240v wattage calcs?

[mojo280]

I should know the answer to this....but I dont. I have a hot tub that is on a 240v/60A breaker. This tub uses a neutral as there are timers and lights that run on 120V. While the heater is on I see roughly 22-24 amps on each of the hots going to the control board. When figuring the total wattage of this hot tub while the heater is on do add the total current draw (22a x 2)?? The reason I ask is I have one of these PG@E smart meters and while its on it says Im using ~5.4kw.

If I calculate 22a x 2 = 44a
240v x 44a = 10,560watts

Im I calculating this incorrectly? Ive been a grunt for too long!

 

[Dennis Alwon]

If the amps is 22amps per phase that means you have 22 amps not 44 amps.

Think of it like this. A dp 200 amp breaker can handle 200 amps on each phase but that is not a 400 amp service.

 

[raberding]

trust the meter

trust the meter

240x 22 is real close to 5.4kw

 

[mojo280]

Ok. So why it would require a 60A breaker if the most its going to draw is ~25A?

 

[qcroanoke]

Ok. So why it would require a 60A breaker if the most its going to draw is ~25A?

Probably the manufacturers instructions or nameplate require it.
Did you have the blowers on when you took the reading?

 

[LarryFine]

Ok. So why it would require a 60A breaker if the most its going to draw is ~25A?

Because there are several motors with their starting currents, and, when joined by the heater element(s), can contribute to voltage drop, which motors abhor.

So, the breaker size helps prevent nuisance tripping, and the wire size helps prevent motor overheating.

 

[mojo280]

If the amps is 22amps per phase that means you have 22 amps not 44 amps.

Think of it like this. A dp 200 amp breaker can handle 200 amps on each phase but that is not a 400 amp service.

ok...22a on the A phase, and 22a on the B phase. how does that not equal a total draw of 44a?? I know this is most likely simple theory but it been a while. I bend pipe...i dont do calcs...lol.

 

[ggunn]

ok...22a on the A phase, and 22a on the B phase. how does that not equal a total draw of 44a?? I know this is most likely simple theory but it been a while. I bend pipe...i dont do calcs...lol.

Think of the current as going around in a circle, in on phase A and out on phase B (half a cycle later it's going out on phase A and in on phase B). The current going around the loop is 22A no matter where you measure it.

 

[rattus]

ok...22a on the A phase, and 22a on the B phase. how does that not equal a total draw of 44a?? I know this is most likely simple theory but it been a while. I bend pipe...i dont do calcs...lol.

If you had two 22A, 120V, parallel loads between L1 and N, for example, I1 would be 44A, L2 would be 0A, and the power would be 44A x 120V.

But, you have in effect, two such loads in series with each other between L1 and L2. Therefore, I1 = 22A = -I2, and the power is 22A x 240V.

 

[LarryFine]

ok...22a on the A phase, and 22a on the B phase. how does that not equal a total draw of 44a??

22a at 240v is twice the power it is at 120v, not twice the current.