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.

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Answer:

W = -3753.8 J

[tex]W_{ideal}[/tex] = - 5163.14 J

[tex]W_{lost} = - 1409.34 \ J[/tex]

[tex]S_G = 4.70 \ \ J/K[/tex]

Explanation:

Given that:

[tex]T_1 = 500 \ K \\T_2 = 371 \ K \\T_{\sigma} = 300 \ K\\P_1 = 6 \ bar\\P_2 = 1.2 \ bar\\[/tex]

[tex]\frac{Cp}{R} = \frac{7}{2} \\n = 1[/tex]

Consider the relation

[tex]\frac{Cp}{R} = \frac{7}{2} \\[/tex]

[tex]Cp = \frac{7*R}{2}[/tex]

[tex]Cp = \frac{7*8.314 \ kJ/kmol.K}{2}[/tex]

[tex]Cp = 29.099 \ kJ/kmol.K[/tex]

Actual work W = ΔH = n Cp (T₂ - T₁)

= 1 × 29.099 (371 - 500)

= -3753.8 J

Change in entropy

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

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

ΔS = 4.6978 J/K

Ideal Work

[tex]W_{ideal} = \delta H - T_{\sigma} \delta S[/tex]

= -3753.8 - 300 ( 4.6978)

= - 5163.14 J

Work lost

[tex]W_{lost} = |W_{ideal}-W|[/tex]

[tex]W_{lost} = |-5163.14 -(-3753.8)|[/tex]

[tex]W_{lost} = - 1409.34 \ J[/tex]

Entropy Generation rate:

[tex]S_G = \frac{W_{lost}}{T_{\sigma}}[/tex]

[tex]S_G = \frac{1409.34}{300}[/tex]

[tex]S_G = 4.70 \ \ J/K[/tex]

The ideal work, work lost and entropy generation rate are; W_ideal = -5163 J; W_lost = 1409.3 J; S_g = 4.698 J/K

What is the change in entropy?

We are given;

Temperature of surrounding; Tσ = 300 Κ

Temperature at initial condition; Τ₁ = 500 Κ

Pressure at initial condition; P₁ = 6 bar

Temperature at final condition; Τ₂ = 371 Κ

Pressure at final condition; P₂ = 1.2 bar

Cp/R = 7/2. where R = 8.314 kJ/mol.k

Thus; Cp = 29.099 kJ/mol

Actual work is;

W_actual = ΔH = n*Cp (T₂ - T₁)

ΔH = 1 × 29.099 × (371 - 500)

ΔH = -3753.8 J

Change in entropy is gotten from the formula;

ΔS = n[Cp In (T₂/T₁) - R In (P₂/P₁)]

Plugging in relevant values and solving gives us;

ΔS = 4.698 J/K

Ideal work is gotten from;

W_ideal = ∆H - Tσ*∆S

W_ideal = -3753.8 - (300 * 4.698)

W_ideal = -5163 J

Work Lost;

W_lost = |W_ideal - W_actual|

W_lost = |-5163 - (-3753.8)|

Solving and taking absolute value gives;

W_lost = 1409.3 J

Entropy generation rate is;

S_g = W_lost/Tσ

S_g = 1409.3/300

S_g = 4.698 J/K

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