Question

Heat exchanger 5 Comb. Compressor Turbine A gas turbine for an automobile is designed with a regenerator. Air enters the compressor of this engine at 100 kPa and 20°C. The compressor pressure ratio is 7.8; the maximum cycle temperature is 865°C; and the cold air stream leaves the regenerator 10°C cooler than the hot air stream at the inlet of the regenerator. Assuming both the compressor and the turbine to be isentropic, determine the rates of heat addition and rejection for this cycle when it produces 130 kW. Use constant specific heats at room temperature Round your final answer to one decimal place Round your final answer to one decimal place in out kW = kW

Answer 1

At compressor inlet

temperature T1 = 20∘C + 273 = 293 K

Maximum cycle temperature

T4 = 865°C + 273 = 1138 K

Compression ratio

p2/p1= 7.8

Gas constants R = 0.287 KJ/kg.k

Specific heat capacity of air at constant pressure

cp=1.005KJ/kg.k

Specific heat capacity of air at constant volume

cv=0.718KJ/kg.k

PV diagram

4 i T 3 2 6 1 Ln

Process 1-2

isentropic compression

T2/T1=(p2/p1)^(k−1/K)

T2/293=(7.8)^(1.4−1/1.4)

T2=526.91 K

process 4-5

isentropic expansion:

T5/T4=(p5/p4)^(k−1/K)

T5/T4=(p1/p2)^(k−1/K)

T5/1138=(1/7.8)^(1.4−1/1.4)

T5 = 632.81 K

At state 3

T3=T5−10

T3=632.81−10 = 622.81 K

T5−T6=T3−T2

T6=T5−(T3−T2)

= 632.81 - (622.81 - 526.91)

= 536.91 K

Net work output

Wnet = Wt−Wc

Wnet = cp(T4−T5) − cp(T2−T1)

Wnet=1.005 kJ/kg-K x (1138−632.81)K − 1.005kJ/K x (526.91−293)K

Wnet = 272.63 kJ/kg

˙Wnet=˙mWnet

130 =˙m(272.63)

˙m=0.4768 Kg/sec

Heat added

˙Qin=˙mcp(T4−T3)

= 0.4768 kg/s x 1.005 kJ/kg-K x (1138 - 622.81)

= 246.88 kJ/s = 246.9 kW

Heat rejected

Qout=˙mcp(T6−T1)

˙Qout=(0.4768 kg/s) x (1.005 kJ/kg-K) x (536.91 - 293)K

= 116.88 kJ/s = 116.9 kW