UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
ASSIGNMENT 1
Fluid Mechanics and Turbomachines – FMM3701
⋄
Module Code: FMM3701
Module Name: Fluid Mechanics and Turbomachines
Student Name: [Student Name]
Student Number: [Student Number]
Assignment No.: Assignment 1
Due Date: 2026
Submitted in partial fulfilment of the requirements for Fluid Mechanics and Turbomachines
at the University of South Africa.
,UNISA | FMM3701 Fluid Mechanics and Turbomachines – Assignment 1
Question 1: Continuity Equation – Tapered Circular Pipe
Question: A circular pipe has a diameter at sections 1 and 2 as 20 cm and 30 cm respec-
tively. 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.
Section 1 Section 2
V1 = 5 m/s V2 =?
D1 = 0.20 m D2 = 0.30 m
Figure 1: Schematic – Tapered Circular Pipe (not to scale)
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
• Fluid: water (incompressible)
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 = 0.03142 m2
4 4 4
Area at section 2:
π(0.30)2 π × 0.09 0.09π
A2 = = = = 0.02250π = 0.07069 m2
4 4 4
Page 1 of 15
, UNISA | FMM3701 Fluid Mechanics and Turbomachines – Assignment 1
1.3 Volume Flow Rate (Discharge) at Section 1
The discharge is given by:
Q = A1 × V1
Q = 0.03142 × 5 = 0.1571 m3 /s
Implementation Insight
Result: The volume flow rate (discharge) through the pipe is Q = 0.1571 m3 /s.
1.4 Velocity at Exit Section 2
Applying the continuity equation for an incompressible fluid:
A1 V1 = A2 V2
Implying that:
A1 V 1 Q
V2 = =
A2 A2
0.1571
V2 = = 2.222 m/s
0.07069
Implementation Insight
Result: The velocity at exit section 2 is V2 = 2.222 m/s. The velocity decreases as the
pipe diameter increases, consistent with the continuity principle.
1.5 Diameter of Equivalent Uniform Pipe
For a pipe of uniform diameter De carrying the same discharge Q at the average velocity, the
conventional approach is to find De such that the cross-sectional area equals the mean of A1
and A2 . However, for an equivalent pipe carrying the same discharge at a mean velocity, the
mean velocity is:
V1 + V2 5 + 2.222
Vmean = = = 3.611 m/s
2 2
Therefore:
Q = Ae × Vmean
Page 2 of 15
College of Science, Engineering and Technology
⋄
ASSIGNMENT 1
Fluid Mechanics and Turbomachines – FMM3701
⋄
Module Code: FMM3701
Module Name: Fluid Mechanics and Turbomachines
Student Name: [Student Name]
Student Number: [Student Number]
Assignment No.: Assignment 1
Due Date: 2026
Submitted in partial fulfilment of the requirements for Fluid Mechanics and Turbomachines
at the University of South Africa.
,UNISA | FMM3701 Fluid Mechanics and Turbomachines – Assignment 1
Question 1: Continuity Equation – Tapered Circular Pipe
Question: A circular pipe has a diameter at sections 1 and 2 as 20 cm and 30 cm respec-
tively. 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.
Section 1 Section 2
V1 = 5 m/s V2 =?
D1 = 0.20 m D2 = 0.30 m
Figure 1: Schematic – Tapered Circular Pipe (not to scale)
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
• Fluid: water (incompressible)
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 = 0.03142 m2
4 4 4
Area at section 2:
π(0.30)2 π × 0.09 0.09π
A2 = = = = 0.02250π = 0.07069 m2
4 4 4
Page 1 of 15
, UNISA | FMM3701 Fluid Mechanics and Turbomachines – Assignment 1
1.3 Volume Flow Rate (Discharge) at Section 1
The discharge is given by:
Q = A1 × V1
Q = 0.03142 × 5 = 0.1571 m3 /s
Implementation Insight
Result: The volume flow rate (discharge) through the pipe is Q = 0.1571 m3 /s.
1.4 Velocity at Exit Section 2
Applying the continuity equation for an incompressible fluid:
A1 V1 = A2 V2
Implying that:
A1 V 1 Q
V2 = =
A2 A2
0.1571
V2 = = 2.222 m/s
0.07069
Implementation Insight
Result: The velocity at exit section 2 is V2 = 2.222 m/s. The velocity decreases as the
pipe diameter increases, consistent with the continuity principle.
1.5 Diameter of Equivalent Uniform Pipe
For a pipe of uniform diameter De carrying the same discharge Q at the average velocity, the
conventional approach is to find De such that the cross-sectional area equals the mean of A1
and A2 . However, for an equivalent pipe carrying the same discharge at a mean velocity, the
mean velocity is:
V1 + V2 5 + 2.222
Vmean = = = 3.611 m/s
2 2
Therefore:
Q = Ae × Vmean
Page 2 of 15
