Mixing of Ideal Gases

MIXING OF IDEAL GASES

When two ideal gases are mixed together, the state of component before mixing is same as the states of both mixture and individual components after mixing. For example, three gases A, B and C at different pressures and temperatures are adiabatically mixed in a closed chamber.

There are two ways of doing adiabatic mixing. The first way is that these three gases, contained in different tanks, are taken by pipe lines and made to meet at one point. The other method is that three different gases are filled in each compartment of fixed volume tank which are made by partitions. After few minutes, the partitions are removed or ruptured to form the mixture.

Some assumptions are made while mixing of gases as follows.

1. Mixing of gases is adiabatic.

2. There is no work done.

According to mass balance,

Total mass of gas after mixing = Sum of masses of different gases before mixing.

mm = MA + MB + MC

where

mm = Total mass of gas mixture

mA, mB, mC = Mass of gases A, B and C respectively.

Similarly, the total volume of gas mixture after mixing is the sum of volume of different gases before mixing.

Vm = VA + VB + VC

where VA, VB and VC = Volume of gas A, B and C respectively.

According to the first law of thermodynamics, the change in internal energy becomes zero.

ΔU = 0     (⸪ Q and W = 0)

Similarly, the total change in internal energy of gas mixture,

where

mm = Mass of mixture

Rm = Gas constant of mixture determined by analysis

Tm = Temperature of gas mixture

Vm = Volume

Similarly, the change in enthalpy can be written as

where

pA, pB and pC are the partial pressure of mixture

pm is the pressure of mixture

TA, TB and TC are the temperature of each constituent

Tm is the temperature of mixture

⸫ Total entropy of gas mixture is given by

ΔSm = ΔSA + ΔSB + ΔSC

For irreversible process,

ΔS > 0

⸫ ΔSm = ΔSA + ΔSB + ΔSC > 0

1. Solved Problem on Mixing of Ideal Gases

Problem 5.8

Methane at 1 bar and 25°C enters an insulated mixing chamber at a rate of 2.5 kg/s. It is mixed with air at 1 bar and 150°C in an air/methane mass ratio of 18. The flow is steady and kinetic energy changes are negligible. The ambient pressure and temperature are 1 bar and 25°C, determine the: (a) temperature of the mixture leaving the chamber and (b) irreversibility of the mixing per kilogram of methane. Take Cv and Cp of methane as 1.7354 kJ/kgK and 2.2537 kJ/kgK respectively.

Given data:

mMe = 2.5 kg/s

pMe = 1 bar = 100 kPa

TMe = 25°C = 273 + 25 = 298 K

pa = 1 bar = 100 kPa

Ta = 150°C = 150 + 273 = 423 K

po = 1 bar = 100 kPa

To = 25°C = 273 + 25 = 298 K

ma / mMe = 18

CpMe = 2.2537 kJ/kgK

CvMe = 1.7354 kJ/kgK

Solution: