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Problem 3.1

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The document provides calculations to determine the pressure of nitrogen in a tank and the minimum required wall thickness. It uses the ideal gas law to calculate the pressure as 3520 lbf/in2. It then sets up a force balance equation considering one hemisphere of the tank to determine the minimum wall thickness is 0.880 inches.

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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 D 3 = p 140 lbm 55.16 ft lbf lbm R 77 460+( ) R 6 2.5 ft( ) 3 ｡ ｢ ｣ ､ ･ ｦ ft 12 in 2 = p 3520 lbf in 2 = cDt pD2 /4 To determine wall thickness, consider a free body diagram for one hemisphere: 裡F 0= p D 2 4 c D t= where c is the circumferential stress in the container Then t p D 2 4 D c = p D 4 c = t 3520 lbf in 2 2.5 ft 4 in 2 30 10 3 lbf = t 0.0733 ft= t 0.880 in=

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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 D 3 = p 140 lbm 55.16 ft lbf lbm R 77 460+( ) R 6 2.5 ft( ) 3 ｡ ｢ ｣ ､ ･ ｦ ft 12 in 2 = p 3520 lbf in 2 = cDt pD2 /4 To determine wall thickness, consider a free body diagram for one hemisphere: 裡F 0= p D 2 4 c D t= where c is the circumferential stress in the container Then t p D 2 4 D c = p D 4 c = t 3520 lbf in 2 2.5 ft 4 in 2 30 10 3 lbf = t 0.0733 ft= t 0.880 in=