Question

A turboprop engine consists of a diffuser, compressor, combustor, turbine, and nozzle. The turbine drives a propeller as well as the compressor. Air enters the diffuser with a volumetric flow rate of 63.7 m³/s at 40 kPa, 240 K, and a velocity of 180 m/s, and decelerates essentially to zero velocity. The compressor pressure ratio is 9 and the compressor has an isentropic efficiency of 85%. The turbine inlet temperature is 1240 K, and its isentropic efficiency is 85%. The turbine exit pressure is 50 kPa and the nozzle exit pressure is 40 kPa. Flow through the diffuser and nozzle is isentropic. Use an air-standard analysis and neglect kinetic energy except at the diffuser inlet and the nozzle exit.Determine the power delivered to the propeller, in MW.

Answer 1

W, = 9.2MW

The detailed solution is given below:

Giveno va = 63.7 mils | Ta=200k [v=o | Me = M = 0.85 Pa= 40kba / Va = 180m/s - 9 / T = 1240K Compressor Combustor o 1 Pq = 50kpa PS = Aokba Turbine 7 Nozzle #state ao Ta=24ok I ha= haatok = 240.02 Pra- Pr@240k = 0.6355 # State 1: hat yang a hot video » h = hatta Ev=o] - 240.02+ 1 КГ 14 L looom?/s2 = 256.22 kJ/kg using interpolation

0. At To =25ok → ho = 250.05 and Port = 0.7329 and TC= 260k → hc = 260.09 and Prc = 0.8405 → Pro = Pro + (re-Pro) x[h-ho] = 0.79902 - The-ho) :: La mia sa-214 #state 2: [ isentropic compression ] 9x 0.79902 - 7.1912 - using interpolation, At Th= 470k h6= 472.24 and Pob= 6.742 and Te= 480k he = 482.49 and Prc = 7.268 has = hot (he-ho) x Poss - Pob] = 480.9 wing . (Poc - Pob) leen Mc = the best has hat have the]

c. b = 2502 + [ato ] seos king # State 3: [P3=] - - To=1240k h ₂ = 1324.93 kilky I paz = 272.3 # Stalen: [isen tropic expansion ] o OB IXO 1 X 272.3 P31 (Peles - alu = (19) 2723 [=] using interpolation = 30.08 At Th=7ook and t = 710k ho= 713.27 and fb = 28.80 he = 724.04 and Prc = 30.38 has = hot (he-hb) x Pras- Pro] = 722 kJikg . (Poe-Pob) Nown n a = h3tha hz has > hy = hz-MChz-has)

= 1324.93 – (0.85) (1324-93–722) 6. h4 = 812.44 kilky # The power delivered to the propellers → Wp = WE Wc = in [[hz ha) -(hz-hi)] Now, m= (vlax Pa = (63.7)x(40) = 37kg 15 Rta (0-287)x (240) 0 => wp= (376gis) (1324-93 – 81244) – (520.65–266-2277 = 9178.22 K5 = 9126.22 KW .: wp= 9.2MW Los 1000 kW = mw]