Roddilla Posts: 145, Reputation: 3 Junior Member #1 Mar 30, 2011, 11:06 AM
Water passes along the hose of diameter 2.5 cm and out through nozzle of end diameter
Water passes along the hose of diameter 2.5 cm and out through nozzle of end diameter 1 cm at a rate of 2kg/s. Density of water is 1000Kgm-3 (a) Calculate volume flow rate (b) Find speeds of water as it passes through pipe and out of nozzle? (c) calculate rate of change of momentum of water along the nozzle
 jcaron2 Posts: 986, Reputation: 204 Senior Member #2 Mar 30, 2011, 03:04 PM
(a) Volume flow rate will be the volume of water that passes through per unit time. We know the mass per unit time (2 kg/s), so all we have to do is convert that mass to volume:

$2 \frac{kg}{s} \cdot \frac{1 m^3}{1000 kg}=0.002 \frac{m^3}{s}$ (a.k.a. 2 L/s)

(b) We now know what volume of water exits the nozzle in 1 second. So if we know how much length that volume of water would occupy in the hose, we know how far the water behind it must be moving each second in order to replace it. Since the hose is cylindrical, we can simply use the equation for the volume of a cylinder:

$V = \pi r^2 L$

$0.002 = \pi (0.0125)^2 L$

$L \; \approx \; 4.1\; m$

so

$v_{hose} \; \approx \; 4.1\;\frac ms$

To calculate the velocity of the water through the nozzle, with its smaller diameter, you can use the same trick (even though the nozzle may only be a few mm thick, the velocity of the water would be the same as if the entire hose had a 1 cm diameter). Just substitute the smaller radius into the equations above. What do you get for an answer?

$v_{nozzle} \; = \; ?$

(c) 2 kg of water flows at 4.1 m in 1 second on one side of the nozzle. Therefore, its momentum is 8.2 kg m/s upstream of the nozzle.

$p_{hose} \; = 2 \; kg \; \cdot \; 4.1 \; \frac ms\; \approx \; 8.2\;\frac {kg \cdot m}{s}$

On the other side of the nozzle, the water is traveling quite a bit faster. So what is the momentum of 1 second's worth of water (i.e. 2 kg) downstream of the nozzle?

$p_{nozzle} \; = \; ?$

The difference between the two is the amount that the water's momentum is changing in one second (a.k.a. the rate of change of momentum). The answer should be in kg m / s^2, otherwise known as Newtons.
 Roddilla Posts: 145, Reputation: 3 Junior Member #3 Mar 31, 2011, 12:04 AM
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But the question doesn't state that the nozzle's volume is 0.002 m3
 jcaron2 Posts: 986, Reputation: 204 Senior Member #4 Mar 31, 2011, 05:16 AM
It doesn't matter what the volume of the nozzle is, only the diameter. As the water leaves the nozzle, it doesn't know or care where it was previously (or what the volume or diameter was of the nozzle, tank, hose, vessel, etc. It just squeezes through a particular orifice at the end of the nozzle (with a diameter of 1 cm in this case), and it goes through at a given flow rate (told to you explicitly in the question). It doesn't matter if the nozzle is just a thin plate of metal with a hole in it at the end of the hose or if it's a hose, itself, with a smaller diameter that's 1 cm long or 2 km long. Regardless, if 2 kg passes out the end in 1 second, that's 0.002 m^3 of water per second no matter what. And, since water is virtually incompressible, in order to squeeze through that orifice, its speed has to increase to the same degree as if the entire hose and everything was that same diameter.

Does that make any more sense?
 Roddilla Posts: 145, Reputation: 3 Junior Member #5 Mar 31, 2011, 07:42 AM
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So you are saying that water takes 1 second to move along nozzle?
 Roddilla Posts: 145, Reputation: 3 Junior Member #6 Mar 31, 2011, 07:47 AM
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The fact that 2 Kg of water are being replaced every second doesn't mean that every second 2kg are having their momentum change from M*4.07ms-1 to M*25.5ms-1
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #7 Mar 31, 2011, 07:54 AM

Please Roddilla, be aware that giving thumbs down requires the answer of the poster you're rating to be either factually incorrect (which is what you can say here) or dangerous to the poster.

However, the post of jcaron2 is nowhere wrong. You have been misunderstanding his post.

"Regardless, if 2 kg passes out the end in 1 second, that's 0.002 m^3 of water per second no matter what."

Does that mean that water takes 1 second to move along the nozzle? Your question "so you are saying that water takes 1 second to move along nozzle?" is wrong, since you are not indicating any volume of water. Just saying that 'water' takes 1 second to move along the nozzle is vague, because it might be 1 drop of water in 1 second, or a thousand litres of water a second and this will still hold true, according to what you asked.

The RATE of flow of water in your question is 2kg/s. 2 kg of water occupies 0.002 m^3.

So, if you have 2kg of water flowing through the nozzle in 1 second, this is the same as saying that you have 0.002 m^3 of water flowing through the nozzle in 1 second.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #8 Mar 31, 2011, 08:07 AM

Originally Posted by Roddila
The fact that 2 Kg of water are being replaced every second doesn't mean that every second 2kg are having their momentum change from M*4.07ms-1 to M*25.5ms-1
Why are you saying that?

Momentum is the product of mass and velocity.

Since the rate of inflow and outflow of water is constant, then saying that the change in momentum is not the same for every 2 kg of water means that some water (mass) is being lost/gained.
 jcaron2 Posts: 986, Reputation: 204 Senior Member #9 Mar 31, 2011, 10:43 AM

The fact that 2 Kg of water are being replaced every second doesn't mean that every second 2kg are having their momentum change from M*4.07ms-1 to M*25.5ms-1
Uh... yes, that's EXACTLY what it means! 2 kg of water approaches the nozzle every second with a momentum of 4.07m/s. After passing through the nozzle, that same 2 kg of water now has a momentum of 25.5 m/s. It's that simple. The fact that the entire 2 kg of water isn't necessarily contained within the nozzle at any one time is irrelevant. The simple fact is that all 2 kg (i.e. 2 liters, i.e. 0.002 m^3) flowed THROUGH the nozzle. I could just as easily have said that 2 grams of water passes through the nozzle every millisecond, 1 microgram passes through every half nanosecond, or 7.2 metric tons per hour. Regardless, the rate of change of momentum (change in momentum over change in time) would have be the same. The question is really asking about the instantaneous change in momentum (i.e. the change in velocity times the mass per time in the limit as t --> 0). However, since the flow rate is constant, no calculus is necessary. The limit as t --> 0 is the same as the value when t = 1s (or when t = 1 ms, or 0.5 ns, or 1 hr as in my other examples).

And since you don't seem to believe me (even though you should), here's a site for fire fighters which gives the basic formulas for computing the force of water leaving the nozzle of a fire hose.

How to Calculate Nozzle Reaction

You'll have to convert to imperial units of pounds and inches, but if you plug the values into their formula, you'll find that their answer agrees perfectly with mine - something in the neighborhood of 9.6 lbs of thrust out the end of the nozzle (which is the same as 42.8N - the answer to part (c) of your question).
 jcaron2 Posts: 986, Reputation: 204 Senior Member #10 Mar 31, 2011, 10:57 AM
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So MY reputation gets negatively impacted because YOU don't get it? Thanks.
 Roddilla Posts: 145, Reputation: 3 Junior Member #11 Mar 31, 2011, 11:20 AM
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Sorry I didn't know it worked like that - but I cannot understand it if you have 2kg of water passing through nozzle per second it doesn't mean that it takes 1 second for momentum to change
 Roddilla Posts: 145, Reputation: 3 Junior Member #12 Mar 31, 2011, 11:28 AM
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I am starting to understand now - thanks for your patience I just didn't know that by saying no I would be harming your reputation

As I am understanding it if the nozzle has a volume of x this is being filled with water and water is getting removed from it at a time y. The volume doesn't count since it is the ratio of x on y which is always constant.
 jcaron2 Posts: 986, Reputation: 204 Senior Member #13 Mar 31, 2011, 12:14 PM
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Yes, you've got it! I'm glad it's making a little more sense now. No problem about the "not helpful" rep. I didn't mean to sound so snarky about it.
 jcaron2 Posts: 986, Reputation: 204 Senior Member #14 Mar 31, 2011, 09:30 PM
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By the way, I noticed the rep changed from a No to a Yes. Thanks, if you were the one that changed it. (I didn't know you could do that). Otherwise, if one of the admins changed it after seeing your reaction, thanks to them.

Either way, it's no big deal and perfectly understandable that you didn't know the implications of a "No".

I'm glad I was able to help. If you still have any more questions about it or need further clarification, don't hesitate to ask.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #15 Apr 1, 2011, 07:31 AM

Yes jcaron, you can change a rep you gave on Go. Somehow weird... while it has its good use in few instances.

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