Borehole pump motor efficiency
Borehole pump motor efficiency
(OP)
I have a pump rated to deliver 75 gpm at a head of 740 feet
coupled to motor rated 20 Hp
The pump currently discharges 60 gpm and draws an average of 47.6 amps on the phases with 230 volts.
Question is how do i compute for the effeciency?
would it be-
[Pout = 14920 (from 20 Hp)] /
[Pin = A (from meter) * V (from meter) * 1.73 *.8 pf]
id get = 98%
or:
from the pump power curve, from the manufacturer, i've plotted the kw, against the current discharge of 60 gpm
ive got a value of 12.4kw = P2
P2 / P1 =
12.4kw / 15kw = 83% = p1 by the way is the pump motor requirement given from the manufacturers curve...
im confused. thanks!
coupled to motor rated 20 Hp
The pump currently discharges 60 gpm and draws an average of 47.6 amps on the phases with 230 volts.
Question is how do i compute for the effeciency?
would it be-
[Pout = 14920 (from 20 Hp)] /
[Pin = A (from meter) * V (from meter) * 1.73 *.8 pf]
id get = 98%
or:
from the pump power curve, from the manufacturer, i've plotted the kw, against the current discharge of 60 gpm
ive got a value of 12.4kw = P2
P2 / P1 =
12.4kw / 15kw = 83% = p1 by the way is the pump motor requirement given from the manufacturers curve...
im confused. thanks!





RE: Borehole pump motor efficiency
RE: Borehole pump motor efficiency
RE: Borehole pump motor efficiency
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RE: Borehole pump motor efficiency
the first equation indeed pertains to motor efficiency.. and ive found out that the correct means would be to get the
eff = (Mechanical Power output) / (Electrical Power input)
if im not mistaken..
my concern now is it possible to calculate a specific efficiency for a pump motor system, like a base data so that if id compute this efficiency, i'd know when im straying too far from that base data...
thanks
RE: Borehole pump motor efficiency
What depth is the pump setting below your measuring point - which I assume is ground level?
RE: Borehole pump motor efficiency
Pump Power Consumed (Hydraulic Power) = flowrate * density * Differential Head
Divide that by your electrical power input to get overall eff.
"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.
http://www.youtube.com/watch?v=hpiIWMWWVco
"Being GREEN isn't easy." Kermit
http://virtualpipeline.spaces.live.com
RE: Borehole pump motor efficiency
RE: Borehole pump motor efficiency
Before doing anything you need to establish accurate flow, discharge head at a convenient point at ground level, power input and the static lift from the water level in the bore to the measuring point with the pump operating.
As an example -
(you fit your own figures to suit the site conditions)
60gpm
200 ft discharge head (gauge reading)
Static lift 400 ft
12.4kW input
This gives a WHP of 60 x 600 / 3960 = 9.1 hp (6.74 kW)
overall efficiency is 6.74 / 12.4kW = 54.3%
Now bear in mind that this is the o/a efficiency of the pump, motor and all losses from the pump inlet to the head measurement point at ground level which includes any discharge pipe, bends, valves plus the static lift to this point plus all the losses on the discharge side of the measuring point.
If you wish to calculate (estimate) the pump / motor efficiency at the pump discharge, you can try and calculate the losses for all the discharge pipe below ground level any valves etc before the head measuring point plus the standing water level below your measuring point with the pump operating, this can be added to the discharge head that you measured and the efficiency re-calculated as follows,
60gpm
200 ft head + (calculated head losses including static lift) lets say 500 ft = 700 ft
re-calculating
60 x 700 / 3960 = 10.6hp = 7.85 kW
Overall pump /motor efficiency is 7.85 / 12.4 = 63%
Of course is all this worth the time and effort which at the end of the day is still only an estimate.
If you need a reliable baseline - just use the flow / head / power input of the installaton - any drop-off over time will be measureable against these figures.