voltage drop on pool heater
- Thread starter Dolfan
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masterinbama
Senior Member
- Location
- Huntsville Alabama
If it's a purely resistive load the drop in voltage will cause an increase in current draw. I would size the wire to minimize voltage drop as much as possible.
masterinbama
Senior Member
- Location
- Huntsville Alabama
It would end up costing more than up sizing the wire. #4 copper or #2 AL will keep you under 3% in Voltage Drop.
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cpinetree
Senior Member
- Location
- SW Florida
Most heaters that we install are 35 amp and use a max 60amp breaker.
#8 wire should be sufficient, that would be 4.4%.
#8 wire should be sufficient, that would be 4.4%.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
121203-0827 EST
masterinbama:
I disagree with your comment.
If a load is a pure constant resistance, then increasing the resistance in series with that load and supplying from a constant voltage source will not increase the current. Rather the more added series resistance the lower the load current.
A pool heater is going to be a resistive device with some sort of modulator. Broadly this could be either a thermostatically controlled switch, a bang-bang servo. Or a temperature controlled phase shift current adjuster, a more conventional servo system.
The peak power capability of the heater should be great enough to provide the desired temperature under the greatest heat loss conditions. Probably this peak power capability, the power when the heater is full on, will really be greater than ever required. Thus, once the water is up to temperature the average power to the heater will be much less than the peak capability.
The major problem with voltage drop between the source and the heater is the initial time to heat the water, it will be somewhat slower, and if under maximum heat loss conditions the desired pool temperature can be maintained.
As long as the heater can do its job of meeting the average power requirements, then the series resistance does not matter. If there was an electronic control at the pool, then its minimum voltage to operate could become a factor.
The site editor messed up and my comments where entered before I was done. Going back and editing I may still be within the timeout limit.
.
masterinbama:
I disagree with your comment.
If a load is a pure constant resistance, then increasing the resistance in series with that load and supplying from a constant voltage source will not increase the current. Rather the more added series resistance the lower the load current.
A pool heater is going to be a resistive device with some sort of modulator. Broadly this could be either a thermostatically controlled switch, a bang-bang servo. Or a temperature controlled phase shift current adjuster, a more conventional servo system.
The peak power capability of the heater should be great enough to provide the desired temperature under the greatest heat loss conditions. Probably this peak power capability, the power when the heater is full on, will really be greater than ever required. Thus, once the water is up to temperature the average power to the heater will be much less than the peak capability.
The major problem with voltage drop between the source and the heater is the initial time to heat the water, it will be somewhat slower, and if under maximum heat loss conditions the desired pool temperature can be maintained.
As long as the heater can do its job of meeting the average power requirements, then the series resistance does not matter. If there was an electronic control at the pool, then its minimum voltage to operate could become a factor.
The site editor messed up and my comments where entered before I was done. Going back and editing I may still be within the timeout limit.
.
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Whatever gives you [1000(5% of 240)]/[400(35A)] = 0.857 ohms or less, per 1000'.
- Location
- Illinois
- Occupation
- retired electrician
A 240 volt heater run at 230 volts will produce 92% of the rated wattage. I doubt that would make any real difference in a pool heater, other than it will have to run longer to get the water up to heat.
masterinbama quote "If it's a purely resistive load the drop in voltage will cause an increase in current draw. I would size the wire to minimize voltage drop as much as possible."
That is not correct. With a resistive load, the amperage will follow ohms law I = E/R. Therefore if the voltage is lower, the current is lower and the
heat output of the heater will be lower. If the heater is rated at 240 volts, you would get about 92% of the rated wattage. It probably will put out enough heat to do the job satisfactorily.
Master, your comment is usually related to motor loads which do have an increase in amperage with lower voltage.
That is not correct. With a resistive load, the amperage will follow ohms law I = E/R. Therefore if the voltage is lower, the current is lower and the
heat output of the heater will be lower. If the heater is rated at 240 volts, you would get about 92% of the rated wattage. It probably will put out enough heat to do the job satisfactorily.
Master, your comment is usually related to motor loads which do have an increase in amperage with lower voltage.
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masterinbama
Senior Member
- Location
- Huntsville Alabama
Please forgive me,I had not yet had my coffee when I replied.
How can a heater have a 60A breaker and nameplate value? Resistance heaters must have a nameplate of 125% the full load current. A pool heat pump will have a max breaker larger than its MCA just like an air conditioner HVAC unit (and the MCA already has a 125% value for the compressor at least).
Calculate the voltage drop from the true running amps, not MCA. If this is a heat pump, I wouldn't skimp on wire size. If a resistance heater, it won't make much difference.
Calculate the voltage drop from the true running amps, not MCA. If this is a heat pump, I wouldn't skimp on wire size. If a resistance heater, it won't make much difference.
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