Flow Through Pipes MCQ Quiz - Objective Question with Answer for Flow Through Pipes - Download Free PDF
Last updated on May 27, 2025
Latest Flow Through Pipes MCQ Objective Questions
Flow Through Pipes Question 1:
Which of the following statements about manometric head is correct?
Answer (Detailed Solution Below)
Flow Through Pipes Question 1 Detailed Solution
Explanation:
Manometric Head:
-
The manometric head is the effective head that accounts for all losses, including friction losses in the system, the velocity head, and the static head.
-
It represents the total energy that a pump delivers to the fluid, factoring in the pumping losses, which include friction losses and other resistance within the system.
-
It is calculated by considering both the suction and discharge pressures as well as the height differences between the inlet and outlet of the pump.
Additional InformationStatic Head:
-
Definition: The static head is the vertical distance between the surface of the fluid in the suction tank and the discharge point.
-
Significance: It represents the potential energy of the fluid at different points in the system due to gravity.
-
Application: Important for calculating the energy required to lift the fluid from the suction side to the discharge side.
Flow Through Pipes Question 2:
The maximum efficiency of transmission of power through a pipe is
Answer (Detailed Solution Below)
Flow Through Pipes Question 2 Detailed Solution
Explanation:
Efficiency of power transmission through pipe is given by,
\(\eta =\frac{H-{{h}_{f}}}{H}\)
Here, H = head available at inlet
hf = frictional head loss
For maximum power transmission through pipe,
\({{h}_{f}}=\frac{H}{3}\)
\(\Rightarrow \eta =\frac{H-\frac{H}{3}}{H}=\frac{2}{3}=0.6667\)
So, ηmax = 66.67%
Flow Through Pipes Question 3:
A horizontal pipeline (diameter = 60 cm) carries oil at the rate of 105 m3/day (specific weight = 9000 N/m3. The frictional head loss of fluid during flow is observed as 8.5 m per 1000 m of pipe run. It is planning to place pumping stations at every 20 km along the pipe, what will be the pressure drop between two pumping stations?
Answer (Detailed Solution Below)
Flow Through Pipes Question 3 Detailed Solution
Concept:
The pressure drop in a pipeline due to frictional head loss can be calculated using:
\(Δ P = γ \cdot h_f\)
Where, ΔP = pressure drop, γ = specific weight of the fluid, hf = head loss over the given length.
Calculation:
Given:
Specific weight of oil, γ = 9000 N/m3
Head loss = 8.5 m per 1000 m of pipe
Distance between pumping stations = 20 km = 20,000 m
Total head loss over 20 km:
\(h_f = \frac{8.5}{1000} \times 20000 = 170~\text{m}\)
Now, calculate pressure drop:
\(Δ P = γ \cdot h_f = 9000 \times 170 = 1,530,000~\text{N/m}^2 = 1.53~\text{MN/m}^2\)
Flow Through Pipes Question 4:
The loss of head at the entrance to pipe (hi) is given by: (where V is the velocity of liquid in pipe and g is the acceleration due to gravity)
Answer (Detailed Solution Below)
Flow Through Pipes Question 4 Detailed Solution
Concept:
Minor losses caused by the disruption of the flow due to the installation of appurtenances, such as valves, bends, and other fittings.
The various losses are given by
1). Sudden expansion loss:
\({\left( {{h_L}} \right)_{exp}} = \frac{{{{\left( {{v_1} - {v_2}} \right)}^2}}}{{2g}}\)
2). Exit loss:
\({\left( {{h_L}} \right)_{exit}} = \frac{{v^2}}{{2g}}\)
3). Entrance loss:
\({\left( {{h_L}} \right)_{ent}} = \frac{{0.5 \;×\; v^2}}{{2g}}\)
Flow Through Pipes Question 5:
Fluid flows through a 100 mm diameter tube at a Reynolds number of 1800. The head loss is 64 m in a 180 m length of tubing. Calculate the discharge in liters per minute (lpm). Assume acceleration due to gravity as 8 ms-2
Answer (Detailed Solution Below)
Flow Through Pipes Question 5 Detailed Solution
Concept:
Fluid Flow in a Pipe: The flow of fluid through a pipe can be described using the Darcy-Weisbach equation for head loss due to friction.
\( h_f = f \cdot \frac{L}{D} \cdot \frac{v^2}{2g} \)
For laminar flow (Re < 2000), the Darcy friction factor is:
\( f = \frac{64}{Re} \)
The discharge (Q) is given by:
\( Q = A \cdot v \)
Calculation:
Given Data:
- Diameter of the tube, D=100 mm = 0.1 m
- Reynolds number, Re=1800
- Head loss, hf = 64 m
- Length of the tube, L=180 m
- Acceleration due to gravity, g=8 m/s²
Calculate the Darcy friction factor (f):
\( f = \frac{64}{Re} = \frac{64}{1800} = 0.03556 \)
Rearrange the Darcy-Weisbach equation to solve for velocity (v):
\( h_f = f \times \frac{L}{D} \times \frac{v^2}{2g} \)
\( 64 = 0.03556 \times\frac{180}{0.1} \times\frac{v^2}{2 \times8} \)
\( 64 = 0.03556 \times1800 \times\frac{v^2}{16} \)
\( 64 = 64.008 \times\frac{v^2}{16} \)
\( v^2 = \frac{64 \cdot 16}{64.008} \)
\( v^2 = 16 \)
\( v = 4 \text{ m/s} \)
Calculate the cross-sectional area (A) of the tube:
\( A = \pi \times\left(\frac{D}{2}\right)^2 \)
\( A = \pi \times\left(\frac{0.1}{2}\right)^2 \)
\( A = \pi \times\left(0.05\right)^2 \)
\( A = \pi \times0.0025 \)
\( A = 0.00785 \text{ m}^2 \)
4. Calculate the discharge (Q):
\( Q = A \cdot v \)
\( Q = 0.00785 \times 4 \)
\( Q = 0.0314 \text{ m}^3/\text{s} \)
5. Convert the discharge to liters per minute (lpm):
\( 1 \text{ m}^3/\text{s} = 1000 \text{ liters/s} \)
\( Q = 0.0314 \text{ m}^3/\text{s} \times1000 \text{ liters/s} \)
\( Q = 31.4 \text{ liters/s} \)
\( Q = 31.4 \text{ liters/s} \times60 \text{ s/min} \)
\( Q = 1884 \text{ lpm} \)
Top Flow Through Pipes MCQ Objective Questions
Head loss due to friction in a circular pipe of diameter D, under laminar flow, is inversely proportional to:
Answer (Detailed Solution Below)
Flow Through Pipes Question 6 Detailed Solution
Download Solution PDFExplanation:
Laminar flow through a circular pipe:
In a constant diameter pipe, the pressure drops uniformly along the pipe length (except for the entrance region)
∵ we know that the average velocity through a circular pipe;\({V_{avg}} = \frac{1}{{8\mu }}\left( { - \frac{{\delta P}}{{\delta x}}} \right){R^2} = \frac{1}{{32\mu }}\left( { - \frac{{\delta P}}{{\delta x}}} \right){D^2}\)
\(\Longrightarrow \frac{1}{{32\mu }}\left( {\frac{{{p_1} - {p_2}}}{L}} \right){D^2} = {V_{avg}}\)
\({p_1} - {p_2} = \frac{{32\mu {V_{avg}}}}{{{D^2}}}\)\(\Longrightarrow \frac{{{p_1} - {p_2}}}{{\rho g}} = \frac{{32\mu {V_{avg}}}}{{\rho g{D^2}}}\)
\(\Longrightarrow \frac{{{p_1} - {p_2}}}{{\rho g}} = \frac{{128\mu {Q}}}{{\pi ×\rho g{D^4}}}\)
Now, ΔP = γ × Hl
Putting ΔP, from the above equation, we get
Hl ∝ \(1 \over D^4\)
From the above expression, it is clear that hydraulic gradient is inversely proportional to D4
Confusion PointsThe above equation is called Hagen–Poiseuille equation, which is valid for only laminar flow in a circular pipe, (as asked in the question), and pressure or head loss is due to the viscous effect of the liquid.
While the Darcy formula, is valid for both laminar and turbulent flow in circular or noncircular sections, pressure loss is due to friction only.
So, when the Darcy formula is available, pressure difference Δ P 1/D5
In a uniform laminar flow through a conduit, the hydraulic gradient varies:
Answer (Detailed Solution Below)
Flow Through Pipes Question 7 Detailed Solution
Download Solution PDFExplanation:
Laminar flow through a circular pipe:
In a constant diameter pipe, the pressure drops uniformly along the pipe length (except for the entrance region)
∵ we know that the average velocity through a circular pipe:
\(Hydraulic \ gradient ={change\ in \ head\over length \ over \ which\ change \ in \ head\ occurs}\)
\(i = {\Delta H\over L}\)
As, per Hagen–Poiseuille, head loss is given by
\(h_f = \dfrac{{{p_1} - {p_2}}}{{ρ g}} = \dfrac{{32μ VL}}{{ρ g{D^2}}} \)
Putting the value of head loss in the above equation we get,
\(i= \dfrac{{32μ VL}}{{ρ g{D^2}}} \times {1\over L}\)
\(i= \dfrac{{32μ V}}{{ρ g{D^2}}} \)
So, from the above equation, we can say that
Hydraulic gradient (i) is directly proportional to velocity (V)
∴ i ∝ V
Confusion Points
- The above equation is called Hagen–Poiseuille equation, which is valid for only laminar flow in a circular pipe, (as asked in the question), and pressure or head loss is due to the viscous effect of the liquid.
- While the Darcy formula, is valid for both laminar and turbulent flow in circular or noncircular sections, pressure loss is due to friction only.
In a pipe of diameter 5 cm, water is flowing at a rate of 8 cm/sec. If the dynamic viscosity of water is 1.6 × 10-2 Pa-s, what type of flow is present?
Answer (Detailed Solution Below)
Flow Through Pipes Question 8 Detailed Solution
Download Solution PDFReynold Number:
- It is the ratio of inertia force to viscous force.
- It determines the type of fluid flow.
The expression for Reynold's number is given as:
\(Re\ =\ \frac{ρVD}{μ}\)
Where ρ = density of the fluid, V = mean velocity, D = diameter of the pipe, and μ = dynamic viscosity of the fluid.
- The flow is laminar if Reynold number < 2000.
- The flow is turbulent if Reynold number > 4000.
Calculation:
Given:
D = 5 cm = 5 × 10-2 m, V = 8 cm/sec = 0.08 m/sec, μ = 1.6 × 10-2 Pa-sec, ρ = 1000 kg/m3
\(Re\ =\ \frac{ρVD}{μ}\)
\(Re\ =\ \frac{1000\;\times\;0.08\;\times\;5\;\times\;10^{-2}}{1.6\times\;10^{-2}}=250\)
The flow is laminar as Reynold number < 2000
Water is flowing through a horizontal pipe of constant diameter and the flow is laminar. If the pipe diameter is increased by 50% keeping the volume flow rate constant, then the pressure drop in the pipe due to friction will be decreased by:
Answer (Detailed Solution Below)
Flow Through Pipes Question 9 Detailed Solution
Download Solution PDFConcept:
Hagen Poiseuille Law (Flow of viscous fluid in circular pipes):
Hagen's poiseuille theory is based on the following assumptions:
- The fluid follows newton's law of viscosity.
- There is no slip of fluid particles at the boundary (i.e. the fluid particle adjacent to the pipe will have zero velocity.)
Pressure drop in the pipe is given by, \(Δ P~=~\frac{32μ VL}{D^2}\)
where ΔP = Pressure drop, μ = Dynamic viscosity, V = average velocity of the fluid stream, L = length of pipe, D = Diameter of pipe
Discharge or volume flow rate, Q = A × V ; A = area of cross-section of pipe, V = average velocity
Q = A × V = \(\frac{\pi}{4}d^2 V\) = \(V~=~\frac{4Q}{\pi d^2}\)
putting the value 'V' in the pressure drop equation, we get:
\(Δ P~=~\frac{128\mu QL}{\pi D^4}\)
Calculation:
Given:
Discharge or volume flow rate is constant, i.e. Q = constant
Pipe diameter is increased by 50 %; so, D2 = 1.5 × D1
where D1 and D2 are pipe diameters in the first and second cases.
∵ \(Δ P~=~\frac{128\mu QL}{\pi D^4}\)
The above equation shows that, \(Δ P~\propto~\frac{1}{D^4}\)
\(\frac{Δ P_1}{Δ P_2}~=~\frac{(D_2)^4}{(D_1)^4}\) ⇒ \(\frac{Δ P_1}{Δ P_2}~=~\frac{(1.5D)^4}{D^4}~=~5.0625\)
⇒ ΔP2 ≃ 0.2 ΔP1
Percentage change in pressure drop = \(\frac{\Delta P_1~-~\Delta P_2}{\Delta P_1} ~\times 100\) = \(\frac{1~-~0.2}{1}~\times~100\) = 80 %
The pressure drop in the pipe due to friction will be decreased by 80 %.
Which of the following represent the Darcy’s friction factor in terms of Reynolds number (Re) for the laminar flow in circular pipes?
Answer (Detailed Solution Below)
Flow Through Pipes Question 10 Detailed Solution
Download Solution PDFDarcy Weisbach Equation for friction losses in circular pipe:
\({h_f} = \frac{{f \times L \times {V^2}}}{{2 \times g \times D}}\)
where,
L = length of the pipe, D = diameter of the circular pipe, V = mean velocity of the flow, f = Darcy’s friction factor = 4 × F’, F’ = coefficient of friction, hf = head loss due to friction
For Laminar Flow
Friction Factor \(f = \frac{{64}}{{{R_e}}}\;and\;F' = \frac{1}{4} \times \frac{{64}}{{{R_e}}} = \frac{{16}}{{{R_e}}}\)
For Turbulent flow
\(f = \frac{{0.316}}{{R_e^{\frac{1}{4}}}}\)
The head loss in a sudden expansion from 8 cm diameter pipe to 16 cm diameter pipe in terms of velocity V1, in the smaller pipe is
Answer (Detailed Solution Below)
Flow Through Pipes Question 11 Detailed Solution
Download Solution PDFConcept:
Equation of continuity: A1V1 = A2V2
A = area of cross section = \(\frac{\pi }{4}{D^2}\)
Head loss due to sudden expansion = \(\frac{{{{\left( {{V_1} - \;{V_2}} \right)}^2}}}{{2g}}\)
Where, V1 = velocity before expansion
V2 = velocity after expansion
g = acceleration due to gravity
Calculation:
Given, D1 = 8 cm, D2 = 16 cm
Using continuity equation:
\(d_1^2{V_1} = d_2^2{V_2}\)
\({V_2} = {\left( {\frac{8}{{16}}} \right)^2}{V_1}\)
\({V_2} = \frac{1}{4}{V_1}\)
Therefore, head loss,
\({H_l} = \frac{{{{\left( {{V_1} - {V_2}} \right)}^2}}}{{2g}} = \frac{{V_1^2{{\left( {1 - \frac{1}{4}} \right)}^2}}}{{2g}} = \frac{{V_1^2\left( {\frac{9}{{16}}} \right)}}{{2g}}\)
\(H_l=\frac{9}{{16}}\left( {\frac{{V_1^2}}{{2g}}} \right)\)
A head loss in a sudden expansion from 6 cm diameter pipe to 12 cm diameter pipe in terms of velocity v1 in the smaller diameter pipe is
Answer (Detailed Solution Below)
Flow Through Pipes Question 12 Detailed Solution
Download Solution PDFConcept:
Loss of head due expansion,
\({h_L} = \frac{{{{\left( {{V_1} - {V_2}} \right)}^2}}}{{2g}}\)
Calculation:
Given:
D1 = 6 cm, D2 = 12 cm
We know,
A1V1 = A2V2
\(\Rightarrow \frac{\pi }{4}D_1^2{V_1} = \frac{\pi }{4}D_2^2{V_2}\)
\(\Rightarrow {V_2} = {\left( {\frac{{{D_1}}}{{{D_2}}}} \right)^2}{V_1}\)
\(\Rightarrow {V_2} = \frac{1}{4}{V_1}\)
Loss of head due expansion,
\(\therefore {h_L} = {\left( {\frac{({{V_1} - {V_2}})^2}{{2g}}} \right)} = {\left( {\frac{({{V_1} - \frac{1}{4}{V_1}})^2}{{2g}}} \right)}\)
\(\Rightarrow {h_L} = \frac{9}{{16}}\frac{{V_1^2}}{{2g}}\)
The pressure drop of water in a pipe is measured by a manometer using a liquid of density 2000 kg/m3. The difference in height of the liquid in the limbs is 10 cm. Then, the pressure drop is
Answer (Detailed Solution Below)
Flow Through Pipes Question 13 Detailed Solution
Download Solution PDFConcept:
Apply Bernoulli’s equation between 1 and 2;
\(\frac{{{p}_{1}}}{\rho g}+\frac{v_{1}^{2}}{\rho g}+{{z}_{1}}=\frac{{{p}_{2}}}{\rho g}+\frac{v_{2}^{2}}{\rho g}+{{z}_{2}}+{{h}_{L}}\)
After a bit of simplification,
\(\left( \frac{{{p}_{1}}}{\rho g}+{{z}_{1}} \right)-\left( \frac{{{p}_{2}}}{\rho g}+{{z}_{2}} \right)= h\left( {\frac{{{S_m}}}{S} - 1} \right)\)
The above expression is representing the piezometric head difference.
For the same reference level,
Pressure head is given by:
\(\frac{{{P_1} - {P_2}}}{{\rho g}} = h\left( {\frac{{{S_m}}}{S} - 1} \right)\)
Here, h represents the height difference between the liquid column in two limbs of the manometer.
Calculation:
Given:
ρm = 2000 kg/m3 ⇒ Sm = 2, S = 1, h = 10 cm = 0.1 m
We know that
\(\frac{{{P_1} - {P_2}}}{{\rho g}} = h\left( {\frac{{{S_m}}}{S} - 1} \right)\)
\({{{P_1} - {P_2}}}{{}} = \rho gh\left( {\frac{{{S_m}}}{S} - 1} \right)\)
ΔP = 1000 × 9.81 × 0.1 × (2 - 1)
ΔP = 981 N/m2
A fluid flows through an orifice of an area 0.4 m2 with an actual discharge of 400 l/s. If the theoretical velocity of flow through the orifice is 2 m/s, what is the coefficient of discharge?
Answer (Detailed Solution Below)
Flow Through Pipes Question 14 Detailed Solution
Download Solution PDFConcepts:
The coefficient of discharge (Cd) is the ratio of the actual discharge (Qa) to theoretical discharge (Qth) .
The actual discharge is the discharge obtained when all the looses through orifice or pipe flow are considered. While, theoretical discharge is the discharge obtained under ideal conditions i.e. no loss is considered.
The theoretical discharge is given as:
Qth = A × Vth
Vth is the theoretical velocity of flow
Calculations:
Given: Vth = 2 m/s; A = 0.4 m2
So, Qth = 0.4 × 2 = 0.8 m3/s or 800 l/s
∴ Cd = 400/800
Cd = 0.5
Other important Coefficients:
1. Coefficient of velocity is the ratio of actual velocity to theoretical velocity.
2. Coefficient of contraction is the ratio of cross-section area at vena-contracta to original cross- sectional area.
For a fully–developed flow of water in a pipe having a diameter 10 cm, velocity 0.2 m/s, and kinetic viscosity 10-5 m2/s, what is the value of the Darcy friction factor?
Answer (Detailed Solution Below)
Flow Through Pipes Question 15 Detailed Solution
Download Solution PDFConcept:
Darcy friction factor is define as,
\({\rm{f}} = \frac{{64}}{{{\rm{Re}}}}{\rm{\;where}},{\rm{\;Re}} = {\rm{Raynolds\;no}}.{\rm{\;}}\)
\({\rm{\;Re}} = \frac{{{\rm{\rho VD}}}}{{\rm{\mu }}} = \frac{{{\rm{VD}}}}{{\rm{\nu }}}\)
where, ρ = density of fluid, V = velocity of fluid, D = Diameter of pipe,
v = kinematic viscosity
If Re > 4000 then the flow become turbulent flow
If Re < 2000 then the flow become laminar flow
Calculation:
Given: D = 10 cm = 0.1 m, v = 0.2 m/s, v = 10-5 m2/s
\({\rm{Re}} = \frac{{0.1 × 0.2}}{{{{10}^{ - 5}}}} = 2000{\rm{\;}}\)
Therefore, it is laminar flow
\({\rm{f}} = \frac{{64}}{{2000}} = 0.032\)
For plate,
If Re > \(5 \times {10^5}\) then the flow become turbulent flow
If Re < \(5 \times {10^5}\) then the flow become laminar flow