Consider acceleration due to gravity = $9.81$ m $s^{-2}$.
Centrifugal Pump Power Calculation at New Speed
This problem involves calculating the power required for a centrifugal pump when its operating speed changes, using the principles of pump affinity laws.
Given Data Summary
- Initial Speed ($N_1$): $1000$ rpm
- Initial Total Head ($H_1$): $30$ m
- Initial Discharge ($Q_1$): $300$ L/min
- Pump Efficiency ($\eta$): $65\%$ or $0.65$
- Final Speed ($N_2$): $1200$ rpm
- Acceleration due to gravity ($g$): $9.81$ m/s²
- Density of water ($\rho$): Assumed as $1000$ kg/m³
Step 1: Calculate Initial Power Required ($P_1$)
The input power ($P$) required by the pump is calculated using the formula:
$P = \frac{\rho \times g \times Q \times H}{\eta}$First, convert the initial discharge ($Q_1$) from Liters per minute to cubic meters per second:
$Q_1 = 300 \text{ L/min} = \frac{300 \times 10^{-3} \text{ m}^3}{60 \text{ s}} = 5 \times 10^{-3} \text{ m}^3/\text{s}$Now, calculate the initial power ($P_1$) in Watts:
$P_1 = \frac{1000 \text{ kg/m}^3 \times 9.81 \text{ m/s}^2 \times 5 \times 10^{-3} \text{ m}^3/\text{s} \times 30 \text{ m}}{0.65}$ $P_1 = \frac{1471.5}{0.65} \text{ W} \approx 2263.85 \text{ W}$Convert $P_1$ to kilowatts (kW):
$P_1 \approx 2.26385 \text{ kW}$Step 2: Calculate Final Power Required ($P_2$) Using Affinity Laws
Pump affinity laws state that the power ($P$) required is proportional to the cube of the pump speed ($N$):
$\frac{P_2}{P_1} = \left(\frac{N_2}{N_1}\right)^3$Calculate the final power ($P_2$) using the initial power ($P_1$) and the speed ratio:
$P_2 = P_1 \times \left(\frac{N_2}{N_1}\right)^3$ $P_2 = 2.26385 \text{ kW} \times \left(\frac{1200 \text{ rpm}}{1000 \text{ rpm}}\right)^3$ $P_2 = 2.26385 \text{ kW} \times (1.2)^3$ $P_2 = 2.26385 \text{ kW} \times 1.728$ $P_2 \approx 3.91353 \text{ kW}$Final Result
Rounding the final power required to two decimal places:
$P_2 \approx 3.91 \text{ kW}$Important Questions from Pump Characteristics
- Impeller speed of a centrifugal pump is increased by $30\%$. The power requirement of the pump increases by $n$ times. The value of $n$ is ________. (Rounded off to two decimal places)
- A centrifugal pump having an overall efficiency of $75\%$ requires $6 \text{ kW}$ power at $1450 \text{ rpm}$ to deliver water against a suction head of $5 \text{ m}$ and a delivery head of $12 \text{ m}$. If the pump runs at $1650$ $\text{rpm}$ and frictional head losses are negligible, the total head developed by the pump in metres will be
- At a speed of 1800 rpm, a centrifugal pump discharges 50 L s$^{-1}$ at its best point of efficiency for a total head of 25 m. The specific speed of the pump in rpm (round off to 2 decimal places) is ________
A six-stage centrifugal pump delivers 120 L $s^{-1}$ against a total head of 510 m. If the design speed of this pump is 1450 rpm, the specific speed of the pump will be ______________.
- The brake power of a centrifugal pump having an impeller diameter of 200 mm is 1.86 kW. If the impeller is replaced with another impeller of 180 mm diameter, the brake power of the pump in kW will be ________.
