Answer:
[tex]D_o = 0.249 m[/tex]
Explanation:
Given data:
Mass of cylinder is 5 kg
Pressure = 200 atm
Temperature = 20 degree celcius = 293 K
wall thickness = 0.5 cm = 0.005 m
we know that Gas constant for nitrogen is calculated as
[tex]\frac{\bar{R}}{M}[/tex]
[tex]R = \frac{8314}{28.01} = 296.82 J/kg.K
FROM IDEAL GAS EQUATION
PV =mRT
[tex]V = \frac{mRT}{P}[/tex]
[tex]= \FRAC{5\times 296.82\times 293}{200\times 101325}[/tex]
V = 0.021457 m^3
internal diameter by using volume formula
[tex]V =\frac{\pi}{4} D_i^2 L[/tex]
[tex]0.021457 = \frac{\pi}{4} D_i^2 (2D_i)[/tex]
solving for [tex]D_i[/tex]
[tex]D_i = 0.239 m[/tex]
calculate the outer diameter
[tex]D_o = D_i + 2t[/tex]
[tex]= 0.239 + 2\times 0.005[/tex]
[tex]D_o = 0.249 m[/tex]
