Question

Steam is compressed to a saturated vapor state at constant pressure. The steam started at 2 MPa and 365 C. What work was required and what was the specific heat transfer?

Please show work so I can actually learn!

Answer 1

We will first refer to the steam tables for the saturation properties of steam at Pressure of 2 MPa:

Pressure (MPa)	Temp (C)	ug (kJ/kg)	vg (m3/kg)
2	212.4	777.784	0.09962

Work done in compressing the superheated steam from superheated state to saturated vapor state:

where

• P = pressure = 2 MPa = 2000 kPa
• v1 = initial specific volume =
  • R = characteristic gas constant =
    • = universal gas constant = 8.314 kg/kmolK
    • M = molecular mass of water = 18 kg/kmol
  • P = 2000 kPa
  • T = initial temp = 365 C = 638 K
• v2 = final specific volume = vg at 2 MPa = 0.09962 m3/kg (from the table above)

Negative sign indicates that work is being done on the system by the surroundings (compression work).

Now, change in internal energy =

where

• u2 = final internal energy = ug at 2 MPa = 777.784 kJ/kg (from the table above)
• u1 = C x T1 = 1.667 x 638 = 1063.546 kJ/kg
  • C = specific heat at constant volume at state 1 = 1.667 kJ/kgK (from the steam table)
  • T1 = 365 C = 638 K

From the first law of thermodynamics, the specific heat transfer =

Negative heat interaction shows that heat is transferred from the system to the surroundings.

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