Francisco,
We can't say suction and discharge pressure are same because there is no meaning of pump head. Inn this regards, power change when pump working point change.
Dear Afshin4451, when I say pressure at suction and discharge to be the same, I'm not referring to PUMP suction and discharge, but to a pumping line suction and discharge, please see image at https://drive.google...UEZZTDR1V1NqTkk .
Like I said, if no friction is considered, there would be free flow. But if you consider friction losses, the pump must compensate for them, in the pumping line the pressure will decrease from suction to pump, increase due to the pump presence, and finally decrease from pump until discharge as I understand.
Best regards
Francisco,
Appreciated if you could express your explanation with using of equation similar article shared by Bobby.
Dear afshin4451, in a case like the one shown in the image, you can say that the "head" (or energy) difference between suction and discharge are due equal to the sum of contributions and losses between those points.
d: discharge
s:suction
z:elevation
v:velocity
g:gravity accelaration
P:pressure
r:density
W:power supplied by the pump dividided by mass flow.
L: losses.
(z+0.5v^2/g+P/(r*g))_s-(z+0.5v^2/g+P/(r*g))_d=W-L
so, if elevation (z), pressure (P) and velocity (v) are the same at suction and discharge, then:
W=L
So, again revisiting my previous assertions, if losses are neglected, L=0 and W=0 , there is no power required from the pump. If losses are:
L=0.5fv^2T/(D*g)
T: equivalent piping lenght
f: friction factor
D: equivalent diameter.
We have:
W=0.5fv^2T/(D*g)
This way you can use an experimental piping circuit to determine a curve of head versus flow for a given pump. The power input from the pump is:
m: mass flow.
P=W*m
Assuming pump power is constant, is flow through the piping is to be increased, the "head" the pump is able to provide decreases. As to how you can determine this curve of head versus flow, using only pump data, like geometry of his internal parts, or characteristics of the associated motor, I don't have that knowledge.
Best regards.