# Lindeburg sample exam problem number 37



## buick455 (Oct 8, 2009)

Being exhausted from work does not help in trying to work problems at night.

Anyway, as I have not worked to many HT and fluids problems lately they are

taking longer than expected. Problem number 37 is one of those problems. I

worked the first part fine except that they got 19.4 degrees C I got 19.4

degrees K which of course changed my final answer. How do you get C from an

equation that only has K in it?? Note: There is no errata on print #3 which is

what I have.

T= (73 W/M) / 2 x pie x (80 W/M^2-K) x (.015m/2)

Change in T = 19.4 C??


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## PKT1106 (Oct 8, 2009)

I assume by "#3", you mean you have the 1st edition, 3rd printing? I have the 2nd edition, 1st printing and #37 in my sample exam is about gravimetric air/fuel ratios.


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## buick455 (Oct 8, 2009)

Yes I have the 1st edition 3 printing. The Gravimetric air/fuel problem in number 36 in my book.


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## buick455 (Oct 8, 2009)

Never mind my question was answered.

Delta T of 19.4 C = delta T of 19.4 K (the equivalency was just not shown by Lindeburg as he just changed units with not explanation)

Thanks


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## JoeysVee (Oct 9, 2009)

Exactly! Also if something is per degree K but you have C, remember the difference in 1 C and 1 K is the same so you can still use something that is per degree K even if you have something that is in C. I hope that makes sense...if not just ignore... :beerchug:


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