Question

Question 15 2 points Save Answer Ar enters the turbocharger compressor (se figure) of an automotive engine at 100 kPa, 27°C (ambient) and exits at 200 kPa. The air is cooled by 50°C in an intercooler before entoring the engine. The intropic efficiency of the compressor is 75% Enter the specific entropy in state 2 (compressor ext) in klikgK Cooler -W Compressor Exhaust Engine 2-3 50 C,

Answer 1

solo Anlet temprature of aur at (1) T = 27°C = 300k Inlet Pressure of aire at (1) Piz 100 kPa los Pa Exit pressure of are at P2 = 200 Kla P 2x 105 Pa Asentropic Efficiency of Comprecor, M = 75 % = 0.75 Now, we Know that for 18entropic Compression of aree in Compressor Jerman Ta Ti P2 (7-1/6) PL for air r=1.4 Ta 1(0-12lr 300 260 2x105 Ta los 300 (220.2857 365.70 K we know that Asentropie feiciency of Compressor, no I sentropic work Actual work

Date.... nc = T2 - To TraTi 0.75 365.7 - 300 & Tza - 300 Tra 387.6 K so Actual temprature of air after compressim State is at 387.6k Now, Know that Specific Entropy of are at State (1) To = 3ook and f S, = 6.692 kJ / kg. K = los pa we Know that specific Change in Entropy of A S=S2-5,= (pln/ Tz Cphn ( T ) - R .In ( 2 ) S2-S, 1.005 lu /387.6 0.287 lu 2x105 300 los S2 6.692 = S2 = so specific 0.0585 ton 6.7505 kJ/kg ok Entropy at State 2 is 6.7505 kJ Kg K