Water Pump problem

[Method]

I posted a message about a waterpump i was using as in inter cooler spray
for a car which was getting very hot with continuous use. I have found if i
put ~30 ohm resistor in parallel with the wires, it takes a lot longer to
heat up but the resistor starts to go brown and smoke. I tried using a 100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

that is what is written on the packet. Are there any other types i can use
which will not heat up?

What am i missing here??

Thanks

 

[Mike Harding]

On Tue, 23 Sep 2003 02:38:02 GMT, "Method" <pmitsaki@bigpond.net.au>
wrote:

I posted a message about a waterpump i was using as in inter cooler spray
for a car which was getting very hot with continuous use. I have found if i
put ~30 ohm resistor in parallel with the wires, it takes a lot longer to
heat up but the resistor starts to go brown and smoke. I tried using a 100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

that is what is written on the packet. Are there any other types i can use
which will not heat up?

Try using a resistor rated at 5 watts or 10 watts.
The one you are using is rated 0.25 of a watt.

And it's not necessary to post to so many newsgroups.

Mike Harding

 

[Andrew Howard]

I think higher wattage values on resistors means they can handle more before
they overheat. Is this right?

Andrew Howard


"Method" <pmitsaki@bigpond.net.au> wrote in message
news:emObb.117740$bo1.48582@news-server.bigpond.net.au...

I posted a message about a waterpump i was using as in inter cooler spray
for a car which was getting very hot with continuous use. I have found if
i
put ~30 ohm resistor in parallel with the wires, it takes a lot longer to
heat up but the resistor starts to go brown and smoke. I tried using a 100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

that is what is written on the packet. Are there any other types i can use
which will not heat up?

What am i missing here??

Thanks

 

[Nirodac]

Ok, if I understand this correctly, your placing a 30 ohm resistor across,
say 14 volts. That's about .46 amps. through the resistor, 14 volts x .46
amps equals about 6.4 watts. Your resistor should be rated, not less than
10 watts at 30 ohms, wire wound would be best. 15 or 20 watt would be even
better (at 30 ohms)
Question? what are you trying to accomplish here, slowing down the water
pump? If your trying to slow the water pump, maybe putting the resistor in
series would work better.

Ray


"Andrew Howard" <andtfoot@hotpop.com> wrote in message
news:nBPbb.117797$bo1.90550@news-server.bigpond.net.au...

I think higher wattage values on resistors means they can handle more
before
they overheat. Is this right?

Andrew Howard


"Method" <pmitsaki@bigpond.net.au> wrote in message
news:emObb.117740$bo1.48582@news-server.bigpond.net.au...
I posted a message about a waterpump i was using as in inter cooler
spray
for a car which was getting very hot with continuous use. I have found
if
i
put ~30 ohm resistor in parallel with the wires, it takes a lot longer
to
heat up but the resistor starts to go brown and smoke. I tried using a
100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

that is what is written on the packet. Are there any other types i can
use
which will not heat up?

What am i missing here??

Thanks

 

[Robert Monsen]

"Method" <pmitsaki@bigpond.net.au> wrote in message
news:emObb.117740$bo1.48582@news-server.bigpond.net.au...

I posted a message about a waterpump i was using as in inter cooler spray
for a car which was getting very hot with continuous use. I have found if
i
put ~30 ohm resistor in parallel with the wires, it takes a lot longer to
heat up but the resistor starts to go brown and smoke. I tried using a 100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

that is what is written on the packet. Are there any other types i can use
which will not heat up?

What am i missing here??

Thanks

Sorry, I flushed the thread, so I'm talking through my hat.

If you use a 30 ohm resistor, and the voltage is 12V, then it'll use up 4.8
watts in heat, effectively demolishing the 1/4 W resistor you are using. (W
= V^2/R)

If you use a 100 ohm resistor, you'll be dissipating 1.44W in the resistor
as heat. Again, a 1/4 W resistor (the 0W25 above) will only last a few
minutes before getting toasty.

If you need to do this, you need some bigger resistors. Resistors come with
a power rating, such as 1/4W, 1/8W, 2W, etc. You can get a big 5W 33 ohm
resistor mailorder (or go to your local electronics store) for a few bucks.
In the US, try www.allcorp.com for some surplus resistors. They'll have what
you need.

Also, if you put resistors in parallel, the effect will be to reduce the
'resistance' of the resistors, but still have the same power rating. So if
you could only find 2W resistors, for example, just buy three 100 ohm 2W
resistors. They will combine in parallel to be 33 ohms total, but will give
you an effective power rating of 6W. For your problem, if you have 12 V, you
could take 20 1k ohm resistors, and put them in parallel to get 50 ohms of
resistance at 5W of power handling capability.

Note that with bigger resistors, they will still get hot. They won't burn
up, though. The smaller ones in parallel won't get as hot, since they aren't
passing as much current each, and probably will get more airflow. The total
heat generated will be the same.

Regards,
Bob Monsen

 

[John G]

How do resistors in parallel with the pump slow it down??
All they should be doing is wasting power unless you are driving the pump
from something else than straight battery.
What is the circuit then maybe someone who has heard of Ohms Law can help
you.
--
John G

Wot's Your Real Problem?

"Method" <pmitsaki@bigpond.net.au> wrote in message
news:emObb.117740$bo1.48582@news-server.bigpond.net.au...

I posted a message about a waterpump i was using as in inter cooler spray
for a car which was getting very hot with continuous use. I have found if
i
put ~30 ohm resistor in parallel with the wires, it takes a lot longer to
heat up but the resistor starts to go brown and smoke. I tried using a 100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

that is what is written on the packet. Are there any other types i can use
which will not heat up?

What am i missing here??

Thanks

 

[The Captain]

"Method" <pmitsaki@bigpond.net.au> wrote in message news:<emObb.117740$bo1.48582@news-server.bigpond.net.au>...

I posted a message about a waterpump i was using as in inter cooler spray
for a car which was getting very hot with continuous use. I have found if i
put ~30 ohm resistor in parallel with the wires, it takes a lot longer to
heat up but the resistor starts to go brown and smoke. I tried using a 100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

that is what is written on the packet. Are there any other types i can use
which will not heat up?

What am i missing here??

Thanks

You need to look at the rating on your pump. If it is overheating
when driven by the nominal 12V from an automotive batery, then it may
be that you have a 6V pump.

There are two solutions to this problem. 1, and this is probably the
better of the two, is to replace the pump with a 12 volt version.
However, there may be other factors at play here, for example, how
exactly you are using the pump. If your pump is designed to probide a
steady flow of water without much back pressure, and you are using it
with a spray fitting, then the motor on the pump, having to work far
harder than its designers intended, will overheat. So replacing your
present pump with a 12 volt high pressure version will solve the
problem of overheating.

If, on the other hand, you want to just make it work as it is, then,
as is mentioned elsewhere in this thread, you really should have the
resistor in series with the pump. To choose the correct resistor you
need to know how much current your pump is drawing at 12 volts. To do
this, you will need a multimeter set to Amps (not milliamps) and
connect in series with the motor. When the motor is running you will
be able to measure the current drawn by the motor.

If, for example, your motor draws 2 amps at 12 volts, then its
internal "resistance" (it's not strictly speaking resistance, but for
practical purposes you can think of it as resistance) will be 12/2 = 6
ohms. Its power disipation will be (12^2)/6 = 24 watts when fed with
12 volts. I can see how this would overheat.

So, to drop the voltage for the motor from 12 to 6 volts, in this
example, you need a series resistor of 6 ohms and 6 watts (explained
later) dissipation capability. Assuming that your motor draws half
the current at half the voltage, which may not be exactly true, but
will be close enough for practical purposes, this will reduce the
voltage over the motor to 6 volts.

The current through the overall setup, 6 ohm resistor in series with a
6 ohm motor, will now be half what it was previously, or 1 amp. The
power, which shows mainly as heat, developed in the motor will be
current squared times resistance (I^2*R)or 12 watts overall; 6 watts
in the motor and 6 watts in the resistor. Hence the call for a
resistor capable of dissipating 6 watts in the previous paragraph. To
be on the safe side, a 10 watt resistor is probably best, since this
will have a greater surface area and will dissipate the heat at a
lower temperature.

At the same time you have reduced the power (heat) produced by the
pump to 6 watts, which is probably enough to prevent overheating.

I hope this helps and that your car goes like a striped ape!

regards

John

 

[Rheilly Phoull]

Uhh John, You might want to reconsider the effects of putting a restriction
on a pump outlet, all the centrifical type which it would be pretty certain
this one would be, will reduce their current when restricted.

--
Regards ............... Rheilly Phoull
"The Captain" <Captain794@yahoo.com> wrote in message
news:7199d521.0309230223.324654d@posting.google.com...

"Method" <pmitsaki@bigpond.net.au> wrote in message
news:<emObb.117740$bo1.48582@news-server.bigpond.net.au>...
I posted a message about a waterpump i was using as in inter cooler
spray
for a car which was getting very hot with continuous use. I have found
if i
put ~30 ohm resistor in parallel with the wires, it takes a lot longer
to
heat up but the resistor starts to go brown and smoke. I tried using a
100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

that is what is written on the packet. Are there any other types i can
use
which will not heat up?

What am i missing here??

Thanks

You need to look at the rating on your pump. If it is overheating
when driven by the nominal 12V from an automotive batery, then it may
be that you have a 6V pump.

There are two solutions to this problem. 1, and this is probably the
better of the two, is to replace the pump with a 12 volt version.
However, there may be other factors at play here, for example, how
exactly you are using the pump. If your pump is designed to probide a
steady flow of water without much back pressure, and you are using it
with a spray fitting, then the motor on the pump, having to work far
harder than its designers intended, will overheat. So replacing your
present pump with a 12 volt high pressure version will solve the
problem of overheating.

If, on the other hand, you want to just make it work as it is, then,
as is mentioned elsewhere in this thread, you really should have the
resistor in series with the pump. To choose the correct resistor you
need to know how much current your pump is drawing at 12 volts. To do
this, you will need a multimeter set to Amps (not milliamps) and
connect in series with the motor. When the motor is running you will
be able to measure the current drawn by the motor.

If, for example, your motor draws 2 amps at 12 volts, then its
internal "resistance" (it's not strictly speaking resistance, but for
practical purposes you can think of it as resistance) will be 12/2 = 6
ohms. Its power disipation will be (12^2)/6 = 24 watts when fed with
12 volts. I can see how this would overheat.

So, to drop the voltage for the motor from 12 to 6 volts, in this
example, you need a series resistor of 6 ohms and 6 watts (explained
later) dissipation capability. Assuming that your motor draws half
the current at half the voltage, which may not be exactly true, but
will be close enough for practical purposes, this will reduce the
voltage over the motor to 6 volts.

The current through the overall setup, 6 ohm resistor in series with a
6 ohm motor, will now be half what it was previously, or 1 amp. The
power, which shows mainly as heat, developed in the motor will be
current squared times resistance (I^2*R)or 12 watts overall; 6 watts
in the motor and 6 watts in the resistor. Hence the call for a
resistor capable of dissipating 6 watts in the previous paragraph. To
be on the safe side, a 10 watt resistor is probably best, since this
will have a greater surface area and will dissipate the heat at a
lower temperature.

At the same time you have reduced the power (heat) produced by the
pump to 6 watts, which is probably enough to prevent overheating.

I hope this helps and that your car goes like a striped ape!

regards

John

 

[Andrew Howard]

So, to drop the voltage for the motor from 12 to 6 volts, in this
example, you need a series resistor of 6 ohms and 6 watts (explained
later) dissipation capability. Assuming that your motor draws half
the current at half the voltage, which may not be exactly true, but
will be close enough for practical purposes, this will reduce the
voltage over the motor to 6 volts.

Do you think it would be possible to use a voltage halver instead?
eg
+ve -------------
|
/
\ R1
/
|
In ------------ +ve
|
/
\ Out
/ R2
|
GND ------------------------ GND

where R1 = R2.

 

[John Popelish]

Method wrote:

I posted a message about a waterpump i was using as in inter cooler spray
for a car which was getting very hot with continuous use. I have found if i
put ~30 ohm resistor in parallel with the wires,

I hope by "in parallel" you mean that you disconnected one of the
wires from the motor and connected the resistor between the motor and
the wire. This is normally called "in series with". This connection
uses up some of the total voltage across the resistor, with the
remainder left for the motor. This will slow the motor, and reduce
the current that passes through it, and the heat it produces.

...it takes a lot longer to
heat up but the resistor starts to go brown and smoke. I tried using a 100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

I am not sure what the 0W5 and 0W25 but the first usually means 0.5
watt (the W acts as both the decimal point and supplies the units),
but 0W25 would mean 0.25 watt. In either case, this is not a large
enough resistor (physical size, not resistance) to radiate the heat
the resistance produces.

I would start with a 10 ohm 2 watt to 5 watt resistor, and measure the
DC voltage across the resistor when the motor is operating. The
wattage being dumped by the resistor is V^2/R. So if the measured
resistor voltage was 2 volts, then the power dissipated would be
2*2/10=0.4 watts, etc. If the calculated power is less than the rated
wattage, the resistor should survive. Of course, you will have to
also find the value of resistance that cools the motor without slowing
it excessively (assuming that this is possible).

that is what is written on the packet. Are there any other types i can use
which will not heat up?

Look for wire wound power resistors that are about as big around as a
pencil and an inch or two long (or bigger). A few 10 ohm units can be
connected in series to produce the total resistance you need.

What am i missing here??

Thanks

You can also buy some 1 amp diodes (assuming your motor can run off
less than an ampere) such as 1N4000 or 3 amp diodes like 1N5400 and
connect them in series and in series with the motor to lower the motor
voltage. You will lose about .7 to 1 volt volt per diode. These may
be cheaper and more available than power resistors.

--
John Popelish

 

[N. Thornton]

So you want to get your 12v screenwash pumps to deliver water spray
continuously. They aren't intended to work for more than say 10secs a
go. To get them to live put your 2 in series, so each gets 6v. You
dont need any resistors.

Now, they'll live but perform dimly. So add another pair of them,
wired the same, and run the water thru one pump after another to build
the pressure up.

Regards, NT

 

[Don Kelly]

"Method" <pmitsaki@bigpond.net.au> wrote in message
news:emObb.117740$bo1.48582@news-server.bigpond.net.au...

I posted a message about a waterpump i was using as in inter cooler spray
for a car which was getting very hot with continuous use. I have found if
i
put ~30 ohm resistor in parallel with the wires, it takes a lot longer to
heat up but the resistor starts to go brown and smoke. I tried using a 100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

that is what is written on the packet. Are there any other types i can use
which will not heat up?

What am i missing here??

Thanks

Ignoring the voltage drop in the leads, putting a resistor in parallel will

not affect the pump. With the normal wire resistance, all the added
resistors do is cause a higher current and voltage drop- this will help
limit the current but is the wrong way to go about it. You are trying to do
the equivalent of dimming a light by connecting a toaster to the same
outlet.
Use a small resistor in <series> with the motor to limit the current taken
by the pump. Some measurements may be needed to determine the size and
rating.

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[Method]

Thanks for all your help, I have picked up a multimeter and will try and
work out a few more things and let you know how it goes.
Thanks again

"Don Kelly" <dhky@peeshaw.ca> wrote in message
news:NO2cb.1027955$3C2.22829137@news3.calgary.shaw.ca...

"Method" <pmitsaki@bigpond.net.au> wrote in message
news:emObb.117740$bo1.48582@news-server.bigpond.net.au...
I posted a message about a waterpump i was using as in inter cooler
spray
for a car which was getting very hot with continuous use. I have found
if
i
put ~30 ohm resistor in parallel with the wires, it takes a lot longer
to
heat up but the resistor starts to go brown and smoke. I tried using a
100
ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

that is what is written on the packet. Are there any other types i can
use
which will not heat up?

What am i missing here??

Thanks

Ignoring the voltage drop in the leads, putting a resistor in parallel
will
not affect the pump. With the normal wire resistance, all the added
resistors do is cause a higher current and voltage drop- this will help
limit the current but is the wrong way to go about it. You are trying to
do
the equivalent of dimming a light by connecting a toaster to the same
outlet.
Use a small resistor in <series> with the motor to limit the current taken
by the pump. Some measurements may be needed to determine the size and
rating.

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[Dimitrij Klingbeil]

John Popelish wrote:

Method wrote:

I posted a message about a waterpump i was using as in inter cooler spray
for a car which was getting very hot with continuous use. I have found if
i put ~30 ohm resistor in parallel with the wires,

I hope by "in parallel" you mean that you disconnected one of the
wires from the motor and connected the resistor between the motor and
the wire. This is normally called "in series with". This connection
uses up some of the total voltage across the resistor, with the
remainder left for the motor. This will slow the motor, and reduce
the current that passes through it, and the heat it produces.

...it takes a lot longer to
heat up but the resistor starts to go brown and smoke. I tried using a
100 ohm resistor and it got very hot but didn't smoke for a while.

The Resistors are
RES 0W5 MET 100R 1% 0W25

I am not sure what the 0W5 and 0W25 but the first usually means 0.5
watt (the W acts as both the decimal point and supplies the units),
but 0W25 would mean 0.25 watt. In either case, this is not a large
enough resistor (physical size, not resistance) to radiate the heat
the resistance produces.

I would start with a 10 ohm 2 watt to 5 watt resistor, and measure the
DC voltage across the resistor when the motor is operating. The
wattage being dumped by the resistor is V^2/R. So if the measured
resistor voltage was 2 volts, then the power dissipated would be
2*2/10=0.4 watts, etc. If the calculated power is less than the rated
wattage, the resistor should survive. Of course, you will have to
also find the value of resistance that cools the motor without slowing
it excessively (assuming that this is possible).

that is what is written on the packet. Are there any other types i can
use which will not heat up?

Look for wire wound power resistors that are about as big around as a
pencil and an inch or two long (or bigger). A few 10 ohm units can be
connected in series to produce the total resistance you need.

What am i missing here??

Thanks

You can also buy some 1 amp diodes (assuming your motor can run off
less than an ampere) such as 1N4000 or 3 amp diodes like 1N5400 and
connect them in series and in series with the motor to lower the motor
voltage. You will lose about .7 to 1 volt volt per diode. These may
be cheaper and more available than power resistors.

This is also a more reliable way to operate the pump since the voltage drop
at the diodes will be nearly constant what cannot be achieved with
resistors. Take into account that a forcibly stopped (for whatever reason)
electrical motor will chew up a lot of an amperage possibly frying the
resistors. Although this is also true for the diodes, the motor voltage
will be higher in this case thus preventing the motor from stopping from a
short overload and making an accidentally stopped motor restart quickly. In
any case, use a fuse rated at less than the maximal current the diodes can
support and take into account the place where the diodes are mounted
(temperature, ventillation, heat sink if any).

 

[Dimitrij Klingbeil]

Andrew Howard wrote:

So, to drop the voltage for the motor from 12 to 6 volts, in this
example, you need a series resistor of 6 ohms and 6 watts (explained
later) dissipation capability. Assuming that your motor draws half
the current at half the voltage, which may not be exactly true, but
will be close enough for practical purposes, this will reduce the
voltage over the motor to 6 volts.


Do you think it would be possible to use a voltage halver instead?
eg
+ve -------------
|
/
\ R1
/
|
In ------------ +ve
|
/
\ Out
/ R2
|
GND ------------------------ GND

where R1 = R2.

Possible, but exceptionally inefficient. A pump is not a couple of LEDs and
it is drawing a lot of current. If you want the +ve-out voltage to remain
stable during operation, the total current through R1 will be something
like 3 times the current drawn by the pump. Although power resistors exist,
why waste several times the required power for nothing.

P.S. Using this for a 'couple of LEDs' is as much inefficient, but with low
currents efficiency problems are negligible more often.