ch03 - Problem 3.1 D = 0.75 m. The gas is at an absolute...

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where σ c is the circumferential stress in the container Σ F0 = p π D 2 4 ⋅σ c π D t = To determine wall thickness, consider a free body diagram for one hemisphere: M6 2 k g = M 25 10 6 N m 2 kg K 297 J × 1 298 K × J Nm × π 0.75 m () 3 6 × = M pV RT = p π D 3 6 = Then the mass of nitrogen is R 297 J kg K = where, from Table A.6, for nitrogen MR T = Assuming ideal gas behavior: Solution Find: Mass of nitrogen; minimum required wall thickness Given: Data on nitrogen tank D = 0.75 m. The gas is at an absolute pressure of 25 MPa and a temperature of 25°C. What is the mass in the tank? If the maximum allowable wall stress in the tank is 210 MPa, find the minimum theoretical wall thickness of the tank. Problem 3.1
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Then t p π D 2 4 π D ⋅σ c = pD 4 σ c = t2 5 1 0 6 N m 2 0.75 m 4 × 1 210 10 6 × m 2 N = t 0.0223m = t 22.3mm =
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h Hg 6.72mm = h Hg 0.909 13.55 999 × 100 × m = SG Hg 13.55 = from Table A.2 h Hg ρ air ρ Hg z = ρ air SG Hg ρ H2O z = Combining p ρ Hg g ⋅∆ h Hg = and also p ρ air g z = We also have from the manometer equation, Eq. 3.7 ρ air 0.909 kg m 3 = ρ air 0.7423 ρ SL = 0.7423 1.225 × kg m 3 = Assume the air density is approximately constant constant from 3000 m to 2900 m. From table A.3 Solution Find: Pressure change in mm Hg for ears to "pop"; descent distance from 8000 m to cause ears to "pop." Given: Data on flight of airplane Ear “popping” is an unpleasant phenomenon sometimes experienced when a change in pressure occurs, for example in a fast-moving elevator or in an airplane. If you are in a two-seater airplane at 3000 m and a descent of 100 m causes your ears to “pop,” what is the pressure change that your ears “pop” at, in millimeters of mercury? If the airplane now rises to 8000 m and again begins descending, how far will the airplane descend before your ears “pop” again? Assume a U.S. Standard Atmosphere. Problem 3.2
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For the ear popping descending from 8000 m, again assume the air density is approximately con constant, this time at 8000 m. From table A.3 ρ air 0.4292 ρ SL = 0.4292 1.225 × kg m 3 = ρ air 0.526 kg m 3 = We also have from the manometer equation ρ air8000 g ⋅∆ z 8000 ⋅ρ air3000 g z 3000 = where the numerical subscripts refer to conditions at 3000m and 8000m. Hence z 8000 ρ air3000 g ρ air8000 g z 3000 = ρ air3000 ρ air8000 z 3000 = z 8000 0.909 0.526 100 × m = z 8000 173m =
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Problem 3.4 (In Excel) When you are on a mountain face and boil water, you notice that the water temperature is 90°C. What is your approximate altitude? The next day, you are at a location where it boils at 85°C. How high did you climb between the two days? Assume a U.S. Standard Atmosphere. Given: Boiling points of water at different elevations Find: Change in elevation Solution From the steam tables, we have the following data for the boiling point (saturation temperature) of water T sat ( o C) p (kPa) 90 70.14 85 57.83 The sea level pressure, from Table A.3, is p SL = 101 kPa Hence T sat ( o C) p/p SL 90 0.694 85 0.573 From Table A.3
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This note was uploaded on 04/15/2008 for the course CHEE 3363 taught by Professor Krishnamoorti during the Spring '07 term at University of Houston.

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ch03 - Problem 3.1 D = 0.75 m. The gas is at an absolute...

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