# Why divide the dimensions of a pressure vessel by 39.37 when calculating discharge capacity?

For a National Board Exam Review:

A 29.53" * 39.37" pressure vessel contains ammonia with f = 0.041. Compute the minimum required discharge capacity of the relief device in $\mathrm{\frac{kg}{hr}}$.

The solution is shown as:

$$C = fDL = 0.041(\frac{29.53}{39.37})(\frac{39.37}{39.37}) = 0.03075 \mathrm{\frac{kg}{s}} = 110.71 \mathrm{\frac{kg}{hr}}$$

What I don't get is why DIVIDE? Isn't it already length and diameter?

• I suppose somebody blindly translated half the data in the original question into customary American units, for some reason. If the size of pressure vessel had been given as 750mm x 1000mm, the arithmetic is a lot easier than messing about with inches. – alephzero Aug 25 '15 at 23:59

## 2 Answers

This is because of units.

Getting the right answer with your units and their approach

What the people have done in your question is converted everything to metres. The division of 39.93 is because that's how you convert inches into metres. L's value is actually 1m.

$$\frac{1000mm}{25.4mm} = 39.37$$ thus: $$39.37\frac{inches}{metre}$$

So the equation that you're using is actually $$C\frac{kg}{sec} = f(metric) \cdot D(metres) \cdot L(metres)$$

The math

According to this report C is either in kg/min or lb/min, while D and L are in ft (or m). The value of f changes depending on whether you've been given the metric or imperial unit one.

The valve you're dealing with is an R717 (look at the second page of the above report) and you've been given the metric value. This valve has an imperial F value of 0.5. So

$$C \frac{lb}{min} = 0.5 \cdot \frac{29.54}{12} \cdot \frac{39.37}{12} = 4.03 \frac{lb}{min}$$ The #/12 is to get the value from inches to ft

Then converting to kg/hr:

$$C \frac{kg}{hr} = C \frac{lb}{min} \cdot \frac{60min}{hour} \cdot \frac{1kg}{2.2lb}$$

which gives:

$$C \frac{kg}{hr} = 4.03 \frac{lb}{min} \cdot \frac{60min}{hour} \cdot \frac{1kg}{2.2lb} = 110.094 \frac{kg}{hr}$$

This is a bit of an unfair question since whenever you're dealing with cross-unit-systems it's hard to keep track of what's going on. Did they at least give you the units of f so that you knew you had been supplied with the metric value?

By dividing the diameter and length dimensions, which are in inches, by 39.37 you are converting them to metres. The problem you are trying to solve has inputs that are in both English and metric units, but the answer needs to be in metric/SI, so you need to convert everything to SI units prior/during calculation.

1 inch = 2.54 cm = 0.0254 m

1/0.0254 = 39.37

If I was doing the calculation I would have multiplied by 0.0254 to convert inches to metres. It's just how I do things. The person who came up with the answer you're quoting divided by 39.37 instead.