Neon gas enters an insulated mixing chamber at 300 K, 1 bar with a mass flow rate of 1 kg/s. A second steam of carbon monoxide enters at 575 K, 1 bar at a mass flow rate of 0.5 kg/s. Assuming the mixture exits at 1 bar and the specific heat ratios for Neon and CO are constant, determine:

(a) The molar composition of the exiting mixture.

(b) The temperature of the exiting mixture, in K.

(c) The rate of entropy production, in kW/K.

a) the molar fraction of neon at the exit is

xₙ = 0.735

and carbon monoxide

xₓ = 0.265

b) the final temperature is

T = 410.55 K

c) the rate of entropy production is

ΔS = 1.83 KW/K

Explanation:

denoting n for neon and x for carbon monoxide:

a) from a mass balance, the molar fraction of neon at the exit is:

outflow mass neon=inflow mass neon

xₙ = outflow mass neon / (total outflow of mass) = inflow mass neon / (total outflow of mass) = (1 kg/seg / 20.18 kg/kmol) / (1 kg/seg / 20.18 kg/kmol + 0.5 kg/seg / 28.01 kg/kmol) = 0.735

and the one of carbon monoxide is

xₓ = 1-xₙ = 1-0.735 = 0.265

b) from the first law of thermodynamics applied to an open system, then

Q - Wo = ΔH + ΔK + ΔV

where

Q = heat flow to the chamber = 0 (insulated)

Wo = external work to the chamber = 0 (there is no propeller to mix)

ΔH = variation of enthalpy

ΔK = variation of kinetic energy ≈ 0 (the changes are small with respect to the one of enthalpy)

ΔV = variation of kinetic energy ≈ 0 (the changes are small with respect to the one of enthalpy)

therefore

ΔH = 0 → H₂ - H₁ = 0 → H₂=H₁

if we assume ideal has behaviour of neon and carbon monoxide, then

H₁ = H ₙ₁ + H ₓ₁ = mₙ₁*cpₙ*Tn + mₓ₁*cpₓ*Tc

H₂ = (m ₙ+mₓ) * cp*T

for an ideal gas mixture

cp = ∑ cpi xi

therefore

mₙ*cpₙ*Tₙ + mₓ*cpₓ*Tₙ₁ = (m ₙ+mₓ) * ∑ cpi xi*T

mₙ / (m ₙ+mₓ) * cpₙ*Tₙ + mₓ / (m ₙ+mₓ) * cpₓ*Tₓ = T ∑ cpi xi

xₙ * cpₙ*Tₙ + xₓ*cpₓ*Tₓ = T * (xₙ * cpₙ+xₓ*cpₓ)

T = [xₙ * cpₙ / (xₙ * cpₙ+xₓ*cpₓ) ]*Tₙ₁ + [xₓ*cpₓ / (xₙ * cpₙ+xₓ*cpₓ) ]*Tₓ₁

denoting

rₙ = xₙ * cpₙ / (xₙ * cpₙ+xₓ*cpₓ)

and

rₓ = xₓ*cpₓ / (xₙ * cpₙ+xₓ*cpₓ)

T = rₙ * Tₙ + rₓ*Tₓ

for an neon, we can approximate its cv through the cv for an monoatomic ideal gas

cvₙ = 3/2 R, R = ideal gas constant=8.314 J/mol K=

since also for an ideal gas: cpₙ - cvₙ = R → cpₙ = 5/2 R

for the carbon monoxide, we can approximate its cv through the cv for an diatomic ideal gas

cvₓ = 7/2 R → cpₓ = 9/2 R

replacing values

rₙ = xₙ * cpₙ / (xₙ * cpₙ+xₓ*cpₓ) = xₙ₁*5/2 R / (xₙ₁*5/2 R+xₓ*9/2 R) =

xₙ₁*5 / (xₙ₁*5 + xₓ*9) = 5xₙ₁ / (5 + 4*xₓ) = 5*0.735 / (5 + 4*0.265) = 0.598

since

rₙ + rₓ = 1 → rₓ = 1-rₙ = 1 - 0.598 = 0.402

then

T = rₙ * Tₙ + rₓ*Tₓ = 0.598 * 300 K + 0.402 * 575 K = 410.55 K

c) since there is no entropy changes due to heat transfer, the only change in entropy is due to the mixing process

since for a pure gas mixing process

ΔS = n*Cp * ln T₂/T₁ - n*R ln (P₂/P₁)

but P₂=P₁ (P=pressure)

ΔS = n*Cp * ln T₂/T₁ = n*Cp*ln T₂ - n*Cp*ln T₁ = S₂-S₁

for a gas mixture as end product

ΔS = (nₓ+nₙ) * Cp*ln T - (nₓ*Cpₓ*ln Tₓ + nₙ*Cpₙ*ln Tₙ)

ΔS = nₙ*Cpₙ ((nₓ+nₙ) * Cp/[nₙ*Cpₙ] * ln T - (nₓ*Cpₓ / (nₙ*Cpₙ) * ln Tₓ + ln Tₙ)

ΔS = nₙ*Cpₙ [ 1/rₙ * ln T - ((rₓ/rₙ) * ln Tₓ + ln Tₙ) ]

replacing values,

ΔS = nₙ*Cpₙ [ 1/rₙ * ln T - ((rₓ/rₙ) * ln Tₓ + ln Tₙ) ] = (1 kg/s / 20.18*10 kg/kmol) * 5/2 * 8.314 kJ/kmol K * [ 1/0.598 * ln 410.55 K - (0.402/0.598 * ln 575 K + ln 300K) ]

= 1.83 KW/K