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I have the following problem.

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My attempt

Initally the efficiency of the supply is 83%, so \$\dfrac{P_{out}}{P_{in}} =83\%\$.

The remaining 17% must be power lost in dissipation in the enclosure \$\dfrac{P_{diss}}{P_{in}} = 17\%\$

If the enclosure is dissipating 25W then we have \$P' = \dfrac{25\text{W}}{17\%} = 1.47 \frac{\text{W}}{\%} \$

So the output power must be \$P_{out} = 1.47\frac{\text{W}}{\%} \cdot 83\% = 122.01\text{W} \$


If the efficiency is increased to 92% we have \$\dfrac{P_{out}}{P_{in}} =92\% \:\$ and \$\dfrac{P_{diss}}{P_{in}} = 8\%\$

If the enclosure is dissipating 25W as before we have \$P' = \dfrac{25\text{W}}{8\%} = 3.125 \frac{\text{W}}{\%} \$

And the output power becomes \$P_{out} = 3.125\frac{\text{W}}{\%} \cdot 92\% = 287.5\text{W} \$

So the percent change of the maximum output power must be $$\text{percent change} = \frac{P_{out,2}-P_{out,1}}{P_{out,1}} \cdot 100\% = 135.54\% $$

That was a lot of calculations (I apologize for that) and it leads me to believe that there is an easier way of solving this. Furthermore, am I really finding the maximum output power if I set \$P_{diss}=25\text{W} \$? Intuitively I would think the maximum output power occured if \$P_{diss} = 0\text{W}\$.

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  • \$\begingroup\$ Maybe it helps if you realize that 3.125 W/% could also be expressed as 312.5 W. That's the input power of the supply (you named it that way in your first equation even). 25 W will go as heat to the case. The rest is the output (easy subtraction). \$\endgroup\$
    – Arsenal
    Feb 28, 2022 at 12:12

1 Answer 1

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it leads me to believe that there is an easier way of solving this

The power output in scenario 1 is \$25\text{ watts}\times \frac{83}{100-83}\$ = 122.06 watts.

The power output in scenario 2 is \$25\text{ watts}\times \frac{92}{100-92}\$ = 287.5 watts.

That's a percentage ratio of 235.54% or an increase of 135.54%.

Intuitively I would think the maximum output power occurred if \$P_{diss} = 0W\$

That's an impossibility because the efficiency is not 100% in either scenario.

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