Question

An oil of density \(870 \, {kg/m}^3\) and viscosity \(0.036 \, {Pa.s}\) flows through a straight pipe of 10 cm diameter and 1.5 km length at the flow rate of 250 liters per minute under the steady and incompressible flow conditions.
To control the flow rate of oil, a valve is fixed at the middle of the pipe causing no change in the total length of the pipe. The total head loss measured across the two ends of the pipe is \(11.60 \, {m}\). Using gravitational acceleration as \(10 \, {m/s}^2\), the minor head loss contributed by the presence of the valve in m {(rounded off to 2 decimal places) is ..........}

Hint:

In fluid mechanics, total head loss = major loss (due to pipe friction) + minor loss (due to fittings like valves).
Use the Darcy–Weisbach formula for major loss and subtract it from total loss to isolate minor contributions.

Answer 1

Step 1: Given data: 
Density \( \rho = 870 \, {kg/m}^3 \), 
Viscosity \( \mu = 0.036 \, {Pa.s} \), 
Diameter \( D = 0.1 \, {m} \), 
Length \( L = 1.5 \, {km} = 1500 \, {m} \), 
Flow rate \( Q = 250 \, {L/min} = \frac{250}{1000 \times 60} = \frac{1}{240} \, {m}^3/{s} \), 
Total head loss = 11.60 m, 
Acceleration due to gravity \( g = 10 \, {m/s}^2 \) 
Step 2: Velocity in the pipe: 
\[ A = \frac{\pi D^2}{4} = \frac{\pi (0.1)^2}{4} = \frac{\pi}{400} \, {m}^2 V = \frac{Q}{A} = \frac{1/240}{\pi/400} = \frac{400}{240\pi} \approx 0.53 \, {m/s} \] 
Step 3: Calculate Reynolds number: 
\[ Re = \frac{\rho V D}{\mu} = \frac{870 \cdot 0.53 \cdot 0.1}{0.036} \approx 1279.17 \] 
Since \( Re<2000 \), the flow is laminar. 
Step 4: Head loss due to pipe (major loss): 
For laminar flow, Darcy's friction factor \( f = \frac{64}{Re} \approx \frac{64}{1279.17} \approx 0.050 \) 
Using Darcy–Weisbach equation: \[ h_f = f \cdot \frac{L}{D} \cdot \frac{V^2}{2g} h_f = 0.050 \cdot \frac{1500}{0.1} \cdot \frac{(0.53)^2}{2 \cdot 10} \approx 10.7 \, {m} \] 
Step 5: Minor head loss due to valve: \[ h_{{minor}} = h_{{total}} - h_{{major}} = 11.60 - 10.7 = 0.90 \, {m} \]