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هوا مثالی از یک مخلوط گازی آرمانی است . Component % by Volume N2 78.10 PowerPoint PPT Presentation


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Chapter 13 Gas Mixtures مخلوط گازها Thermodynamics: An Engineering Approach , 7th edition by Yunus A. Çengel and Michael A. Boles. مخلوط های گازی که در این فصل بررسی می گردند تنها شامل گازهایی می باشند که میل به واکنش با یکدیگر ندارند و واکنش ناپذیر می باشند.

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هوا مثالی از یک مخلوط گازی آرمانی است . Component % by Volume N2 78.10

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Component by volume n2 78 10

Chapter 13Gas Mixtures Thermodynamics: An Engineering Approach, 7th editionby Yunus A. engel and Michael A. Boles


Component by volume n2 78 10

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  • Component % by Volume

  • N2 78.10

  • O2 20.95

  • Argon 0.92

  • CO2 + trace elements 0.03


Component by volume n2 78 10

k gases

T = TmV = Vm

P = Pmm = mm

  • .

  • V k P T .

  • :gravimetric analysis

  • :molar analysis

  • mm Nm :


Component by volume n2 78 10

mfi yi .

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k V P T . k P T .

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vfi .

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i (partial pressure) .

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k . (component pressure) Pi' .

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partial pressure of gas i = component pressure of gas i


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Component by volume n2 78 10

Example 13-1

An ideal-gas mixture has the following volumetric analysis

Component % by Volume

N2 60

CO2 40

(a)Find the analysis on a mass basis.

For ideal-gas mixtures, the percent by volume is the volume fraction. Recall


Component by volume n2 78 10

Comp. yiMiyiMimfi = yiMi/Mm

kg/kmolkg/kmol kgi/kgm

N2 0.60 28 16.8 0.488

CO2 0.40 44 17.6 0.512

Mm = yiMi = 34.4

(b) What is the mass of 1 m3 of this gas when P = 1.5 MPa and T = 30oC?


Component by volume n2 78 10

(c) Find the specific heats at 300 K.

Using Table A-2, Cp N2 = 1.039 kJ/kgK and Cp CO2 = 0.846 kJ/kgK


Component by volume n2 78 10

(d) This gas is heated in a steady-flow process such that the temperature is increased by 120oC. Find the required heat transfer.

The conservation of mass and energy for steady-flow are

The heat transfer per unit mass flow is


Component by volume n2 78 10

(e) This mixture undergoes an isentropic process from 0.1 MPa, 30oC, to 0.2 MPa. Find T2.

The ratio of specific heats for the mixture is

Assuming constant properties for the isentropic process

(f) Find Sm per kg of mixture when the mixture is compressed isothermally from 0.1 MPa to 0.2 MPa.


Component by volume n2 78 10

But, the compression process is isothermal, T2 = T1. The partial pressures are given by

The entropy change becomes

For this problem the components are already mixed before the compression process. So,

Then,


Component by volume n2 78 10

Why is sm negative for this problem? Find the entropy change using the average specific heats of the mixture. Is your result the same as that above? Should it be?

(g) Both the N2 and CO2 are supplied in separate lines at 0.2 MPa and 300 K to a mixing chamber and are mixed adiabatically. The resulting mixture has the composition as given in part (a). Determine the entropy change due to the mixing process per unit mass of mixture.


Component by volume n2 78 10

Take the time to apply the steady-flow conservation of energy and mass to show that the temperature of the mixture at state 3 is 300 K.

But the mixing process is isothermal, T3 = T2 = T1. The partial pressures are given by

The entropy change becomes


Component by volume n2 78 10

But here the components are not mixed initially. So,

and in the mixture state 3,

Then,


Component by volume n2 78 10

NOZZLE

C = 500 m/s

T = 500oC

Mixture

N2 and CO2

Then,

If the process is adiabatic, why did the entropy increase?

Extra Assignment

Nitrogen and carbon dioxide are to be mixed and allowed to flow through a convergent nozzle. The exit velocity to the nozzle is to be the speed of sound for the mixture and have a value of 500 m/s when the nozzle exit temperature of the mixture is 500C. Determine the required mole fractions of the nitrogen and carbon dioxide to produce this mixture. From Chapter 17, the speed of sound is given by

Answer: yN2 = 0.589, yCO2 = 0.411


Component by volume n2 78 10

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