Eng-Tips is the largest forum for Engineering Professionals on the Internet.
Members share and learn making Eng-Tips Forums the best source of engineering information on the Internet!
-
Congratulations dmapguru on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Liquid Height above outlet nozzle 2
- Thread starter jkate
- Start date
Bernouli's equation. Take the velocity at the surface of the tank as zero. Take the pressure at the outlet of that last pipe as atmospheric. Account for minor losses and frictional losses in the pipe as appropriate.
If the outlet pipe isn't submerged enough, you can get flow limited by vortex or weir effects.
If that last pipe doesn't just dump out somewhere, but goes into a distribution system, you'd either need to assume a pressure at some point or have some means of analyzing the whole system.
If the outlet pipe isn't submerged enough, you can get flow limited by vortex or weir effects.
If that last pipe doesn't just dump out somewhere, but goes into a distribution system, you'd either need to assume a pressure at some point or have some means of analyzing the whole system.
- Thread starter
- #4
Just look at the last tank. The outflow from the nozzle must be 350 gpm. The only thing driving that outflow is the height of water above the nozzle's centerline. Find the height of water that you need to give you a velocity through the nozzle resulting in a flow rate of 350 gpm and you've got your target level. You must then get the level of that tank to that height above the nozzle and keep it there by doing the same thing from the second tank. Now do it again from the first. Flow through all nozzles in all tanks must stabilize at a steady state of 350 gpm.
This web page should help you figure it all out.
This web page should help you figure it all out.
- Thread starter
- #6
BigInch, thanks for that link.
The equations given are:
Q = C*Aspout*Vjet
Solving for Vjet
Vjet = Q/C*Aspout
Vjet2 = 2g*z
Using:
Q = 350 gpm (0.779 ft3/sec)
Aspout = 0.2007 ft2 (flow area of 6" SCH40 pipe)
Vjet = 3.96 ft/sec
(3.96 ft/sec)2 = 64.4*z
z = 3.96 ft/sec)2/64.4
z = 0.24ft = 2.88in above centerline
Does this seem correct? Also, you mention to do this same calculation for the second tank and the first tank. If those tanks also have 6" connections wouldn't the height above the nozzle be the same (2.88in)?
Thanks.
The equations given are:
Q = C*Aspout*Vjet
Solving for Vjet
Vjet = Q/C*Aspout
Vjet2 = 2g*z
Using:
Q = 350 gpm (0.779 ft3/sec)
Aspout = 0.2007 ft2 (flow area of 6" SCH40 pipe)
Vjet = 3.96 ft/sec
(3.96 ft/sec)2 = 64.4*z
z = 3.96 ft/sec)2/64.4
z = 0.24ft = 2.88in above centerline
Does this seem correct? Also, you mention to do this same calculation for the second tank and the first tank. If those tanks also have 6" connections wouldn't the height above the nozzle be the same (2.88in)?
Thanks.
Jkate:
You would do well to post your attachment/sketches as PDF’s, we can see them and print them much easier in that format. The same applies to your flat plate problem (Struct. forum, thread507-379753), which is far from a flat plate in your latest iteration. You hadn’t addressed the flat plate problem correctly either, before it so radically changed. Maybe you should be asking these questions of your boss, so he knows what you know and what you don’t know, and he can help guide you and keep you out of trouble. That would be good for you and the company. E-Tips is a wonderful place to come for help and discussion on engineering problems, but it shouldn’t be a substitute for material you should have covered in your earlier engineering course work. Dig out you text books and class notes, and do a little self study.
You would do well to post your attachment/sketches as PDF’s, we can see them and print them much easier in that format. The same applies to your flat plate problem (Struct. forum, thread507-379753), which is far from a flat plate in your latest iteration. You hadn’t addressed the flat plate problem correctly either, before it so radically changed. Maybe you should be asking these questions of your boss, so he knows what you know and what you don’t know, and he can help guide you and keep you out of trouble. That would be good for you and the company. E-Tips is a wonderful place to come for help and discussion on engineering problems, but it shouldn’t be a substitute for material you should have covered in your earlier engineering course work. Dig out you text books and class notes, and do a little self study.
The height in each tank is set by the headloss of the outlet pipe.
You can use the orifice equation.
Use 0.6 as the Cd coefficient. g = Gravitational constant (9.8 m/s).
There is an online calculator here:
The orifice equation will only work for full pipe flow through the orifice. Weird things happen if you have air/water or a vortex situation.
A 6-Inch nozzle should have approximately 6-Inches of headloss
You can use the orifice equation.
Use 0.6 as the Cd coefficient. g = Gravitational constant (9.8 m/s).
There is an online calculator here:
The orifice equation will only work for full pipe flow through the orifice. Weird things happen if you have air/water or a vortex situation.
A 6-Inch nozzle should have approximately 6-Inches of headloss
Not sure what you are doing here, but some observations.
The outlet nozzle does not have to be at the top of the tank. You can pump off the bottom.
If you are using the nozzle to control the level in the tank, then think about putting some type of weir or overflow box. A nozzle on the side of the tank will not do a good job of keeping a constant level in the tank.
There needs to be some thought to prevent air from entering the outlet nozzle. A nozzle at the water surface is not a good idea as it will allow air to enter the discharge pipe.
A nozzle near the surface may also cause a vortex and pull air into the outlet.
The outlet nozzle does not have to be at the top of the tank. You can pump off the bottom.
If you are using the nozzle to control the level in the tank, then think about putting some type of weir or overflow box. A nozzle on the side of the tank will not do a good job of keeping a constant level in the tank.
There needs to be some thought to prevent air from entering the outlet nozzle. A nozzle at the water surface is not a good idea as it will allow air to enter the discharge pipe.
A nozzle near the surface may also cause a vortex and pull air into the outlet.
-
1
- #11
LittleInch
Petroleum
Also be aware that the equations used are for your nozzle to be open to atmosphere and hence just pouring all over the ground. in that case 6" sounds about right. Your diagram looks like some sort of multi chamber interceptor / settling system and the last section is your final overflow weir??
you don't say or have included in any of the calcs the friction losses in the interconnecting pipes, valves etc or the friction losses from nozzle A to the centrifugal pump you mention in a post above. In that instance it is highly likely that the pressure at the end of the pipe will fall below atmospheric pressure unless you very carefully match your pump flow to the pipe and hence getting 350 gpm will be quite difficult.
you essentially have two halves to your sytem here from what I can see - the inlet half comprising the incoming flow which appears to be above the water line- OK - and the first 2 1/2 tanks and then the final chamber in the interceptor which is connected to your pump (not shown). The critical item in the last half is the pump and its discharge system and the friction losses in the pipe feeding the pump which will determine what head you need in the final chamber. Then its a matter of whether there is any flow control on your discharge to get it equal to the inlet flow?
Not quite as simple as it looks?
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
you don't say or have included in any of the calcs the friction losses in the interconnecting pipes, valves etc or the friction losses from nozzle A to the centrifugal pump you mention in a post above. In that instance it is highly likely that the pressure at the end of the pipe will fall below atmospheric pressure unless you very carefully match your pump flow to the pipe and hence getting 350 gpm will be quite difficult.
you essentially have two halves to your sytem here from what I can see - the inlet half comprising the incoming flow which appears to be above the water line- OK - and the first 2 1/2 tanks and then the final chamber in the interceptor which is connected to your pump (not shown). The critical item in the last half is the pump and its discharge system and the friction losses in the pipe feeding the pump which will determine what head you need in the final chamber. Then its a matter of whether there is any flow control on your discharge to get it equal to the inlet flow?
Not quite as simple as it looks?
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
-
1
- #12
LittleInch is correct - you cannot solve this problem without a lot more detail on the pump (you need the pump curve), the outlet piping (size, length, routing and elevation) and the control system. Any calculations done upstream of Nozzle A are a waste of time without knowing what the pressure is at A, and you can only know this pressure if you have the details of the pump, outlet piping and control system. You have to start at the point of known pressure, which is presumably the discharge of the outlet piping and work backwards until you get to point A.
The chances of getting all these calculations accurate enough to confidently predict the level in the final compartment of the tank are very slim. I would not try to convince anybody that I could do such a calculation. And if the flow rate varies even slightly you could rapidly move from an overflowing tank to sucking air into the pump.
The usual way to handle a flow problem of this nature is to control the level in the final compartment by modulating the valve on the discharge side of the pump. You know 350 gpm is coming in to the tank and the valve will ensure that 350 gpm leaves the tank by enforcing a "conservation of volume" strategy.
Katmar Software - AioFlo Pipe Hydraulics
"An undefined problem has an infinite number of solutions"
The chances of getting all these calculations accurate enough to confidently predict the level in the final compartment of the tank are very slim. I would not try to convince anybody that I could do such a calculation. And if the flow rate varies even slightly you could rapidly move from an overflowing tank to sucking air into the pump.
The usual way to handle a flow problem of this nature is to control the level in the final compartment by modulating the valve on the discharge side of the pump. You know 350 gpm is coming in to the tank and the valve will ensure that 350 gpm leaves the tank by enforcing a "conservation of volume" strategy.
Katmar Software - AioFlo Pipe Hydraulics
"An undefined problem has an infinite number of solutions"
- Thread starter
- #13
- Status
Similar threads
- Question
- Question
- Question