Originally Posted by

**rww88**
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 [tex]V =175 \left(\dfrac{1}{\rho}\right)^{0.43}[/tex] 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 ?

Nice example!

If we define new variables D = density in lb/ft^{3}, and U = velocity in ft/s, we would replace [tex]\rho[/tex] with

[tex]\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[/tex]

and V with

[tex]V = U \dfrac{ft}{s} \times \dfrac{1 m}{3.28 ft}= \dfrac{U}{3.28}m/s[/tex]

making the new equation

[tex]\dfrac{U}{3.28} =175 \left(\dfrac{1}{16D}\right)^{0.43}[/tex]

which, solving for U to make it a formula, becomes

[tex]U =574 \left(\dfrac{1}{16D}\right)^{0.43}[/tex]

Or at least I think so. Check it out with some specific numbers.

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