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### Brayton

Course: MAE 4263, Fall 2008
School: Oklahoma State
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Word Count: 717

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Cycles: Brayton From the ...rst law, the maximum transfers for component SSSF control volumes are w1 q2 w3 q4 2 3 4 1 th = (h1 = (h3 = (h3 = (h1 wnet \$ q2 3 h2 ) = h2 ) = h4 ) = h4 ) = =1 negative positive positive negative T1 T2 Generally the maximum tolerable temperature T3 Tmax is given. For the simple reversible Brayton cycle of given pressure ratio rp \$ (p2 =p1 ) = (p3 =p4 ) [compare Example 9.4, pp. 417...

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Cycles: Brayton From the ...rst law, the maximum transfers for component SSSF control volumes are w1 q2 w3 q4 2 3 4 1 th = (h1 = (h3 = (h3 = (h1 wnet \$ q2 3 h2 ) = h2 ) = h4 ) = h4 ) = =1 negative positive positive negative T1 T2 Generally the maximum tolerable temperature T3 Tmax is given. For the simple reversible Brayton cycle of given pressure ratio rp \$ (p2 =p1 ) = (p3 =p4 ) [compare Example 9.4, pp. 417 in 5th Ed.] we can use Equation 6.43 [p. 256] and Table A-22, or the constant-k approximation Equation 6.45 [p. 257], to compute the states after adiabatic compression and expansion w1 q2s w3 q4s 2s 3 = (h1 = (h3 = (h3 = (h1 \$ h2s ) = cp T1 1 h2s ) = cp (T3 h4s ) = cp T3 h4s ) = cp (T1 1 1 1 (k rp 1)=k T2s ) 1 rp T4s ) ! (k 1)=k 4s ! 1 th wnet = q2 3 (k 1)=k rp The closed-form approximation shows that the e ciency creasing pressure ratio rp [Figure 9.11 on p. 419] th increases with in- where we take k = (7=5) = 1:4. Increasing pressure ratio is not entirely practical, because the work output per pound of air throughput becomes very small at high values of rp [Figure 9.12 on p. 420] wnet = (h1 h2 ) + (h3 h4 ) = cp [T1 T2 + T3 T4 ] h = cp T1 1 p(2=7) + T3 1 r = cp T1 1 p(2=7) + r T3 T1 pr (2=7) 1 pr (2=7) i 1 wnet approaches zero when T2 reaches T3 at rp = (T3 =T1 ) . To maximize the speci...c work, we need an intermediate pressure ratio [pp. 420f] which depends on the temperature ratio (T3 =T1 ) rp = T3 T1 (7=4) (7=2) For example, if the environmental temperature is 80 o F (540 o R) and the maximum temperature is 1700 o F (2160 o R), then (T3 =T1 ) = 4, and the speci...c work (in BTU/lbm of air throughput) goes to zero at rp = 1 and at rp = 43:5 = 128, and in between has a maximum at rp = 41:75 = 11:3 where the e ciency is th = 50%, compared to the Carnot e ciency of 75% which theoretically could be approached at that temperature ratio. This is an isentropic calculation to explain the trade-o between speci...c work (which aects turbine size and cost) on the one hand, and e ciency (which aects fuel consumption and expense) on the other hand. When we consider irreversibilities, there is another, important more factor limiting high pressure ratios: compressor and turbine e ciency. The back-work ratio of the reversible Brayton cycle is 1 w1 2s (h1 h2s ) T 1 pr = = (2=7) w3 4s (h3 h4s ) T 3 1 pr which we can plot for our example case of (T3 =T1 ) = 4 (2=7) With turbine and compressor losses this changes to w1 2a (h1 h2 ) = = w3 4a (h3 h4 ) com p turb = 1 turb com p 1 T 1 pr T3 1 pr (2=7) (2=7) 2 which we can plot for example for turb = 0:70 and com p = 0:80 which exceeds unity at pressure ratios of 17 and beyond-- in other words, the net work becomes negative! When we take losses into account [Example 9.6 on pp. 422] w1 2a = 1 com p w1 2s h2a q2a 3 w3 4a h4a q4a 1 = h1 w1 2a = h3 h2a = turb w3 4s = h3 w3 4a = h1 h4a the thermal e ciency has a fairly wide peak at a modest pressure ratio-- for our example, in the range of rp 6 to 10, with a thermal e ciency approaching 30%. The big improvement to the Brayton cycle comes with regenerative heating, which is only possible if T4 > T2 ; reverting to a reversible calculation, this (7=4) requires that rp < (T3 =T1 ) and w1 2 q2 2:5 q2:5 3 w3 4 q4 4:5 q4:5 1 th = (h1 = q4 = (h3 = (h3 = q2 = (h1 \$ h2 ) = cp (T1 4:5 T2 ) T4 ) T4 ) T2 ) r(2=7) T3 =T1 h4 ) = cp (T3 h4 ) = cp (T3 2:5 h2 ) = cp (...

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