Hi, I\'m studying for my exam that\'s over four days, and so far allthe exercice

Question

Hi, I'm studying for my exam that's over four days, and so far allthe exercices have gone pretty good. Until now. Problem 4.24 (inthe SI edition) is so bizzar that I just don't know what to do. Theonly thing I can think about it is: huh? And that'spretty sad. Given is that air expands through a turbine from 900K to 500K.The pressure changes from 10 to 1 bar. The inlet velocity is smallcompared to the exit velocity of 100 m/s. The turbine operates atsteady state and develops a power output of 3200 kW. Find the massflow rate of the air (in kg/s) and the exit area (in m^3) if thepotential energy effects and heat trasfer between the turbine andits surroundings are negligible. I hope you can help me. Thanks in advance, Saartje Hi, I'm studying for my exam that's over four days, and so far allthe exercices have gone pretty good. Until now. Problem 4.24 (inthe SI edition) is so bizzar that I just don't know what to do. Theonly thing I can think about it is: huh? And that'spretty sad. Given is that air expands through a turbine from 900K to 500K.The pressure changes from 10 to 1 bar. The inlet velocity is smallcompared to the exit velocity of 100 m/s. The turbine operates atsteady state and develops a power output of 3200 kW. Find the massflow rate of the air (in kg/s) and the exit area (in m^3) if thepotential energy effects and heat trasfer between the turbine andits surroundings are negligible. I hope you can help me. Thanks in advance, Saartje

Explanation / Answer

For air inlet: P1 = 10bar T1 = 900K V1<<V2 At outlet: Wcv = 3200kW P2 = 1bar T2 = 500K V2 = 100m/s Begin with a steady state energy balance: 0 = Qcv - Wcv + m [ (h1 -h2) + (V12 -V22)/2 + g(z1 -z2)] Here, Qcv = V12 = z = 0so the energy balance reduces and we can solve for m sincem1 = m2 = m Rearranging the energy balance you get: m = Wcv / [(h1 - h2) -V22/2] From Table A-22; h1 = 932.93kJ/kg, h2 =503.03kJ/kg Using the formula, m = 3200kW*1kJ/s/kW / [(932.93-503.03)kJ/kg- (1002m2/s2/2)*1N/kgm/s2*1kJ/103Nm                                 =7.53 kg/s The exit area is: A2 = v2m/V2 = RT2m/ P2V2      =    [(8.314/28.97)kJ/kg K * 500K * 7.53kg/s / 1 bar * 100m/s] * 1bar/105N/m2 * 103Nm/1kJ      = 0.108 m2 Hope that helps ;-) Don't forget to rate! Hope that helps ;-) Don't forget to rate!

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