) T 3 T 2 , (10.126) = ((1100 K ) (725 . 4 K )) ((539 . 5 K ) (280 K )) (1100 K ) (539 . 6 K ) , (10.127) = 0 . 205 . (10.128) If we had been able to employ a Carnot cycle operating between the same temperature bounds, we would have found the Carnot efficiency to be η Carnot = 1 T 1 T 3 = 1 280 K 1100 K = 0 . 745 > 0 . 205 . (10.129) A plot of the T s diagram for this Brayton cycle is shown in Fig. 10.17. Note that from 1 to 2 (as well as 3 to 4) there is area under the curve in the T s diagram. But the process is adiabatic! Recall that isentropic processes are both adiabatic and reversible. The 1-2 process is an example of a process that is adiabatic but irreversible. So the entropy change is not due to heat addition effects but instead is due to other effects. CC BY-NC-ND. 2011, J. M. Powers.
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302 CHAPTER 10. CYCLES s T P = 3 5 k P a P = 1 4 0 k P a 1 2 3 4 2s 4s Figure 10.17: T s diagram of Brayton power cycle for spacecraft with turbine and com- pressor inefficiencies. Example 10.6 We are given a turbojet flying with a flight speed of 300 m/s . The compression ratio of the compressor is 7. The ambient air is at T a = 300 K , P a = 100 kPa . The turbine inlet temperature is 1500 K . The mass flow rate is ˙ m = 10 kg/s . All of the turbine work is used to drive the compressor. Find the exit velocity and the thrust force generated. Assume an air standard with a CPIG; k = 1 . 4 , c P = 1 . 0045 kJ/kg/K . A plot of the T s diagram for this Brayton cycle is shown in Fig. 10.18. We first calculate the ram compression effect: h 1 + 1 2 v 2 1 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright 0 = h a + 1 2 v 2 a . (10.130) We typically neglect the kinetic energy of the flow once it has been brought to near rest within the engine. So we get h 1 h a = 1 2 v 2 a , (10.131) c P ( T 1 T a ) = 1 2 v 2 a , (10.132) T 1 = T a + v 2 a 2 c P , (10.133) = (300 K ) + ( 300 m s ) 2 2 parenleftBig 1 . 0045 kJ kg K parenrightBig kJ 1000 m 2 s 2 , (10.134) = 344 . 8 K. (10.135) CC BY-NC-ND. 2011, J. M. Powers.
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10.2. BRAYTON 303 T s v 2 /2 q in w t w c a v 2 /2 1 2 3 4 5 Figure 10.18: T s diagram of Brayton cycle in a turbojet engine.
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  • Spring '10
  • Powers
  • Thermodynamics, Heat engine, Carnot cycle, Gas turbine, Thermodynamic cycles, J. M. Powers

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