induction stove efficiency 2

[vij36]

Hi ,

Bit confused with a simple calculation. Recently done an induction stove testing.

It is an 12 kW induction stove. Have set it to 10 kW. heats 30 liters of water from 28 degree celsius to 90 degree celsius in 25 minutes.

Wanted to calculate the system efficiency.

Q = 30 kg * 4184 J/kg·K * 62 K = 7,947,520 J

Jouls to kilocalories:
7,947,520 J / 4184 J/kcal = 1890.5 kcal.

kcal to kW = 1890/860 = 2.2 kW

Energy required for water heating = 2.2 kW

Induction stove:
Power drawn by the induction stove :

1.732 * 415 volts * 13.3 amps * (PF = 0.8) / (1000)

this gives 4.24 kW.
For PF 0.9 it is 4.77 kW

Induction stove efficiency = (energy required to heat water/ power drawn by stove)
 
= 2.2/ 4.24 = 51.88%.

But induction supposed to be 85% efficient .

calculation File attached

Please point if any mistakes in calculation or the stove is inefficient ...

Kindly throw light

Sincerely,

 

[IRstuff]

The spreadsheet cheated a bit with the 4.2 instead of 4.184...

But, yes, the remaining power has to go somewhere, and that somewhere is heat, which is dissipated into the air. You can get a cheap IR thermometer to see what's getting hot inside the heater, and those are all heat sources that do nothing to heat the water. All induction heaters have coils of some sort, which have finite resistance, which consume power and do nothing to heat the water.

As mentioned above, evaporative losses, convection losses, etc., all contribute to a lower efficiency.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

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[vij36]

In the midst of this discussion i got this fundamental doubt ...

in the kcal calculation "time" factor is considered (m*c*dt/ Time). But in kW calculation i.e. V*I*(3)^.5*Cos(Phi) same "time" factor is missing.

How does the this "time" factor embedded and hidden in the kW calculation ?
How to confirm that the "time" factor playing its role in this formula .. V*I*(3)^.5*Cos(Phi)? because i don't see any T value in the formula...

Kindly explain

 

[MintJulep]

Look at your electricity bill. What are you paying for?

 

[3DDave]

Electrical current, I, is a measure of charge transfer per time.

"Electric current is measured in units of amperes; the symbol for the ampere is A. One ampere is equal to one coulomb passing a point in a wire in one second. We can calculate current, 𝐼 , using the formula 𝐼 = 𝑄 𝑡 , where 𝑄 represents an amount of charge passing a point in an amount of time, 𝑡 ."

https://www.nagwa.com/en/explainers/980189686841/

 

[IRstuff]

How does the this "time" factor embedded and hidden in the kW calculation ?

It's not hidden; power is the time rate of energy, i.e., energy/time = power

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

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[vij36]

understood ... back to basics with the help of the members here .. thank you all