Power drawn by a water pump to project water through nozzle

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1. Dec 4, 2016

omoplata

1. The problem statement, all variables and given/known data

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.

2. Relevant 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.

3. 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?

2. Dec 4, 2016

TSny

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.