Question #213398

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.


1
Expert's answer
2021-07-07T11:41:57-0400

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;

mCO2+mN2=mmm_{CO_2}+m_{N_2} =m_m

Where mm is the mass of mixture

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

Energy balance;

Qin+mCO2hCO2Q_{in}+m_{CO_2}h_{CO_2}+mN2hN2_{N_2}h_{N_2}=mmhm_mh_m

Qin 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;

Cp=72×RC_p =\frac 7 2 \times R

Where R= R M\frac {R^~}M

R=8.314

M is Molar mass for each gas.

CpC_p for Nitrogen;

RN2R_{N_2} =8.31428\frac {8.314}{28}

Cp=8.31428×72C_p=\frac {8.314} {28} \times \frac 7 2 =1.03925kJ/kgK

CpC_p for Carbon dioxide;

RCO2=8.31444R_{CO_2}=\frac {8.314} {44}

Cp=8.31444×72C_p =\frac {8.314}{44} \times \frac 7 2=0.6613kJ/kgK

CpC_p for the mixture;

mm =5kg/s

Mass fraction mf for each gas in mixture:

mfCO2mf_{CO_2} =25\frac 2 5 =0.4

mfN2=35mf_{N_2}=\frac 3 5 =0.6

Number of moles in mixture nmn_m ;

n=mM\frac m M

nm=nCO2+nN2n_m =n_{CO_2} +n_{N_2}

nm=244+328n_m=\frac 2 {44} +\frac 3{28} =0.1526

Mole fractions y;

yCO2=0.045450.1526=0.2979y_{CO_2}=\frac {0.04545} {0.1526}=0.2979

yN2=0.10710.1526=0.7121y_{N_2}=\frac {0.1071}{0.1526}=0.7121

Molar mass of mixture Mm;

Mm=yCO2MCO2+yN2MN2M_m=y_{CO_2}M_{CO_2} +y_{N_2}M_{N_2}

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

Mm=32.77

Rm=8.31432.77=0.2537\frac {8.314}{32.77}=0.2537

Cp=72×0.2537\frac 7 2\times 0.2537 =0.8880kJ/kgK

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

Since h=CpT

100+2(0.6613×\times308)+3(1.03925×\times323)=5×0.8880×Tm

Tm=341.04K or 68.04°C

Partial pressures for individual gases;

P=y×Pm

PCO2=0.2979×4=1.1916P_{CO_2}=0.2979×4=1.1916 bar

PN2=0.7021×4=2.8084P_{N_2}=0.7021×4=2.8084 bar







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