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Fluid Flow in Pipes, pressures and flow rates help needed 4

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carla00

Mechanical
Jun 16, 2005
22
Hi

Could anyone point me in the right direction ?

I am trying to work out an inlet flow rate (or pressure, whichever is easiest)required for a water sprinkler system.

All the info I have is the outlet flow rate and pressure of the nozzles, and pipe specifications and layout.

I can't use Q1 = Q2 (sqr p1/p2) because I dont have P1 or Q1. I tried guessing P1 as a pressure higher than that of the nozzle (assuming that the pressure in the pipework is higher than that of the nozzle) but the resulting flow rate was greater than that of the nozzle. Surely it should be less ?

I'm probably going about this completely wrongly so would appreciate any assistance.

Cheers

Carla
 
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Hi

I was going to look at head loss once I've found the inlet pressure or flowrate at the inlet to the nozzle. Basically, I'm trying to find the minimum pressure or flow required at the inlet of the nozzles.

Once I know the above, I'm going to work it backwards, looking at the flow required in the pipes, and then accounting for head loss to determine the total inlet flow or pressure.

Basically, the system is a pressurised tank feeding several sprinkler nozzles and I'm trying to work out the required flow rate coming out of the tank.
 
Not exactly. For full cone nozzle, the exponential is generally 0.46 and for full cone wide spray nozzles it is 0.46.

Just note that if you already have the system and now trying to increase the flow by increasing pressure, it is not possible as the things are different at pump end. As the head increases(i.e as the nozzle pressure increases) the flow decreases.

But if you want to just know the flowrate for a given nozzle you can use the above approximations.

Get the engineer's guide for spray technology from the link below. It is very useful.


Regards,
 
You are correct; you can't use Q1 = Q2 (sqr p1/p2) because it is a pump similarity law, and does not apply to your situation.

You don't want the required flow rate coming out of the tank, Q1 = Q2. You want to determine if the tank pressure (P1) is sufficient to deliver Q at P2 within the conveyance system.

Use Hazen-Williams to solve for hf at Q, as wjb suggested. Use Bernoulli's and solve for P1, v1 is 0, v2 is Q/A.

If P1 (derived) is less than P1 (actual), you must change something: P1, the pipe network, Q or P2 (via the sprinkler heads). Sounds like everything else is already spec'd, so P1 it is.


Remember: The Chinese ideogram for “crisis” is comprised of the characters for “danger” and “opportunity.”
-Steve
 
Hi, and thanks for the advice, I'm actually looking at a catalogue that is quite similar to the link.

For the nozzles I am using, the flow rate does increase as pressure increases. I'm looking at a table showing flow/pressure info at the mo.

Hang on a min, I think I'm being stupid here, so please humor me, I am very new to this ! If I know the pressure in the pipe, and the flow through the nozzle (given on the nozzle spec sheet), how can I work that back to get a flow rate in the pipe ?

And surely the eqn Q1=Q2 (sqr p1/p2) cannot be used as I only have one pressure.

 
Iha, I've just read your reply. Please ignore my previous post.

You're right, I do want to find a tank pressure that would be sufficient to supply the nozzles.

I'm not familiar with Hazen Williams so I will get reading over the weekend. Thanks for that
 
"If I know the pressure in the pipe, and the flow through the nozzle (given on the nozzle spec sheet), how can I work that back to get a flow rate in the pipe ?"

Tips on solving for flow/pressure drop:

- The flow through the nozzle is the flow through the pipe. ("Mass continuity")

- The pressure drop from the tank to the atmosphere is the sum of the pressure drop through the piping and the pressure drop through the nozzle. However you calculate the relationship between flow and pressure drop differently for the nozzle and the pipes. The pipes you use a Moody chart or the Hazen-Williams equation, the nozzle you use a chart provided by the manufacturer. You'll have to iterate to reach a solution.
 
Hi Carla,

You don't say so explicitly, but I guess you have a ladder network of branches with nozzles on individual branches? If this is so, and the network is not perfectly symmetrical you will have to solve this by trial and error.

Basically what you need to do is guess flowrates for each nozzle and then work out the pressure drop from each nozzle all the way back to the supply. Then based on the differences you get for the pressure wherever branches meet and for the pressure at the source you have to revise all your nozzle flowrate guesses and do it all again. You will have to repeat this until your guesses for the flowrates are varying from trial to trial by a "sufficiently small" amount.

This is a problem best solved by a computer program. The chances are good that the supplier of the nozzles will be able to do this for you. Network software is fairly expensive, but you can get a free version of some software that will handle a limited number of branches from

When I tested this software it seemed to work in most cases but it did do strange things now and then. Let it solve the problem for you and then test the final solution by hand to make sure that it is correct - actually this is a good procedure to follow no matter how good the software is.
 
Thanks for the advice.

Yes it is a pipe network (fairly basic) that branches out to various different nozzles. Luckily the branches will be more or less the same length and I have the info on pressure losses in the nozzles, and I think I even have info on the pipe and connectors.

Right, I'm going to have a look at the Hazen-Williams equation and begin from there.

The objective is to come up with the required pressure, which I will use to find out the force needed to mechanically push the liquid out of the tank.

I will keep you all posted !



 
You are mixing up the things here. First, you should find out what flowrate you require through your nozzles. You can have x pressure drop for y flowrate for one nozzle size and z pressure drop for y flowrate. If x>z then you have higher size nozzle with pressure drop z. So your nozzle cost increases. However, your pumping cost decreases. Get a break even.

Once you fix up the proper nozzle size, you know the flowrate and you know the pressure at the inlet of the nozzle at the expense of which you get the required y flowrate. (i.e pressure head gets converted to velocity head). Your pump discharge pressure should be the sum of frictional drop in the network+residual pressure(I know this is a crude term to say) at the nozzle inlet.

Regards,


 
Hi Quark,

The nozzles are ones that we already use, and vary (generally between 4 types of nozzle)depending upon the application. What I am going to do is begin by looking at several applications and specifying the nozzles before doing any calculations.

Basically, I don't want to make any changes to the system itself (nozzles, pipework, fittings etc). I have all the the data required for the above and was hoping to work backwards from there. The only thing I am looking to change is the method of pressurising the tank, while keeping the pipes and nozzles exactly the same.

What I'm planning to do is....(please tell me if I am wrong, I haven't even had the chance to look into it properly yet!) I will be starting at the nozzles, using the specified flowrate at a given pressure (I will use the current average system pressure to begin with, in order to get an approximate idea of the flowrate out of each nozzle - this is given in the supplier catalogue).

Then I will work back from each nozzle, using Hazen-Williams to get the head loss, hopeflly to get an idea of the pressure of water coming out of the tank.

Aw no, I've just realised I can do this for each branch, but how would I go about calculating the loss in the main feed pipe from which all the branches come off ?

I've a feeling it's going to be a late night tonight !
 
You should better use Hardy Cross analysis for calculating pressure drop in a pipe network. The pressure drop in the mains can be calculated by either Darcy's formula or Hazen's formula(I prefer the first one) in a similar way you do it for branch piping. This link gives you a description of the procedure.


Regards,
 
Sorry for butting in - as a civil engineer I should not be here - but this is one of the most interesting threads I have followed for some time !!. I guess you have a dendritic pipe network with a nozzle on the end of each branch (perhaps some along the branches ?). The flow through wach nozzle is dependent on the prssure.

And you have not fixed any boundary conditions - pressure or flow so you are iterating around in circles ??.

You have to either 1) fix something somewhere in the system e.g flow through a nozzle (which fixes the pressure at that nozzle) or pressure at the tank which will eventually allow you to determine flows through the system or 2) plot curves of different pressures at the tank against flows through the system (System pressure -v- flow curves). I have a feeling that is what you really want to do.

Whatever the solution is iterative.

I suggest you set up a spread sheet.

1) First of all set up a table of pressure v flow for the nozzle

2) For each branch pipeline prepare a curve of flow v headloss.

3) Combine the two curves to give flow v pressure at the head of the branch pipe.

4) At each node prepare a curve of Flow v pressure – (the pressure is the same at the node in each pipe so you add the flows from the curves you prepared at 3) above.

5) Work your way back up the system repeating step 4) at each node.

6) you will eventually end up with a system curve at the tank

QED
 
PS I should have given you also step 0) - set up a curve of flow V head loss per m (ft) length of pipe (for each different pipe diameter) using Hazen Williams, (or whatever head loss formulae you like).

You can use these curves to establish the curves in step 2) and step 5).

The whole process shouldn't take more than 20 minutes!
 
or 2 hours for a recently graduated engineer, especially after a year out of the industry !!

Thanks for the advice everyone, I will have a look tonight and then let you know how it went.

Carla
 
Hi Again,

I worked out the head loss using Hazen-Williams for some branches of the system, and because the system is quite small it turns out the loss is negligible. So back to square one :(

What I am doing now is starting from the tank end. Basically there is a pipe in the tank which is reduced to a smaller diameter as it comes out of the tank. I have set the pressure and flow rate in the system and have used the continuity eqn to get the velocity in pipe 1.

Now I am using bernoulli's eqn to get the pressure in the pipe at the tank end.

The only problem is I'm stuck on the units and can't find anything on the internew that gives me them !

The units I've got are... Q-m^3 / sec, A - M^2, V - m/s.

Any idea what units of pressure I need to be working in ? I think it's Pa, but I'm not sure !
 
If the head loss in the pipes is small then you have nothing left to solve - The pressure at each nozzle is the same as the pressure at the tank and the flow is the sum of the flows through each nozzle. Once you have done that you may check that the head loss is negligible.

I doubt that the accuracy of your calculation warrants you worrying about Barnoulli!!
 
Are you sure about this ? according to Bernnoulli's equation you get an increase in flow rate and a drop in pressure at a pipe contraction (off the top of my head !)
Baring mind that there is a large contraction at the entrance to the pipe and also another a bit further along the pipe, I thought there would be a bit of a pressure change.

I did use bernoulli's eqn to work it out and found that there is a change in pressure. I just didn't know how significant it is as I wastn't sure of the units. I'll assume it is a very small change for now then, and can aways go back and check it later.

I guess it's simple then - use flow rate coming out of the tank to work out the speed of the piston ! I think I can manage that one ! and then just look at the forces (i.e. tank pressure and friction in piston rings) to work out the force required to push the piston.
 
There is a reduction in pressure with increasing velocity (I think that is why planes fly !) but there is also a recovery in pressure with a reduction in velocity. If you are neglecting head losses through the pipes the Energy at the nozzle is the same as the Energy at the tank and you don't need to work all these changes in velocity and pressure through the system.

The gauge pressure head upstream of the nozzle (neglecting losses) is the pressure head at the tank - V^2/2g where v is the velocity in the pipe at the nozzle. But for a first approximation we are neglecting losses and assuming v^2/2g is small and also negligible.

If you are neglecting losses it is that simple.


 
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