relationship between the pump speed and the flowrate/flowability

[raysheild]

dear friends,

There is this problem we were facinng when i was doing my project for the completion of my national diploma in Yaba tech. before i come across this site.

i and my project-mate were working on an hydraulic press then,we've constructed and mounted all our eqipment i.e the pump,valve,tank and the hydraulic press.the problem we faced then is for the pump to pump the oilwhich it doesn't and we were sure the pump is o.k

but accidentally we realise that the problem was that because we multiply the speed of the pump i.e runing twice the speed of the primer.but when by accident we reduced the sped of the pump to half the speed of the primer, it started workink smoothly and we submited it.

knowing that i could get an answer to my question in this forum can someone please tell me the relationship between the operating speed and the flowability of the fluid

 

[hero123]

will u explain what do u mean by flowability?

 

[BobPE]

Look in your fluids text books for the affinity laws. These laws will allow you to look at change in speed vrs flow mathmatically.

BobPE

 

[raysheild]

by the flowability i mean when the pump was working at twice the speed of the pump,it was pumping less fliud with air which couldn't push the hydraulic press. but when it was operating at a speed that is half the speed of the motor,it pumps the fluid not even minding the air and save time for bleeding.

 

[hero123]

see, only solution that comes to my mind is the affinity laws as stated earlier:
the laws are as follows:

at constant efficiency,

and at constant rpm (N)
Flow (Q) goes directly as impellar Diameter (D)
Head (H) goes directly as square of impellar Diameter (D)
BHP (P) goes directly as cube of impellar Diameter (D)

and at constant diameter (D)

Flow (Q) goes directly as rpm(N)
Head (H) goes directly as square of rpm(N)
BHP (P) goes directly as cube of rpm(N)


hope this helps u work out ur problem

 

[dadfap]

The affinity laws are expressed mathematically as follows
New rate =Old rate(N2/N1)
New head = Old head(N2/N1)^2

N = speed.

The only explanation that I can think of is a serious loss of efficency in the pump