UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
TDA3701 APPLIED THERMODYNAMICS
Assignment 01 — May 2026
⋄
Module Code: TDA3701
Module Name: Applied Thermodynamics
Assignment No.: Assignment 01
Due Date: May 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for TDA3701
at the University of South Africa.
,UNISA | TDA3701 Applied Thermodynamics — Assignment 1
Question 1: Reversible Adiabatic Expansion of Steam
Question: Steam expands reversibly in a cylinder behind a piston from 7 bar dry saturated,
to a pressure of 0.75 bar. Assuming that the cylinder is perfectly thermally insulated, calcu-
late the work done during the expansion per kilogram of steam. Calculate:
1.1 The Change in Internal Energy [12 marks]
Given:
P1 = 7 bar, x1 = 1 (dry saturated), P2 = 0.75 bar
Since the cylinder is perfectly thermally insulated:
Q=0
For a reversible adiabatic (isentropic) process, entropy is conserved:
s1 = s2
Step 1: Read steam table values at 7 bar.
s1 = sg = 6.694 kJ/kgK, u1 = ug = 2573 kJ/kg
Step 2: Read steam table values at 0.75 bar.
sf = 1.213 kJ/kgK, sf g = 6.243 kJ/kgK
uf = 384 kJ/kg, uf g = 2094 kJ/kg
Step 3: Determine the dryness fraction at state 2.
Using the entropy relation for a wet mixture:
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,UNISA | TDA3701 Applied Thermodynamics — Assignment 1
s2 = sf + x2 sf g
Since s1 = s2 = 6.694 kJ/kgK:
6.694 = 1.213 + x2 (6.243)
6.694 − 1.213 5.481
x2 = =
6.243 6.243
x2 = 0.878
Step 4: Determine internal energy at state 2.
u2 = uf + x2 uf g
u2 = 384 + (0.878)(2094)
u2 = 384 + 1838.53
u2 = 2222.53 kJ/kg
Step 5: Compute the change in internal energy.
∆U = u2 − u1 = 2222.53 − 2573
∆U = −350.47 kJ/kg
The negative sign confirms that internal energy decreases as the steam does work on the pis-
Page 2 of 14
,UNISA | TDA3701 Applied Thermodynamics — Assignment 1
ton.
1.2 The Work Done [3 marks]
Given:
Q = 0, ∆U = −350.47 kJ/kg
Applying the first law of thermodynamics:
Q = ∆U + W
0 = ∆U + W
W = −∆U
W = −(−350.47)
W = 350.47 kJ/kg
The work done by the steam on the piston during adiabatic expansion equals 350.47 kJ per
kilogram of steam.
1.3 The Change in Enthalpy [6 marks]
Step 1: Read enthalpy at state 1 from steam tables at 7 bar.
h1 = hg = 2763 kJ/kg
Step 2: Read enthalpy values at 0.75 bar.
Page 3 of 14
, UNISA | TDA3701 Applied Thermodynamics — Assignment 1
hf = 384 kJ/kg, hf g = 2278 kJ/kg
Step 3: Compute enthalpy at state 2 using x2 = 0.878.
h2 = hf + x2 hf g
h2 = 384 + (0.878)(2278)
h2 = 384 + 1999.08
h2 = 2383.08 kJ/kg
Step 4: Compute change in enthalpy.
∆H = h2 − h1 = 2383.08 − 2763
∆H = −379.92 kJ/kg
The drop in enthalpy reflects the combined reduction in internal energy and the pressure-
volume work transferred out of the steam.
1.4 The Change in Specific Volume [5 marks]
Step 1: Read specific volume at state 1 from steam tables at 7 bar.
v1 = vg = 0.273 m3 /kg
Step 2: Read specific volume values at 0.75 bar.
Page 4 of 14
College of Science, Engineering and Technology
⋄
TDA3701 APPLIED THERMODYNAMICS
Assignment 01 — May 2026
⋄
Module Code: TDA3701
Module Name: Applied Thermodynamics
Assignment No.: Assignment 01
Due Date: May 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for TDA3701
at the University of South Africa.
,UNISA | TDA3701 Applied Thermodynamics — Assignment 1
Question 1: Reversible Adiabatic Expansion of Steam
Question: Steam expands reversibly in a cylinder behind a piston from 7 bar dry saturated,
to a pressure of 0.75 bar. Assuming that the cylinder is perfectly thermally insulated, calcu-
late the work done during the expansion per kilogram of steam. Calculate:
1.1 The Change in Internal Energy [12 marks]
Given:
P1 = 7 bar, x1 = 1 (dry saturated), P2 = 0.75 bar
Since the cylinder is perfectly thermally insulated:
Q=0
For a reversible adiabatic (isentropic) process, entropy is conserved:
s1 = s2
Step 1: Read steam table values at 7 bar.
s1 = sg = 6.694 kJ/kgK, u1 = ug = 2573 kJ/kg
Step 2: Read steam table values at 0.75 bar.
sf = 1.213 kJ/kgK, sf g = 6.243 kJ/kgK
uf = 384 kJ/kg, uf g = 2094 kJ/kg
Step 3: Determine the dryness fraction at state 2.
Using the entropy relation for a wet mixture:
Page 1 of 14
,UNISA | TDA3701 Applied Thermodynamics — Assignment 1
s2 = sf + x2 sf g
Since s1 = s2 = 6.694 kJ/kgK:
6.694 = 1.213 + x2 (6.243)
6.694 − 1.213 5.481
x2 = =
6.243 6.243
x2 = 0.878
Step 4: Determine internal energy at state 2.
u2 = uf + x2 uf g
u2 = 384 + (0.878)(2094)
u2 = 384 + 1838.53
u2 = 2222.53 kJ/kg
Step 5: Compute the change in internal energy.
∆U = u2 − u1 = 2222.53 − 2573
∆U = −350.47 kJ/kg
The negative sign confirms that internal energy decreases as the steam does work on the pis-
Page 2 of 14
,UNISA | TDA3701 Applied Thermodynamics — Assignment 1
ton.
1.2 The Work Done [3 marks]
Given:
Q = 0, ∆U = −350.47 kJ/kg
Applying the first law of thermodynamics:
Q = ∆U + W
0 = ∆U + W
W = −∆U
W = −(−350.47)
W = 350.47 kJ/kg
The work done by the steam on the piston during adiabatic expansion equals 350.47 kJ per
kilogram of steam.
1.3 The Change in Enthalpy [6 marks]
Step 1: Read enthalpy at state 1 from steam tables at 7 bar.
h1 = hg = 2763 kJ/kg
Step 2: Read enthalpy values at 0.75 bar.
Page 3 of 14
, UNISA | TDA3701 Applied Thermodynamics — Assignment 1
hf = 384 kJ/kg, hf g = 2278 kJ/kg
Step 3: Compute enthalpy at state 2 using x2 = 0.878.
h2 = hf + x2 hf g
h2 = 384 + (0.878)(2278)
h2 = 384 + 1999.08
h2 = 2383.08 kJ/kg
Step 4: Compute change in enthalpy.
∆H = h2 − h1 = 2383.08 − 2763
∆H = −379.92 kJ/kg
The drop in enthalpy reflects the combined reduction in internal energy and the pressure-
volume work transferred out of the steam.
1.4 The Change in Specific Volume [5 marks]
Step 1: Read specific volume at state 1 from steam tables at 7 bar.
v1 = vg = 0.273 m3 /kg
Step 2: Read specific volume values at 0.75 bar.
Page 4 of 14
