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

no he didnt use it

explanation:

he said no