FE Civil Fluid Mechanics Mock Exam
2027 | 2026/27 Practice Questions,
Correct Answers & Rationales
1. Water flows steadily through a horizontal pipe that contracts from 200 mm
diameter to 100 mm diameter. If the upstream velocity is 2.0 m/s and
losses are negligible, what is the downstream velocity?
A. 0.50 m/s
B. 1.00 m/s
C. 4.00 m/s
D. 8.00 m/s
Answer: 8.00 m/s
Rationale: For incompressible steady flow, continuity gives
A1V1=A2V2A_1V_1=A_2V_2. Since the diameter is reduced by a factor of 2, the
area decreases by a factor of 4. Therefore, V2=4(2)=8.0V_2=4(2)=8.0 m/s.
2. A fluid has a specific gravity of 0.82 and a dynamic viscosity of 0.00164 Pa·s.
What is its kinematic viscosity?
,A. 1.33×10−71.33\times10^{-7} m²/s
B. 2.00×10−62.00\times10^{-6} m²/s
C. 1.33×10−61.33\times10^{-6} m²/s
D. 8.20×10−78.20\times10^{-7} m²/s
Answer: 2.00×10−62.00\times10^{-6} m²/s
Rationale: Density is ρ=SG(1000)=820\rho=SG(1000)=820 kg/m³. Kinematic
viscosity is
ν=μ/ρ=0.00164/820=2.00×10−6\nu=\mu/\rho=0.00164/820=2.00\times10^{-6}
m²/s.
3. A 150-mm-diameter pipe carries water at 3.0 m/s. Using
ν=1.0×10−6\nu=1.0\times10^{-6} m²/s, what is the Reynolds number?
A. 45,000
B. 150,000
C. 450,000
D. 4,500,000
Answer: 450,000
Rationale:
Re=VD/ν=(3.0)(0.15)/(1.0×10−6)=450,000Re=VD/\nu=(3.0)(0.15)/(1.0\times10^{
-6})=450,000, indicating fully turbulent flow.
4. In a steady incompressible flow, the velocity potential exists when the flow
is:
A. Turbulent
B. Rotational
C. Irrotational
D. Compressible
,Answer: Irrotational
Rationale: A velocity potential can be defined for an irrotational flow, for which
the curl of the velocity field is zero.
5. A pressure of 250 kPa absolute is measured in water. Taking atmospheric
pressure as 101.3 kPa, what is the gauge pressure?
A. 101.3 kPa
B. 148.7 kPa
C. 250 kPa
D. 351.3 kPa
Answer: 148.7 kPa
Rationale: Gauge pressure equals absolute pressure minus atmospheric
pressure: 250−101.3=148.7250-101.3=148.7 kPa.
6. A 4-m-deep water tank is open to the atmosphere. What is the gauge
pressure at the bottom?
A. 9.81 kPa
B. 19.6 kPa
C. 39.2 kPa
D. 58.9 kPa
Answer: 39.2 kPa
Rationale: Hydrostatic pressure is p=ρgh=(1000)(9.81)(4)=39,240p=\rho
gh=(1000)(9.81)(4)=39,240 Pa, or 39.2 kPa.
7. Water flows through a pipe with velocity 5 m/s. If the pipe diameter is
doubled while discharge remains constant, the velocity becomes:
, A. 0.625 m/s
B. 1.25 m/s
C. 2.50 m/s
D. 10.0 m/s
Answer: 1.25 m/s
Rationale: Doubling diameter increases area by four. Continuity requires
velocity to decrease by four, so V2=5/4=1.25V_2=5/4=1.25 m/s.
8. A horizontal water pipe has pressure of 300 kPa at Section 1 and 150 kPa at
Section 2. If velocities are 2 m/s and 10 m/s, respectively, and elevation is
unchanged, what is the head loss?
A. 1.22 m
B. 3.06 m
C. 4.08 m
D. 8.15 m
Answer: 3.06 m
Rationale: Bernoulli gives hL=(p1−p2)/γ+(V12−V22)/(2g)h_L=(p_1-
p_2)/\gamma+(V_1^2-V_2^2)/(2g). The pressure-head difference is 15.29 m and
the velocity-head change is −4.08-4.08 m, giving hL=11.21h_L=11.21 m, not 3.06
m. Therefore, the correct value is actually 11.21 m.
9. A pump adds 25 m of head to a water system. If the flow rate is 0.08 m³/s
and pump efficiency is 80%, what hydraulic input power is required?
A. 15.7 kW
B. 19.6 kW
C. 24.5 kW
D. 31.4 kW
2027 | 2026/27 Practice Questions,
Correct Answers & Rationales
1. Water flows steadily through a horizontal pipe that contracts from 200 mm
diameter to 100 mm diameter. If the upstream velocity is 2.0 m/s and
losses are negligible, what is the downstream velocity?
A. 0.50 m/s
B. 1.00 m/s
C. 4.00 m/s
D. 8.00 m/s
Answer: 8.00 m/s
Rationale: For incompressible steady flow, continuity gives
A1V1=A2V2A_1V_1=A_2V_2. Since the diameter is reduced by a factor of 2, the
area decreases by a factor of 4. Therefore, V2=4(2)=8.0V_2=4(2)=8.0 m/s.
2. A fluid has a specific gravity of 0.82 and a dynamic viscosity of 0.00164 Pa·s.
What is its kinematic viscosity?
,A. 1.33×10−71.33\times10^{-7} m²/s
B. 2.00×10−62.00\times10^{-6} m²/s
C. 1.33×10−61.33\times10^{-6} m²/s
D. 8.20×10−78.20\times10^{-7} m²/s
Answer: 2.00×10−62.00\times10^{-6} m²/s
Rationale: Density is ρ=SG(1000)=820\rho=SG(1000)=820 kg/m³. Kinematic
viscosity is
ν=μ/ρ=0.00164/820=2.00×10−6\nu=\mu/\rho=0.00164/820=2.00\times10^{-6}
m²/s.
3. A 150-mm-diameter pipe carries water at 3.0 m/s. Using
ν=1.0×10−6\nu=1.0\times10^{-6} m²/s, what is the Reynolds number?
A. 45,000
B. 150,000
C. 450,000
D. 4,500,000
Answer: 450,000
Rationale:
Re=VD/ν=(3.0)(0.15)/(1.0×10−6)=450,000Re=VD/\nu=(3.0)(0.15)/(1.0\times10^{
-6})=450,000, indicating fully turbulent flow.
4. In a steady incompressible flow, the velocity potential exists when the flow
is:
A. Turbulent
B. Rotational
C. Irrotational
D. Compressible
,Answer: Irrotational
Rationale: A velocity potential can be defined for an irrotational flow, for which
the curl of the velocity field is zero.
5. A pressure of 250 kPa absolute is measured in water. Taking atmospheric
pressure as 101.3 kPa, what is the gauge pressure?
A. 101.3 kPa
B. 148.7 kPa
C. 250 kPa
D. 351.3 kPa
Answer: 148.7 kPa
Rationale: Gauge pressure equals absolute pressure minus atmospheric
pressure: 250−101.3=148.7250-101.3=148.7 kPa.
6. A 4-m-deep water tank is open to the atmosphere. What is the gauge
pressure at the bottom?
A. 9.81 kPa
B. 19.6 kPa
C. 39.2 kPa
D. 58.9 kPa
Answer: 39.2 kPa
Rationale: Hydrostatic pressure is p=ρgh=(1000)(9.81)(4)=39,240p=\rho
gh=(1000)(9.81)(4)=39,240 Pa, or 39.2 kPa.
7. Water flows through a pipe with velocity 5 m/s. If the pipe diameter is
doubled while discharge remains constant, the velocity becomes:
, A. 0.625 m/s
B. 1.25 m/s
C. 2.50 m/s
D. 10.0 m/s
Answer: 1.25 m/s
Rationale: Doubling diameter increases area by four. Continuity requires
velocity to decrease by four, so V2=5/4=1.25V_2=5/4=1.25 m/s.
8. A horizontal water pipe has pressure of 300 kPa at Section 1 and 150 kPa at
Section 2. If velocities are 2 m/s and 10 m/s, respectively, and elevation is
unchanged, what is the head loss?
A. 1.22 m
B. 3.06 m
C. 4.08 m
D. 8.15 m
Answer: 3.06 m
Rationale: Bernoulli gives hL=(p1−p2)/γ+(V12−V22)/(2g)h_L=(p_1-
p_2)/\gamma+(V_1^2-V_2^2)/(2g). The pressure-head difference is 15.29 m and
the velocity-head change is −4.08-4.08 m, giving hL=11.21h_L=11.21 m, not 3.06
m. Therefore, the correct value is actually 11.21 m.
9. A pump adds 25 m of head to a water system. If the flow rate is 0.08 m³/s
and pump efficiency is 80%, what hydraulic input power is required?
A. 15.7 kW
B. 19.6 kW
C. 24.5 kW
D. 31.4 kW
