Question

2. One lb of air undergoes a power cycle consisting of the following processes: 1->2: constant volume from p1=20 psi, T1=500°R to T2=820°R 2->3: adiabatic expansion to v3 = 1.4 v2 3->1: constant pressure compression Sketch the cycle on a p-v diagram. Assuming ideal gas behavior, determine (a) The pressure at state 2 in psi. (b) The temperature at state 3 in °R. (c) The thermal efficiency of the cycle.

Answer 1

Refer to the diagram:

PA T2 AN P. Ti Tz V = V₂ V3

a) 1-> 2: isochoric process

P1 = initial temperature = 20 psi = 137895 Pa

T1 = initial temperature = 500 degree R = 277.78 K

T2 = final temperature = 820 degree R = 455.56 K

P2 = final pressure = ?

As the volume is constant,

P2 T2 P1 T

$\Rightarrow P_{2}=\frac{P_{1}}{T_{1}}*T_{2}$

$\Rightarrow P_{2}=\frac{137895}{277.78}*455.56=226148.2 Pa = 32.8 psi$ [answer]

b) Given:

For the process from 3 -> 1: isobaric process

T1 = initial temperature = 500 degree R = 277.78 K

V1 = initial volume

V3 = final volume

T3 = final temperature

As the pressure is constant:

$\frac{V_{3}}{T_{3}}=\frac{V_{1}}{T_{1}}$

$\Rightarrow T_{3}=\frac{V_{3}T_{1}}{V_{1}}$

[answer]

T3 1.4V1 * 277.78 V1 = 1.4 * 277.78 = 388.89K = 700'R

c) As the gas is air, it is a diatomic gas, therefore, $C_{V}=\frac{5R}{2}, C_{P}=\frac{7R}{2}, \gamma=\frac{7}{5}$ [where R = universal gas constant]

As the temperature is increased only from state 1 to state 2.

Heat provided = Q = . [since the process is isochoric]

5 nCv (T2 – T1) = 5nR(T2 – T1)

Now, work done from 1 to 2 : $W_{12}=0$ [since volume is constant]

work done from 2 to 3 : $W_{23}=\frac{nR(T_{3}-T_{2})}{1-\gamma}$ [since the process is adiabatic]

work done from 3 to 1 : [since the process is isobaric]

Total work done = $W=W_{12}+W_{23}+W_{31}$

  $=0+\frac{nR(T_{3}-T_{2})}{1-\gamma}+( -0.4*nRT_{1}) = \frac{nR(T_{3}-T_{2})}{1-\gamma} -0.4*nRT_{1}$

The thermal efficiency of the cycle = total work done / heat provided

$\eta = \frac{W}{Q} = \frac{\frac{nR(T_{3}-T_{2})}{1-\gamma} -0.4*nRT_{1}}{\frac{5}{2}nR(T_{2}-T_{1})}$

  $= \frac{\frac{(388.89-455.56)}{1-\frac{7}{5}} -0.4*277.78}{\frac{5}{2}(455.56-277.78)}=0.125$

Therefore, the efficiency of the cycle is 0.125 = 12.5%. [answer]