self priming slurry pump chart

Solved: A water pump is used to deliver the water at constant rate. By what factor power of pump s [Others]

Answer

The correct option is B. $$n^3$$n 3

Explanation

To determine the factor by which the power of a pump must be increased to raise the delivery rate by a factor of $$n$$n, we can analyze the relationship between power, mass flow rate, and velocity.

1. Understand the power equation. The power $$P$$P delivered by the pump can be expressed as: $$P = \vec{F} \cdot \vec{V}$$P=F⋅V where $$\vec{F}$$F is the force exerted by the pump and $$\vec{V}$$V is the velocity of the water being pumped.

2. Relate power to mass flow rate. The power can also be expressed in terms of mass flow rate $$\dot{m}$$m˙ and velocity $$v$$v: $$P = \dot{m} \cdot v$$P=m˙⋅v If we denote the initial mass flow rate as $$\dot{m}$$m˙ and the initial velocity as $$v$$v, then the power becomes: $$P = \dot{m} v$$P=m˙v

3. Increase the delivery rate. If we want to increase the delivery rate by a factor of $$n$$n, the new mass flow rate $$\dot{m}'$$m˙′ becomes: $$\dot{m}' = n \cdot \dot{m}$$m˙′=n⋅m˙ Assuming the velocity also increases by a factor of $$n$$n (i.e., $$v' = n \cdot v$$v′=n⋅v), we can express the new power $$P'$$P′ as: $$P' = \dot{m}' \cdot v' = (n \cdot \dot{m}) \cdot (n \cdot v) = n^2 \cdot \dot{m} \cdot v = n^2 P$$P′=m˙′⋅v′=(n⋅m˙)⋅(n⋅v)=n 2⋅m˙⋅v=n 2 P

4. Final power relationship. The power required to achieve the increased delivery rate is: $$P' = n^3 P$$P′=n 3 P This shows that the power must be increased by a factor of $$n^3$$n 3

Thus, the correct answer is that the power of the pump should be increased by a factor of $$n^3$$n 3