Question

Given $N$ the pump power, $\rho$ density of pumped water, $H$ total head and $Q$ pump capacity, the overall efficiency of pump is

• A. $\dfrac{Q \rho H g}{1000 N}$
• B. $\dfrac{N \rho H g}{1000 Q}$
• C. $\dfrac{Q \rho N g}{1000 H}$
• D. $\dfrac{1000 Q \rho N g}{H}$

Hint:

Pump efficiency calculations are crucial for optimizing energy usage in mining operations, especially for water management systems.

Answer 1

The Correct Option is A

The overall efficiency of a pump is defined as the ratio of the hydraulic power output to the input power. Hydraulic power is given by $Q \rho H g$, where $Q$ is the pump capacity (flow rate), $\rho$ is the density of the pumped water, $H$ is the total head, and $g$ is the acceleration due to gravity. The input power is $N$. Efficiency $\eta$ is thus $\dfrac{\text{Hydraulic Power}}{N} = \dfrac{Q \rho H g}{N}$. The factor of 1000 in the denominator adjusts for unit consistency (e.g., converting watts to kilowatts). Therefore, the correct formula is $\dfrac{Q \rho H g}{1000 N}$.
Thus, the correct answer is $\dfrac{Q \rho H g}{1000 N}$.