hi, can you help me please??

Question

Air in a piston/cylinder setup goes through a car not cycle with the BV diagram shown in figure the high and low temperatures are 600k and 300K respectively the heat added at the high temperature is 250 kilojoule per kilogram and the lowest pressure in the cycle is 75 kg find the specific volume and pressure after heat rejection and the network per unit Mass

Answer 1

We have:

q_H = 250 kJ/kg , T_H = 600 K, T_L = 300 K, P₃ = 75 kPa 

The states as

1 : 600 K , 2: 600 K, 3: 75 kPa, 300 K 4: 300 K

Since this is a Carnot cycle and we know the temperatures the efficiency is

η = 1 − T_L/T_H = 1 - 300/600 = 0.5

And the net work becomes 

 w_NET = ηq_H = 0.5 × 250

= 125 kJ/kg

The heat rejected is

 q_L = q_H – w_NET = 125 kJ/kg

After heat rejection 

q_L = RT_L ln (v₃/v₄)

v₃ = RT₃ / P₃ = 0.287 x 300 / 75 = 1.148 m³/kg

v₄ = v₃ exp(-q_L/RT_L) = 1.148 exp(−125/0.287 × 300) = 0.2688 m³/kg

P₄ = RT₄ / v₄ = 0.287 × 300 / 0.2688 = 320 kPa