Respuesta :
Answer:
The answer is "2074.2 KW"
Explanation:
The mole part of [tex]CO_2[/tex]=0.1
The mole part of [tex]H_2O[/tex]= O.19
The mole part of [tex]N_2[/tex]=0.71
At temperature ([tex]T_1[/tex]) mixture receives from turbine =720K
At pressure ([tex]p_1[/tex]) Â mixture receives from turbine =0.35 Mpa
Flow rate volumetric V = 3.2 [tex]\frac{m^3}{s}[/tex]
A turbine leaves the blend at temperature ([tex]T_2[/tex]) = 380K
The solution comes out of a pressure ([tex]p_2[/tex])= 0.11 Mpa
Decreasing healthy mass balance:
[tex]W_{cv}= m(h_1-h_2)\\\\W_{cv}= m\frac{(\bar{h_1}-\bar{h_2})}{M_{mix}}\\\\W_{cv}= m\frac{Y_{co_2}(\bar{h_1}-\bar{h_2})_{co_2}+Y_{H_2o}(\bar{h_1}-\bar{h_2})_{H_2o}+Y_{N_2}(\bar{h_1}-\bar{h_2})_{N_2}}{M_{mix}}\\\\\ Mass \ flow \ rate:\\[/tex]
[tex]m= \frac{(AV)_1}{V_1}\\\\m= \frac{(AV)_1 P_1}{(\bar{\frac{R}{M}})_{mix}V_1}\\\\\frac{m}{M_{mix}}= \frac{(AV_1)(P1)}{\bar{R}T_1}[/tex]
by increasing value we get:
[tex]\frac{m}{M_{mix}}= \frac{8.2 \frac{kg}{s}(0.35 \times 10^6 \frac{N}{m^2})}{8.314 \frac{N.M}{kmol}(720 K)}[/tex]
After solve we get = 0.1871 [tex]\frac{Kmol (mix)}{s}[/tex]
[tex]W_{cv}= m\frac{Y_{co_2}(\bar{h_1}-\bar{h_2})_{co_2}+Y_{H_2o}(\bar{h_1}-\bar{h_2})_{H_2o}+Y_{N_2}(\bar{h_1}-\bar{h_2})_{N_2}}{M_{mix}}[/tex]
[tex]T_1 =720K \ \ \bar h_1=28,211 \frac{KJ}{Kmol} \ \ co_2\\\\T_2 =380K \ \ \bar h_2=12,552 \frac{KJ}{Kmol} \ \ co_2\\\\T_1 =720K \ \ \bar h_1=24,840 \frac{KJ}{Kmol} \ \ H_2o\\\\T_2 =380K \ \ \bar h_2=12,672 \frac{KJ}{Kmol} \ \ H_2o\\\\[/tex]
[tex]T_1 =720K \ \ \bar h_1=21,220 \frac{KJ}{Kmol} \ \ N_2\\\\T_2 =380K \ \ \bar h_2=11,055 \frac{KJ}{Kmol} \ \ N_2\\\\[/tex]
put the value in above given formula it will give [tex]W_{cv}=[/tex]2074.2 KW
