I got a 128.7kW,50Hz,415V delta connected motor with "Trickle voltage heating" method installed. The motor totally enclosed, have very little room available for installation of heater. The heating system consist of Single phase 500VA,415/24VAC transformer connected to two of three winding. A 16A fuse installed at the secondary side of transformer. When the circuit energised, fuse blown immediately and we found out the winding resistance was very low (0.03ohm). I am thinking of connect a external resistor in series to limit the current, will this method work and where can i get the resistor?
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Trickle Voltage Heating 1
- Thread starter tiewrc
- Start date
500VA / 24V = 20.8A
Unfortunately 24V/.03oHms = 800A
You need something on the order of 24V/20A = 960oHms
I don't know what the inductive contribution the the motor's impedance would be.
Putting a resistor in the circuit will drop all the power across the resistor... Which is not the result you want.
P = V[sup]2[/sup] / R
so V = SQRT(R x P)
V = SQRT(0.03 x 500)
V = 3.87 Volts seems closer to where you should be.
Keith Cress
Flamin Systems, Inc.-
Unfortunately 24V/.03oHms = 800A
You need something on the order of 24V/20A = 960oHms
I don't know what the inductive contribution the the motor's impedance would be.
Putting a resistor in the circuit will drop all the power across the resistor... Which is not the result you want.
P = V[sup]2[/sup] / R
so V = SQRT(R x P)
V = SQRT(0.03 x 500)
V = 3.87 Volts seems closer to where you should be.
Keith Cress
Flamin Systems, Inc.-
- Thread starter
- #3
what if i add a resistor rated at 1.6ohm,400W connected in series with following assumption:
1)To maintain existing installed equipment with additional external resistor to limit the operating current up to 75% of rated transformer secondary current(~20A)
2)R=24/(0.75x20)=1.6ohm
3)P=(15x15)x1.6=360W
May i know where can i get such a resistor? or connect a same rating of lamp will work?
1)To maintain existing installed equipment with additional external resistor to limit the operating current up to 75% of rated transformer secondary current(~20A)
2)R=24/(0.75x20)=1.6ohm
3)P=(15x15)x1.6=360W
May i know where can i get such a resistor? or connect a same rating of lamp will work?
Hang on, as others will probably have some suggestions.
Keith Cress
Flamin Systems, Inc.-
Keith Cress
Flamin Systems, Inc.-
We should be using impedance for current and resistance for heating.
16 amps through 0.03 ohms will develop 7.8 watts.
I would suggest 25 to 50 watts.
I would estimate the full load current to be about 210 amps.
600% of 210 = 1260 locked rotor amps.
415V/1260A = .4 ohms impedance.
For 25 watts out of .03 ohms resistance we need about 30 amps.
To push 30 amps through .4 ohms impedance we need 12 volts.
12 volts x 30 amps is 360 VA.
If you are really lucky,the transformer has a 12:24 volt secondary and you can reconnect it for 12 volts.
If you have to buy a transformer, go for a 500VA or 750 VA. There are too many assumptions in my calculations to size the transformer precisely at 360 VA.
Or, you could wire two 50 watt strip heaters in series and strap or bolt one low on each side of the motor. Connect them to rated voltage so that each gets only one half of rated voltage. The total combined heat output of the heaters will be 25 watts. Strip heaters are cheap and long lasting, especially at 50% voltage.
Bill
--------------------
"Why not the best?"
Jimmy Carter
16 amps through 0.03 ohms will develop 7.8 watts.
I would suggest 25 to 50 watts.
I would estimate the full load current to be about 210 amps.
600% of 210 = 1260 locked rotor amps.
415V/1260A = .4 ohms impedance.
For 25 watts out of .03 ohms resistance we need about 30 amps.
To push 30 amps through .4 ohms impedance we need 12 volts.
12 volts x 30 amps is 360 VA.
If you are really lucky,the transformer has a 12:24 volt secondary and you can reconnect it for 12 volts.
If you have to buy a transformer, go for a 500VA or 750 VA. There are too many assumptions in my calculations to size the transformer precisely at 360 VA.
Or, you could wire two 50 watt strip heaters in series and strap or bolt one low on each side of the motor. Connect them to rated voltage so that each gets only one half of rated voltage. The total combined heat output of the heaters will be 25 watts. Strip heaters are cheap and long lasting, especially at 50% voltage.
Bill
--------------------
"Why not the best?"
Jimmy Carter
- Thread starter
- #6
Thanks Bill.
The transformer rated at 500VA, 415/24VAC installed at site. We are not able to install the strip heater as its an existing submersible pump and the client asking for the heating for the pump(on standby) recently. The only way we can think of is to use the winding as heater. When we switch on the heating circuit, the current reading at primary side of transformer=28A (~480A at 24V side). The motor has pf of 0.76.
I believe there should be some impedance on the transformer and motor itself to limit the current to low value, not sure what cause it to fail. We megger our circuit and everything seem to be ok.
The transformer rated at 500VA, 415/24VAC installed at site. We are not able to install the strip heater as its an existing submersible pump and the client asking for the heating for the pump(on standby) recently. The only way we can think of is to use the winding as heater. When we switch on the heating circuit, the current reading at primary side of transformer=28A (~480A at 24V side). The motor has pf of 0.76.
I believe there should be some impedance on the transformer and motor itself to limit the current to low value, not sure what cause it to fail. We megger our circuit and everything seem to be ok.
It can be done. The rule is to have about 25-35% of the rated current going through the windings. If the winding resistance is low like this and you have to use an external resistor, so much voltage drop will take place at the resistor that the resistor will put out a LOT of heat, possibly dangerous amounts.
A better method is to use an electronic controller which uses phase-angle firing to reduce the effective voltage to the windings and automatically adjusts to the eventual resistance changes with temperature rise. There are 3 or 4 standard products that are designed to retrofit to existing starters on the market, but also, many Soft Starters and VFDs are providing this as "free" built-in features as well now. Some people also do it themselves with DC supplies, such as small welders, but they require manual adjustments as the motor heats up.
Other possibly useful information may be found in the FAQs for this site: faq237-1157
A better method is to use an electronic controller which uses phase-angle firing to reduce the effective voltage to the windings and automatically adjusts to the eventual resistance changes with temperature rise. There are 3 or 4 standard products that are designed to retrofit to existing starters on the market, but also, many Soft Starters and VFDs are providing this as "free" built-in features as well now. Some people also do it themselves with DC supplies, such as small welders, but they require manual adjustments as the motor heats up.
Other possibly useful information may be found in the FAQs for this site: faq237-1157
LionelHutz
Electrical
I would agree with jraef that you are looking for about 30% of rated current to heat the motor. Since this is a submersible under water it may even require more but start at 30%.
At any rate, now that you have provided the data that when you supply 24VAC the motor draws 28A you can figure out where to go. The current should be around 50A so I would use about 48VAC with the transformer sized at about 4kVA. This should provide a little over 50A of current.
At any rate, now that you have provided the data that when you supply 24VAC the motor draws 28A you can figure out where to go. The current should be around 50A so I would use about 48VAC with the transformer sized at about 4kVA. This should provide a little over 50A of current.
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1
- #9
I would go with jraef's analysis. His comments remind me of another possibility. Some softstart manufacturers also make a stand-alone DC injection brake unit for stopping with a mag starter. I've seen these used for motor heating by leaving a few amps DC flowing thru the motor when the starter is open. The output current and, as a result, the motor heating, is adjustable.
This same DC injection equipment is sometimes used to prevent reverse rotation of fans in air handlers, also.
Check with Benshaw in Pennsylvania for an example of this type of equipment.
This same DC injection equipment is sometimes used to prevent reverse rotation of fans in air handlers, also.
Check with Benshaw in Pennsylvania for an example of this type of equipment.
Skogsgurra
Electrical
If you need to heat the winding, you should use DC.
A transformer with a very low voltage plus a rectifier will do the trick. Only DC resistance in play - no impedance to push current through. You can easily use a 100 or 200 VA transformer if you do not need more than 50 W heating.
Bridge rectifiers for 35 A are cheap and standard.
Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
A transformer with a very low voltage plus a rectifier will do the trick. Only DC resistance in play - no impedance to push current through. You can easily use a 100 or 200 VA transformer if you do not need more than 50 W heating.
Bridge rectifiers for 35 A are cheap and standard.
Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
- Thread starter
- #13
Skogsgurra
Electrical
Answer for tiewrc,
if one winding has 30 milliohms then you need sqrt(25/.03) = 29 A for 25 W. Not difficult at all. The transformer needs to output enough voltage to cover rectifier drop (two or three volts) plus winding drop, which is close to 1 V. So transformer will be something like 4 - 5 V at 40 A. A 250 VA transformer does it easily.
Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
if one winding has 30 milliohms then you need sqrt(25/.03) = 29 A for 25 W. Not difficult at all. The transformer needs to output enough voltage to cover rectifier drop (two or three volts) plus winding drop, which is close to 1 V. So transformer will be something like 4 - 5 V at 40 A. A 250 VA transformer does it easily.
Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
I have an old write-up from US Motors that recommends ac current be used at about 25% of the nameplate current rating (single-phase).
It says this normally requires about 7 to 10% of nameplate voltage and they recommend a small step-down transformer be used that has +/- 5% and +/- 10% voltage taps for final adjustment.
The advantage given for ac current is that transformer action will heat the rotor as well as the stator and this heat can travel along the shaft by conduction to help heat the bearings more effectively than convection heating from the stator.
Seems like this trickle heating used to be an available option from GE and others in MCC starters.
It says this normally requires about 7 to 10% of nameplate voltage and they recommend a small step-down transformer be used that has +/- 5% and +/- 10% voltage taps for final adjustment.
The advantage given for ac current is that transformer action will heat the rotor as well as the stator and this heat can travel along the shaft by conduction to help heat the bearings more effectively than convection heating from the stator.
Seems like this trickle heating used to be an available option from GE and others in MCC starters.
LionelHutz
Electrical
I like Gunnar's solution which keeps the size of the transformer down but it does add a rectifier and running 30A through a bridge rectifier requires a decent heatsink. Also, I'd expect picking the transformer to maintain 30A is hard to do when the output voltage is so low. Any small transformer output change will affect the current more compared to using a higher voltage AC voltage.
Overall, it may be easier to just use a solid-state current controlled solution as already suggested.
Overall, it may be easier to just use a solid-state current controlled solution as already suggested.
I checked GE MCC options - they use an SCR for theirs.
Another consideration that often comes up with this application:
Even with the motor OFF there is still voltage at the motor terminals, so appropriate signage is important.
Skogsgurra
Electrical
Yes, you are right LH. Motor itself can be used for the heat sink. Makes the losses in rectifier useful for heating, too. Done that in a large ABB motor with lots of room in the peckerhead. Even put the transformer and contactor there. Could use an AC contactor with all three contacts in parallel, so it was a small one.
Most of the circuit impedance is in the transformer. By having a transformer with bad regulation helps keeping current fairly constant. Since absolute amperage isn't critical, same variation usually doesn't hurt.
Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
Most of the circuit impedance is in the transformer. By having a transformer with bad regulation helps keeping current fairly constant. Since absolute amperage isn't critical, same variation usually doesn't hurt.
Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
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