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

The answer is "2074.2 KW"

Step-by-step 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
`(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= ((AV)_1)/(V_1)\\\\m= \frac{(AV)_1 P_1}{(\bar{(R)/(M)})_(mix)V_1}\\\\(m)/(M_(mix))= \frac{(AV_1)(P1)}{\bar{R}T_1}`

by increasing value we get:

`(m)/(M_(mix))= (8.2 (kg)/(s)(0.35 * 10^6 (N)/(m^2)))/(8.314 (N.M)/(kmol)(720 K))`

After solve we get = 0.1871
`(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 (KJ)/(Kmol) \ \ co_2\\\\T_2 =380K \ \ \bar h_2=12,552 (KJ)/(Kmol) \ \ co_2\\\\T_1 =720K \ \ \bar h_1=24,840 (KJ)/(Kmol) \ \ H_2o\\\\T_2 =380K \ \ \bar h_2=12,672 (KJ)/(Kmol) \ \ H_2o\\\\`

`T_1 =720K \ \ \bar h_1=21,220 (KJ)/(Kmol) \ \ N_2\\\\T_2 =380K \ \ \bar h_2=11,055 (KJ)/(Kmol) \ \ N_2\\\\`

put the value in above given formula it will give
`W_(cv)=`2074.2 KW