Question

"A gas turbine receives a mixture having the following molar analysis: 10% CO2, 19% H2O, 71% N2, at 720 K, 0.35 MPa and a volumetric flow rate of 3.2 m3 /s. The mixture exits the turbine at 380 K, 0.11 MPa. For adiabatic operation with negligible kinetic and potential energy effects, determine the power developed at steady state, in kW." NOTE: the process is NOT isentropic.

Answer 1

Answer:

The answer is "2074.2 KW"

Explanation:

The mole part of $CO_2$=0.1

The mole part of $H_2O$= O.19

The mole part of $N_2$=0.71

At temperature ($T_1$) mixture receives from turbine =720K

At pressure ($p_1$)  mixture receives from turbine =0.35 Mpa

Flow rate volumetric V = 3.2 $\frac{m^3}{s}$

A turbine leaves the blend at temperature ($T_2$) = 380K

The solution comes out of a pressure ($p_2$)= 0.11 Mpa

Decreasing healthy mass balance:

$W_{cv}= m(h_1-h_2)\\\\W_{cv}= m\frac{(\bar{h_1}-\bar{h_2})}{M_{mix}}\\\\W_{cv}= m\frac{Y_{co_2}(\bar{h_1}-\bar{h_2})_{co_2}+Y_{H_2o}(\bar{h_1}-\bar{h_2})_{H_2o}+Y_{N_2}(\bar{h_1}-\bar{h_2})_{N_2}}{M_{mix}}\\\\\ Mass \ flow \ rate:$

$m= \frac{(AV)_1}{V_1}\\\\m= \frac{(AV)_1 P_1}{(\bar{\frac{R}{M}})_{mix}V_1}\\\\\frac{m}{M_{mix}}= \frac{(AV_1)(P1)}{\bar{R}T_1}$

by increasing value we get:

$\frac{m}{M_{mix}}= \frac{8.2 \frac{kg}{s}(0.35 \times 10^6 \frac{N}{m^2})}{8.314 \frac{N.M}{kmol}(720 K)}$

After solve we get = 0.1871 $\frac{Kmol (mix)}{s}$

$W_{cv}= m\frac{Y_{co_2}(\bar{h_1}-\bar{h_2})_{co_2}+Y_{H_2o}(\bar{h_1}-\bar{h_2})_{H_2o}+Y_{N_2}(\bar{h_1}-\bar{h_2})_{N_2}}{M_{mix}}$

$T_1 =720K \ \ \bar h_1=28,211 \frac{KJ}{Kmol} \ \ co_2\\\\T_2 =380K \ \ \bar h_2=12,552 \frac{KJ}{Kmol} \ \ co_2\\\\T_1 =720K \ \ \bar h_1=24,840 \frac{KJ}{Kmol} \ \ H_2o\\\\T_2 =380K \ \ \bar h_2=12,672 \frac{KJ}{Kmol} \ \ H_2o$

$T_1 =720K \ \ \bar h_1=21,220 \frac{KJ}{Kmol} \ \ N_2\\\\T_2 =380K \ \ \bar h_2=11,055 \frac{KJ}{Kmol} \ \ N_2$

put the value in above given formula it will give $W_{cv}=$2074.2 KW