Question

Supercharging of an engine is used to increase the inlet air density so that more fuel can be added, the result of

which is an increased power output. Assume that ambient air, 100 kPa and 27°C, enters the supercharger at a rate

of 250 L/s. The supercharger (compressor) has an isentropic efficiency of 75%, and uses 20 kW of power input.

Assume that the ideal and actual compressor have the same exit pressure. Find the ideal specific work and verify

that the exit pressure is 175 kPa. Find the percent increase in air density entering the engine due to the

supercharger and the entropy generation

Answer 1

m_in = m_ex = m = V. /vin = 0.29 kg/s  

Energy: m_hin - W = mh_ex

Assume: Q. = 0

Entropy: m. s_in + S_gen = m

Inlet state:

v_in = RT_in/P_in = 0.8614 m³ /kg, P_{r in} = 1.1167

?_c = w_{C s}/w_{C ac} => - W_S = - W_AC × ?_c = 15 kW

-w_{C s} = - . W_S/m = 51.724 kJ/kg, -w_{C ac} = 68.966 kJ/kg

from table A7

h_{ex s} = h_in - w_{C s} = 300.62 + 51.724 = 352.3 kJ/kg  

? T_{ex s} = 351.5 K, P_{r ex} = 1.949  

P_ex = P_in × P_{r ex}/P_{r in} = 100 × 1.949 / 1.1167 = 174.5 kPa

The actual exit state is  

h_{ex ac} = h_in - w_{C ac} = 369.6 kJ/kg ? Tex ac = 368.6 K

v_ex = RT_ex/P_ex = 0.606 m3 / kg

?_ex / ?_in = v_in/v_ex = 0.8614/0.606 = 1.42 or 42 % increase

s_gen = s_ex - s_in = 7.0767 - 6.8693 - 0.287 ln(174/100)] = 0.0484 kJ/kg K