UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
FMM3701: Fluid Mechan-
ics and Turbomachines
Assignment 1 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
FMM3701
Module Code:
Fluid Mechanics and Turbomachines
Module Name:
Assignment 1
Assignment:
2026
Due Date:
Submitted in partial fulfilment of the requirements for FMM3701 — UNISA 2026
,UNISA | FMM3701 Assignment 1 – Fluid Mechanics
Question 1: Continuity Equation – Circular Pipe Flow
Question (Rewritten): A circular pipe has a diameter at sections 1 and 2 as 20 cm and 30
cm respectively. Determine the volume flow rate (discharge) through the pipe if the velocity
of water at section 1 is given as 5 m/s. Determine the velocity of fluid at the exit section 2.
Determine the diameter of an equivalent pipe if the tapered pipe is to be replaced with a pipe
of uniform diameter for the same discharge.
1.1 Given Information
Diameter at section 1: D1 = 20 cm = 0.20 m
Diameter at section 2: D2 = 30 cm = 0.30 m
Velocity at section 1: V1 = 5 m/s
1.2 Cross-sectional Areas
The cross-sectional area of a circular pipe is:
πD2
A=
4
Area at section 1:
π(0.20)2 π × 0.04 0.04π
A1 = = = = 0.01π
4 4 4
A1 = 0.03142 m2
Area at section 2:
π(0.30)2 π × 0.09 0.09π
A2 = = = = 0.0225π
4 4 4
A2 = 0.07069 m2
1.3 Volume Flow Rate (Discharge)
The volume flow rate Q is given by:
Q = A1 × V1
Page 2 of 16
, UNISA | FMM3701 Assignment 1 – Fluid Mechanics
Q = 0.03142 × 5
Q = 0.1571 m3 /s
1.4 Velocity at Section 2
Applying the continuity equation (conservation of mass for incompressible flow):
A1 V1 = A2 V2
Solving for V2 :
A1 V 1 Q
V2 = =
A2 A2
0.1571
V2 =
0.07069
V2 = 2.222 m/s
Verification using ratio method:
2 2
A1 D1 0.20 4
V2 = V1 × = V1 × =5× =5× = 2.222 m/s✓
A2 D2 0.30 9
1.5 Diameter of Equivalent Uniform Pipe
For the same discharge Q through a uniform pipe, the average velocity Veq must be deter-
mined. Since no specific equivalent velocity is stated, the mean velocity of the tapered pipe is
used as the design velocity. The mean velocity between sections 1 and 2 is:
V1 + V2 5 + 2.222 7.222
Veq = = = = 3.611 m/s
2 2 2
The cross-sectional area of the equivalent pipe:
Q 0.1571
Aeq = = = 0.04350 m2
Veq 3.611
Therefore, the equivalent diameter:
2
πDeq 2 4Aeq 4 × 0.04350 0.17400
Aeq = =⇒ Deq = = = = 0.05538 m2
4 π π 3.1416
Page 3 of 16
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
FMM3701: Fluid Mechan-
ics and Turbomachines
Assignment 1 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
FMM3701
Module Code:
Fluid Mechanics and Turbomachines
Module Name:
Assignment 1
Assignment:
2026
Due Date:
Submitted in partial fulfilment of the requirements for FMM3701 — UNISA 2026
,UNISA | FMM3701 Assignment 1 – Fluid Mechanics
Question 1: Continuity Equation – Circular Pipe Flow
Question (Rewritten): A circular pipe has a diameter at sections 1 and 2 as 20 cm and 30
cm respectively. Determine the volume flow rate (discharge) through the pipe if the velocity
of water at section 1 is given as 5 m/s. Determine the velocity of fluid at the exit section 2.
Determine the diameter of an equivalent pipe if the tapered pipe is to be replaced with a pipe
of uniform diameter for the same discharge.
1.1 Given Information
Diameter at section 1: D1 = 20 cm = 0.20 m
Diameter at section 2: D2 = 30 cm = 0.30 m
Velocity at section 1: V1 = 5 m/s
1.2 Cross-sectional Areas
The cross-sectional area of a circular pipe is:
πD2
A=
4
Area at section 1:
π(0.20)2 π × 0.04 0.04π
A1 = = = = 0.01π
4 4 4
A1 = 0.03142 m2
Area at section 2:
π(0.30)2 π × 0.09 0.09π
A2 = = = = 0.0225π
4 4 4
A2 = 0.07069 m2
1.3 Volume Flow Rate (Discharge)
The volume flow rate Q is given by:
Q = A1 × V1
Page 2 of 16
, UNISA | FMM3701 Assignment 1 – Fluid Mechanics
Q = 0.03142 × 5
Q = 0.1571 m3 /s
1.4 Velocity at Section 2
Applying the continuity equation (conservation of mass for incompressible flow):
A1 V1 = A2 V2
Solving for V2 :
A1 V 1 Q
V2 = =
A2 A2
0.1571
V2 =
0.07069
V2 = 2.222 m/s
Verification using ratio method:
2 2
A1 D1 0.20 4
V2 = V1 × = V1 × =5× =5× = 2.222 m/s✓
A2 D2 0.30 9
1.5 Diameter of Equivalent Uniform Pipe
For the same discharge Q through a uniform pipe, the average velocity Veq must be deter-
mined. Since no specific equivalent velocity is stated, the mean velocity of the tapered pipe is
used as the design velocity. The mean velocity between sections 1 and 2 is:
V1 + V2 5 + 2.222 7.222
Veq = = = = 3.611 m/s
2 2 2
The cross-sectional area of the equivalent pipe:
Q 0.1571
Aeq = = = 0.04350 m2
Veq 3.611
Therefore, the equivalent diameter:
2
πDeq 2 4Aeq 4 × 0.04350 0.17400
Aeq = =⇒ Deq = = = = 0.05538 m2
4 π π 3.1416
Page 3 of 16
