Question

A steady flow adiabatic turbine accepts gas at conditions T1, P1 and discharges at conditions T2 and P2. Assuming ideal gas, determine (per mole of gas) W, Wideal, Wlost and SG for the following. Take Tσ = 300 Κ, Τ1 = 500 Κ, P1 = 6 bar, Τ2 = 371 Κ, P2 = 1.2 bar, and Cp/R = 7/2.
Chemical Engineering (thermodynamics) Please answer as soon as possible.

Answer 1

Answer:

W = -3753.8 J

$W_{ideal}$ = - 5163.14 J

$W_{lost} = - 1409.34 \ J$

$S_G = 4.70 \ \ J/K$

Explanation:

Given that:

$T_1 = 500 \ K \\T_2 = 371 \ K \\T_{\sigma} = 300 \ K\\P_1 = 6 \ bar\\P_2 = 1.2 \ bar$

$\frac{Cp}{R} = \frac{7}{2} \\n = 1$

Consider the relation

$\frac{Cp}{R} = \frac{7}{2}$

$Cp = \frac{7*R}{2}$

$Cp = \frac{7*8.314 \ kJ/kmol.K}{2}$

$Cp = 29.099 \ kJ/kmol.K$

Actual work W = ΔH = n Cp (T₂ - T₁)

= 1 × 29.099 (371 - 500)

= -3753.8 J

Change in entropy

ΔS = $n[Cp In \frac{T_2}{T_1} - R In \frac{P_2}{P_1}]$

ΔS =  $1*[29.099 In(\frac{371}{500}) - 8.314 In (\frac{1.2}{6})]$

ΔS = 4.6978 J/K

Ideal Work

$W_{ideal} = \delta H - T_{\sigma} \delta S$

= -3753.8 - 300 ( 4.6978)

= - 5163.14 J

Work lost

$W_{lost} = |W_{ideal}-W|$

$W_{lost} = |-5163.14 -(-3753.8)|$

$W_{lost} = - 1409.34 \ J$

Entropy Generation rate:

$S_G = \frac{W_{lost}}{T_{\sigma}}$

$S_G = \frac{1409.34}{300}$

$S_G = 4.70 \ \ J/K$