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Pump Thrust 2
- Thread starter civil21
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2
- #2
Why do you need to know axial thrust? All properly designed pumps have zero axial thrust at the pump shaft becuase all of the thrust developed by the impeller is taken up in thrust bearings. If you have a double suction split case pump, then the axial thrust is balanced and the pump therefore does not need thrust bearings.
(As a side note, I always laugh when I see specs calling for Kingsbury thrust bearings on double suction axially split-case pumps....)
As far as radial thrust is concerned, that is much more complicated because you need to know the shaft design that you are dealing with, how and where it is loaded, what type of impeller you use, the duty point, and the fluid being pumped. The impeller being used is really important becuase you need to know the area of the outlet channels of the impeller. Also, is the entire rotating assembly statically and dynamically balanced because an off-balance (and misalignment) will impart more force to the rotating impeller.
There are formulas you can use with overhung shafts, but they assume constant (or near-constant) cross section of the shaft. With shafts commonly consisting of several different steps, this is not valid for any but the simplest of pump designs.
(As a side note, I always laugh when I see specs calling for Kingsbury thrust bearings on double suction axially split-case pumps....)
As far as radial thrust is concerned, that is much more complicated because you need to know the shaft design that you are dealing with, how and where it is loaded, what type of impeller you use, the duty point, and the fluid being pumped. The impeller being used is really important becuase you need to know the area of the outlet channels of the impeller. Also, is the entire rotating assembly statically and dynamically balanced because an off-balance (and misalignment) will impart more force to the rotating impeller.
There are formulas you can use with overhung shafts, but they assume constant (or near-constant) cross section of the shaft. With shafts commonly consisting of several different steps, this is not valid for any but the simplest of pump designs.
- Thread starter
- #3
I realize pumps come with thrust bearings to account for the axial thrust loads and radial bearings for radial load, I am just curious how one would calculate it. I also tend to agree with tsdead regarding the split case pumps in theory. However, it is always good practice to require thrust bearings because "Murphy" WILL find thrust which can be significant. My question comes about due to curiosity. I have seen some elaborate equations to calculate axial and radial thrust and was curious if there was a quick and easy (substitute approximate) way to determine the magnitude of the unbalanced pressure between the front and back of the impeller.
Well, the axial thrust is the change in momentum of the fluid along the x-axis (with that axis defined as the axis of rotation of the shaft) plus the pressure differential between the back of the impeller and the front of it - corrected for the area of thrust - that being the cross section of the impeller minus any balance holes.
Just do a force calc similar to that of a turbine blade, assume a blade angle (which is not constant, but for quick and dirty, just try). For pressure behind the impeller, assume a few pounds over suction pressure (5 psig is probably sufficient for quick and dirty) or assume its the same as suction pressure (not a bad assumption, really) behind the impeller to be more conservative. Remember, in overhung end suction type pumps, the shaft seal sees a fluid pressure just slightly above suction (it depends on several factors, but this is generally the case).
As far as Murphy rearing his ugly head with split-case pumps, I can point to hundreds and hundreds of such installations where the pumps have standard single anti-friction radial bearings and do not overthrust the motor (which cannot take any thrust for the most part.) Yes, if designed and built incorrectly, you will not have a balanced radial thrust - but then, you get what you pay for.
Now, if you are using sleeve bearings (not recommended with split case pumps for the most part) then yes, you will need a limited end float coupling. Also, if you use a gear coupling, then the coupling itself will transmit axial thrust to the pump and/or motor.
Unfortunately, I don't think you can come up with a good and meaningful thrust value by any "quck and dirty" method. You can come up with something, but then the question arises, what use is that number????
As far as radial thurst, you can use this method:
R = K/Kso * H/Hso * Rso
where:
R = radial force, in pounds, at the operating condition
K = Kso * [ 1- (Q/Qn)^3.3]
Q = operating flowrate, in gpm
Qn = flowrate at BEP
Kso = thrust factor at shutoff head and is equal to:
Specific speed (Nss) Kso
2000 0.34
2200 0.355
2400 0.365
>2400 0.37
Hso = total shutoff head, in feet
H = total oeprating head, in feet
Rso = radial thrust, in pounds, at shutoff
Rso = Kso * Hso / (2.31 * D2 * B2)
where:
D2 = impeller diameter, in inches
B2 = impeller width, in inches, at discharge (including shrouds)
As you see, you need to assume values (or measure them if you have the impeller) of the impeller dimensions. If you want to find deflection since radial thrust in and of itself is a useless value, then use the standard beam equations and use the method of superposition with a stepped shaft. This is compounded if you have wear rings because they act like bearings, so you wind up with a shaft that looks kinda like an S. If you know the clearence of the wear rings, then you can assume them to be rigid motion limiters and use that boundary condition in your analysis. Also, assume anti-friction bearings to be rigid connections.
I hope I haven't muddied the waters too much.
Regards,
Tim
Just do a force calc similar to that of a turbine blade, assume a blade angle (which is not constant, but for quick and dirty, just try). For pressure behind the impeller, assume a few pounds over suction pressure (5 psig is probably sufficient for quick and dirty) or assume its the same as suction pressure (not a bad assumption, really) behind the impeller to be more conservative. Remember, in overhung end suction type pumps, the shaft seal sees a fluid pressure just slightly above suction (it depends on several factors, but this is generally the case).
As far as Murphy rearing his ugly head with split-case pumps, I can point to hundreds and hundreds of such installations where the pumps have standard single anti-friction radial bearings and do not overthrust the motor (which cannot take any thrust for the most part.) Yes, if designed and built incorrectly, you will not have a balanced radial thrust - but then, you get what you pay for.
Now, if you are using sleeve bearings (not recommended with split case pumps for the most part) then yes, you will need a limited end float coupling. Also, if you use a gear coupling, then the coupling itself will transmit axial thrust to the pump and/or motor.
Unfortunately, I don't think you can come up with a good and meaningful thrust value by any "quck and dirty" method. You can come up with something, but then the question arises, what use is that number????
As far as radial thurst, you can use this method:
R = K/Kso * H/Hso * Rso
where:
R = radial force, in pounds, at the operating condition
K = Kso * [ 1- (Q/Qn)^3.3]
Q = operating flowrate, in gpm
Qn = flowrate at BEP
Kso = thrust factor at shutoff head and is equal to:
Specific speed (Nss) Kso
2000 0.34
2200 0.355
2400 0.365
>2400 0.37
Hso = total shutoff head, in feet
H = total oeprating head, in feet
Rso = radial thrust, in pounds, at shutoff
Rso = Kso * Hso / (2.31 * D2 * B2)
where:
D2 = impeller diameter, in inches
B2 = impeller width, in inches, at discharge (including shrouds)
As you see, you need to assume values (or measure them if you have the impeller) of the impeller dimensions. If you want to find deflection since radial thrust in and of itself is a useless value, then use the standard beam equations and use the method of superposition with a stepped shaft. This is compounded if you have wear rings because they act like bearings, so you wind up with a shaft that looks kinda like an S. If you know the clearence of the wear rings, then you can assume them to be rigid motion limiters and use that boundary condition in your analysis. Also, assume anti-friction bearings to be rigid connections.
I hope I haven't muddied the waters too much.
Regards,
Tim
- Thread starter
- #5
dear
in centifugal pumps axial thrust happen in Q=0 and calculated as following:
Fa=(0.7-0.9)p.g.h.An where:
Fa= total axila thrus
p= specefic density
g=gravity
h= total head
An= unbalanc area between back and front of impeller
do u want more..?
bye now
in centifugal pumps axial thrust happen in Q=0 and calculated as following:
Fa=(0.7-0.9)p.g.h.An where:
Fa= total axila thrus
p= specefic density
g=gravity
h= total head
An= unbalanc area between back and front of impeller
do u want more..?
bye now
Allow me to expand a bit. There are many "approximate" methods to calculate the total axial thrust on a pump shaft, but so far there has been no analytical method proposed that gives any results worth a dam becuase they lack repeatable accuracy.
I must point out that the above equation only takes into account the static pressure difference between the front and back of the impeller - it does not account for the axial thrust component of the momentum change of the fluid.
Not only is the pressure different on both sides of the impeller (which will place an axial load on the shaft) but you also have a fluid impinging on the impeller vanes. Resolving that force in all three dimensions you will see that it is dependant upon the vane angle. Therefore, the axial thrust is a function of MORE than just the static pressure difference between the front and back of the impeller.
Stepanoff proposed a formula in 1957 for axial thrust esitmation, but his formula has been shown to be grossly in error at off-deign situations.
Lobanoff and Ross presented some formulas in 1985 which more closely approximates the axial thrust in several different pump designs but still lacks in accuracy. They generated many forumlas for use in pumps but even they mention that their equations are only estimates and are subject to significant error. "The location of the impeller relative to the stationary walls, impeller shroud geometry, surface roughness of the walls, wear ring clearance, and balance hole geometry affect the pressure distribution, and hence, the axial thrust."
If you want a quick and dirty answer, go ahead and use the pressure differential equaltion for thrust - but be prepared at being off by as much as 900% or more.
Regards,
Tim S.
I must point out that the above equation only takes into account the static pressure difference between the front and back of the impeller - it does not account for the axial thrust component of the momentum change of the fluid.
Not only is the pressure different on both sides of the impeller (which will place an axial load on the shaft) but you also have a fluid impinging on the impeller vanes. Resolving that force in all three dimensions you will see that it is dependant upon the vane angle. Therefore, the axial thrust is a function of MORE than just the static pressure difference between the front and back of the impeller.
Stepanoff proposed a formula in 1957 for axial thrust esitmation, but his formula has been shown to be grossly in error at off-deign situations.
Lobanoff and Ross presented some formulas in 1985 which more closely approximates the axial thrust in several different pump designs but still lacks in accuracy. They generated many forumlas for use in pumps but even they mention that their equations are only estimates and are subject to significant error. "The location of the impeller relative to the stationary walls, impeller shroud geometry, surface roughness of the walls, wear ring clearance, and balance hole geometry affect the pressure distribution, and hence, the axial thrust."
If you want a quick and dirty answer, go ahead and use the pressure differential equaltion for thrust - but be prepared at being off by as much as 900% or more.
Regards,
Tim S.
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