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Consider a pipe in which a fluid flows such that the pipe is at a higher temperature than the fluid. At any cross section of the pipe we get a varying temperature profile. We can define a mean temperature at this cross section, which is basically the value that would give us the same energy flow rate through the cross section as the varying temperature profile would. Using this we can come up with a relation for the mean temperature $T_m$

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In this derivation, the enthalpy h of the fluid is substituted by $c_p T$, which means this temperature $T$ is in Kelvins. I say this because we can write enthalpy $h= c_p T$ only when we have given the value of enthalpy at $T=0K $ as $0$.

My question is -

Since the temperature T is in K, does that mean I have to always remember that whenever I substitute the temperature profile (temperature variation relation) in the formula for $T_m$ to determine $T_m$ I would have to do so in kelvins?

For instance in this practice problem they have given the temperature profile in Kelvins

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Engineers work in Kelvin. Note that the question does also reference air temperature in degrees C.

We were often given temperatures in Kelvin, Fahrenheit, Celcius or Rankine.

We got in the habit of converting to K as the very first step. The most common error was using deg C and not adding 273.13...

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  • $\begingroup$ Ohh. What if the temperature profile in the question was given in Celsius. Could I substitute it in the formula for $T_m$ to get $T_m$ in degree Celsius? Or would I have to convert that profile first in kelvins and then substitute? $\endgroup$
    – Harshit Rajput
    Apr 19, 2022 at 15:02
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    $\begingroup$ @HarshitRajput what if the profile was in Celcius with negative numbers... Which is why K is used. $\endgroup$
    – Solar Mike
    Apr 19, 2022 at 15:09
  • $\begingroup$ Engineers not in the dumb USA. I will add that the absolute heat conent is seldom useful - it's the changes that matter, and delta-T doesn't need K or R. $\endgroup$
    – Tiger Guy
    Apr 19, 2022 at 16:51
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    $\begingroup$ @TigerGuy it's temperature and o-rings that get them... But the profile in the example is not about delta-T. $\endgroup$
    – Solar Mike
    Apr 19, 2022 at 16:53

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