Axolotl writes:

Andrew Burgess wrote in
:


No. The wattage is a maximum value. Throttling the output, for
example, will reduce the power draw.

I do not think this is true. Two elements, pressure and volume, influence
the power consumption of any pump. If you look at Mark's Standard
Handbook for Mechanical Engineers 14-22 thru 14-25 you will find the
formula for centrifugal pump output.

I looked for something like this on the web and was unsuccessful.
(I only own EE books ;-)

Stripped of its more complicated
bits the formula shows that the power required, p = Q*H, where p = power,
Q=volume and H = head. Throttling reduces the volume output but increases
the internal pressure. Essentially throttling the pump has the same
effect within the pump of increasing the head.

I cannot think of a refutation for this, so maybe I'm misremembering
the 'minimum power consumed at no output' thing.

Unless you've oversimplified and the relationship isn't linear? Aren't
head-flow curves non-linear?

I'm going to email my friend with the power meter right now and borrow
his meter and do some measurements...

I have a question about centrifugal pumps. Jan recommended putting the pump in
the filter, which would mean the pump is pulling water out of the pond not
pushing it into the filter.

No. It's pulling water out of the filter and pumping to the pond. The pond
to filter flow is by gravity.

There's a device at Radio Shack for like $30.00 that you plug stuff into and
it'll tell youi how much power something is drawing. It'll also count it as
time goes on. So you could plug your pump into it, let it run for a month,
look at your enegry bill to get the kilowatt per hour cost and figure it all
out. Here's the device on the net:

http://store.yahoo.com/ahernstore/p4400.html

Sam

"RROLD1" wrote in message
...
Hello all,

I have purchased a pump with a rating that does not make any sense at all
(trusted the knowlegable sales staff). I am trying to figure out my cost
per
month but the only specification provided is output watts. Now that I have
compared it to the next pump size down, I am really confused. I don't want
to
install it if a smaller one will do the job.

Here are the specs:

My pump:
Rated 3800 GPH
Output Watts: 736

Next pump down:
Rated 3300 GPH
Output Watts: 330

After reading the specs I called the dealer and asked the average monthly
cost
to run each pump. They replied that the smaller pump will cost about
$28.00 and
the other (the one I have) about $64.00. There is only a 500 GPH
difference! Do
they know what they are talking about?

After searching the newsgroups, I have found a lot of similar questions
but no
difinitive answers (at least none that I understand).
Why don't the pump companies offer a table of actual killowats used per
feet of
head?. Its the answer everyone seems to be after.

Can anyone help me sort this out?

Additional Info: The pump is submersible, I have a 2' head, and a bio
filter/falls.

Thanks,

Steve

"Sam Hopkins" writes:

There's a device at Radio Shack for like $30.00 that you plug stuff into and
it'll tell youi how much power something is drawing. It'll also count it as
time goes on. So you could plug your pump into it, let it run for a month,
look at your enegry bill to get the kilowatt per hour cost and figure it all
out. Here's the device on the net:

http://store.yahoo.com/ahernstore/p4400.html

OK I ordered one. Thanks Sam! Easier then driving to my friends
to borrow his...

fyi ,,, P=U x I aka Watt = Voltage x Current ,,, measure how much Current
the pump is taking ,, and mutliply it with your voltage . For a 550 watts
pump the current should be mesured ~5 Amps on a 110 Vac leader.

However Measuing the Current is not going to be a easy task with that
instrument as you have to measure Current (Ampere) in serie with the Pump ,
there is however Ampere meters that you can simply clamp around the wire to
measure the Amps.

Best regards

Rune S.

"Andrew Burgess" wrote in message
...
"Sam Hopkins" writes:

There's a device at Radio Shack for like $30.00 that you plug stuff into
and
it'll tell youi how much power something is drawing. It'll also count it
as
time goes on. So you could plug your pump into it, let it run for a
month,
look at your enegry bill to get the kilowatt per hour cost and figure it
all
out. Here's the device on the net:

http://store.yahoo.com/ahernstore/p4400.html

OK I ordered one. Thanks Sam! Easier then driving to my friends
to borrow his...

Second thought this meter might be suited with that clamp to hang on the
wire =) Me bad ,, humble appologies.

R.stolen
"R.stol" wrote in message
...
fyi ,,, P=U x I aka Watt = Voltage x Current ,,, measure how much Current
the pump is taking ,, and mutliply it with your voltage . For a 550 watts
pump the current should be mesured ~5 Amps on a 110 Vac leader.

However Measuing the Current is not going to be a easy task with that
instrument as you have to measure Current (Ampere) in serie with the Pump
,
there is however Ampere meters that you can simply clamp around the wire
to
measure the Amps.

Best regards

Rune S.

"Andrew Burgess" wrote in message
...
"Sam Hopkins" writes:

There's a device at Radio Shack for like $30.00 that you plug stuff
into
and
it'll tell youi how much power something is drawing. It'll also count
it
as
time goes on. So you could plug your pump into it, let it run for a
month,
look at your enegry bill to get the kilowatt per hour cost and figure
it
all
out. Here's the device on the net:

http://store.yahoo.com/ahernstore/p4400.html

OK I ordered one. Thanks Sam! Easier then driving to my friends
to borrow his...

"R.stol" writes:

fyi ,,, P=U x I aka Watt = Voltage x Current

That's for DC. AC requires considering the power factor.

,,, measure how much Current
the pump is taking ,, and mutliply it with your voltage.

Times the power factor gives the power..

For a 550 watts
pump the current should be mesured ~5 Amps on a 110 Vac leader.

However Measuing the Current is not going to be a easy task with that
instrument as you have to measure Current (Ampere) in serie with the Pump ,
there is however Ampere meters that you can simply clamp around the wire to
measure the Amps.

It has a series resistor.

Yes, I have simplified the formula the original is
n=lQH/3960P US or n=lQh/270P metric.

Where
n = efficiency coefficient,
l (lambda) = specific gravity of the material being pumped for water this
is 1,
Q = the volume in gallons/minute or M3/h
H = the head or pressure in ft or meters
P = the power in BHP, 1 BHP = 746W (if I remember correctly).

As the formula has no exponents in it, I believe that it is linear.
Having taken one of my pond pumps apart I would guess that the value for
n is very low, probably not much above 0.1 and maybe even lower.
AXO