At the beginning of compression in an airstandard Diesel cyc

At the beginning of compression in an air-standard Diesel cycle, p_1 = 170 kPa, V_1 = 0.016 m^3, and T_1 = 315 K. The compression ratio is 15 and the maximum cycle temperature is 1400 K. Determine the mass of air, in kg. the heat addition and heat rejection per cycle, each in kJ. the network, in kJ, and the thermal efficiency.

Solution

solution:for given air standard cycle

1)compression ratio=v1/v2

v2=1.066*10^-3 m3

2)isentropic compression processes

T2/T1=(v1/v2)^(1.4-1)

then we get T2=930.58 k

T2/T1=(P2/P1)^(1.4-1/1.4)

we get P2=7532.98 kpa

P2=P3

3)cut of ratio=k

T3=T1(r^1.4-1)k

k=1.504

where

k=v3/v2

v3=1.6032*10^-3 m3

4)from isentropic processes we get

T4=T3(r/k)^(1.4-1)

T4=558.72 k

5)T4/T3=(P4/P3)^(1.4-1/1.4)

P4=302.481 kpa

6) work done in compression a ndin expansion is given by

W=P1v1-P2v2/1.4-1

Wc=-13.275 kj

We=P3v3-P4v4/1.4-1

We=18.092 kj

net work=we+Wc=18.092-13.275=4.817 kj

7)heat supplied at constant pressure is given by

q23=m*1.005*(1400-930.56)

q41=m*.718*(558.72-315)

here net work=q23-q41

we get

m=.01623 kg

8) total heat supplied is

q23=.01623*1.005*(1400-930.56)=7.658 kj

9)heat rejected

q41=.01623*.718*(558.72-315)=2.84 kj

10)efficiecy=(net work/heat supplied)=4.817/7.658=.6293=62.93%