Question

Two kilograms of steam initially at 200 degrees Celsius and 0.1 MPa are to be reversibly and adiabatically compressed to 10 bar. Determine the final temperature and the work required.

Answer 1

An adiabatic reversible process is an isoentropic process:

You have to use the next equation:

Temperature must be in Kelvin

this is true for a reversible adiabatic compression or expansion

K is the ratio of CP/CV of the gas it is tipically used a value of 1.4, you can find values of K (sometimes called ) of different substances on books or internet.

Looking for values of steam (water in gas phase) i get the next values for differente pressures:

k = 1.28 for 14.5 psi or 100,000 Pa or 1 bar

k = 1.32 for 150 psi or 1,000,000 (aproximately) or 10 bar

If we take the average value of this 2 we have

k = 1.3

Let´s calculate

T₂ = 804 K = 804 - 273 = 531.69 Celsius

Now to calculate work you apply the equation:

Now we have to multiply the work by the molecular weight of water (18)

w = 11,950 * 18 = 215,100 J/Kg

Now multiply that by the Kg of water (2)

w = 430,200 Joules