TITLE: Conversion factor
QUESTION [0 upvotes]: I have two equations that evaluate effective leakage area (ELA). In one equation, the ELA is tested at 25 pascals, in the other its tested at 50 pa, so these equations become ELA25 and ELA50 respectively. The same is true for the units for cubic feet per minute - CFM25 and CFM50.
The equation to find ELA25 is:
2(0.05x)^1/2; x = CFM25
The equation to find ELA50 is:
x/10; x = CFM50
Lastly, there is an CFM50 is approximately 1.57 times the value of CFM25.
My question is this: Can someone come up with an equation that evaluates ELA25 using CFM50?
I don't know enough about algebra to problem solve this without help. Thanks in advance. Also, I probably have the wrong tag in place, I'm not sure what to even title this question.

REPLY [1 votes]: Let $F_{25}$ be the flow rate in $\frac {\text{ft}^3}{\text{min}}$ for $25$ Pa (don't you just love intermixing U.S. units with S.I.?). And $F_{50}$ is the flow rate at $50$ Pa. And let $A_{25}, A_{50}$ be the Effective Leak Areas for each.
You say you have the following equations:
$$A_{25} = 2\sqrt{(0.05)F_{25}}$$
$$A_{50} = \frac{F_{50}}{10}$$
$$F_{50} = (1.57)F_{25}$$
Which is really curious because 25 and 50 Pa are extremely low pressures (less than 0.01 lbs/in^2), so I cannot imagine what would cause effective area to vary directly with flow rate near 50 Pa, but vary instead with the square root of flow rate at 25 Pa.
But your mathematical question is simple:
$$\frac{F_{50}}{1.57} = F_{25}$$
$$A_{25} = 2\sqrt{(0.05)\left(\frac{F_{50}}{1.57}\right)}$$
$$A_{25} = (0.357)\sqrt{F_{50}}$$
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