Question

Steam enters an adiabatic turbine at 20 MPa and 700° C. Steam leaves the turbine at a temperature of 160° C with a quality of 88%. Determine the mass flow rate of steam necessary to produce 18 MW power output. Neglect the effect of elevation and velocity changes (20) 5. An air standard Otto cycle compresses air from 14.7 psia and 80°F at the beginning of the compression process. It adds 155 Btu/(Ibun cycle) of heat during the combustion process raising the maximum temperature to 1600°F. Draw the P-v diagram and determine: (25) a. Temperature and pressure at the end of compression process b. Work done per (Ibe cycle)during compression process c. Work done per (Ib cycle during expansion process d. Compression ratio e. Overall efficiency of the cycle.

Do 4 and 5: Show full workings and unit conversions

Answer 1

Dear learner as per the HOMEWORKLIB POLICY, we are told to do only 1st question in case of multiple questions but, I am doing some extra just for you! Good luck.

solution : steam Turbine V Q. 4 given that P = 20MPa ; T = 700°c Pout = 18 Mhe T2 = 1609; X, = 88% or 0.88 Neglect the effect of elevation and velocity changes. - Cie. Pel & KE changes) Now from steam table of superheated rapour - at p, and I we w have hi = 3807.8 kJ/kg also from Steam table of Satweated steam at - T2 = 160°C , we have ht=675.53 kalng , hyqe 2082.6 kulkg het in be the mass flow rate of steam necessary to produce power i p = m (hi ha) , here - ha = hf + R2 hfgz 18816f="" (1907. 8-2506.218)X103 = 675.53+ 0.68 (2062-6) = * = 13.95 kg/sec fue a hu = 2506.218 kolky

Q. 5 solution & Unit conversions : 14,7 psiq = po 1,013 bar = 1,013X105 Pa 1 Btu/lom = 2326 T/kg ; • 10 = 273 % 17-32 - °C -> (1°F–3) X5 - _ v Given data þ. = 14.7 psiq = 1,013 bax = 1,013X105 Pa Ti = 80°F = 26167€ Þ = 299.67K Qadd = 155 Bitullom Cycle = 360.53 kuleq ayole Tz = 1600°F = 87.11€ = 1144,11 K we knowo air properties at atm pressure ad 6 = 1.005 kJ/kyki Grz 0.748 kilkgk ; V=14 - R = 0.287 kJ / kg k from ideal gas equation Þiv= MRT, → 1,013 X10*XV, = 1X0.287X109x299,67 | V = V4 = 0.849 m3

9 Tempis pressure at the end of compression - Qadd = Cu(T3-T2) => 360.53X103= 0.718810*x (1144. –T2) =T2 = 641.979K1 AM For sale isengropic compression - (2) You Do (641.929) >b2 = 14.57 bar = 14.57x105Pal Ary B) compression cook per (ebm cycle) Ę Wlcomp = - bv, -bvs -MR (Ts –Tu) e-1X0.287810°¥(299.67–641979) Wecamp = 24560k5|1g cycle I m ry 24-1 © Expansion wook :- . P 3 V = Pq vur or or The at point 2 P2 V2 = MRT2 14.57X105X V2 = 1X0.287X1094041979 TV2 = V2 = 0,1264 m3 PLLC 34 1140.17 = ( 0.1264104-1) = 10.849)

=> 1Ty = 534,076k kle kelement z MR (T3-Ty) 1x0,287X10% (1144.1/-534.07) 104-1 Exp I hear = 437.699 kJ/kg you (e) overall eff: - klepp - Klcomp Q add = (437.699-245.60)x103 360.53X103 = 0-5328 B2 12 = 53.2011 con compression Ratio - rc = V2 = 0.1264 = 0.1988 Am 0.849