# Power drawn by a water pump to project water through nozzle

• omoplata
In summary, the problem involves finding the power consumption of a pump necessary to lift water to a certain height and give it a certain velocity. The solution involves using various equations, including ##v_f^2 = v_i^2 + 2 a S##, ##P = x \rho g##, ##F = P A##, and ##Q = F v##. After calculations, the answer given in the book is ##25\rm{kW}##, while the calculated answer is ##2.77\rm{kW}##. The discrepancy is due to not considering the kinetic energy of the water in the calculations.
omoplata

## Homework Statement

[/B]
This is problem 10.27 in the Exercises for the Feynman Lectures on Physics, or Problem C-2 in Chapter 10 of Leighton and Vogt's Exercises in Introductory Physics.

## Homework Equations

Using ##v_f^2 = v_i^2 + 2 a S## for the water in the vertical direction, we can find the initial velocity ##v_i## of the water. ##v_f## is the final velocity, ##a## is the acceleration (or in this case the deceleration), ##S## is the distance traveled.

I can use ##P = x \rho g## to find the pressure difference the pump has to overcome, where ##x## is the height the water has to climb, ##\rho## is the density of water, and ##g## is the gravitational field.

##F = P A## to find the force necessary, where ##A## is the cross sectional area of the nozzle.

##Q = F v## to find the rate of work done on the water, where ##v## is the velocity of the water.

Then I can divide by 60% to find the actual power consumption of the motor.

## The Attempt at a Solution

The initial vertical velocity is ##v_i = v \sin(30^{\circ}) = v/2##. The vertical velocity at the apex is 0. Acceleration is ##-g##. Vertical distance is ##h##. Therefore,
$$0 = \left( \frac v 2 \right) ^2 - 2\ g\ h \\v = \sqrt{8\ g\ h}$$
So I can find the velocity ##v## of the water. To find the pressure difference,
$$P = x\ \rho\ g$$
To find the force,
$$F = P\ A = x\ \rho\ g\ A$$
To find the power expended on the water,
$$Q = F\ v = x\ \rho\ g\ A\ \sqrt{8\ g\ h}$$
I'm going to use the following conversion factors.
$$1\ \rm{sq\ in} = 6.45 \times 10^{-4}\ \rm{m^2} \\1\ \rm{ft} = 3.05 \times 10^{-1}\ \rm{m} \\1\ \rm{lb\ ft^{-3}} = 1.60 \times 10^1 \rm{kg\ m^{-3}}$$
After calculation, I get ##Q = 1.66 \times 10^3 \rm{W}##. After dividing by 60%, I get ##2.77 \times 10^3 \rm{W}##, or ##2.77\rm{kW}##. The answer given in the book is ##25\rm{kW}##. Where did I go wrong?

The pump needs to give the water its kinetic energy as well as lift the water. Rather than relate power to force, it might be easier to relate power to energy.

## 1. What factors affect the power drawn by a water pump?

The power drawn by a water pump is affected by several factors, including the flow rate of the water, the pressure difference between the pump inlet and outlet, the efficiency of the pump, and the viscosity of the fluid being pumped.

## 2. How does the flow rate of water affect the power drawn by a water pump?

The flow rate of water is directly proportional to the power drawn by a water pump. This means that as the flow rate increases, the power required to pump the water through the nozzle also increases. This is because more energy is needed to overcome the resistance of the fluid and maintain a constant flow rate.

## 3. Why does the pressure difference between the pump inlet and outlet affect the power drawn by a water pump?

The pressure difference between the pump inlet and outlet, also known as the head, is a measure of the energy required to move the water through the pump. The higher the head, the more power is needed to overcome the resistance of the fluid and maintain a constant flow rate through the nozzle.

## 4. How does the efficiency of a water pump impact the power drawn?

The efficiency of a water pump is a measure of how well it converts the input power into useful work, which in this case is pumping water through a nozzle. A more efficient pump will require less power to achieve the same flow rate, while a less efficient pump will require more power to achieve the same flow rate.

## 5. Does the viscosity of the fluid being pumped affect the power drawn by a water pump?

Yes, the viscosity of the fluid being pumped can have a significant impact on the power drawn by a water pump. This is because more energy is needed to overcome the resistance of a thicker fluid, resulting in a higher power requirement. Therefore, the power drawn by a water pump will be greater when pumping thicker fluids compared to thinner fluids.

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