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