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I am working out the temperature for an Ideal diesel cycle:

Based on the isentropic properties for an ideal gas. $$T_f = T_i * r^{k-1}$$ Where $T_i = 50 ºF = 510ºR$

If I use ºF, then $T_f = 172.9ºF = 632.9 ºR$

If I use ºR, then $T_f = 1763 ºR = 13030 º F$

It gives me completely different temperatures; how am I supposed to determine which one is correct?

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  • $\begingroup$ You need to clarify R(degrees) in the expression, is it "Rankine", or it is the ideal gas constant? I think there is a mistake in your question/calculations. $\endgroup$
    – r13
    Apr 11 at 19:02
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We cannot obtain negative temperatures in thermodynamics calculations. Only the absolute scales prevent this. The two absolute scales are Kelvin and degrees Rankin.

Here are the two calculations to show that both give the same result. I take that $r^{k-1} = 3.45$ based on using your Rankin values.

$$T_f = (50 + 459.67)(3.45) = 1763\ ^o\mathrm{R}$$

$$T_i = ((50 - 32)/1.8) + 273.15 = 283.15\ \mathrm{K}$$

$$T_f = (283.15)(3.45) = 979.44\ \mathrm{K}$$

$$T_f = 1.8(979.44 -273.15) + 32 + 459.67 = 1763\ ^o\mathrm{R}$$

In summary, the K and $^o$R scales are interchangeable in thermodynamic calculations because they are absolute scales. The $^o$C and $^o$F scales are not thermodynamic temperature scales.

This of course presumes that, for whichever absolute temperature scale you use, you also use the appropriately scaled constants. Your equation is unitless when divided by $T_i$, so no conversion is needed. But, for a simple yet effective counter, consider that the value of the gas law constant $R$ is not the same in the Kelvin unit scale as it is in the Rankin unit scale.

So, always use absolute temperature scales in thermodynamics calculations. Convert back to either of the other two "colloquial" (common usage) scales when desired. And always check that units on all other factors are consistent with the absolute temperature scale that you choose.

As a side note, some move is afoot to use degrees Kelvin interchangeably with Kelvin. This would make all temperature scales prefaced by the word "degrees". I find the practice a bit hard to accept. Cementing the notion that the former is an absolute scale while the latter is a difference helps distinguish these two confusions.

$$ 273.15\ K = 0\ ^oC \ \ \mathrm{but}\ \ 273.15\ ^oK = 273.15\ ^oC$$

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  • $\begingroup$ Thanks for clarifying what happens with the Imperial system. I took the liberty of proposing a correction to your initial "As a side note, some move is afoot to use degrees Celcius interchangeably with Kelvin". $\endgroup$
    – NMech
    Apr 13 at 6:16
  • $\begingroup$ @NMech The move is not to use degrees Celsius with degrees Kelvin. The move is to eliminate Kelvin and call it degrees Kelvin. $\endgroup$ Apr 13 at 13:09
  • $\begingroup$ Ok, apologies for that. I was not aware of it. $\endgroup$
    – NMech
    Apr 13 at 13:12
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    $\begingroup$ @NMech The change is rarely comments on openly and seems to be hidden rather obliquely. I ran across it a year or so ago as I was developing some lecture notes to emphasize the distinctions on the four temperature scales. Added to this is the poor habit to use the Celsius symbol without the degrees notation. Thanks for the edit in any case as a way to help me improve the example. $\endgroup$ Apr 13 at 13:30
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To my knowledge when you are doing thermodynamic calculations you should always use K (Kelvin), unless stated otherwise in the textbook you are using.

Otherwise the results from those relationships (with powers) will always be all over the place.

Another way to put it (probably more correct) is, "Always use the units in the textbook. When in doubt use Kelvin".

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    $\begingroup$ or you're in industry in the US, in which case you'll be stuck with imperial units almost certainly. $\endgroup$
    – Tiger Guy
    Apr 11 at 17:19
  • $\begingroup$ @TigerGuy, You are right, I have no clue what they do in the US, in that respect. I suspect Rankine degrees can be used in the state equation of ideal gas (without significant changes, since both Kelvin and Rankine are absolute temperatures), but I don't think that this would work on the isentropic process because of the exponent. $\endgroup$
    – NMech
    Apr 11 at 18:57
  • $\begingroup$ @NMech Yes, the isentropic equation of state as stated by OP will still work if temperatures are expressed in degrees Rankine (muliplying both $T_{\textrm{f}}$ and $T_{\textrm{i}}$ by $9/5$ doesn't make any difference to the validity of the equation, because it applies the same operation to each side), but not in degrees Celsius or degrees Fahrenheit (adding $273.15$ or $523.7$ to each temperature value would apply different operations to the two sides of the equation, thus invalidating it). $\endgroup$ Apr 12 at 9:16

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