Answer to Question #273273 in Mechanics | Relativity for luca

Question

An air-standard cycle is executed in a closed system and is composed of the following four processes:

1-2 Isentropic compression from 100 kPa and 27°C to 1 MPa

2-3 P = constant heat addition in the amount of 2800 kJ/kg

3-4 v = constant heat rejection to 100 kPa

4-1 P = constant heat rejection to initial state

(a) Show the cycle on P-v and T-s diagrams.

(b) Calculate the maximum temperature in the cycle.

(c) Determine the thermal efficiency.

Assume constant specific heats at room temperature.

Answer 1

Given ,

• P ₁ = 100 KPa
• T ₁ = 27 ^* C = 300 K.
• P ₂ = P ₃ = 1 MPa =1000 KPa.
• Q ₂₃ = 2800 KJ/kg.
• V ₃ = V ₄ .
• P ₄ = 100 KPa.

a)

b) For process 1 - 2

$\frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{γ-1}{γ}} \\ T_2=300(\frac{1000}{100})^{\frac{1.4-1}{1.4}}$

T₂=579.2K

For heat addition 2-3 -

$Q_23=C_p(T_3 -T_2)$

2800=1.005(T₃ - 579.2)

T₃ =3365.26K=3092.25C

c) For process 3-4 

$\frac{T_4}{T_3}=\frac{P_4}{P_3} \\ T_4=3365.26 \times \frac{100}{1000} \\ T_4=336.526K$

Total heat rejected in the cycle -

$Q_r=Q_34 +Q_41 \\ Q_r=C_v (T_3 -T-4) + C_p (T_4 - T_1) \\ Q_r= 0.718(3365.26 - 579.2) - 1.005(336.526 - 300) \\ Q_r=2037.009 Kj/kg$

Now,

Work done in the cycle will be -

W=Qsupplied − Qrejected

W=2800−2037.099

W=762.901KJ/kg.

Efficiency of the cycle will be -

$n = \frac{Workdone}{heatsuppied} \\ n = \frac{762.901}{2800} \\ n=0.2724= 27.24$ %