Question

A mixture of water at 500 kPa and a quality of 20% is heated in an isobaric process until its volume becomes 0.52261 m^3/kg. Find:

a. the change in specifc volume

b. the work/unit mass

c. the change in specific internal energy

d. the heat transferred/unit mass

e. the change in specific enthalpy

f. the final pressure

g. the final temperature

Answer 1

From saturated steam table at 500 kPa,

Tsat = 151.836 C

Hl = 640.185 kJ/kg ; Hv = 2748.11 kJ/kg

Ul = 639.639 kJ/kg ; Uv = 2560.7 kJ/kg

Vl = 0.00109256 m³/kg ; Vv = 0.374804 m³​/kg

Given: Quality of steam or vapour mass fraction x = 0.2

H1 = xHv + (1-x)Hl = 0.2*2748.11 + (1-0.2)*640.185 = 1061.77 kJ/kg

Similarly,

U1 = 1021.45 kJ/kg

V1 = 0.0758348 m³/kg

Now, after isobaric heating, V2 = 0.52261 m³/kg > Vv = 0.374804 m³​/kg

This implies that steam is superheated.

Now, from superheated steam table, find the corresponding temperature for P = 500 kPa and V = 0.52261m³​/kg

we get, T = 300 C

H2 = 3064.6 kJ/kg

U2 = 2803.29 kJ/kg

a) Change in specific volume = V2 -V1 = 0.52261 - 0.0758348 = 0.446775 m³/kg

b) For work,

By 1st Law of thermodynamics,

U = Q + W

For constant pressure process, Q = H2 - H1 = 3064.6 - 1061.77 = 2002.83 kJ/kg

.: U2 - U1 = Q + W

.: 2803.29 - 1021.45 = 2002.83 + W

.: W = -220.99 kJ/kg ...........(-ve sign implies that work is done by the system (steam) )

c) U = 2803.29 - 1021.45 = 1781.84 kJ/kg

d) Q = H2 - H1 = 2002.83 kJ/kg

e) H2 - H1 = 2002.83 kJ/kg

f) P2 = P1 = 500 kPa ........since isobaric

g) T = 300 C