Question

Water in a piston-cylinder assembly is heated from saturated liquid to saturated vapor at a constant pressure of 40 bar in a reversible process. How much work per unit mass is done? How much heat transfer per unit mass (via 푇 d푠 relationship)? [194.11 kJ kg−1 , 1704.2 kJ kg−1 ] How much heat transfer per unit mass (via internal energy)? [1714.1 kJ kg−1 ]

Answer 1

a. W = 194.11 KJ/kg

b. Q = 1713.5 KJ/kg

c. Q = 1713.41 KJ/kg

Step-by-step explanation:

a.

Since, this is a constant pressure process. Therefore, the work done will be:

W = PΔV

where,

W = work done per unit mass

P = constant pressure = 40 bar = 4000 KPa

ΔV = change in volume = Vg - Vf

At 4000 KPa, from saturated water table:

Vf = 0.001252 m³/kg

Vg = 0.049779 m³/kg

Therefore,

ΔV = (0.049779 - 0.001252) m³/kg

ΔV = 0.048527 m³/kg

Now, the work done will be:

W = (4000 KPa)(0.048527 m³/kg)

W = 194.11 KJ/kg

b.

Now for heat transfer, using the relation:

Q = ΔH

where,

Q = Heat Transfer per unit mass

ΔH = change in enthalpy of water = Hg - Hf

At 4000 KPa, from saturated water table:

Hf = 1087.4 KJ/kg

Hg = 2800.8 KJ/kg

Therefore,

Q = (2800.8 - 1087.4) KJ/kg

Q = 1713.5 KJ/kg

c.

Heat transfer per unit mass with the help of internal energy can be found out by using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Transfer per unit mass

ΔU = change in internal energy of water = Ug - Uf

W = work done per unit mass = 194.11 KJ/kg

At 4000 KPa, from saturated water table:

Uf = 1082.4 KJ/kg

Ug = 2601.7 KJ/kg

Therefore,

ΔU = (2601.7 - 1082.4) KJ/kg = 1519.3 KJ/kg

Hence,

Q = (1519.3 + 194.11) KJ/kg

Q = 1713.41 KJ/kg