hw 12 thermo.pdf

# hw 12 thermo.pdf - 10.93 A diesel engine has a compression...

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Unformatted text preview: 10.93 A diesel engine has a compression ratio of 20:] with an inlet of 95 kPa= 290 K. state 1= with volume 0.5 L. The maximum cycle temperature is 1300 K. Find the maximum pressure= the net speciﬁc work. the cut off ratio= see problem 10.92, and the thermal efﬁciency. Solution: Compression process (isentropic) from Eqs. 624-25 72 = 71671 7 9-2)“ = 290 x 20"—4 = 961 K P2 = 95 x (20) 1-4 = 6297.5 kPa; v2 = 171720 = RT17'(20 P1) = 0.043 305 m37kg 1672 = 62 - 1114 12...,(12 — T1) = 0.717 (961 - 290) = 431.1 1:ng Combustion at constant P which is the maximtun pressure P3 = P2 = 6298 kPa; V3 = 02 T3 7T2 = 0.043305 x 13007961 = 0.03205 m3.-'1;g Cut-off ratio: vs 1' v2 = T3 .-'"1'2 = 1800 961 = 1.8?3 2W3 = P (03 — 02) = 6293 >4 (0.03215 — 0.043305) = 241.5 131.03g 293:113'112+2W3=h3-h2=cpoCT3-T2} = 1.004(1800 - 961) = 842.4 kl-"kg Expansion process [isentropic) from Eq.6.24 T4 = T3( 93 W04 = 1300 (0.03205 0.3761) 0-4 = 693 K 3W4 = 63 - 64 747 Cmes - T4) = 0.717 (1300 — 693) = 790.1 k1..-kg Cycle net work and efﬁciency,r “7w: 21473 + 3w4+ 11792 = 241.5 + 790.1- 481.1: 550.5 kJJ'kg n = “PM qH = 550.57 342.4 = 0.653 10.99 A diesel engine has air before compression at 280 K, 85 kPa. The highest temperature is 2200 K and the highest pressure is 6 MP3. Find the volumetric compression ratio and the mean effective pressure using cold air properties at 300 K. Solution: Compression (192031] = (VII'Vﬂk = CRk CR = le'vg = (92291)“ = (6000135)”‘4 = 20.92 T2 = T1(Pg.r"91}k'm = 230 x (6000135) ”357 = 944.3 K Combustion. Highest temperature is after combustion. 1:12 = Cprg —T 2) = 1.004 kJIkg—K (2200 — 944.8} K = 1260.2 ka'kg Expansion (v4 = CR x V: and P2 = P3} _ . k—l _ . k—l _ ,. k—l T4 — T3 (V3194) — T3 [V3 -' (CR X Vﬂ] — T3 i Ts" (CR X T2)] = 2200 K x (2200.-"20.92 x 944.3) "'4 = 914.2 K qL = 114 — u1= C3411); - T1) = 0.117 kJ.-"kg-K (914.2 — 280) K = 454.7 kJ.-1cg v1 = RTl..-"P1 = 0.28? kJIkg-K x 230 [(\$85 kPa = {3.9454 m3.-'kg Displacement and mep ﬁom net work v1 - v2 = vl- Vlr'CR = v1[1 — (L’CRH = 0.9002 m31'kg Pmeff = “’netf'tvl - V2) = (914 - qu-T V1 - V2) = (1260.2 — 454.7}-'0.9002 = 894.8 kPa ...
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