help on reducing wattage

[buju357]

help on reducing wattage

Hi all

I need some help on this:

I have a appliance that is rated at 4000 watts (220vac) , I want to hook it up to a timer that is only rated at 3500 watts , so I would like to reduce the amount of wattage that this appliace can get.

my calculations are as follows:
3000 watts / 220 vac = 13.6 amps

now according to this website:
Ohm's Law / Watt's Law Calculator

voltage 220 v
current 14 amps
resistance 15.7 ohms
power 3080 watts

does this mean I should use a 15.7 ohm resistor ?

 

[davenn]

if the appliance draws 3500 Watts to function correctly, then thats what it uses, nothing you can do about it. unless its a heating element for example that you can turn it down to a lower heat setting with a rehostat (Oven, range top element etc).

any resistor you put inline is only going to try and dissipate the additional wattage and get VERY hot and probably burn up

What is the appliance, it mite help if you told us that

Dave

 

[Reloadron]

If the load (appliance) was purely resistive like a heating element then yes, a resistor could be placed in series with the load to limit the current and subsequently the power. However, there are a few problems.

Initially you have a 4000 watt load running on 220 VAC so we get 4000 / 220 = 18.18 amps of current. With that we can calculate the load resistance as 220 / 18.18 = 12.1 ohms.

We want to limit the current to the load so we have a 3000 watt load. A 3000 watt load would be 3000 / 220 = 13.64 amps. With that we can calculate the load resistance would be 220 / 13.64 = 16.140 ohms.

So we need to get the load resistance up from 12.1 ohms to about 16.1 ohms. That gives us 16.1 - 12.1 = 4 Ohms. Looks like if we add a 4 ohm resistor we are where we want to be give or take a little. We can't change the load resistance but we can add resistance. However, this seems too easy doesn't it?

With our 3000 watt load the circuit current will be about 13.63 amps. The voltage drop across our 4 ohm resistor will be 4 * 13.63 = ~55 volts. The voltage drop across our original load will be 12.1 * 13.63 = ~165 volts. Our series resistor will need to easily dissipate 750 watts or so but figure 1000 watts minimum so it runs cool. You are looking at very expensive planar type heat sinked resistor in a 4 or 5 ohm flavor. Not very practical but it would work.

There is another problem. All of the above assumes a purely resistive load like heating elements. It does not consider for example a heating element with any type of motor like a fan blower. Since our 12.1 ohm load now only has about 165 volts across it (remember our 4 ohm resistor is dropping 55 volts) that means any motors or fans designed for 220 VAC now only sees 165 volts. The motor may not like this lower working voltage and burn up. Additionally even if it did run it would run much slower than normal moving much less air and now we stand a very good chance of burning up our load. Of course all of that assumes heater elements but I am sure you get the idea.

The best approach would be to have a contactor supply the load and use the timer to drive the contactor. Contactor or SSR (Solid State Relay) would likely be a much more viable solution.

More info on the load as was mentioned would sure be helpful.

Ron

 

[davenn]

I still think tho, the real point is ...

it doesnt matter where the extra 500 Watts is being dissipated in the load only or load + a resistor, the total dissipation is still going to be more than what the timer switch can handle (3kW) !!!

Dave

 

[buju357]

thanks for the replies.
the appliance is a hot water geyser , one of these fancy things where the element is deep inside it and it would have to be opened up and the element replaced using new seals etc , its a new geyser so still under warranty so I don't want to mess around with it.

Reloadron , your explanation is excellent and very easily understood - thanx .

I think the rheostat is a good idea , but correct me if I'm wrong here , but wont the rheostat switch on /off and still let the heating element draw its 4000 watts ?

 

[Nigel Goodwin]

You're making this FAR more complicated than it is - simply use your timer to switch a suitable relay, and the relay to switch the heater - job done.

 

[Reloadron]

I still think tho, the real point is ...

it doesnt matter where the extra 500 Watts is being dissipated in the load only or load + a resistor, the total dissipation is still going to be more than what the timer switch can handle (3kW) !!!

Dave

Hi Dave

The original post states in part:

I have a appliance that is rated at 4000 watts (220vac) , I want to hook it up to a timer that is only rated at 3500 watts , so I would like to reduce the amount of wattage that this appliace can get.

The switch in the timer is rated at 3500 watts. That was my take from the original post.

The load is a 4 KW load running on 220 VAC. That works out to be a current draw of about 18 amps. For a purely resistive load, like heating elements for example, that would be a load resistance of about 12 ohms. The load resistance and applied voltage are fixed so nothing can be done about it.

However, if I place a 4 ohm resistor in series with my load the total resistance increases from 12 ohms to 16 ohms. With the same 220 volts applied my circuit current drops from 18 amps to about 13.75 amps from the original 18 amps (give or take). The overall power dissipated in the circuit drops from the original 4 KW to about 3 KW which is about 500 watts less than the switch is rated for.

The big problem doing this, as I mentioned, is the 4 ohm resistor will be looking to dissipate over 700 watts of power (quite toasty) so you would need a heck of a large resistor well heat sinked. I also listed some other consequences of simply placing a resistor in series with the load. Anyway, we still don't know what the load actually is.

Finally I suggested the use of a contactor or SSR as what I saw as a more viable solution.

Ron

 

[Reloadron]

I think the rheostat is a good idea , but correct me if I'm wrong here , but wont the rheostat switch on /off and still let the heating element draw its 4000 watts ?

Nope. Placing a resistance in series with the load will reduce the total current and thus the power overall. When the time comes, consider a relay or solid state relay driven by the timer.

Ron

 

[Diver300]

You could run the heater from a lower voltage. It would heat up much more slowly if running from 120 V but it would work fine, and not overload the timer.

 

[MikeMl]

You could run the heater from a lower voltage. It would heat up much more slowly if running from 120 V but it would work fine, and not overload the timer.

Running a heater designed for 240V on 120V means that it will take four times longer to heat a given volume of water. If the heated water is being used (as in a residential water heater), then it couldn't keep up unless the flow is very small. Not a good idea. Just get a timer and a relay.

 

[buju357]

Ive put in the timer and a 25 amp (220v) relay - problem solved.

 

[gary350]

A choke coil will work if it is AC but if the appliance has an electric motor surge current is typically = to locked rotor current which is several times more than full load current so the motor will never start. The motor will try to start over heat and burn up.

You could use a timer to make the appliance come off less often.

 

[KeepItSimpleStupid]

I need to warn everyone, that make sure you use the nameplate and make sure you know the voltage in your country. 240 VAC is common in the US. Way back when it seemed to be 120/220, now it's 120/240. If you had a heater rated at 220 and were using it on 240, there will be other issues. Timer and a contactor is the way to go. I got a 7 day programmable DIN rail timer for about $15 on ebay. They have them with 110 and 220 inputs, so you would need a contactor with a 220/240 V coil.

 

[KeepItSimpleStupid]

So, you just connect the timer such that it activates a relay/contactor. You now have contacts rated for the new appliance. So, you have to package the gizmo.

 

[RODALCO]

As already said by other posters, get a 25 Amp contactor which can be controlled via the timeclock.
You could get a three phase 16 Amp contactor ( they are relatively cheap ) and put the three contacts in parallel which effectively gives you a 45 Amp contactor. You are switching a resistive load which is not hard on the contacts.
We often run streetlight contactors in this mode with the three contacts paralleled up.
Ensure that the contactor coil is rated for the correct supply voltage.

 

[kubeek]

Six years later?