Flow Through Pipes MCQ Quiz - Objective Question with Answer for Flow Through Pipes - Download Free PDF

Explanation:

Manometric Head:

1. 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.

2. 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.

3. 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:

1. Definition: The static head is the vertical distance between the surface of the fluid in the suction tank and the discharge point.

2. Significance: It represents the potential energy of the fluid at different points in the system due to gravity.

3. Application: Important for calculating the energy required to lift the fluid from the suction side to the discharge side.

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%

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, h_f = 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:

\(Δ P = γ \cdot h_f = 9000 \times 170 = 1,530,000~\text{N/m}^2 = 1.53~\text{MN/m}^2\)

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:

\({\left( {{h_L}} \right)_{ent}} = \frac{{0.5 \;×\; v^2}}{{2g}}\)

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, h_f = 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):

\( 1 \text{ m}^3/\text{s} = 1000 \text{ liters/s} \)

\( Q = 0.0314 \text{ m}^3/\text{s} \times1000 \text{ liters/s} \)

\( Q = 31.4 \text{ liters/s} \)

\( Q = 31.4 \text{ liters/s} \times60 \text{ s/min} \)

\( Q = 1884 \text{ lpm} \)

Explanation:

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;\({V_{avg}} = \frac{1}{{8\mu }}\left( { - \frac{{\delta P}}{{\delta x}}} \right){R^2} = \frac{1}{{32\mu }}\left( { - \frac{{\delta P}}{{\delta x}}} \right){D^2}\)

\(\Longrightarrow \frac{1}{{32\mu }}\left( {\frac{{{p_1} - {p_2}}}{L}} \right){D^2} = {V_{avg}}\)

\(\Longrightarrow \frac{{{p_1} - {p_2}}}{{\rho g}} = \frac{{32\mu {V_{avg}}}}{{\rho g{D^2}}}\)

\(\Longrightarrow \frac{{{p_1} - {p_2}}}{{\rho g}} = \frac{{128\mu {Q}}}{{\pi ×\rho g{D^4}}}\)

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

Explanation:
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.

Reynold 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

Concept:

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 %.

Darcy 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, h_f = 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

\(f = \frac{{0.316}}{{R_e^{\frac{1}{4}}}}\)

Concept:

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,

\({H_l} = \frac{{{{\left( {{V_1} - {V_2}} \right)}^2}}}{{2g}} = \frac{{V_1^2{{\left( {1 - \frac{1}{4}} \right)}^2}}}{{2g}} = \frac{{V_1^2\left( {\frac{9}{{16}}} \right)}}{{2g}}\)

\(H_l=\frac{9}{{16}}\left( {\frac{{V_1^2}}{{2g}}} \right)\)

Concept:

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,

\(\therefore {h_L} = {\left( {\frac{({{V_1} - {V_2}})^2}{{2g}}} \right)} = {\left( {\frac{({{V_1} - \frac{1}{4}{V_1}})^2}{{2g}}} \right)}\)

\(\Rightarrow {h_L} = \frac{9}{{16}}\frac{{V_1^2}}{{2g}}\)

Concept:

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:

ρ_m = 2000 kg/m³ ⇒ S_m = 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

Concepts:

The coefficient of discharge (C_d) is the ratio of the actual discharge (Q_a) to theoretical discharge (Q_th) .

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:

Q_th = A × V_th

V_th is the theoretical velocity of flow

Calculations:

Given: V_th = 2 m/s; A = 0.4 m²

So, Q_th = 0.4 × 2 = 0.8 m³/s  or 800 l/s

∴ C_d = 400/800

C_d = 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.

Concept:

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