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Centrifugal Pump Head 1
- Thread starter Max1976
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A gardening problem... hmmmz
try connecting a gardenhose and hold up (1st or 2nd floor) the other end. when no more water comes out, the difference in height from the pump to the top of liquid in the hose is your shutoff head (in meters water kolom (0.9881 bar/m). than by timing the outflow (buckets per minute) at lower hieght, find the rest of your curve.
Have fun
try connecting a gardenhose and hold up (1st or 2nd floor) the other end. when no more water comes out, the difference in height from the pump to the top of liquid in the hose is your shutoff head (in meters water kolom (0.9881 bar/m). than by timing the outflow (buckets per minute) at lower hieght, find the rest of your curve.
Have fun
- Thread starter
- #6
dPh = rho * g * h for shutof head. this means your liquid is standing stil and no pressure is lost to pipe-(hose) friction.
Once water comes out the other end you can find the pump discharge pressure by:
P2 = P1 - dPf - dPh.
P2 = the pressure at the other end (1.014bara)
P1 = your pump discharge pressure (the point you actually want to fill in on your pump graph).
dPf = a bit more difficult. but can be approximated from dPf = 1/2 * rho * v^2 if v is low, like 0.5 m/s. So take a short lenght of the biggest hose you can find. Now measure flow by stopwastching your "buckets" (extrapolate to m3/h)and deviding that number by 0.25*pi*inside-hose-diameter^2 to find v. When your hose is long and thin, you'll have to use Reynolds, moody and frictionfactors (nasty bit of calculating which you realy want macros in excel to do for you).
Once water comes out the other end you can find the pump discharge pressure by:
P2 = P1 - dPf - dPh.
P2 = the pressure at the other end (1.014bara)
P1 = your pump discharge pressure (the point you actually want to fill in on your pump graph).
dPf = a bit more difficult. but can be approximated from dPf = 1/2 * rho * v^2 if v is low, like 0.5 m/s. So take a short lenght of the biggest hose you can find. Now measure flow by stopwastching your "buckets" (extrapolate to m3/h)and deviding that number by 0.25*pi*inside-hose-diameter^2 to find v. When your hose is long and thin, you'll have to use Reynolds, moody and frictionfactors (nasty bit of calculating which you realy want macros in excel to do for you).
typo: pump graph should be pump curve.
in a pump curve you set out flow against pump head.
now I come to think of it my previous answer can be done simpler if you can connect two hoses to the discharge end. read of waterkolemn height of one that is held to high for outflow (discharge pressure)and measure flow on the other one. Use a valve on the second hose to get more points on your pump curve. Now you are circumventing pressure loss from gardenhose.
in a pump curve you set out flow against pump head.
now I come to think of it my previous answer can be done simpler if you can connect two hoses to the discharge end. read of waterkolemn height of one that is held to high for outflow (discharge pressure)and measure flow on the other one. Use a valve on the second hose to get more points on your pump curve. Now you are circumventing pressure loss from gardenhose.
Unless you resort to experiments, what you estimate won't be very accurate.
But if you know the pump impeller diameter and specific speed, you can generate a "generic" curve shape.
If you want to get more accurate than that you need to take the pump apart and study the number of impeller vanes and their outlet angle to get a better idea of the expected head coefficient.
But if you know the pump impeller diameter and specific speed, you can generate a "generic" curve shape.
If you want to get more accurate than that you need to take the pump apart and study the number of impeller vanes and their outlet angle to get a better idea of the expected head coefficient.
@ Artisi: measuring static water head in a tube is a valid way of calibration that is more reliable and precise than any common pressure guage or theoretical approach. By the way, when peeing up a wall factor in velocity head and crosswind
@ bradshsi: Euler is fun, I'll give you that, but this involves a lot of guessing approach angles, front and back blade eddie loss, re circulation leakage loss, turbulence zone loss and assuming material friction factors and so on and so on.
My advice is this: If you want to know whats realy happening, get a T-piece, a tube, a valve, a bucket and a stopwatch.
@ max could you tell us a bit more about your problem?
@ bradshsi: Euler is fun, I'll give you that, but this involves a lot of guessing approach angles, front and back blade eddie loss, re circulation leakage loss, turbulence zone loss and assuming material friction factors and so on and so on.
My advice is this: If you want to know whats realy happening, get a T-piece, a tube, a valve, a bucket and a stopwatch.
@ max could you tell us a bit more about your problem?
Euler is fun. But if you work for a pump company (as I do), you have access to significantly better experimental correlations than just that.
If you know the general construction of the pump you can make a decent stab at the curve shape. Lets take your list of items in turn:
Approach angles. Not really an issue in predicting a HQ curve unless the suction pipework is especially bad.
Front & back eddie loss. I presume you are referring to the expected vortex strength, which is reasonably predictable when the geometry is known.
Recirculation leakage loss. Again this can be predicted from the pump geometry. Note that leakage losses occur at all flow rates not just when the pump is operating in recirculation.
Turbulence zone loss. Not sure what you are referring to here. Most centrifugal pumps operate in the turbulent regime.
Friction factors. Provided the casting is not extremely rough or smooth these can be taken into account.
I'm not trying to be argumentative here. By far the easiest way to confirm the pump curve is by experimentation or a pressure gauge.
However if those are not an option, prediction from the geometry is a possibility.
If you know the general construction of the pump you can make a decent stab at the curve shape. Lets take your list of items in turn:
Approach angles. Not really an issue in predicting a HQ curve unless the suction pipework is especially bad.
Front & back eddie loss. I presume you are referring to the expected vortex strength, which is reasonably predictable when the geometry is known.
Recirculation leakage loss. Again this can be predicted from the pump geometry. Note that leakage losses occur at all flow rates not just when the pump is operating in recirculation.
Turbulence zone loss. Not sure what you are referring to here. Most centrifugal pumps operate in the turbulent regime.
Friction factors. Provided the casting is not extremely rough or smooth these can be taken into account.
I'm not trying to be argumentative here. By far the easiest way to confirm the pump curve is by experimentation or a pressure gauge.
However if those are not an option, prediction from the geometry is a possibility.
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- #13
What kind of impeller is it? for instance: retracted(vortex pump) or closed or open... And can you find out what the impeller speed or range of speed is, that might be of interest when constructing a correlation.
@Bradshsi: do you also have experience with experimental correlations for centrifugal air blowers? Ive got a problem with one and am thinking of starting a new thread if it's not to much to aks to share this kind of data. I'd certainly understand if youre reluctant to do so.
@Bradshsi: do you also have experience with experimental correlations for centrifugal air blowers? Ive got a problem with one and am thinking of starting a new thread if it's not to much to aks to share this kind of data. I'd certainly understand if youre reluctant to do so.
Is this a real pump or an imaginary one?
No speed given, no driver size, no real data only very vague info lacking any real engineering input.
"it could be considered a gardening problem"
"Supposing the following partial further information available:"
No speed given, no driver size, no real data only very vague info lacking any real engineering input.
"it could be considered a gardening problem"
"Supposing the following partial further information available:"
- Thread starter
- #17
Deear All,
Thank you for the contributes;
the problem was a real gardening problem, however data are not available;
If you like, consider the following pump as reference:
Output: 1 HP,
Outlet: 2"
Rated: H 7.5 (M), Q 220 (L/M)
Maximum: H 12 (M), Q 450 (L/M)
Dimension: 350 L x 162 W x 180 H (mm)
Weight: 11 Kgs
Thanks again for the help, best regards
Thank you for the contributes;
the problem was a real gardening problem, however data are not available;
If you like, consider the following pump as reference:
Output: 1 HP,
Outlet: 2"
Rated: H 7.5 (M), Q 220 (L/M)
Maximum: H 12 (M), Q 450 (L/M)
Dimension: 350 L x 162 W x 180 H (mm)
Weight: 11 Kgs
Thanks again for the help, best regards
In your attachment, you have the performance curve for this pump. From the rated data you give I am assuming it is the JKH-750. Just check follow that black line to your axis's and that will give you your flow vs head pressure.
That is assuming that chart is for your pump, and no motor or impellar changes have been made.
Am I missing something here?
That is assuming that chart is for your pump, and no motor or impellar changes have been made.
Am I missing something here?
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