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 Chapter 1. Basics 2. Fluid Statics 3. Kinematics 4. Laws (Integral) 5. Laws (Diff.) 6. Modeling/Similitude 7. Inviscid 8. Viscous 9. External Flow 10. Open-Channel Appendix Basic Math Units Basic Equations Water/Air Tables Sections Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Chean Chin Ngo Kurt Gramoll ©Kurt Gramoll

FLUID MECHANICS - CASE STUDY SOLUTION

Compressed Tanks

Table: Gas Constants for the Problem
 Gas Constant, R (kJ/kg-K) Air 0.2870 Ammonia 0.4882 Carbon Dioxide 0.1889 Helium 2.0771 Hydrogen 4.1243 Nitrogen 0.2968 Oxygen 0.2598 R-12 0.06876 R-134a 0.08149

The conditions of the gases are given as follows:
 Number of Tanks Temperature (K) Absolute Pressure (MPa) Oxygen 5 290 1.00 Nitrogen 4 290 0.85 Carbon Dioxide 6 290 1.25

The mass of each gas can be determined by the ideal gas law:

m = pV/RT

For oxygen, the gas constant R is 0.2598 kJ/kg-K. The mass of the oxygen in a single tank is

m = (1000)(0.25)/(0.2598)(290)
= 3.318 kg

For nitrogen, the gas constant R is 0.2968 kJ/kg-K. The mass of the nitrogen in a single tank is

m = (850)(0.25)/(0.2968)(290)
= 2.469 kg

For carbon dioxide, the gas constant R is 0.1889 kJ/kg-K. The mass of the carbon dioxide in a single tank is

m = (1,250)(0.25)/(0.1889)(290) = 5.705 kg

The total weight of the tanks is then given by

Wtotal = weight of empty tanks + weight of gases
= (15)(5)(9.8) + (5)(3.318)(9.8)
+ (4)(2.469)(9.8) + (6)(5.705)(9.8)
= 1,330 N

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