Thank you; your statements make complete sense. Here is an example equation: The maximum gas velocity V (in m/s) in a pipe is given by V =175 (1/density)**0.43 where the units of gas density are kg/m**3. What would this equation transform to if the gas density was in lb/ft**3 and the desired velocity units were ft/s ?
I assume you are using "**" to mean an exponent, so the problem is this:
The maximum gas velocity V (in m/s) in a pipe is given by \(\displaystyle V =175 \left(\dfrac{1}{\rho}\right)^{0.43}\) where the units of gas density are kg/m3. What would this equation transform to if the gas density was in lb/ft3 and the desired velocity units were ft/s ?
Nice example!
If we define new variables D = density in lb/ft
3, and U = velocity in ft/s, we would replace \(\displaystyle \rho\) with
\(\displaystyle \rho = D \dfrac{lb}{ft^3} \times \dfrac{0.4536 kg}{1 lb} \times \left(\dfrac{3.28 ft}{1 m}\right)^3 = 16 D kg/m^3\)
and V with
\(\displaystyle V = U \dfrac{ft}{s} \times \dfrac{1 m}{3.28 ft}= \dfrac{U}{3.28}m/s\)
making the new equation
\(\displaystyle \dfrac{U}{3.28} =175 \left(\dfrac{1}{16D}\right)^{0.43}\)
which, solving for U to make it a formula, becomes
\(\displaystyle U =574 \left(\dfrac{1}{16D}\right)^{0.43}\)
Or at least I think so. Check it out with some specific numbers.