The Stovebolt Site Forums The Shop Area The Electrical Bay Heater Switch is HOT!!!

No the switch should not get so hot that it burns you. check all connection it mke sure they are making good.

Wow! we topped 40! Not bad for my 2nd post ever! Everyones been real informative (whether they agree or not... )and again thanks for the help.

-Jeff

and yer up to 8 already!!
what was the problem?

Bill

I have my '55 2nd series GMC heater out of the truck for a rebuild so I did a little measuring and a little figuring. This is a 6 volt system and has a single resistor in the fan speed control to give two fan speeds.

The resistance of the motor measures out at around 0.9 ohms. The single resistor measures about 3.7 ohms.

Since the resistors job is to drop the voltage to the motor to something less than 6 volts, I calculated the current flowing in the resistor for varying voltages across the resistor:

E = volts
R = resistance (ohms)
I = current (amps)

P = IxIxR (power in watts)
I = E/R (current through resistor

1 volt drop: 1/3.7 = 0.27 amps ; .269 watts
2 volt drop: 2/3.7 = 0.54 amps ; 1.07 watts
3 volt drop: 3/3.7 = 0.81 amps ; 2.43 watts
4 volt drop: 4/3.7 = 1.08 amps ; 4.32 watts
5 volt drop: 5/3.7 = 1.35 amps ; 6.74 watts
6 volt drop: 6/3.7 = 1.62 amps ; 9.71 watts

Lets take the case with the 4 volt drop and calculate the voltage dropped in the motor.

I = 1.08 amps throughout the circuit.
Motor resistance = .9 ohm (measured)

Motor voltage drop due to resistance
Mvd = 1.08 amp x .9 ohm = .972 volt

If we add up all of the voltage drops in the circuit, the should all sum to zero.

Battery = + 6.0 volts
Resistor = - 4.0 volts
Motor = - .972 volts
-----------------------
?????? = + 1.028 volts

Looks like we have a little voltage left over!

We know the current is steady at 1.08 amps lets
redo this calculation in power Watts

Battery = 6.0 volts x 1.08 amps = 6.48 watts (energy in)
Resistor = 4.0 volts x 1.08 amps = 4.4 watts (heat from resistor)
Motor = 0.972 volts x 1.08 amps = 1.05 watts (heat in motor windings)

Looks like we have 1.028 volts x 1.08 amps = 1.11 watts of energy actually turning the fan!

Stuart, in your experiment, your assumption that the fan would always draw 6v/.95ohm = 6.3 amps, only applies when the rotor is locked and the torque produced in the shaft is at it's maximum.

If you release the shaft and allow it to rotate, the current draw and the torque in the shaft will drop as the motor speeds up. The more mechanical power used by the driven device, the higher the current draw.

John

looks like ya better check the battery, must be overcharged

if 'the circuit' is the 3.7ohm resistor, wouldn't the current always be 1.62 amps, so 'drop' 6 volts? isn't the circuit actually 4.6ohms? [.9 + 3.7], making the complete circuit current 1.3amps?
where's the current draw of the motor?
the same as the resistor?
do the same amps create the heat from the resistor and the motion in the motor??
what's the 'equality of currents' in this?

does the measured resistance of the motor take into account the 'non-linear' characteristic in the calculations?

of course we also know that in the real world the "ideal" values necessary for the formulas to be accurate don't exist, but it still appears it's current that makes the world go round, and the theories just let us pick a "looks like" that matches our out-look

I hadn't thought about the current required to move the motor from rest being higher than that needed to maintain a steady speed but I do make the assumption folks knew that starting a motor is what jumps up the electric bill, not running it

Bill

As Bill pointed out, there is a flaw in the above calculations. One can't just choose an arbitrary voltage drop to assign to the resistor. The actual drop across each resistance will depend on the current, which must be calculated using the total resistance of the circuit (including the motor, not just the resistor alone) and the battery voltage.

Electrical concepts are often explained using the flow of water through a hose as an analogy (as if fluid dynamics were simpler). A certain amount of water will flow through a hose (current) at a given source water pressure (voltage) and a given amount of friction or restriction (resistance) within the hose. No matter how many restrictions there are to the flow (e.g. crimping the hose or putting a valve in the hose or running a hydraulic motor), the amount of water coming out the end must be the same as the amount going in at the spigot. The water can't build up or disappear anywhere within the hose, so the flow has to be constant throughout the hose. That's the "equality of currents".

That same flow creates heat due to friction at the crimp or valve in the hose and produces the motion of the hydraulic motor. Both the valve and the motor also create pressure drops within the hose at their exits, since some of the energy was used in creating the heat and motion. The pressure at the end of the hose, at the point it meets the atmosphere, is zero. One can also picture a battery (or generator) as a water pump in this analogy.

Does this help explain the concept, or am I all wet and tempting fate by mixing water and electricity?

Good Evening Bill, Curt and anyone else still following this thread.

Curt, your water analogy is spot on.

Bill, you ask a lot of good questions so let's see if I can give you qood answers.

"if 'the circuit' is the 3.7ohm resistor, wouldn't the current always be 1.62 amps, so 'drop' 6 volts? isn't the circuit actually 4.6ohms? [.9 + 3.7], making the complete circuit current 1.3amps?"

Part 1a. If the circuit was JUST the 3.7ohm resistor the total current flow would be 1.62 amps, and the voltage dropped acrossed the resistor would be equal (and opposite) to the battery voltage.

Part 1b. If the circuit was the 3.7ohm resistor + the .9ohm resistance of the motor, circuit current would be 1.3 amps. The resistor would then drop 1.3amps x 3.7ohms = 4.8 volts and the motor would drop 1.3amps x .9ohms = 1.2 volts.

Kirchoff's voltage law says the sum of all of the voltages in a circuit must equal zero.

+6v (battery) - 4.8v (resistor) - 1.2v (motor) = zero and Kirchoff's voltage law is satisfied

Kirchoff's current law says that the sum of the currents entering and leaving a node (device) in a circuit must equal zero.

The Battery has 1.3 amps out from (+) terminal to the resistor.
The Resistor has 1.3 amps in from battery, 1.3 amps out to the motor.
The Motor has 1.3 amps in from resistor, 1.3 amps out to the battery (-) terminal.

Each node (device) has a sum of zero amps (In - Out) and Kirchoff's current law is satisfied.

This scenario requires one specific condition to remain true. That condition is that the shaft of the motor MUST BE MECHANICALLY RESTRAINED FROM MOVING (called a locked rotor). As soon as the motor begins to turn, Ohm's law is useless as far as helping us determine current flow and voltage drop in the motor. As I did in the last post, you can make an assumption about the current and calculate information about the motor but it is impossible to calculate motor current based on what we know at this point.

Rotating electrical machines are energy CONVERSION devices. They convert electrical energy to mechanical energy (motor) and vice versa (generator).

In a motor, we supply electrical power in (Volts * Amps)= (Watts) and have mechanical power (Torque x Angular Momentum) = (HorsePower) coming out on the shaft. We also lose some of the electrical power to the resistance of the windings and the brushes/commutator.

In a generator, we supply mechanical power in on the shaft and take electrical power out on the terminals. Again, some of the mechanical power that is converted to electrical power is lost to the resistance of the windings and the brushes/commutator.

DC motors and generators are essentially the same machine. The 1948-1951 Chevy Truck shop manual has instructions on how to run your generator as a motor.

In the next post, I'll attempt to explain what's going on in these DC machines that gives Ohm's law (and many of us Stovebolters) such heartburn.

John

olblue, doesn't the motor/generator also create back emf that would tend to add to or at least offset the apparent current that the motor converts to heat in order to turn?

Any time we have an electrical conductor moving through a magnetic field, an electrical current will be induced in the wire. That's one of the science lessons from about middle school as I recall.

If we take our heater motor, hook the wires to a voltmeter and hook a variable speed drill to the shaft and turn the drill on, what happens? Yup, we get a voltage on the voltmeter. If you turn the drill faster, the voltage goes higher, if you stop the drill, the voltage disappears. If you reverse the drill, the voltage reverses. This voltage is called the 'speed voltage' and is caused by the movement of the conductors through the magnetic field left in the steel parts of the motor.

The magnitude of this speed voltage depends on two things, how fast the motor is turning and how strong the magnetic fields are. Faster motor -> more voltage, stronger magnetic field -> more voltage.

This should all seem somewhat intuitive as far as a generator goes. Increase the engine speed and the ammeter shows more charge or have your voltage regulator fail and supply to much field voltage to the generator and the generator will overcharge your battery.


In the case of a motor where we supply the voltage and current to move the rotor what happens to the speed voltage? It's still there! Not only that, it still behaves just like it does in the generator with one new wrinkle. The 'speed voltage' OPPOSES the voltage supplied from the outside source.

If we apply full voltage to our heater motor, at the moment the switch is thrown, the current through the motor will be in accordance with Ohm's law and limited by the non-rotating resistance of the motor.

As the motor speeds up, the 'speed voltage' will increase and act to 'reduce' the voltage available to push current through the motor. As the motor continues to speed up, the current flow through it will drop off as the speed voltage increases. At some point, the motor will reach an equilibrium point where there is not enough current flowing to continue to accelerate the motor and it runs at a steady speed.

The motor is using just enough current to spin the unloaded rotor and is running at it's maximum speed. What's the resistance of the motor at this point?

The same as it was before, around .9 ohm. Why is there MUCH less current flowing through it than expected? Because the 'speed voltage' created by the motion of the motor armature is opposing the voltage supplied by the battery and effectively reducing the voltage available to push current through the motor.

What happens if we put a load on the motor? The shaft will slow down, the 'speed voltage' will drop and the battery will push more current through the motor. Since the resistance of the motor is quite small, it takes only a small drop in motor speed to create a substantial increase in current and mechanical power delivered.

What happens if we supply mechanical power to the shaft of our motor and increase the speed and the associated 'speed voltage' above the supply voltage? Our motor has now become a generator and will PUSH current in the opposite direction back into the battery and charge it.

My brain is strained, and if you got this far, I'll bet yours is to. I hope this long-winded blurb helps a few people get a better grasp on 'juice'.

The original question about why the resistors on the back of the switch get hot? They're supposed to, that's how the motor speed is lowered. Newer cars place the blower resistors inside the air duct to help keep them cool and I suspect to keep wires and such from touching them.

If they seem way to hot, then something is overloading the motor and causing it to draw to much current, the oilite bushings my be dried out and gummed up or there may be trash rubbing on the fan wheel.

John

and where's the energy [volts or current or especially watts] that is lost to heat at the resistor? doesn't Kirchoffs equality see the heat lost as current out? kinda like the pinhole in the hose where the water out the business end = the water in less the loss from the pin hole - like the canvas hoses fire depts usta use, weeping alla way along

I was always a bit iffy on them lil electrons leapin from hole to hole til they fell into the 'earth', but it seems some of'em must commit hari-kari when they hit the resistor

I'm not sure I understand the theories of hydraulics either, but my hydraulic pressing equipt has been working reliably since I put it together 20+ years ago, and tho there's slight weeping around the rams, I haven't added anything to the reservoir in that time!

and ol_blue, I can testify that a 2 & 3 horse motors do convert quite a few of my dollars to heat - doesn't the back emf have more to do w/ the generator thing and the variability of the resistance of the running motor - folks are always asking why I leave'em runnin idle

question is, considering that the electrical circuit is really a closed one, ground being the completion of the loop, and the tires insulating the whole thing from the earth, the fact that batteries run outta amps pretty fast indicates the 'equality' is a one way street - the amps are going someplace other than "back to minus", and the motor requires very little to keep it turnin once started, so the actual energy must nearly all be going into heat, one place or other - the outlet of the 'hose' isn't going back into the reservoir like my hydraulic pumps

Bill

Hi Bill,

At the resistor, on the battery side the power coming in is 6v x 1.3a = 7.8watts.

We drop 4.8 volts acrossed the resistor which means we've converted 4.8v x 1.3a = 6.24watts converted to heat.

At the output of the resistor, the power remaining in the circuit is (6-4.8)v x 1.3a = 1.56watts

What's lost in the resistor is the voltage.

It's like a hydroelectric dam in the river. In the upper resevoir, we have a lot of potential work in the water. When we let the water run through the turbines, we extract energy from the water as it drops. Down stream, we still have the same amount of water moving downriver but we have used up the potential energy.

as much as i love talking about timing valves with the engine running (never tried it but seems like it might work) and undercoating and such, this thread has really been educational for me as it deals with one of my 'other' hobbies; electricity. i also love electronics. maybe we can get a lenthy debate in the radio section going. now if i can just find somewhere to discuss guitars and remote controlled cars, i'm set. :-)
tony

Well, what happens when you cross the blue wire from the Dohicky to the blue wire on the Thingamabob and short it against the Johnson rod? And does this effect the spare tire pressure in the storage shed?

This is the longest discussion I have ever seen on a very simple topic. I think that there were some inappropriate terms used earlier on that caused confusion. Here is a sample:

red58
Extreme Gabster
Stovebolter # 2136

posted 02-13-2006 12:47 AM
--------------------------------------------------------------------------------
easy: I [current draw of motor]=E [12V]/R [the resistance of the motor windings] translates to P=IxE
power in watts [like yer portable electric heater, which is just a resistance wire element], is what makes heat, and that's determined by the current the motor uses to run times the voltage you have - the slower you run the motor, the more current has to be dumped someplace else, which is the resistance unit on the back of [or inside] the switch, and that piece is quite small, hence can get red hot
-------------------------------------------------
Obviously, current is not"being dumped someplace else". This is a series circuit here and the current through the control is the same as through the motor.

Also, the total current does go down as the control is changed to slow down the motor. The voltage drop across the motor also goes down and the control now dissapates energy as heat. This energy exists because the control now has inserted resistance into the circuit and current is passing through it.

I also take exception to the term "juice" because it does not define anything and people use it to describe voltage, current or power flow.

On another note, it was good to see some information on rotating DC machines. Although the affects shouldn't be considered too significant in this discussion, the current flow through the motor with a stalled rotor is different than with a spinning one and even blocking the air intake or output will change this. If you look at the of the specifications for the Chev starters of the fifties you will see a current draw test with the rotor blocked.

Finally, yes the resistors in these controls get hot enough to easily burn skin.

after reading all this i have forgoten all that i have ever learned, (about electricity) thik about the voltage as pressure pushing the ampers through the wires to turn the motor. Your switch puts more resistance in the wire, need more pressure to push the amps. don't have more presure, 6 or 12 volts steady. so not as many electrons (amps) go to the motor. the resitance slowing the amps creates heat. Engergy can only be changed, not destroyed. volts are only pressure. amps are flow.

If your'e floatin down the river in a cement canoe and all 4 wheels fall off how many pancakes does it take to shingle a doghouse? E=MC2?

I spent 4 years in the USAF too as an engine mech and all I have to show for it is tinnitis from the darn GE-J47s.

Weeds

With all being said above,"IT'S NORMAL"

An old post, but nicely done! Really informative, and just what I was searching for. Thanks, if you're still around.