# Thread: Assistance with re-arranging equations

1. ## Assistance with re-arranging equations

Hi,

Im after a little assistance in re-arranging what i understand to be quite a basic equation used in ventilation engineering shown below:

Q = CdA √2ΔP/ρ

Im trying to re-arrange this equation in order to find the value of 'CdA'. I had presumed this to be simply 'CdA = Q / √2ΔP/ρ' - however my results dont seem to compare with those achieved in notes.

Can anyone re-arrange this equation to find 'CdA' and if possible please attempt to explain how this is done.

James

2. Originally Posted by peach49
Hi,

Im after a little assistance in re-arranging what i understand to be quite a basic equation used in ventilation engineering shown below:

Q = CdA √2ΔP/ρ

Im trying to re-arrange this equation in order to find the value of 'CdA'. I had presumed this to be simply 'CdA = Q / √2ΔP/ρ' - however my results dont seem to compare with those achieved in notes.

Can anyone re-arrange this equation to find 'CdA' and if possible please attempt to explain how this is done.

James

1. Is $\displaystyle dA$ meant to be a differential?

2. If NOT:

$\displaystyle Q=\dfrac{CdA \cdot \sqrt{2} \cdot \Delta P}{\rho}~\implies~\dfrac{Q\cdot \rho}{\sqrt{2} \cdot \Delta P}=CdA$

dA is not referring to a differential. However the way you have shown the original equation above doesnt seem to correspond to how its shown elsewhere, as the 'ρ' is shown within the square root (ie the square root is calculating the sum of '2ΔP/ρ') Does this affect the rearrangement of the equation?

Also could you explain as to how you reached this conclusion? Are there any websites anyone can recommend that clearly show the process behind rearranging equations?

Thanks again

4. Originally Posted by peach49

dA is not referring to a differential. However the way you have shown the original equation above doesnt seem to correspond to how its shown elsewhere, as the 'ρ' is shown within the square root (ie the square root is calculating the sum of '2ΔP/ρ') Does this affect the rearrangement of the equation?

Also could you explain as to how you reached this conclusion? Are there any websites anyone can recommend that clearly show the process behind rearranging equations?

Thanks again
The only reason you are finding this to be opaque is, at a guess, that you are not used to solving problems with just variables. If I asked you to solve
$\displaystyle 3x = 9$
you would immediately divide by 3.

In this case
$\displaystyle Q = CdA \sqrt{\frac{2 \Delta P}{\rho}}$

Divide both sides by the square root:
$\displaystyle CdA = \frac{Q}{\sqrt{\frac{2 \Delta P}{\rho}}}$

The last step might be a little less common for you. We need to clear the fraction under the square root as that puts a fraction in the denominator. Multiply the numerator and denominator by $\displaystyle \sqrt{\rho}$ and you finally get
$\displaystyle CdA = Q\sqrt{\frac{\rho}{2 \Delta P}}$

-Dan

5. I see this as an equation for determining the flow of gas thru a square edged orifice where Csubd is a combined discharge coefficent and A is the orifice area. Delta P /p would be under the radical.Q = Cd A rad delta P/p

bjh