Electrical puzzle
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I need a little help from some of you mechanical/electrical genious'. And I know you're out there! Lol.
Ok, my '36 (stock engine with the old, stock, 6 volt electrical and generator) runs a little on the hot side in the hot summer months. 'Specially when I'm in a parade or something that makes me go particularly slow. I installed one of those new-fangled 6-volt electric pusher fans in front of the radiator and it works great. Gets enough air flow through that beautiful little grill to really make a difference.
Here's the problem. With the fan running, I don't get enough electric spark to make the car run right! No kidding, with the fan on, I get a miss-especially during acceleration. I can shut off the fan and it runs good. With it on it misses. New, good battery, so that aint it. Any ideas? I'd be greatful...slim
:)
Ok, my '36 (stock engine with the old, stock, 6 volt electrical and generator) runs a little on the hot side in the hot summer months. 'Specially when I'm in a parade or something that makes me go particularly slow. I installed one of those new-fangled 6-volt electric pusher fans in front of the radiator and it works great. Gets enough air flow through that beautiful little grill to really make a difference.
Here's the problem. With the fan running, I don't get enough electric spark to make the car run right! No kidding, with the fan on, I get a miss-especially during acceleration. I can shut off the fan and it runs good. With it on it misses. New, good battery, so that aint it. Any ideas? I'd be greatful...slim
:) 0
Comments
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What I have done on my Hudson was to fit a six blade fan, never have a heating problem with it, trouble is getting it hot enough.LOL.0
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Sounds to me like you have the fan wired up through the ignition switch, which is lowering the voltage, causing weak spark. Run it through a heavy duty relay and see what happens.0
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Geoff C., N.Z. wrote:Sounds to me like you have the fan wired up through the ignition switch, which is lowering the voltage, causing weak spark. Run it through a heavy duty relay and see what happens.
Alright! Now we are getting somewhere. I now have it straight to the battery. Will I be better off through a relay?
I don't get this @%&*...slim0 -
If it works, leave it. However, if you wanted to run it through the ignition switch, you would need a relay. Simply put, the ignition switch would energise the relay, which would be hooked up to the battery terminal on the regulator. A fan would run at least 10 amps I would imagine, and the switch cannot handle such a current and would overheat, hency your problem. The relay coil only takes about half an amp, if that, so will not affect anything. I assume you have a manual switch to control the fan?0
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The aux fan should have come with a thermal controller with relay and external switch for dash over-ride control. Just installed one of these on my 75 Jeep DJ-5D. The controler uses a thermostat tucked into the radiator fins close to the upper inlet. The controller has a temp setting for automatic start. Also has a wire to a dash switch to force turn-on.
The fan can draw up to 20amps @12v - 6V probably higher. The main fan power is via the controller operated relay then right to the batt + through an inline fuse; the fan neg to frame ground. The controller is powered by ignition +12 with a couple amp inline fuse.
Like Geoff said, a relay is used to switch the high current to the fan. The relay/controller power is less than a 1/2amp and powered off the ignition switch.
The fan will come on automatically when the controller temp setting is exceeded or the dash over ride switch is enabled.
Hope this helps,
Jack0 -
forjack842 wrote:The aux fan should have come with a thermal controller with relay and external switch for dash over-ride control. Just installed one of these on my 75 Jeep DJ-5D. The controler uses a thermostat tucked into the radiator fins close to the upper inlet. The controller has a temp setting for automatic start. Also has a wire to a dash switch to force turn-on.
The fan can draw up to 20amps @12v - 6V probably higher. The main fan power is via the controller operated relay then right to the batt + through an inline fuse; the fan neg to frame ground. The controller is powered by ignition +12 with a couple amp inline fuse.
Like Geoff said, a relay is used to switch the high current to the fan. The relay/controller power is less than a 1/2amp and powered off the ignition switch.
The fan will come on automatically when the controller temp setting is exceeded or the dash over ride switch is enabled.
Hope this helps,
Jack
The fan can draw up to 20amps @12v = 240 watts. This is 40 amps at 6 volts!!! No wonder you are running out of puff!! Perhaps you need to look for a less thirsty fan.0 -
I doubt very much if a 12v fan is going to draw anywhere near 20A.
You might also confirm that the generator is putting out the specified 7.2 volts at light load and 1,000 rpm. If lower than that, the VR may need adjusting.0 -
ESSX28-1 wrote:The fan can draw up to 20amps @12v = 240 watts. This is 40 amps at 6 volts!!! No wonder you are running out of puff!! Perhaps you need to look for a less thirsty fan.
not too bust you too much cause i know your trying to help but he is not drawing 40 amps....the idea you used of power=power only works for transformers if you would like to know how you do it here it is: first find the resistance of the windings of your fan: voltage=resistance x current we know voltage and maximum current so 12v=resistance x 20amp means a winding resistance of .6 ohms next we know we want to find out current for a 6v system so again voltage=resistance x current therefore 6v=.6 ohms x current and that gives us 10 amps so at most the fan would be drawing 10 amps...which is still high for a fan but i agree with the other post about using a relay or connecting direct to battery....WARNING!!!! if connecting right to battery or even through and ignition switch make sure to use an inline fuse....last thing you want is a fire under the hood of the beautiful machines :-)
Devon0 -
teddy1221 wrote:not too bust you too much cause i know your trying to help but he is not drawing 40 amps....the idea you used of power=power only works for transformers if you would like to know how you do it here it is: first find the resistance of the windings of your fan: voltage=resistance x current we know voltage and maximum current so 12v=resistance x 20amp means a winding resistance of .6 ohms next we know we want to find out current for a 6v system so again voltage=resistance x current therefore 6v=.6 ohms x current and that gives us 10 amps so at most the fan would be drawing 10 amps...which is still high for a fan but i agree with the other post about using a relay or connecting direct to battery....WARNING!!!! if connecting right to battery or even through and ignition switch make sure to use an inline fuse....last thing you want is a fire under the hood of the beautiful machines :-)
Devon
Thanks for the update, it's over 50 years since I last studied any of this stuff!!0 -
Geoff C., N.Z. wrote:If it works, leave it. However, if you wanted to run it through the ignition switch, you would need a relay.
Geoff...Do you have a source and part # for a 6v relay? And would it work with 8v?
Thanks.0 -
Unfortunately you're wrong.
The actual power required to move the same amount of air by the exact same fan physical design requires the same power whether 12v or 6 v. Using your example, if the correct voltage fan is used in both cases then the 12v fan has an internal resistance of 0.6 Ohms and requires (P=I^2 x R) 240 Watts of power to move its rated cubic feet of air per minute. To achieve the same air movement power at 6 volts we'll take 240W / 6V (Power is also is defined as P= Volts x Amps) or 40 Amps. The fan motor would have a resistance of (R=P / I^2) 0.15 Ohms. The above defines the ideal without losses and the like.
The voltage drop in the wire run for a 6 volt system and a fan at 40amps would be considerable if the fan were to be connected at the ignition switch. However, if connected at the battery the loss would be virtually limited to the fan supply circuit leaving just the battery terminal and internal battery resistance to be passed to the rest of the car systems. Simply put, the voltage drop at the battery would be slight with the fan running and the ignition system would virtually see the entire battery supply less its own designed in losses.
The use of a inline fuse is always the correct way to protect your precious valuables.
Jack
teddy1221 wrote:not too bust you too much cause i know your trying to help but he is not drawing 40 amps....the idea you used of power=power only works for transformers if you would like to know how you do it here it is: first find the resistance of the windings of your fan: voltage=resistance x current we know voltage and maximum current so 12v=resistance x 20amp means a winding resistance of .6 ohms next we know we want to find out current for a 6v system so again voltage=resistance x current therefore 6v=.6 ohms x current and that gives us 10 amps so at most the fan would be drawing 10 amps...which is still high for a fan but i agree with the other post about using a relay or connecting direct to battery....WARNING!!!! if connecting right to battery or even through and ignition switch make sure to use an inline fuse....last thing you want is a fire under the hood of the beautiful machines :-)
Devon0 -
junkcarfann wrote:Geoff...Do you have a source and part # for a 6v relay? And would it work with 8v?
Thanks.
No I don't sorry, as I'm in New Zealand, and don't know the availability of this stuff in the U.S. However, a standard headlamp relay should cope with it okay, and I'm sure these are available and most of them have fuses inbuilt. Good luck. And my two-bits worth on the fan current, if you are using a 12 volt fan on 6 volts, then it will in fact only draw half the current ( 10 amps) according to ohms law. The resistance is fixed, therefore the wattage will be halved if the voltage is halved. Of course the fan will only turn at half the speed as well most likely. And even 10 amps through a standard ignition switch is likely to overload the contacts and cause reduced voltage through the circuit.
Geoff.0 -
junkcarfann wrote:Geoff...Do you have a source and part # for a 6v relay? And would it work with 8v?
Thanks.
Here's a few:
Ebay # 280477652010, 300420192021, 110537387512,
8VDC is no problem with these.0 -
Just to carify the original question:
Quote:
"I installed one of those new-fangled 6-volt electric pusher fans in front of the radiator and it works great. Gets enough air flow through that beautiful little grill to really make a difference."0 -
forjack842 wrote:Unfortunately you're wrong.
The actual power required to move the same amount of air by the exact same fan physical design requires the same power whether 12v or 6 v. Using your example, if the correct voltage fan is used in both cases then the 12v fan has an internal resistance of 0.6 Ohms and requires (P=I^2 x R) 240 Watts of power to move its rated cubic feet of air per minute. To achieve the same air movement power at 6 volts we'll take 240W / 6V (Power is also is defined as P= Volts x Amps) or 40 Amps. The fan motor would have a resistance of (R=P / I^2) 0.15 Ohms. The above defines the ideal without losses and the like.
The voltage drop in the wire run for a 6 volt system and a fan at 40amps would be considerable if the fan were to be connected at the ignition switch. However, if connected at the battery the loss would be virtually limited to the fan supply circuit leaving just the battery terminal and internal battery resistance to be passed to the rest of the car systems. Simply put, the voltage drop at the battery would be slight with the fan running and the ignition system would virtually see the entire battery supply less its own designed in losses.
The use of a inline fuse is always the correct way to protect your precious valuables.
Jack
hey jack i always like this kind of discisson :-) it always helps other learn aswell as ourselves...but the only thing i have to disagree with what you said is that the resistance of the fan changes...resistance is a constant determined by the amount of wire used to create the coil inside the fan. its not a variable. therefore the fan under a 12v system would push more then the same fan on a 6v system...power is also a variable...resistance is the constance unless you were to rewind the fan with more wire.
what you think? 0 -
Teddy,
We initially started the discussion using a 6v fan. I mentioned a 12v fan and 20A because I installed that fan in my jeep and that's what it was rated at.
A 6v aux fan would draw upwards of 40A to be as air flow effective as the 12v 20A fan.
With 2 separate fans in mind,
Power is the constant in this debate. The power(torque) to move a specific amount of air through a restriction within a specific amount of time.
This is exactly the same as Horse Power moving a specific weight, a specific distance, in a specific amount of time. All participating variables must deliver the exact power or horse power to meet the need.
Be it volts/amps or gasoline/air the combination plus all losses from conversion/transmission must deliver the specific power(torque) to accomplish the task.
In the 6v fan, the power delivered is a relationship between volt/amp input and the air flow. If the voltage is held constant at 6v and the air flow is constant, then the current must be adjusted until the power required to move the air is met. This is accomplished by lowering the resistance of the fan coils but maintaining the magnetic flux rotational power between the coils and the armature.
If the voltage changes from 6v to 12v the required air flow delivered does not change, the rotational torque transfered to move that air does not change therefore the current must decrease to reach the same power delivered in the 6v circuit. To accomplish the currect reduction the field/armature wire size is reduced while still maintaining the required flux for the required rotational torque.
To reduce this to simple terms, Power = Volts x Amps (Ohm's Law for Power)
240w = 12v x 20A or 6v x 40A or
To solve for Resistance: Power / Current ^2 (Ohm's Law for Resistance)
12V circuit: 240Watts / (20A x 20A) = 0.6 ohms
6v circuit: 240Watts / (40A x 40A) = 0.15 ohms
Go to this web page and verify the above using the online calculator.
http://www.onlineconversion.com/ohms_law.htm
Therefore the real difference the user is concerned with between the 6v and 12v fans is the current. The ignition circuit in the 6v car is way under rated to support the fan. So the fan must be directly battery connected via properly rated relay and wire. This is also the exact wiring technique used in the 12v circuit simply because the 12v ignition system is also under rated for an additional 20 amp draw.
In regards to the fan, the power delivered is a relationship between volt/amp input and the air flow through the blades. If the voltage is held constant and the air flow through the blades is a constant, then the current must be increased until the power required to move the air is met. This is accomplished by lowering the resistance of the fan coils but maintaining the magnetic flux rotational power between the coils and the armature.
If the voltage changes from 6v to 12v the required air flow delivered does not change, the rotational torque transfered to the blades does not change therefore the current must decrease to reach the same power delivered in the 6v circuit. To accomplish the currect reduction the field wire size is reduced while still maintaining the required flux for the required rotational torque.0 -
It would be interesting to know exactly what the 6 volt fan is rated in watts. Theoretically you are correct, but I suspect the 6 volt fan would not shift as much air as a 12 volt. No 6 volt system will survive for long with a 40 amp discharge!0
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Hey guys, there are a couple of misperceptions floating around in this discussion. First, the "rating" of a device may not actually indicate the current it actually draws at the specified voltage. Sometimes the amperage or wattage rating indicates what it can handle without damage. Second, for many types of motors, measuring the resistance doesn't give you a valid value, unlike when measuring a simple resistive load. Best to connect it to the correct voltage and measure the current.0
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You are of course correct Park. Like trying to measure the resistance of a bulb, it measures zero, because it doesn't have any resistance until it gets voltage through it which heats it up.0
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You are both right. The 6v fan would definately be a "PIG" and its actual power would be measured via the current it drew in its actual configuration.
The axiom I presented was the ideal case for the motor and all of its inherent losses. And if actually implemented would require a system power re-evaluation before being worthy of everyday use .
Even in the 12v system the battery and alternator are working a bit to provide 20A just for the coolant fan. In the 6v system the 29 Hud/Essex would require a definite power redesign. Or you would surely have a cool radiator for there would be no battery left to run the ignition!0 -
I have been watching this thread from the start and some "electrical wizards" are having fun with it. I have a '50 Super Six with a Scott 6 volt fan mounted on the front of the rad. The fan is hooked to the ign switch thru a toggle switch. No relay. I checked the voltage at the ign. side of the coil - engine idleing - no charge - ( gen. light on )- 5.6 volts. I turned the fan on - the voltage went down to 5 volts. Of course the voltage goes up when the eng. speed is raised. It sounds like his origional problem is bad wireing or a weak battery. Norm0
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Well a 50's gen will handle the fan draw nicely. The 29 has a gen output max of 15 or 20A with ignition wire of about 10g. I isn't designed to handle your fan.
Jack0 -
Jack - Denverslim's is a '36. My generator is not puttng out anything at idle speed. Norm0
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That's typical at idle, the battery cuts in and manages power needs while the genny is under RPM'd.0
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