UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
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TDA3701: Applied Thermodynamics
Assignment 1 — May 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
TDA3701
Module Code:
Applied Thermodynamics
Module Name:
Assignment 1
Assignment:
May 2026
Due Date:
56
Total Marks:
Submitted in partial fulfilment of the requirements for TDA3701 – UNISA 2026
, UNISA | TDA3701 Applied Thermodynamics – Assignment 1
Question 1: Reversible Adiabatic Steam Expansion
Steam expands reversibly in a cylinder behind a piston from 7 bar dry saturated to a pres-
sure of 0.75 bar. The cylinder is perfectly thermally insulated. The work done per kilogram of
steam is calculated below.
1.1 The Change in Internal Energy [12 marks]
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 change in internal energy per kilogram of steam.
Given information:
P1 = 7 bar (dry saturated, so x1 = 1)
P2 = 0.75 bar
Q = 0 (thermally insulated)
∴ s1 = s2 (reversible adiabatic process)
Step 1: Extract steam table values at State 1 (7 bar, dry saturated).
s1 = sg = 6.694 kJ/kgK
u1 = ug = 2573 kJ/kg
Step 2: Extract steam table values at State 2 (0.75 bar).
sf = 1.213 kJ/kgK
sf g = 6.243 kJ/kgK
uf = 384 kJ/kg
uf g = 2094 kJ/kg
Page 2 of 15
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
TDA3701: Applied Thermodynamics
Assignment 1 — May 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
TDA3701
Module Code:
Applied Thermodynamics
Module Name:
Assignment 1
Assignment:
May 2026
Due Date:
56
Total Marks:
Submitted in partial fulfilment of the requirements for TDA3701 – UNISA 2026
, UNISA | TDA3701 Applied Thermodynamics – Assignment 1
Question 1: Reversible Adiabatic Steam Expansion
Steam expands reversibly in a cylinder behind a piston from 7 bar dry saturated to a pres-
sure of 0.75 bar. The cylinder is perfectly thermally insulated. The work done per kilogram of
steam is calculated below.
1.1 The Change in Internal Energy [12 marks]
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 change in internal energy per kilogram of steam.
Given information:
P1 = 7 bar (dry saturated, so x1 = 1)
P2 = 0.75 bar
Q = 0 (thermally insulated)
∴ s1 = s2 (reversible adiabatic process)
Step 1: Extract steam table values at State 1 (7 bar, dry saturated).
s1 = sg = 6.694 kJ/kgK
u1 = ug = 2573 kJ/kg
Step 2: Extract steam table values at State 2 (0.75 bar).
sf = 1.213 kJ/kgK
sf g = 6.243 kJ/kgK
uf = 384 kJ/kg
uf g = 2094 kJ/kg
Page 2 of 15
