redfun
Mechanical
- Jun 12, 2011
- 4
We are having a motor-pump set up in our lab, that over heats after pumping water for 2 hours.( Simply because it is was built as a pneumatic set up initially).We want to run the set up over months non stop to check a polymer heart valve. So a new motor was ordered. Now I just need to cross check if the order was done correctly by my predecessor.
Calculation of shaft power of present pump ( single cylinder positive displacement pump)
Input
Present motor power rating: 1/15 HP=0.0667 HP= 49.6915 W
Power of new motor( brush less DC motor whose speed can be varied between 8-400rpm ): 0.1667 HP= 124.1915 W
Suction head, Hs= 0.44m
Discharge head, Hd= 0.44m (Hs=Hd as all that pump does is recirculation)
Density of glycerin, ?= 1261 kg/m^3 ( planning to shift to glycerin as fluid so calculations are done using glycerin density)
Gravity= 9.81 m/s^2
Requirement
Discharge, Q= 5 L/min= 0.00008333 m^3/s (Heart discharge rate)
Equations
Pump shaft power, Ps= Hydraulic power, Hp / Pump efficiency, ?
Hydraulic power, Hp (W) = Q (m^3/s) x (Hd + Hs in m) x ? (kg/m^3) x g (m/s^2)
Assumptions
I have assumed positive suction head as the height above pump center line to water source.
And the discharge head as the height from pump center line to water destination.
Calculations
Hydraulic power= 0.00008333x0.88x1261x9.81= 0.9071W
Observation
According to this calculation the power requirement of the pump is lesser than the power provided by the motor- new and old, even if a reasonable loss factor is considered. But we know that the heating could not be avoided in the present motor unless a heat removal source is provided ( well we ran the set up overnight after we kept a table fan in front of the motor –so in and around the valve endured 25 hours of testing before failing). So the questions that may be raised are would the new motor have the necessary torque-my guess is it should as it has a higher power rating than the present motor. And the second question is would it endure months of operation without overheating. I have presently not found a way of predicting if this would happen.
Presently the pump is horizontally oriented, but can I orient it vertically to avoid air trapping problem ?Now how could I do calculations that would show the difference in power requirements of vertical and horizontal positioning of pump? ( May be the effect not considering the leakage would be inconsequential because of the fact that the pump is very small, and hence the work that needs to be done against gravity is low)
I hope to achieve the 5L/min discharge by running the motor at 120 rpm, (this is the same number of reciprocations I want from the pump piston per minute.) And the discharge per stroke to be adjusted would be 42 cc.
"Please correct me if I am wrong in application of formulas or assumptions .Any observations or suggestions are welcome"
Calculation of shaft power of present pump ( single cylinder positive displacement pump)
Input
Present motor power rating: 1/15 HP=0.0667 HP= 49.6915 W
Power of new motor( brush less DC motor whose speed can be varied between 8-400rpm ): 0.1667 HP= 124.1915 W
Suction head, Hs= 0.44m
Discharge head, Hd= 0.44m (Hs=Hd as all that pump does is recirculation)
Density of glycerin, ?= 1261 kg/m^3 ( planning to shift to glycerin as fluid so calculations are done using glycerin density)
Gravity= 9.81 m/s^2
Requirement
Discharge, Q= 5 L/min= 0.00008333 m^3/s (Heart discharge rate)
Equations
Pump shaft power, Ps= Hydraulic power, Hp / Pump efficiency, ?
Hydraulic power, Hp (W) = Q (m^3/s) x (Hd + Hs in m) x ? (kg/m^3) x g (m/s^2)
Assumptions
I have assumed positive suction head as the height above pump center line to water source.
And the discharge head as the height from pump center line to water destination.
Calculations
Hydraulic power= 0.00008333x0.88x1261x9.81= 0.9071W
Observation
According to this calculation the power requirement of the pump is lesser than the power provided by the motor- new and old, even if a reasonable loss factor is considered. But we know that the heating could not be avoided in the present motor unless a heat removal source is provided ( well we ran the set up overnight after we kept a table fan in front of the motor –so in and around the valve endured 25 hours of testing before failing). So the questions that may be raised are would the new motor have the necessary torque-my guess is it should as it has a higher power rating than the present motor. And the second question is would it endure months of operation without overheating. I have presently not found a way of predicting if this would happen.
Presently the pump is horizontally oriented, but can I orient it vertically to avoid air trapping problem ?Now how could I do calculations that would show the difference in power requirements of vertical and horizontal positioning of pump? ( May be the effect not considering the leakage would be inconsequential because of the fact that the pump is very small, and hence the work that needs to be done against gravity is low)
I hope to achieve the 5L/min discharge by running the motor at 120 rpm, (this is the same number of reciprocations I want from the pump piston per minute.) And the discharge per stroke to be adjusted would be 42 cc.
"Please correct me if I am wrong in application of formulas or assumptions .Any observations or suggestions are welcome"