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Volume and head pressure Question
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Posted by: GJensen

08/22/2006, 08:51:34

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Good Morning,
I am attempting to calculate what volume and height of water is required to provide a minimum .4 gallons per minute pressure for at least 15 minutes to two nozzles. There is no auxillary pump. The only available pressure is that which is created by the volume of water above the nozzles. The total area of water release in the nozzles is approximately 0.044 sq. in.(very small). Pressure would obviously start higher and gradually decline as water is depleted from storage vessel.

What is the approximate volume and height required to maintain the .4gpm for the full 15 minutes? (6 gallons is not the correct answer.)

Thanks for any and all insight.
GJ








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Re: Volume and head pressure
Re: Volume and head pressure -- GJensen Post Reply Top of thread Forum
Posted by: zekeman

08/28/2006, 11:20:49

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Homework problem? I'll assume not
You have to start with a vessel area, A as large as practical to get the minimum vol.
Next you assume the height x which is to be solved. First, you get the minimum height,xmin you need to deliver 0.4gpm or about .05ft3/minute or .000833ft3/sec. if the nozzle area is "a"
vmin=.00083/a and from Bernouli
xmin=vmin^2/2g=(.00083/.044/144)^2/2g=2.716^2/2g=.115 ft
Now invoking
d/dt(Ax)=-av=-a*sqrt(2gx) which says tha the rate of change of volume is the rate at which it is spewed out of the nozzles, and so
(A/a)/sqrt(2g)*dx/sqrt(x)=-dt
Integrating both sides with the x limits of x to xmin, I get
A/a/sqrt(2g){sqrt(x)-sqrt(xmin}=t=15*60sec
I'll let you solve this for x after assuming a value for A. The container volume is xA







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Re: Volume and head pressure
Re: Volume and head pressure -- GJensen Post Reply Top of thread Forum
Posted by: Scooter

08/22/2006, 11:26:01

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I'm not sure volume & height are going to "keep" your flow at
a true .4GPM's. Height will give you Head Pressure, but as the
Height decresses, so does Head Pressure. Is there some reason
you can't just feed this nozzle from,,,like a 55 gal drum or some other container ?







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Re: Re: Volume and head pressure
Re: Re: Volume and head pressure -- Scooter Post Reply Top of thread Forum
Posted by: GJensen

08/22/2006, 12:47:51

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KNOWN:
flow: Minimum 0.4gpm
duration: 15 minutes
min. total volume: (.4 X 15minutes)= 6 gallons
Area of opening: .044 sq. in. (24 holes @ .0018 sq. in. each)

Problem
The storage vessel must be above the nozzles, but as close as possible and part of the assembly. The vessel must be as small as possible and still meet the .4gpm minimum flow at the 15 minute mark.

The two nozzles each have 12 openings, totaling .044 sq. in. of area where fluid will dispense.

Flow will obviously be higher when vessel is full but how much fluid is necessary to satisfy the 15 minute requirement.

Thanks








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Re: Re: Re: Volume and head pressure
Re: Re: Re: Volume and head pressure -- GJensen Post Reply Top of thread Forum
Posted by: ipajewsk

08/23/2006, 12:19:48

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First: Calculate the required velocity of the water flowing through the "nozzle".

(.4 GPM)*(231in^3/gal)/[(60sec/min)*(.044in^2)] = 35in/sec

Next, use Bernoulli's equation to find pressure needed to generate that flow.
For first iteration, assume no elevation change, insignificant losses through nozzle, on streamline, etc:

Pressure = 1/2 x (.0361 lbm/in^3) x (35in/sec)^2
which I think gives a Pressure of 22.1 psi. (Might want to double check my unit conversion)

Now, that is the minimum pressure you must maintain for 15 minutes. Now, how much water you need to achieve that pressure is going to depend on the design of your tank. Its geometry will affect how fast your pressure head changes. That being said, I don't think that this flowrate is very reasonable for your nozzle size. You would need a static pressure head of 50 ft! Without using a pump, you will probably need to increase the size of your nozzles.

Hope this helped!








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Re: Re: Re: Volume and head pressure
Re: Re: Re: Volume and head pressure -- GJensen Post Reply Top of thread Forum
Posted by: Scooter

08/22/2006, 13:49:59

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Looks like we must first change sq. in's to Cub. in's.
To do this we must know the configuration of the nozzles.
(round x length) (Sq. x length),etc. Then:

Total area in Cub.In's divided into 1728 cub. in. per gal.
This should give us the volume in gal's.

Then we can go from there. What do you think ?








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