Need help about reactor (inductor)

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SUMMARY

The discussion centers on calculating the necessary dimensional scaling of a reactor to increase its capacity from 1 kVA to 81 kVA. The established conclusion is that the dimensions must be multiplied by a factor of 3. This is derived from the relationship between capacity and dimensions, where capacity scales with the fourth power of the dimension scaling factor (m). The calculations confirm that if the capacity increases by a factor of 81, the dimensional scaling must be 3, as shown by the equation E'I' / EI = m^4.

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MissP.25_5
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Can someone help me solve this? I don't really understand how to do this.

A 1kvA reactor's capacity is needed to be enlargedto 81kvA. How much does its dimension needs to be multiplied?

Hint:
when its dimension is m times enlargened, surface A'=m'A, flux ∅'=m^2∅, current I'=m^2I, voltage E'=m^2E, Capacity E'I'=m^4EI.

Loss Pe'=m^3Pe
Temperature θ'=loss/surface area = (m^3Pe)/m^2A = mθ

My attempt:
Capacity E'I'=m^4EI. So,
E'I' / EI = m^4
81/1 = m^4
m = 3.

Is the answer 3?
 
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Looks like a straight scaling to me too.
It helps if you lay it out formally:

If capacity is given by ##\small C=EI##
And you know the E and I scale by: ##\small E'=m^{\tiny 2}E## and ##\small I'=m^{\tiny 2}I##; then $$C'=E'I'=m^4EI=m^4C$$
The only way you can get this wrong is if any of these relations are wrong, or if you hit the wrong buttons on your calculator.

Your answer is saying you need to increase the characteristic dimension of the reactor to 3x it's current size to get an 81x increase in capacity.
 
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