Reconsider Prob. 12-76 . Using EES (or other) software, extend the problem to compare the soluti

step 1 Hi friends here it is given two tanks connected by a valve. In one container there is O2 of mass

Reconsider Prob. 12-76 . Using EES (or other) software,...

Key Concepts

Substance-Specific Thermodynamic Properties

Different substances, such as methane, have unique thermodynamic properties that affect their behavior under various conditions. This concept emphasizes the importance of using accurate data and appropriate equations of state when analyzing specific substances, as variations in molecular structure lead to differences in properties like compressibility, heat capacity, and phase behavior.

Thermodynamic Software Tools

Software tools like Engineering Equation Solver (EES) are used to model and simulate thermodynamic processes. They provide built-in functions and property databases which allow for the analysis of both ideal and real fluid behavior, offering a more efficient and accurate means to solve complex problems in thermodynamics.

Real Fluid Behavior

Real fluids exhibit non-ideal behavior due to intermolecular forces and the finite size of molecules. This concept takes into account deviations from ideality by using experimentally derived data or advanced equations of state, providing more accurate results in scenarios where ideal gas assumptions fail, such as high pressures or low temperatures.

Generalized Chart Data

Generalized charts compile experimental data in a format that can be used to interpolate the properties of substances under various conditions. These charts are useful in cases where direct measurement is complicated or when hand calculations are required, providing a visual and pragmatic tool for estimating fluid properties beyond the ideal gas approximation.

Ideal-Gas Assumption

This concept involves assuming that a gas behaves in a perfectly elastic and non-interacting manner, where its pressure, volume, and temperature are related by the ideal gas law (PV = nRT). It simplifies many calculations by ignoring intermolecular forces and finite molecular volume, making it useful for many engineering applications under low-pressure and high-temperature conditions.

Transcript

00:01 Hi friends here it is given two tanks connected by a valve.

00:20 In one container there is o2 of mass 1 kg at temperature 15 degrees celsius or 288 kelvin and pressure 300 kiosk in other tank having nitrogen of volume 2 meter cube at temperature 50 degrees celsius or 323 kelvin and temperature 500 kilo pascal.

01:56 They are isolated with each other using the table.

02:14 Molecular mass of oxygen 31 .99 kg per kilo mole.

02:23 Gas constant, 02598 kilojoules.

02:31 Per kg per kelvin for nitrogen molecular mass is 0 .2968 kg per kilo mole and gas constant to be sorry molecular mass is 28 .013 and gas constant is 0 .2968 kilojoules per kaj per kelvin gas constant for real gas 8 .315 kilo.

03:23 Kilo j.

03:23 Per k .g.

03:25 Per kelvin.

03:28 Specific heat at constant pressure of oxygen.

03:32 0 .918 kilojoule per k .g.

03:36 Per kelvin.

03:45 Specific heat of oxygen at constant volume, 0658 kilojoule per kg per kelvin.

04:03 And that for nitrogen at constant volume, 0 .743 kilojoule per kege per kelvin.

04:12 Kelvin and at constant pressure it is to be 1 .039 kilojoules per kg .g.

04:21 Per kelvin now let us start solving it first we calculate a specific volume of both the components of o2 and n2 the b -o2 you will get ro2 temperature o2 first condition upon pressure so 0598 into 288 divided by 300 so it is to be equal to 0 .24941 meter cube per k.

05:36 Similarly for nitrogen specific volume 0 .2968 into 323 upon 500 kilo -pascal so it is to be 0 .19173 meter cube per kg.

06:00 Volume of first chamber volume of oxygen m .o2 into specific volume 1 into 0 .294241 so it is to be 2 .24941 so it is to be 2 .24941 meter cube mass of nitrogen volume upon specific volume 2 .2 .2 .4941 meter cube mass of nitrogen volume volume.

06:44 1 .19173 so 10 .43 kg mass of mixture mass of oxygen plus mass of nitrogen 1 plus 10 .4 3 that is 11 .43 kg volume of mixture volume of oxygen plus volume of nitrogen that is 2 .4941 plus 2 .4941 plus 2...

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