air handler name plates

[sparky333]

I have been an electrician for over 20 years and a journeyman for 16 .But I have been confused over the name plates on these new units i have been wiring. this is just an example.
208/240 volts then the ampacitys will say 35/40 for whatever heat strips are in that unit.
This is just an example with the ampacitys >my question is How Come the smaller Number is on the left Of the slash?.My knowledge has always been that at 208 Volts the amps will be larger then at 240v and vice versa. the more volts the less amps.Right ?

Just like a name plate on a motor is right 120/240v then the amps is like 20/10a just an example
or a three phase 208/480 = more/less amps

could some on please help me out with this?

 

[boboelectric]

Look at it, and you'll get the hang of it.

 

[jeremysterling]

With a 240/480 motor the resistance of the windings changes according to how you configure the t-leads.

The resistance of the heater strip does not change even if you apply 208 or 240. Simple ohm's law. Divide the KW of the heater by voltage you are using to get the current of the heater.

The same goes for water heaters. You can run a 240V water heater with 208V supply, you will use less current and you will also get less hot water!

 

[sparky333]

i diid the calculation?

i diid the calculation?

5 kw heat strip
5000 w divided by 208 = 24.03 a
5000 w divided by 240 = 20 a which is less simple ohms law?

still confused sorry?

 

[augie47]

what jeremy says!
The key is that the only constant you are working with is the resistance of the element.
The wattage is not constant.
If you know the element is rated say 9600 watts @240 it will pull 40 amps.
Using ohms law R=E/I you find the resistance is 6 ohms.
Similary I=E/R, so on 208 it's 34.6 (35) amps and W=E?/R or 7210 watts.
Many of the nameplates will also show that in a slash wattage rating ie: 9.6/7.2 kw

 

[iwire]

This is just an example.
208/240 volts then the ampacitys will say 35/40 for whatever heat strips are in that unit.
This is just an example with the ampacitys >my question is How Come the smaller Number is on the left Of the slash?

When you run the same heat strip at a lower voltage it will draw less current.

Lets look at a heat strip rated 40 amps at 240 volts.


In watts that would be 240 volts x40 amps =9600 watts, now what would happen if we ran that same exact heat strip on 208?

First we need to find the resistance of the heat strip, to do that we divide the voltage by the current

240 volts / 40 amps =6 ohms of resistance

Now that we know the resistance we can use ohms law again to find out what that heat strip will draw at 208 volts

208 volts / 6 ohms =34.66 amps.

So the exact same heating strip that needed 40 amps at 240 volts will only need 34.66 amps at 208 volts.

HOWEVER ... at 240 volt that heat strip was putting out 9600 watts of heat, at 208 it is not, it is now putting out less heat.

208 volts x 34.66 amps = 7209 watts

So by dropping the voltage the current went down but so did the amount of heat produced.

My knowledge has always been that at 208 Volts the amps will be larger then at 240v and vice versa. the more volts the less amps.Right ?

Now, lets talk about what happens if we need the heat output of 9600 watts to remain the same at either voltage.

In the first example we had a 240 volt heat strip that put out 9600 watts with a resistance of 6 ohms and a current of 40

If we have to maintain the same wattage output at a lower voltage we will need a different heat strip with a lower resistance and that will draw more current.

9600 watts / 208 volts = 46.15 amps and the internal resistance of the heater would have to drop from 6 ohms to 4.5 ohms.

Has this made sense?

 

[sparky333]

I got it now

I got it now

I need to go but thanks and i need to get back into therory and the books thanks again

 

[augie47]

I need to go but thanks and i need to get back into therory and the books thanks again

or trust the nameplate

 

[infinity]

I need to go but thanks and i need to get back into therory and the books thanks again

The theory is simple if you remember that the current will be different depending on the type of load. A resistive load will have the current and voltage directly proportional. Increase the voltage you increase the current. An inductive load, such as a motor, will be the opposite or inversely proportional. Increase the voltage and you'll decrease the current.

 

[iwire]

An inductive load, such as a motor, will be the opposite or inversely proportional. Increase the voltage and you'll decrease the current.

I cannot agree with that as written.

If the motors mechanical output remains the same that above is true, but reducing the voltage to lightly loaded motors will also reduce current. (To a point)

 

[infinity]

I meant for a motor rated for more than one voltage, i.e- 208-240 rated.

 

[Dennis Alwon]

Well I got ready to post but I see Bob has explained it well enough. Since I went thru the work I am going to post anyway. The key here is that an element at 240 volts has a different wattage then at 208V.

Normally you would see two wattages at the different voltage. Thus a 5000 watt element at 240 is not really 5000 watts at 208. Let?s look at it

5000 watt element @ 240

Using ohms law we have to find resistance
R= V^2/P, V= Voltage, P= Wattage, I= amperage, R= Resistance

R= 240^2/ 5000
R= 11.52

Amps of 5000 watts on 240V
I= V/R
I= 240/11.52
I= 20.83 amps

So what is the wattage of the same element at 208V.
We know R= 11.52 and V= 208 so we look for a formula with those known items
P=V^2 / R
P= 208^2/ 11.52
P= 3756 watts

So
I= P / V
I= 3756 / 208
I= 18.06

 

[defears]

Think of a heating element as a fixed resistance. Motors try to compensate for slightly lower voltages.

 

[jeremysterling]

...Divide the KW of the heater by voltage you are using to get the current of the heater.

Sorry, I had not had coffee yet when I wrote that. Dennis and Iwire explained it correctly. I should have written:

Divide the square of the rated voltage by the KW rating to find the resistance of the heater.

Then follow Dennis' math and Ohm's law!

5 kw heat strip
5000 w divided by 208 = 24.03 a
5000 w divided by 240 = 20 a which is less simple ohms law?

still confused sorry?

Not anymore. Thanks, Dennis.

 

[ronaldrc]

5 kw heat strip
5000 w divided by 208 = 24.03 a
5000 w divided by 240 = 20 a which is less simple ohms law?

still confused sorry?

The thing is if the heater generates 5000 watts at 240 at 208 it will be less than 5000 watts at 208 volts.

Go back to ohms law and study it.

 

[augie47]

sparky333, does it make sense to you now ?

 

[ActionDave]

With a 240/480 motor the resistance of the windings changes according to how you configure the t-leads.

I think this is the part that throws a lot of guys. I know it did me for a while till I learned what Jeremy said above.

 

[Dennis Alwon]

sparky333, does it make sense to you now ?

You need to yell louder. Try this

sparky333, does it make sense to you now ?

 

[ronaldrc]

Bob

I agree thats the first discussion I got involved in on a electrical forum in 1995.

Here a Calculator for converting Wattage when the voltage is higher or lower.
Bottom two calculators on this web page.

http://home.comcast.net/~ronaldrc/wsb/calculators2.htm

 

[LarryFine]

There's a big difference between using simple Ohm's Law math for a varying voltage across a constant-impedance load, and making an equipment change, which means varying the impedance, for different design voltages.

When we say that voltage varies inversely with current for a given power level, we're talking about changing the load's characteristics to suit a change in supply voltage, because the goal is to be able to deliver a given load.

The design power output is the constant, and the load impedance is the direct variable. Then, the voltage we actually supply, and the resultant current, still follow Ohm's law. Note that most motor types don't follow this behavior.

A motor attempts to deliver the necessary torque the load demands, and will use more current as it's slowed by the load in an attempt to speed back up. That's why overloaded motors get hot; the increased current.

When we're talking about what happens when we vary the voltage across a given resistance, with no equipment change, then the load impedance is the constant; the supply voltage, and the resultant current, are variables.