Question

A piston–cylinder device initially contains 0.22 kg of steam at 200 kPa and 300°C. Now, the steam is cooled at constant pressure until it is at 150°C. Determine the volume change of the cylinder during this process using the compressibility factor, and compare the result to the actual value.

Answer 1

Explanation:

Defining compressibility factor (Z), which is also known as the gas deviation factor, is a correction factor which describes the deviation of a real gas from ideal gas behaviour.

Z = p/(density * Rs *T)

Where Rs is specific gas constant =R/M

Deriving the volume values from steam tables,

At 200kPa and 300°C, V1 = 1.31623 m3/kg

At 200kPa and 150°C, V2 = 0.95986 m3/kg

Change in volume, DV = m(V1 - V2)

= 0.2*(1.31623 - 0.95986)

= 0.07128 m3/kg

Also from steam tables,

The critical pressure, Pc = 22.06MPa

The critical temperature, Tc = 647.1K

Remember, gas constant, R = 0.4615 kPa.m3/kg.K

Therefore, reduced temperature at initial state, Tr = T1/Tc

= (360 + 273)/647.1

= 0.886

Also, reduced pressure at initial state, Pr = P1/Pr

= 200 x 10^-3/22.06

= 0.0091

But compressibility factor at initial state, Z1 = 0.9956

Ideal volume, Vi = m*R*T/P1

= (0.2*0.4615*(300+273))/200

= 0.2644m3

Compressibility volume, Vc = Z1*Vi

= 0.9956* 0.2644

= 0.2633m3

Reduced temperature at final state, Tr = T2/Tc

= (150+273)/647.1

= 0.65

Reduced pressure at final state, Pr = P2/Pc

= 200 x 10^-3/22.06

= 0.0091

Compressibility factor at final state, Z2 = 0.9897