Please help me solve this equation (Dimensional Analysis)

Question

I have an equation I need help solving. This is an mechanical engineering question about calculating leak rate. If possible could you walk me through the process to solve this because I want to learn how to do it and not just get an answer.

Here is the equation:

L = 0.7 x 0.000000134std. cc cm/cm^2 sec bar x 1.256in x 449lb/in^2 x 1.35 x 0.20

(L needs to be in std. cc/sec)

Here is the equation explained if you are interested to see what the equation is solving for: http://www.engineersedge.com/fluid_flow/oring_leak_rate_13605.htm

Any help with this would be appreciated! Please let me know if you have any questions.

Walker

Answer 1

Comments

Firstly the question is especially confusing as you have used a variety of metric and imperial units including non-standard abbreviations of some. All this compounds the difficulty of the dimensional analysis and getting an answer.

Lets look at the underlying formula:

$$L=0.7F\times D\times P\times Q\times(1-S)^2$$

$F$ is listed as measured in "std. cc cm/cm$^2$ sec bar". This is a very ugly way of writing it. Firstly "cc" is short for cubic centimetre, i.e. cm$^3$. Next bar would be better written as 10$^5$ Pascal (Pa). So putting these bits together gives:

$$\frac{cc\cdot cm}{cm^2\cdot sec\cdot bar}\times10^{-8}=\frac{cm^4}{cm^2\cdot s\cdot 10^5Pa}\times10^{-8}=\frac{cm^2}{s\cdot 10^5Pa}\times10^{-8}=\frac{(10^{-2}m)^2}{s\cdot 10^5Pa}\times10^{-8}=\frac{m^2}{s\cdot Pa}\times10^{-17}$$

$D$ needs to be converted to metres. $1.256in=0.0319024m$.

$P$ has its problems too as $lb/in^2$ is a terrible unit for pressure as $lb$ can mean both mass and force which leads to plenty of confusion. Normally in the US measurement system this is written as $lbf/in^2$ where $lbf$ means pound-force to avoid the confusion. $449lb/in^2=3095740Pa$

$Q$ and $S$ are dimensionless coefficients (between 0 and 1) so they don't play any role in performing dimensional analysis. Only in determining the answer.

Dimensional Analysis Ignoring the dimensionless values in the equation gives: $$L=F\times D\times P$$ $$L=\frac{m^2}{s\cdot Pa}\times m \times Pa$$ $$L=\frac{m^3}{s}$$ Dimensionally this is equivalent to $cc/sec$.

Calculation

$$L=0.7 \times 0.000000134std. cc cm/cm^2 sec bar \times 1.256in \times 449lb/in^2 \times 1.35 \times 0.20$$ becomes: $$L=0.7\times 0.000000134\times \frac{m^2}{s\cdot Pa}\times10^{-17}\times0.0319024m\times3095740Pa\times1.35\times0.20$$ $$L=2.5\times10^{-20}\frac{m^3}{s}$$ Converting this to $\frac{cm^3}{s}$ gives: $2.5\times10^{-14} std. cc/s$

I have no idea about what sort of values are realistic when working with O-rings but this sounds on the small side suggesting that some of the values supplied or the formula is incorrect. Given the horrendous mixture of units there could be many points of error in its derivation or application.

Answer 2

The units are explained in this paragraph from the original source of the formula:

NOTE: For convenience, the formula contains mixed units. It was set up this way because in the United States O-ring diameters are usually given in inches, and pressures in pounds per square inch while permeability figures are usually shown in metric units. The 0.7 factor/constant resolves these inconsistencies.

In other words, the makers of the formula recognized that the units in the formula are a heterogeneous mess; but they find them convenient to use in the context of engineering in the United States, where these are the units in which you would acquire the data. I suspect that a physicist (even in the US) would take a different approach.

If the formula had been stated consistently in all SI units, and users of the formula received their data in mixed SI and common US units, the users of the formula would have to convert each set of data to all SI units before applying the formula. Instead of requiring that, the necessary conversion factors have all been incorporated into the leading factor of $0.7$.

By the way, I think the factor $10^{-8}$ in the formula in the original source is meant to indicate that your $F$ quantity, which you wrote as $0.000000134$, should be written as $13.4$ instead (that is, use the fact that $0.000000134 = 13.4 \times 10^{-8}$, and then consider the factor of $10^{-8}$ to be part of the "units" of the formula). Hence

$$ L = 0.7 \times 13.4 \times 1.256 \times 449 \times 1.35 \times 0.20\ \text{std. cc/sec}$$

if there are no other errors in the numbers.