Wattage Drop?

[rakesdl]

Earlier I posted a thread concerning changing a three phase electric heater to single phase, after some modifications to the wiring in the heater I have been able to make it run on 240 single phase. My question is this, The heater originally produced 15000 watt, since i have changed it to run on 240 single phase how much of a wattage drop/btu if any has this created. And is there a way to find out how much it is putting out now?

 

[mmg440]

Not knowing how your rewire affected the unit. Are You still using all the elements? Are they at the designed voltage per element? It is hard to predict what the results of such a modification would be.

The get what you are drawing for watts now.
Get or borrow a AC clamp on amp meter. Take a reading on one of the two power lines into the unit(they both should be carring the same current) then multiply the amps times the volts (240 but you may want to verify with a meter) and you will get the wattage the unit is drawing.

 

[Oldram__Newram]

If wired correctly power (watts) should stay the same. So per ohms law as voltage goes down, current goes up. You should be drawing more amperage now so make sure wiring (both interal and external of the unit) is sized correctly.

http://www.csgnetwork.com/ohmslaw2.html

 

[rinny_tin_tin]

If the design 3-phase voltage is the same as your 1-phase voltage, in this case 240, there is no reason to see any "wattage drop", or increased losses owed to i^2R heating. The likely reason the unit was originally made 3 phase was for a an application where 3-phase was available and to balance out the loading across three phases. The heaters are resistance elements and thus no power factor concerns. However, if the original heaters were designed as 440 V and you went to 240, then your current draw would be double, and your i^2R losses would also double. If the design was 240 and you stayed 240 - your current would stay the same. At 240 V for a 15 kW heater, your current should be 62.5 Amperes.

 

[rinny_tin_tin]

I need to correct (my tendency to quickly respond without thinking)

The 3-phase amperes at 240 VAC (15 kW heater) is 36.2 Amperes while the single phase amperes at 250 VAC would be 62.5 (about 1.73 times more!) So, the i^2 losses would be about double for single vice 3-phase. So, the heater would still produce 15kW@ 1-phase - however, the cable would tend to get hotter and a larger cable would be needed 0 6 AWG vice 10 AWG, for example

 

[timgr]

You guys sure about this? I thought polyphase was used because the power dissipation per amp is higher than with single phase. I don't think there's an easy way to reconfigure the circuit to single phase without a loss in power dissipation, or else require a change in voltage.

(linked from http://www.allaboutcircuits.com/vol_2/chpt_10/2.html )

Looks to me like you make this a single phase load by removing load #3 from the circuit. Then your power dissipation will drop by 1/3 (giving a heat output of 10 kW), but I expect the wire size won't need to change.

I don't see any other easy way to rewire a 3-phase circuit like this and get the same power dissipation with single phase, other than running each load at 120V and putting them all in parallel (ie 120V single phase). Then you'd need significantly heavier wire. The loads aren't designed to bridge 230V, so you can't arbitrarily put one load in parallel with the other two.

No experience with this, but I do know a little about circuits...

 

[mmg440]

Using the clamp on ac amp meter and finding out what the actual amps are when the heater is in operation Would most likely disclose how he wired the unit. There are to many variables to know if the unit is using all the loads. Maybe if rakesdl post a schematic of what the original heater wiring was and what he changed to make it a single phase unit. Then might be able to figure. I still think the simplest way at this point is to make a measurement. You will also have the information to make sure the circuit it is on can handle the amp load.

 

[rakesdl]

I have read the replies and am sending a photo of the wiring diagram. I had one heating element working and attempted to get two of them working today, and I did but the 10/2 wire got very hot and had to disconnect. It seems to be working well with one element on, but would like to get two working. As far as the changes I have made I eliminated L2, placing one wire in L1 and one in L3, I have two hot wires coming in, one on T1 and T3 with a ground. As you probably have determined I am not very good at electricity, just trying to save $ by making this work. Any help would be appreciated

 

[rinny_tin_tin]

I have read the replies and am sending a photo of the wiring diagram. I had one heating element working and attempted to get two of them working today, and I did but the 10/2 wire got very hot and had to disconnect. It seems to be working well with one element on, but would like to get two working. As far as the changes I have made I eliminated L2, placing one wire in L1 and one in L3, I have two hot wires coming in, one on T1 and T3 with a ground. As you probably have determined I am not very good at electricity, just trying to save $ by making this work. Any help would be appreciated

Yeah - your heaters are configured in delta for three phase @ 208/220. For single phase, reconfigure L1, L2, & L3 in parallel and connect with at least No 6 or larger across to 240/250 to handle 60+ amperes.

To determine yourself:

kW for single phase is the Volt-Ampere product (i.e., Volts * Amperes)
For three phase, the formula is 1.73 * Voltage * Current * Power Factor (PF)
For resistance heating elements, PF is unity = 1

So, for 15kW: 1.73 * 250 = 432.5

Current = P/V = 15000/432.5 = 34 Amperes - Thus why it was wired with 10 AWG

Now that you are running single phase, you will have 1.7 times more current, or simply 15000/250 = 60 Amperes = 6 AWG

6 AWG should work fine, but I'd run 4 AWG fed from a 75 A breaker, etc.

Yes - you would have greater heat losses in the conductor feeding the heater elements (I^2R) - and perhaps greater voltage drop as yoru current demand is higher, and when you first hit that switch, you will likely notice
a voltage sag/dip - however, in theory, it should work - and nonetheless, you will still have 15kW of heat if that is what it was originally rated at.

 

[rakesdl]

Thanks to all who responded, and thanks to rinny tin tin for the directions to wire the unit. From reading his reply, I get the feeling that feeding this thing (electricity) is going to be like providing fuel for a F-16 fighter jet, please tell me if this is wrong? I mentioned that the 10-2 wire got hot, that was the wire that I temporally used to feed the heater from the fuse panel, the wiring inside the heater did not get hot. Also rinny tin tin you said to wire the heater in parallel, that means replacing the wires back to original configuration right? And I don't understand how that I have three wires coming into the heater from the breaker box when the 75A breaker only has two connections going out? Again thanks for all the help.

 

[Charles (in GA)]

The problem with this whole thing is that article 424 of the NEC, "Fixed Electric Space Heating Equipment", specifically 424.22 (A) and (B), restrict you to a max of 60 amps circuit protection per element "shall be protected at not more than 60 ampres." and "Equipment rated more than 48 amps and employing such resistance elements shall have the heating elements subdivided, and each subdivided load shall not exceed 48 ampres."

In addition, the branch circuit wiring supplying the heater, if under 50kW has to be sized at not less than 125% of the rated load.

When this heater was three phase, you had three breakers protecting the various elements (one for each phase), now you have reduced this, and by combining the loads, it sounds as if you have exceed code limitations.

If by chance I am reading this wrong, feel free to correct me. It is fairly simple reading and all seems to make sense.

Charles

 

[Charles (in GA)]

Thanks to all who responded, and thanks to rinny tin tin for the directions to wire the unit. From reading his reply, I get the feeling that feeding this thing (electricity) is going to be like providing fuel for a F-16 fighter jet, please tell me if this is wrong?

The power company calls and thanks me every time I turn on my 6000 watts of lighting. I'll bet they send a limo to your house bearing gifts when you fire up the heater. I cannot imagine paying the bill on this. It costs about a dollar an hour for me to run my lights.

Charles

 

[TRDon]

The power company calls and thanks me every time I turn on my 6000 watts of lighting. I'll bet they send a limo to your house bearing gifts when you fire up the heater. I cannot imagine paying the bill on this. It costs about a dollar an hour for me to run my lights.

Charles

Holy ****! KWs are about half that where I live

 

[Mr_fixit]

So you have either a 5000 , or 10000 or 15000 watt heater, depending on how you hook it up.... Maybe 10,000 watts is enough?

 

[rakesdl]

I think given the expert information i have received here concerning the safety issues and the amount of electricity this thing is going to burn, 10000 is enough. It will only be used occasionally as I only work there as an escape from the spouse or when something needs repairing. The temperatures here are not that bad most of the time and the 24x32 is insulated well with triple pane windows and insulated garage doors. I currently use a 115000btu kerosene heater that heats the building in only minutes, but also creates deadly fumes. I was considering purchasing something like a Dayton G73 (5000 Watts) but think that was to small. Then came across this heater for free and seemed to be determined to make it work. If i can get 10000 watts from it safely without supporting the power company that seems the be the way to go. On a side note I found this forum about six months ago and have been luring (just learning) since, I would like to say thanks again to all who helped with this problem and to all a HAPPY NEW YEAR!

 

[timgr]

According to my quick research, the voltage across each leg of the delta in 3-phase is sqrt(3)*120 or 207V. If each leg supplies 5000W, then R=(207)^2/5000 = 8.6 ohms.

If you run two elements in series and one element in parallel with the series pair with single-phase 240, you'll have a resistance that's 2/3 the normal R, ie 5.7 ohms. V^2/R = (240)^2/5.7 = 10,105 watts. This is assuming that the single element across 240V can stand the increased dissipation from 5000 to 6700 watts. The remaining 3400 watts will come from the two elements in series.

Note that you need 42 amps to generate 10,100 watts at 240 volts ... if you disconnect one of the elements in the delta, you can connect the heater to make either 6700 watts (28 amps, one element) or 3400 watts (2 elements in series, 14 amps).

 

[rinny_tin_tin]

Thanks to all who responded, and thanks to rinny tin tin for the directions to wire the unit. From reading his reply, I get the feeling that feeding this thing (electricity) is going to be like providing fuel for a F-16 fighter jet, please tell me if this is wrong? I mentioned that the 10-2 wire got hot, that was the wire that I temporally used to feed the heater from the fuse panel, the wiring inside the heater did not get hot. Also rinny tin tin you said to wire the heater in parallel, that means replacing the wires back to original configuration right? And I don't understand how that I have three wires coming into the heater from the breaker box when the 75A breaker only has two connections going out? Again thanks for all the help.

Hey Rake - yes - sort of like feeding fuel to an F15, but 15kW is 15 kW - so whether you feed it 3-phase or 1 phase - 15 kW is 15 kW and will about cost you the same.

To wire in parallel: You need to disconnect the three heater element from Delta (Triangle configuration) and put the elements in parallel (think of a ladder configuration : sort of like this | | |

So that one leg of the 240 connects all the tops of the three legs. while the other leg connects all the bottoms. Think of how you would wire three car batteries in parallel such that you still get 12 V but three times the cranking current, etc.

When you configure your heaters as above - you only have two wires to connect. A three phase supply has three wires - which is why the heaters were configured in delta. This should work for you if you use the fatter conductor - 6 AWG or 4 AWG - whatever it came out to

Let me know how you do!

Best

 

[rinny_tin_tin]

If you only want 10 kW - then only wire/connect two elements in parallel and keep all other numbers/details same (e.g., 6 AWG, 75A breaker, etc)