Question

An oil of density $870 \,\text{kg/m}^3$ and viscosity $0.036 \,\text{Pas}$ flows through a straight pipe of 10 cm diameter and 1.5 km length at the flow rate of 250 liters per minute under 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 $10 \,\text{m/s}^2$, the minor head loss contributed by the presence of the valve in m (rounded off to 2 decimal places) is ...............

Hint:

For laminar pipe flow, always use Hagen–Poiseuille relation: $h_f = \dfrac{32 \mu V L}{\rho g D^2}$. Minor losses are obtained by subtracting theoretical major loss from total measured head loss.

Answer 1

Correct Answer: 0.85

Step 1: Extract given data.
- Density of oil: $\rho = 870 \,\text{kg/m}^3$. - Viscosity: $\mu = 0.036 \,\text{Pas}$. - Diameter: $D = 0.1 \,\text{m}$. - Length: $L = 1500 \,\text{m}$. - Flow rate: $Q = 250 \,\text{L/min} = \dfrac{250}{1000 \times 60} = 0.004167 \,\text{m}^3/s$. - Total head loss (measured): $h_{total} = 11.60 \,\text{m}$. - $g = 10 \,\text{m/s}^2$. 

Step 2: Compute velocity.
Area of pipe: \[ A = \frac{\pi D^2}{4} = \frac{\pi (0.1)^2}{4} = 7.854 \times 10^{-3} \,\text{m}^2. \] Velocity: \[ V = \frac{Q}{A} = \frac{0.004167}{7.854 \times 10^{-3}} = 0.531 \,\text{m/s}. \] 

Step 3: Reynolds number.
\[ Re = \frac{\rho V D}{\mu} = \frac{870 \times 0.531 \times 0.1}{0.036}. \] \[ Re = \frac{46.2}{0.036} \approx 1283. \] Since $Re<2000$, the flow is laminar. 

Step 4: Theoretical (major) head loss for laminar flow.
Darcy–Weisbach equation for laminar flow: \[ h_f = \frac{32 \mu V L}{\rho g D^2}. \] Substitute values: \[ h_f = \frac{32 (0.036)(0.531)(1500)}{870 (10)(0.1^2)}. \] Numerator: $32 \times 0.036 \times 0.531 \times 1500 = 918.9$. Denominator: $870 \times 10 \times 0.01 = 87$. \[ h_f = \frac{918.9}{87} = 10.56 \,\text{m}. \] 



Step 5: Minor loss due to valve.
\[ h_{minor} = h_{total} - h_f = 11.60 - 10.56 = 1.04 \,\text{m}. \] Final Answer:
\[ \boxed{1.04 \,\text{m}} \]