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,

 

[vij36]

I am unable to see the link the attached file.

 

[pierreick]

Hi,
With your figures should be 67.78%.
Q=m*cp*Delta T , P= Q/time
Pe= U*I*sqrt(3)* cos(phi)
eta= P/Pe
my 2 cents
Pierre

 

[MintJulep]

Does the heat from the stove go only into the water?

Probably not.

 

[1503-44]

Efficiency looks great to me, but I suspect your power factor is off.
Water Mass 30 kg
T2 90 C
T1 28 C
dT 62 C
Tavg 59 C
Heat Cap 4184 J/Kg-K
Heat 7782240 J
Time m 25 min
Time s 1500 sec
No conversion to kcal is required
1 Joule/sec = 1 Watt = 1 Kg-m/s
7782240J/1500s = 5188.16 Watts
5.18816 kW

Conversion to kcal
4184.000633 J/kcal
1859.999719 kcal
4463.999325 kcal/h
1 kcal/h = 1.163 W
Note the time element.
J & Cal is energy, Watt is power
4464 kcal/h/W
5191 watts
5.191 kW

Other
Did you have the lid on the pot?
How much of the water was evaporated? How much water mass remained in the pot at 90°C?
Add around an average of 2300 kJ/kg additional heat of vaporisation x water lost to evaporation.

https://www.engineeringtoolbox.com/water-properties-d_1573.html

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[MortenA]

I think MinyJulep and 1503-44 is on to something!

You heat the pot as well - and water may evaporate.

--- Best regards, Morten Andersen

 

[1503-44]

Yes, Pot (and lid) should be insulated.
But now I think the only elephant in the room is the electronics.
I'll let the EEs work that out.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[IRstuff]

Hypothetically, power factor is very high, since the energy transfer is mostly in phase.

That said, is your induction heater still at the efficiency it's supposed to be when you throttle the power?

Is your pot stainless steel or contains stainless steel? It's unusual to find 30 liter stainless steel pots. An aluminum pot would have significantly less efficiency.

Finally, note that some people claim a noticeably lower efficiency for induction, more like 76%

https://www.aceee.org/files/proceedings/2014/data/papers/9-702.pdf

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

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

faq731-376 forum1529 Entire Forum list

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[MintJulep]

How does a heater set to 10 kW draw only 4 or 5 kW from the line?

Magic!

Energy into the water = m * c * dt

Energy used = power draw * time

Efficiency = energy to water / energy in

 

[IRstuff]

How does a heater set to 10 kW draw only 4 or 5 kW from the line?

That's not what we're saying; 5.2 kW required to heat the water to 90C, but heater set to 10kW, hence 52% efficiency

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

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

faq731-376 forum1529 Entire Forum list

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[MintJulep]

See first post.

 

[pierreick]

Hi,
my calculation :

Mass 30 kg
water

from 28 C
to 90 C
in 25 minutes


Q=m*cp*Delta T 1860 Kcal

cp 1 kca/kg/C

P =Q/time 5183.2 Watts

P elec=U*I*(3)^.5*Cos(Phi)

U 415 V
I 13.3 A
Cos (phi) 0.8

Pe 7648.043546 Watts

ղ =P/Pe 0.677715807

Something is weird in the "stars" calculation above!

[COLOR=]EDIT[/color]:

https://www.aceee.org/files/proceedings/2014/data/papers/9-702.pdf

Pierre

 

[IRstuff]

I think the OP was confused, since his Joule heating calculation should have resulted in 5.188 kW, which actually means his efficiency was 5.188/4.24 = 120%, so that's pretty cool.

However, his math is incorrect. 415V * 13.3A * 0.8 * 1.732 = 7.647 kW, which results in an efficiency of 67.8%, which is within the calculation uncertainty of the paper I linked.

It's interesting, though, that his calculated efficiency was 51.88%, which is 5.118kW/10kW.

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

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

faq731-376 forum1529 Entire Forum list

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[1503-44]

This shows there ere other factors to consider. Apparently all induction systems are not equal.

https://ultraflexpower.com/learn-about-induction-heating/induction-heating-calculation-to

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[MintJulep]

However, his math is incorrect. 415V * 13.3A * 0.8 * 1.732 = 7.647 kW,

Ok, 10 kW out for 7.647 in. Still magic.

Or poor calibration.

Anyway, if we trust the volts, amps, pf and time then I agree with 67.8%

 

[1503-44]

Is it a DC system with an inverter somewhere?
5188/ 5519 = 94%
I know < 0 about E
I'm probably proving that right now.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[IRstuff]

An induction heater has to run AC

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

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

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[1503-44]

Thought so. I'll go away now.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[vij36]

yes i have made a mistake in the calculation.

Inquired the vendor and got to know PF is 0.92. Now the efficiency comes up to 60%.
So where the remaining power converted ? Is it dissipated to atmosphere ?
Asked vendor to give us test certificate or any calculation for the efficiency.

Please find the attached sheet.

 

[MintJulep]

Is it dissipated to atmosphere ?

Yes.

Also, if the PF gets higher then the efficiency will decrease. Because it is drawing more real power to produce the same end result.

Energy into the water = m * c * dt

Energy used = power draw * time

Efficiency = energy to water / energy in

When the denominator gets bigger a fraction gets smaller.