Force of water out of a hose

[zaguar]

The question states: A firemans hose has a nozzle of area 5cm^2 and the hose has a cross sectional area of 360cm^2. The hose ejects water at 25m/s and at a rate of 12L/s What is the force that the fireman must exert to keep the hose steady?

My attempt:

What I have done so far is to calculate the speed of water as it travels through the parts of the hose.

In the nozzle it travels at 24m/s (12000/500) and in the hose it is traveling at (12000/36000) = 0.33ms^-1, and as it is ejected, it is traveling at 25m/s.

So it goes from traveling at 0.33m/s to 24m/s to 25m/s. Hence, the force is

m(v-u)/t = 12kg/s * (25-0.33) = 296N​

So the fireman exerts a force of 296N in the direction of the flow of water.

Is this correct? It seems to simplistic to ignore the 24m/s in the nozzle, yet it seems to work.

 

[Kurdt]

This question is to do with momentum. Try and use the conservation of momentum to find the force the firefighter must exert on the hose to steady himself. You know the volume of water being ejected so its easy to find the mass. Also think of the time dependence of the mass.

 

[zaguar]

Sorry, but I don't understand what you're getting at. What was wrong with the original approach?

 

[siddharth]

Sorry, but I don't understand what you're getting at. What was wrong with the original approach?

How is m=12kg/s?
The mass flow rate will depend on the cross section of the hose and the velocity. So, it won't be constant.

More specifically $$\dot{m} = \rho A v$$, where $$\dot{m}$$ is the mass flow rate.

What Kurdt says, is to find the difference in the momentum entering and leaving the hose. You can find the force from that.

 

[zaguar]

Isn't that what I did?

Isn't the flow rate always going at 12L/s and the velocity changing, so that there is always 12 litres of water going through the pipe per second, but the speed is different in different parts of the pipe?

So the water is traveling at 0.33m/s in the hose, and exiting at 25m/s. If it is doing that, F=(mv-mu)/t=12*(25-0.33)=296N