Answer to Question #213398 in Molecular Physics | Thermodynamics for Yomna

Question

In a mixer enters two streams: CO2 at a rate of 2kg/s and 35oC as well as N2 at 

a rate of 3 kg/s and 50oC. During mixing, 100kW of heat is added while all inlet 

and outlet streams are at 4 bar. Obtain outlet mixture temperature, partial 

pressures of each gas.

Answer 1

Question

In a mixer enters two streams :C02 at a rate of 2kg/s and 35°c as well as N2 at a rate of 3kg/s and 50°c.During mixing,100kW of heat is added while all inlet and outlet streams are at 4 bar. Obtain outlet mixture temperature ,partial pressures of each gas.

Solution

Mass balance;

$m_{CO_2}+m_{N_2} =m_m$

Where m_m is the mass of mixture

m_m=2kg/s +3kg/s=5kg/s

Energy balance;

$Q_{in}+m_{CO_2}h_{CO_2}$+m$_{N_2}h_{N_2}$=m$_mh_m$

Q_in is the heat added during mixing.

Since both Nitrogen and Carbon dioxide are diatomic gases, their specific heat capacity at constant pressure is give as follows;

$C_p =\frac 7 2 \times R$

Where R= $\frac {R^~}M$

R=8.314

M is Molar mass for each gas.

$C_p$ for Nitrogen;

$R_{N_2}$ =$\frac {8.314}{28}$

$C_p=\frac {8.314} {28} \times \frac 7 2$ =1.03925kJ/kgK

$C_p$ for Carbon dioxide;

$R_{CO_2}=\frac {8.314} {44}$

$C_p =\frac {8.314}{44} \times \frac 7 2$=0.6613kJ/kgK

$C_p$ for the mixture;

m_m =5kg/s

Mass fraction mf for each gas in mixture:

$mf_{CO_2}$ =$\frac 2 5$ =0.4

$mf_{N_2}=\frac 3 5$ =0.6

Number of moles in mixture $n_m$ ;

n=$\frac m M$

$n_m =n_{CO_2} +n_{N_2}$

$n_m=\frac 2 {44} +\frac 3{28}$ =0.1526

Mole fractions y;

$y_{CO_2}=\frac {0.04545} {0.1526}=0.2979$

$y_{N_2}=\frac {0.1071}{0.1526}=0.7121$

Molar mass of mixture M_m;

$M_m=y_{CO_2}M_{CO_2} +y_{N_2}M_{N_2}$

$M_m=(0.2979\times{44} )+(0.7021\times{28})$

M_m=32.77

R_m=$\frac {8.314}{32.77}=0.2537$

C_p=$\frac 7 2\times 0.2537$ =0.8880kJ/kgK

Use the energy equation to find the Temperature of the mixture;

Since h=C_pT

100+2(0.6613$\times$308)+3(1.03925$\times$323)=5×0.8880×T_m

T_m=341.04K or 68.04°C

Partial pressures for individual gases;

P=y×P_m

$P_{CO_2}=0.2979×4=1.1916$ bar

$P_{N_2}=0.7021×4=2.8084$ bar