Block Heater Questions

[dboyw]

Couple questions regarding the block heater.

I bought the marine plug, can't remember name now, and I was going the put on front of the truck behind the license plate. Now I think I want to put it in the rear of the truck. I am going to buy a holder that you put a seven pin trailer plug into and use it the the new heater plug.

OK, questions:

What gauge wire do I need for the plug to run to the rear of the truck?

Should I cut the original plug and wire off close to the plug or close to the heater part?

Has anyone ever put two plugs on a truck? One in front and one in rear. I thought of doing this but is there really any advantage to doing this?

Would this LED http://www.radioshack.com/product/i...2058&kwCatId=2032058&kw=led&parentPage=search
work to verify that power has been established when plugged in to the truck?

Thanks for the help. And if there is anything else you think I should know, please tell me. Thanks

 

[whatabudro]

I would personally not run a cord to the back of your truck. There are too many possibilities of the wiring getting chafed. You're not dealing with 12V here either. One minor nick in the wire could mean death. You can change the connector on the cord if you want to, just make sure it is rated for 20A 120V IMO. Here's some more info for you.

Voltage drop:

VD = 2xLxKxI divided by CM

L = length
K = Constant ( copper = 12.9 ohms ) (aluminum = 21.2 ohms)
I = amps
CM = circular mills
18awg = 1620
16awg = 2580
14awg = 4110
12awg = 6530
10awg = 10380


I'm going to quote DAVEBEN (for formula reasons only) and say the block heater is 1000 watts.

P=IxE E=IxR (PIE & EIR)

P= power/watts
I= amps
E= voltage

E= voltage
I= amps
R= resistance

Now if you know any two variables you can figure the third.

Using P=IxE:

1000p/ 120e = 8.3i (amps)

Now back to the voltage drop formula using a 100 foot 18 gauge copper wire cord:

VD = 2xLxKxI / CM

2x100x12.9x8.33 / 1620 = 13.27%

Now we must go back and figure out the amperage again since our voltage is now 120volts minus 13.27%

100% - 13.27% = 86.73%

120volts x .87 = 104.4volts

1000p/104.4e = 9.61i

So by using that cord you raised amp draw from 8.33 to 9.61

Here is where the problem is:

18awg cord is rated at 7 amps in most cases (voltage drop doesn't matter its too small anyway)
16awg cord is rated at 10 amps in most cases

So you would think 16 awg would be fine. I was estimating your voltage at the receptacle at 120 volts to start with. Most do not have 120 volts, some are lower than 110 volts. So if we started with actual and figured the voltage drop, then the 16 gauge cord would not be large enough.

 

[stroke_smoke]

i knew i sucked at math but geeez . i'd just use a big ole camper extension cord. wait, i do that anyway. oh well.

 

[Rex-a-FORD]

Couple questions regarding the block heater.

OK, questions:

What gauge wire do I need for the plug to run to the rear of the truck?

I'd suggest 12 AWG extension cord 12-2/W Ground. Also solder all splice connections. Use heatshrink tubing over the splices to keep moisture out.

Should I cut the original plug and wire off close to the plug or close to the heater part?

If the original cord is still good I'd cut it as long as you can to make it easy to work with.

Has anyone ever put two plugs on a truck?

This will be dangerous, since the prongs on the plug you aren't using will energized at 120 volts AC. Shock hazard.

One in front and one in rear. I thought of doing this but is there really any advantage to doing this?

Would this LED http://www.radioshack.com/product/i...2058&kwCatId=2032058&kw=led&parentPage=search
work to verify that power has been established when plugged in to the truck?

The LED you linked to is only good for 12 Volts. You're working with 120 Volt AC for the block heater.

Thanks for the help. And if there is anything else you think I should know, please tell me. Thanks

My reponses are in RED within your quoted text above.

Please be very careful working with the block heater. There have been several trucks burned up by the block heater plug or cord failing and starting a fire.

If in doubt just go buy a new cord for the block heater.

 

[RenoF250]

I would personally not run a cord to the back of your truck. There are too many possibilities of the wiring getting chafed. You're not dealing with 12V here either. One minor nick in the wire could mean death. You can change the connector on the cord if you want to, just make sure it is rated for 20A 120V IMO. Here's some more info for you.

Voltage drop:

VD = 2xLxKxI divided by CM

L = length
K = Constant ( copper = 12.9 ohms ) (aluminum = 21.2 ohms)
I = amps
CM = circular mills
18awg = 1620
16awg = 2580
14awg = 4110
12awg = 6530
10awg = 10380


I'm going to quote DAVEBEN (for formula reasons only) and say the block heater is 1000 watts.

P=IxE E=IxR (PIE & EIR)

P= power/watts
I= amps
E= voltage

E= voltage
I= amps
R= resistance

Now if you know any two variables you can figure the third.

Using P=IxE:

1000p/ 120e = 8.3i (amps)

Now back to the voltage drop formula using a 100 foot 18 gauge copper wire cord:

VD = 2xLxKxI / CM

2x100x12.9x8.33 / 1620 = 13.27%

Now we must go back and figure out the amperage again since our voltage is now 120volts minus 13.27%

100% - 13.27% = 86.73%

120volts x .87 = 104.4volts

1000p/104.4e = 9.61i

So by using that cord you raised amp draw from 8.33 to 9.61

Here is where the problem is:

18awg cord is rated at 7 amps in most cases (voltage drop doesn't matter its too small anyway)
16awg cord is rated at 10 amps in most cases

So you would think 16 awg would be fine. I was estimating your voltage at the receptacle at 120 volts to start with. Most do not have 120 volts, some are lower than 110 volts. So if we started with actual and figured the voltage drop, then the 16 gauge cord would not be large enough.

I am afraid your math is incorrect, the smaller cord will not result in more current. The heater is just a passive resistor and will simply operate at a lower voltage and produce less heat with smaller gauge wire. I think the wire in the cord on the truck is 16 gauge. Bigger is better though so I would probably use 12 gauge and perhaps fuse it for the max current the heater should draw to lower the risk of fire.

 

[whatabudro]

I am afraid your math is incorrect, the smaller cord will not result in more current. The heater is just a passive resistor and will simply operate at a lower voltage and produce less heat with smaller gauge wire. I think the wire in the cord on the truck is 16 gauge. Bigger is better though so I would probably use 12 gauge and perhaps fuse it for the max current the heater should draw to lower the risk of fire.

You're right, the smaller cord does not result in more current, directly, but it is the cause. The smaller cord results in in voltage drop. The length of wire and size of wire determine the the resistance of the circuit. The increased resistance through the conductor results in lower voltage. The lower the voltage is the greater the amperage. This is not something that I made up, it's called ohms law. The cord that comes on the truck is not considered a part of the circuit when making calculations. It is considered fixture wiring and is the responsiblity of the company and engineer that designed the product to be properly sized, also UL for giving them a label. Running a 120V cord on your truck from front to rear, or even making changes to the cord on the front of the truck, fall into your wallet. There are a lot of people in this world that think they can wire things just fine. I know because I repair them daily.

 

[RenoF250]

You're right, the smaller cord does not result in more current, directly, but it is the cause. The smaller cord results in in voltage drop. The length of wire and size of wire determine the the resistance of the circuit. The increased resistance through the conductor results in lower voltage. The lower the voltage is the greater the amperage. This is not something that I made up, it's called ohms law. The cord that comes on the truck is not considered a part of the circuit when making calculations. It is considered fixture wiring and is the responsiblity of the company and engineer that designed the product to be properly sized, also UL for giving them a label. Running a 120V cord on your truck from front to rear, or even making changes to the cord on the front of the truck, fall into your wallet. There are a lot of people in this world that think they can wire things just fine. I know because I repair them daily.

No, you are missing the point, the small wire increases the total resistance of the circuit and lowers the overall current. A heater element designed to create 1kW at 120V is going to be 120 * 120/1000 = 14.4 ohms and a current of 8.33A. Now lets say you add a bunch of wire that has a resistance of 1 ohm making a total for the circuit 15.4 ohms. Now the current will be 7.8A * 120V = 935W with 112.2V across the heater and 7.8V lost across the wire.

 

[whatabudro]

No, you are missing the point, the small wire increases the total resistance of the circuit and lowers the overall current. A heater element designed to create 1kW at 120V is going to be 120 * 120/1000 = 14.4 ohms and a current of 8.33A. Now lets say you add a bunch of wire that has a resistance of 1 ohm making a total for the circuit 15.4 ohms. Now the current will be 7.8A * 120V = 935W with 112.2V across the heater and 7.8V lost across the wire.

I am not missing the point, I know them all, I am a master electrician with 12 years in the trade. The point I made was explaining to the original poster what is considered safe and exceptable by NEC (national electric code) I stated that installing a 120V cord on his vehicle would not be safe. I said this because he is not qualified to make such an installation or he would not have asked. It can be done, but it should be done by someone who is qualified to do so. The guy at home depot in the electrical section is not considered quailified.

Your formula above does nothing but prove my point. The NEC allows no more than 3% voltage drop across a branch circuit, with no more than a total of 5% drop from the service feeders to the outlet NEC 210.19(A)(1)FPN4. Your formula will work fine on a circuit board, but not on a 120V circuit. You came up with a 6.5% voltage drop, on that circuit alone without including the feeders. The reason for this code, is fires caused in the past. The NEC is written by the NFPA ( national fire protection association) not by electricians. They are MINIMUM guidelines for what is acceptable. The formula I used and the way I used it are industry standards. The unnecessary resistance causes heat which in turn causes fires. Increasing the resistance on a circuit lowers the voltage. Once the voltage of a circuit is known, THEN the formula for finding the amperage is calculated. The circuit resistance is not calculated in with the amperage/watts of the load for determining the final load.

 

[whatabudro]

dboyw:

You should just install the plug you purchased on the front of you vehicle, if that plug is rated for the proper voltage and amperage. Then buy a cord from your local home improvment center, that is 12ga and has a lighted cord end on it and is as short as possible, that should solve your concern of knowing that the circuit is energized.

 

[Crumm]

I bought the marine plug, can't remember name now.

It is probably a Marinco plug.