UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
THT3701: Applied Thermo-
dynamics and Heat Transfer
Assignment 01 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
THT3701
Module Code:
Applied Thermodynamics and Heat Trans-
Module Name:
fer
Assignment 01
Assignment:
2026
Due Date:
20
Total Marks:
Submitted in partial fulfilment of the requirements for THT3701 — UNISA 2026
,UNISA | THT3701 Assignment 01 – 2026
Question 1: Spacecraft Surface Temperature
The outer surface of a spacecraft in space has an emissivity of 0.8 and a solar
absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of
950 W/m2 , determine the surface temperature of the spacecraft when the radia-
tion emitted equals the solar energy absorbed. (5)
Assumptions
1. Steady operating conditions exist, implying that the surface temperature remains con-
stant.
2. Thermal properties of the spacecraft surface are constant.
3. The temperature of outer space is taken as Tspace = 0 K (deep space, negligible back-
ground radiation).
Given Information
• Solar absorptivity: α = 0.3
• Surface emissivity: ε = 0.8
• Incident solar irradiation: q̇solar = 950 W/m2
• Stefan-Boltzmann constant: σ = 5.67 × 10−8 W/m2 · K4
Step-by-Step Solution
Step 1: State the Energy Balance Condition
The condition given is that radiation emitted by the surface equals the solar energy absorbed
by the surface. Therefore:
Q̇solar, absorbed = Q̇rad, emitted (1)
Step 2: Express the Absorbed Solar Energy
The rate of solar energy absorbed per unit area is:
Page 2 of 15
,UNISA | THT3701 Assignment 01 – 2026
Q̇solar, absorbed = α · As · q̇solar (2)
Substituting the known values:
Q̇solar, absorbed = 0.3 × As × 950 W/m2 (3)
Step 3: Express the Radiation Emitted by the Surface
By the Stefan-Boltzmann law, the radiation emitted by the surface into space is:
Q̇rad, emitted = ε · σ · As · Ts4 − Tspace
4
(4)
Since Tspace = 0 K:
Q̇rad, emitted = ε · σ · As · Ts4 = 0.8 × 5.67 × 10−8 × As × Ts4 (5)
Step 4: Apply the Energy Balance
Setting absorbed equals emitted and cancelling As from both sides:
0.3 × 950 = 0.8 × 5.67 × 10−8 × Ts4 (6)
285 = 4.536 × 10−8 × Ts4 (7)
Step 5: Solve for Ts4
Dividing both sides by 4.536 × 10−8 :
285
Ts4 = = 6.284 × 109 K4 (8)
4.536 × 10−8
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,UNISA | THT3701 Assignment 01 – 2026
Step 6: Solve for Ts
Taking the fourth root of both sides:
1
0.3 × 950 4
Ts = (9)
0.8 × 5.67 × 10−8
0.25
Ts = 6.284 × 109 (10)
Ts = 281.5 K (11)
Implementation Insight
The surface temperature of 281.5 K (≈ 8.4◦ C) is relatively mild. This results from the
low absorptivity (α = 0.3), meaning only 30% of incident solar radiation is absorbed,
while the high emissivity (ε = 0.8) allows efficient re-radiation. Spacecraft are often
coated with specialised paints or multi-layer insulation to control precisely this thermal
balance.
Page 4 of 15
, UNISA | THT3701 Assignment 01 – 2026
Question 2: Convection Heat Transfer Coefficient of a Solid Plate
A solid plate, with a thickness of 15 cm and a thermal conductivity of 80 W/m·K,
is being cooled at the upper surface by air. The air temperature is 10◦ C, while
the temperatures at the upper and lower surfaces of the plate are 50 and 60◦ C,
respectively. Determine the convection heat transfer coefficient of air at the upper
surface and discuss whether the value is reasonable or not for forced convection of
air. (5)
Assumptions
1. Steady-state, one-dimensional heat conduction through the plate.
2. Thermal conductivity k is constant and uniform.
3. Heat flows upward (from lower surface to upper surface) and is then removed by convec-
tion at the upper surface.
Given Information
• Plate thickness: L = 15 cm = 0.15 m
• Thermal conductivity: k = 80 W/m · K
• Air (fluid) temperature: T∞ = 10◦ C
• Upper surface temperature: Ts,1 = 50◦ C
• Lower surface temperature: Ts,2 = 60◦ C
Step-by-Step Solution
Step 1: Determine the Conduction Heat Flux Through the Plate
By Fourier’s law of heat conduction (one-dimensional, steady-state), the heat flux conducted
through the plate from the lower surface to the upper surface is:
Ts,2 − Ts,1
q̇cond = k · (12)
L
Substituting values:
Page 5 of 15
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
THT3701: Applied Thermo-
dynamics and Heat Transfer
Assignment 01 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
THT3701
Module Code:
Applied Thermodynamics and Heat Trans-
Module Name:
fer
Assignment 01
Assignment:
2026
Due Date:
20
Total Marks:
Submitted in partial fulfilment of the requirements for THT3701 — UNISA 2026
,UNISA | THT3701 Assignment 01 – 2026
Question 1: Spacecraft Surface Temperature
The outer surface of a spacecraft in space has an emissivity of 0.8 and a solar
absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of
950 W/m2 , determine the surface temperature of the spacecraft when the radia-
tion emitted equals the solar energy absorbed. (5)
Assumptions
1. Steady operating conditions exist, implying that the surface temperature remains con-
stant.
2. Thermal properties of the spacecraft surface are constant.
3. The temperature of outer space is taken as Tspace = 0 K (deep space, negligible back-
ground radiation).
Given Information
• Solar absorptivity: α = 0.3
• Surface emissivity: ε = 0.8
• Incident solar irradiation: q̇solar = 950 W/m2
• Stefan-Boltzmann constant: σ = 5.67 × 10−8 W/m2 · K4
Step-by-Step Solution
Step 1: State the Energy Balance Condition
The condition given is that radiation emitted by the surface equals the solar energy absorbed
by the surface. Therefore:
Q̇solar, absorbed = Q̇rad, emitted (1)
Step 2: Express the Absorbed Solar Energy
The rate of solar energy absorbed per unit area is:
Page 2 of 15
,UNISA | THT3701 Assignment 01 – 2026
Q̇solar, absorbed = α · As · q̇solar (2)
Substituting the known values:
Q̇solar, absorbed = 0.3 × As × 950 W/m2 (3)
Step 3: Express the Radiation Emitted by the Surface
By the Stefan-Boltzmann law, the radiation emitted by the surface into space is:
Q̇rad, emitted = ε · σ · As · Ts4 − Tspace
4
(4)
Since Tspace = 0 K:
Q̇rad, emitted = ε · σ · As · Ts4 = 0.8 × 5.67 × 10−8 × As × Ts4 (5)
Step 4: Apply the Energy Balance
Setting absorbed equals emitted and cancelling As from both sides:
0.3 × 950 = 0.8 × 5.67 × 10−8 × Ts4 (6)
285 = 4.536 × 10−8 × Ts4 (7)
Step 5: Solve for Ts4
Dividing both sides by 4.536 × 10−8 :
285
Ts4 = = 6.284 × 109 K4 (8)
4.536 × 10−8
Page 3 of 15
,UNISA | THT3701 Assignment 01 – 2026
Step 6: Solve for Ts
Taking the fourth root of both sides:
1
0.3 × 950 4
Ts = (9)
0.8 × 5.67 × 10−8
0.25
Ts = 6.284 × 109 (10)
Ts = 281.5 K (11)
Implementation Insight
The surface temperature of 281.5 K (≈ 8.4◦ C) is relatively mild. This results from the
low absorptivity (α = 0.3), meaning only 30% of incident solar radiation is absorbed,
while the high emissivity (ε = 0.8) allows efficient re-radiation. Spacecraft are often
coated with specialised paints or multi-layer insulation to control precisely this thermal
balance.
Page 4 of 15
, UNISA | THT3701 Assignment 01 – 2026
Question 2: Convection Heat Transfer Coefficient of a Solid Plate
A solid plate, with a thickness of 15 cm and a thermal conductivity of 80 W/m·K,
is being cooled at the upper surface by air. The air temperature is 10◦ C, while
the temperatures at the upper and lower surfaces of the plate are 50 and 60◦ C,
respectively. Determine the convection heat transfer coefficient of air at the upper
surface and discuss whether the value is reasonable or not for forced convection of
air. (5)
Assumptions
1. Steady-state, one-dimensional heat conduction through the plate.
2. Thermal conductivity k is constant and uniform.
3. Heat flows upward (from lower surface to upper surface) and is then removed by convec-
tion at the upper surface.
Given Information
• Plate thickness: L = 15 cm = 0.15 m
• Thermal conductivity: k = 80 W/m · K
• Air (fluid) temperature: T∞ = 10◦ C
• Upper surface temperature: Ts,1 = 50◦ C
• Lower surface temperature: Ts,2 = 60◦ C
Step-by-Step Solution
Step 1: Determine the Conduction Heat Flux Through the Plate
By Fourier’s law of heat conduction (one-dimensional, steady-state), the heat flux conducted
through the plate from the lower surface to the upper surface is:
Ts,2 − Ts,1
q̇cond = k · (12)
L
Substituting values:
Page 5 of 15
