Question

Two reservoirs located at the same altitude are connected by a straight horizontal pipe of length 120 m and inner diameter 0.5 m. A pump transfers a liquid of density 800 kg m$^{-3}$ at a flow rate of 1 m$^{3}$ s$^{-1}$ from Reservoir-1 to Reservoir-2. The liquid levels in Reservoir-1 and Reservoir-2 are 2 m and 10 m, respectively. The friction factor is 0.01. The acceleration due to gravity is 9.8 m s$^{-2}$. The required power of the pump is \(\underline{\hspace{2cm}}\) kW (rounded off to one decimal place).

Hint:

Total pump head is the sum of static head and friction losses; velocity heads cancel when reservoirs are large.

Answer 1

Correct Answer: 87 - 88

Step 1: Calculate Flow Velocity in Pipe

$$v = \frac{Q}{A} = \frac{Q}{\pi D^2/4} = \frac{1}{\pi (0.5)^2/4} = \frac{1}{0.1963} = 5.093 \text{ m/s}$$

Step 2: Apply Bernoulli's Equation with Pump

Taking the liquid surface of Reservoir-1 as reference (datum = 0):

Point 1 (surface of Reservoir-1):

• Elevation: $z_1 = 2$ m (above pipe centerline)
• Pressure: $P_1 = P_{atm}$
• Velocity: $v_1 \approx 0$

Point 2 (surface of Reservoir-2):

• Elevation: $z_2 = 10$ m (above pipe centerline)
• Pressure: $P_2 = P_{atm}$
• Velocity: $v_2 \approx 0$

The energy equation (Bernoulli with pump and friction):

$$\frac{P_1}{\rho g} + \frac{v_1^2}{2g} + z_1 + H_p = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} + z_2 + h_f$$

Since $P_1 = P_2 = P_{atm}$ and $v_1 = v_2 \approx 0$:

$$z_1 + H_p = z_2 + h_f$$

$$H_p = (z_2 - z_1) + h_f = (10 - 2) + h_f = 8 + h_f$$

Step 3: Calculate Friction Head Loss

Using the Darcy-Weisbach equation:

$$h_f = f \frac{L}{D} \frac{v^2}{2g}$$

$$h_f = 0.01 \times \frac{120}{0.5} \times \frac{(5.093)^2}{2 \times 9.8}$$

$$h_f = 0.01 \times 240 \times \frac{25.94}{19.6}$$

$$h_f = 2.4 \times 1.324 = 3.178 \text{ m}$$

Step 4: Calculate Total Head Required

$$H_p = 8 + 3.178 = 11.178 \text{ m}$$

Step 5: Calculate Pump Power

The hydraulic power required by the pump:

$$P = \rho g Q H_p$$

$$P = 800 \times 9.8 \times 1 \times 11.178$$

$$P = 7840 \times 11.178 = 87,635.52 \text{ W}$$

$$P = 87.6 \text{ kW}$$

Answer

The required power of the pump is 87.6 kW (rounded to one decimal place).