Question

Give the appropriate values and units for these quantities:

a) The specific internal energy of steam at 250oC and 5 atm.

b) The temperature at which water vapor and liquid are in equilibrium when the pressure is 0.11 bar.

c) The enthalpy of 6 kg of water in equilibrium with its vapor at 40 oC.

Answer 1

a) h = u + pv
from steam tables (saturated) At t = 250^oC h = 1085.8 kJ/kg ; v = 0.00125 m³/kg
  1085.8 = u + 5 * 0.00125
specific internal energy u = 1085.79375 kJ/kg
b) water vapor and liquid are equilibrium at 100^oC temperature and 1 atm pressure
100^oC ---------------------- 1 atm
? ----------------------- 0.1085616 atm (0.11 bar)
cross multiply to get answer , T = 10.85^oC
Therefore at 0.11 bar pressure the temperature at which water vapor and liquid are in equilibrium is at 10.85^o C
c) H = mC_pT
  C_p = 4.179 kJ/K kg ( heat capacity of liquid water at 40⁰C)
H = 6 kg* 4.179 kJ/K kg * (40+273.15) K = 7851.9231 kJ
   enthalpy of 6 Kg water in equilibrium with its vapor at 40^oC is 7851.9231 kJ