![Problem 3.1 [Difficulty: 2]
Given: Data on nitrogen tank
Find: Pressure of nitrogen; minimum required wall thickness
Assumption: Ideal gas behavior
Solution:
Ideal gas equation of state: p V⋅ M R⋅ T⋅=
where, from Table A.6, for nitrogen R 55.16
ft lbf⋅
lbm R⋅
⋅=
Then the pressure of nitrogen is p
M R⋅ T⋅
V
= M R⋅ T⋅
6
π D
3
⋅
⎛
⎜
⎝
⎞
⎟
⎠
⋅=
p 140 lbm⋅ 55.16×
ft lbf⋅
lbm R⋅
⋅ 77 460+( )× R⋅
6
π 2.5 ft⋅( )
3
×
⎡
⎢
⎣
⎤
⎥
⎦
×
ft
12 in⋅
⎛
⎜
⎝
⎞
⎟
⎠
2
×=
p 3520
lbf
in
2
⋅=
σcπDt
pπD2
/4
To determine wall thickness, consider a free body diagram for one hemisphere:
ΣF 0= p
π D
2
⋅
4
⋅ σc π⋅ D⋅ t⋅−=
where σc is the circumferential stress in the container
Then t
p π⋅ D
2
⋅
4 π⋅ D⋅ σc⋅
=
p D⋅
4 σc⋅
=
t 3520
lbf
in
2
⋅
2.5 ft⋅
4
×
in
2
30 10
3
× lbf⋅
×=
t 0.0733 ft⋅= t 0.880 in⋅=](https://image.slidesharecdn.com/problem3-190914005354/85/Problem-3-1-1-320.jpg)
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Problem 3.1
- 1. Problem 3.1 [Difficulty: 2] Given: Data on nitrogen tank Find: Pressure of nitrogen; minimum required wall thickness Assumption: Ideal gas behavior Solution: Ideal gas equation of state: p V⋅ M R⋅ T⋅= where, from Table A.6, for nitrogen R 55.16 ft lbf⋅ lbm R⋅ ⋅= Then the pressure of nitrogen is p M R⋅ T⋅ V = M R⋅ T⋅ 6 π D 3 ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= p 140 lbm⋅ 55.16× ft lbf⋅ lbm R⋅ ⋅ 77 460+( )× R⋅ 6 π 2.5 ft⋅( ) 3 × ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × ft 12 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ×= p 3520 lbf in 2 ⋅= σcπDt pπD2 /4 To determine wall thickness, consider a free body diagram for one hemisphere: ΣF 0= p π D 2 ⋅ 4 ⋅ σc π⋅ D⋅ t⋅−= where σc is the circumferential stress in the container Then t p π⋅ D 2 ⋅ 4 π⋅ D⋅ σc⋅ = p D⋅ 4 σc⋅ = t 3520 lbf in 2 ⋅ 2.5 ft⋅ 4 × in 2 30 10 3 × lbf⋅ ×= t 0.0733 ft⋅= t 0.880 in⋅=