# PhysicsPhysics Engine Pump problem

#### Jerome

##### New member
An engine pumps water from a river 10m below its own level and discharges it through a nozzle of diameter 10cm with a speed of 50ms^-1. Find the power required assuming (a) no losses. (b) 70% efficiency. Water weighs 10^3kgm^-3. (g = 10ms^-2)

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
An engine pumps water from a river 10m below its own level and discharges it through a nozzle of diameter 10cm with a speed of 50ms^-1. Find the power required assuming (a) no losses. (b) 70% efficiency. Water weighs 10^3kgm^-3. (g = 10ms^-2)

Hi Jerome!

I am not sure where you are stuck, so I'll just give some hints.

Suppose we discharge water for 1 second.
What is the volume that comes out of the nozzle during this 1 second?
What is its mass?
How much energy does it take to transport that mass 10 meters up?

#### Jerome

##### New member
how to go about the diameter thing!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
how to go about the diameter thing!
Well, what is the area of a circle disk with the given diameter?

#### Jerome

##### New member
Here....
d = 2 * r
10 = 2 * r
r = 10/2
r = 5cm
Area = PI*r*r
Area = (22/7) * 25
Area = 550/7
= 78.57 ~
am i correct?

#### SuperSonic4

##### Well-known member
MHB Math Helper
Here....
d = 2 * r
10 = 2 * r
r = 10/2
r = 5cm
Area = PI*r*r
Area = (22/7) * 25
Area = 550/7
= 78.57 ~
am i correct?
That's a close estimate (and the right way of working) but in order to avoid rounding errors it's better to leave your answer in terms of $$\displaystyle \pi$$ at this stage and say $$\displaystyle A = 25\pi \text{ cm}^2$$

Now that you have the area of the nozzle can you work out the volume of water that passes through it per second? If you don't know the equation remember that volumetric flow rate has dimensions of $$\displaystyle \text{[length]}^3 \cdot \text{[time]}^{-1}$$

$$\displaystyle Q_V = Av$$

#### Jerome

##### New member
i don't know how to work it out

#### SuperSonic4

##### Well-known member
MHB Math Helper
i don't know how to work it out
As I said in my last post the volumetric flow rate has dimensions of $$\displaystyle \text{[length]}^3 \cdot \text{[time]}^{-1}$$

Area has dimensions of $$\displaystyle \text{[length]}^2$$ and velocity has dimensions $$\displaystyle \text{[length]} \cdot \text{[time]}^{-1}$$

Can you work out how to get the volumetric flow rate using these dimensions?

edit: When working out be wary of units

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