Question

Air in a piston/cylinder goes through a Carnot cycle. The high and low temperatures are 600 K and 300 K, respectively. The heat added at the high temperature is 250 kJ/kg, and the lowest pressure in the cycle is 75 kPa. Find the specific volume and pressure after heat rejection and the net work per unit mass.

Answer 1

Volume = 0.2688
`m^(3)`/kg

Pressure = 320kPa

Net work/unit mass = 125kJ/kg

Step-by-step explanation:

qH=250 (heat at high temp)

TH= 600K (high temp)

TL= 300K (low temp)

Pl= 75kPa (lowest pressure)

Efficiency = η = 1 -
`(TL)/(TH)` = 0.5 (Plugging the values from above)

So, net work done per unit mass = η*qH = 0.5*250 = 125kJ/kg

qL is the heat rejected which would be = qH - net work done = 250 - 125 = 125kJ/kg

VL (volume low) =
`(RTL)/(Pl)` = 0.287*300/75 = 1.148kJ/kg

Specific volume = VL
`*e^{(-qL)/(R*TL) }`

= 1.148
`*e^{(-125*300)/(.287) }` = .02688
`m^(3)/kg`

PH = R*TH/VH = 0.287*300/0.2688 = 320kPa