Conversion confusion: Rankine in ideal gas constant

[Undoubtedly0]

Hi all. I have ran into a problem in converting from imperial to SI units, as follows.

The ideal gas constant for air is often given in imperial units as

$$R = 1716 \frac{ft*lbf}{slug*°R}$$

where

$$1 ft*lbf = 1.356(10^{-3}) kJ$$

$$1 slug = 14.59 kg$$

$$°R = \frac{9}{5}K$$

Thus making these substitutions gives

$$R = 1716 \frac{ft*lbf}{slug*°R} = 1716 \frac{(1.356(10^{-3}) kJ)}{(14.59 kg)(\frac{9}{5}K)} = 0.0886 \frac{kJ}{kg*K}$$

Yet it is quite common knowledge that in reality,

$$R = 0.287 \frac{kJ}{kg*K}$$

Somewhere, then, in my conversion, there must be an error. I found that if I instead reversed the temperature conversion to be °R = K*5/9, the desired result is given - but how could this be? Surely it is must be true that °R = K*9/5.

Thanks for the help.

 

[Chestermiller]

Hi all. I have ran into a problem in converting from imperial to SI units, as follows.

The ideal gas constant for air is often given in imperial units as

$$R = 1716 \frac{ft*lbf}{slug*°R}$$

where

$$1 ft*lbf = 1.356(10^{-3}) kJ$$

$$1 slug = 14.59 kg$$

$$°R = \frac{9}{5}K$$

Thus making these substitutions gives

$$R = 1716 \frac{ft*lbf}{slug*°R} = 1716 \frac{(1.356(10^{-3}) kJ)}{(14.59 kg)(\frac{9}{5}K)} = 0.0886 \frac{kJ}{kg*K}$$

Yet it is quite common knowledge that in reality,

$$R = 0.287 \frac{kJ}{kg*K}$$

Somewhere, then, in my conversion, there must be an error. I found that if I instead reversed the temperature conversion to be °R = K*5/9, the desired result is given - but how could this be? Surely it is must be true that °R = K*9/5.

Thanks for the help.

$$1\ degree\ R=\frac{5}{9}\ degree\ K$$so there are 5/9 degree K/degree R