# (30 points) Consider the turbocharger of an internal combustion engine. The exhaust gases enter the turbine at 450C at a rate of 0.02 kg/s and leave at 400C. Air enters the compressor at 708C and 95 kPa at a rate of 0.018 kg/s and

(30 points) Consider the turbocharger of an internal combustion engine. The exhaust gases enter the turbine at 450C at a rate of 0.02 kg/s and leave at 400C. Air enters the compressor at 708C and 95 kPa at a rate of 0.018 kg/s and leaves at 135 kPa. The mechanical efficiency between the turbine and the compressor is 95 percent (5 percent of turbine work is lost during its transmission to the compressor). Using air properties for the exhaust gases, determine (a) the air temperature at the compressor exit and (b) the isentropic efficiency of the compressor.

by (100 points)
(a) Tâ‚‚ = 126Â°C , Tâ‚‚s = 106Â°C(b) 64%Explanation:we will begin with a step by step process;using the expression WÑ‚ = m(ex)Cp(Tâ‚ - Tâ‚‚)where WÑ‚ = turbine power outputm(ex) = mass of the exhaust gasCp = specific heat at constant pressure Tâ‚ = exhaust gas temperature at the turbine inletTâ‚‚ = exhaust gas temperature at the turbine exitusing from the gas table, the specific heat of exhaust gas at avg temperature of 425Â°C as 1.075kJ/kg.Kmaking substitutions we have;Tâ‚ = 450Â°C, Tâ‚‚ = 400Â°C, m(ex) = 0.02kg/s, Cp = 1.075kJ/kg.KSubstituting values into formula gives;WÑ‚ = m(ex)Cp(Tâ‚ - Tâ‚‚) WÑ‚ = [(0.02kg/s)(1.075kJ/kg.K)(450+273)K-(400+273)K]WÑ‚ = 1.075 kWThe relationship of mechanical efficiency between turbine and compressor is given as;Ð›Ð¼ = Wc/WÑ‚ (1)where WÑ‚ = turbine output Â  Â  Â  Â  Â  Â Wc = compressor output Â  Â  Â  Â  Â  Â  Ð›Ð¼ = mechanical efficiency Wc = Ð›Ð¼ Ã— WÑ‚ Wc = 0.95 Ã— 1.075(kW)Wc = 1.021 kWcompressor power:Wc = m(air)Cp(Tâ‚‚ - Tâ‚)where m(air) = mass of airTâ‚ = air temp at the compressor inlet Tâ‚‚ = air temp at the compressor exit Cp = specific heat at constant pressure Tâ‚‚ = Tâ‚ + Wc/m(air)Cp (2)the specific heat of air at the expected average temperature of 100Â°C is 1.011 kJ/kg.Ksubstituting Tâ‚ = 70Â°C, Wc = 1.021 kW, m(air) = 0.018 kg/sm Â Cp = 1.011 kJ/kg.KTâ‚‚ = 70 Â°C + (1.021)kW/ (0.018kg/s)(1.011 kJ/kg.K)Tâ‚‚ = 126 Â°C Considering an isentropic process, the air temperature at the compressor exit; Tâ‚‚s = Tâ‚ (PÂ²/PÂ¹)âˆ§(k-1)/(k)where Tâ‚ = 70Â°C, Pâ‚= 95kPa, Pâ‚‚ = 135kPa, k = 1.397Tâ‚‚s = (70+273)K (135/95 kPa) âˆ§ (1.397-1)/(1.397)Tâ‚‚s = 379 K = 106 Â°C(b). the isentropic efficiency of the compressor Â  Â  Â Ð›c = (Tâ‚‚s - Tâ‚ / Tâ‚‚ - Tâ‚) Â  Â  Â  Â  Ð›c = (106Â°C - 70Â°C) / (126Â°C - 70Â°C) Â  Â  Â  Â  Â  Â Ð›c = 0.642 = 64%the isentropic efficiency of the compressor is 64%cheers i hope this helps