Explanation The van der Waals equation is: P = R T V _ − b − a V _ 2 (1) Here, gas constant is R, temperature is T , constants are a, b, and molar volume is V _ . For a reversible turbine the entropy balance is as follows: 0 = S _ i n − S _ o u t Here, inlet and outlet molar entropy is S i n and S o u t respectively. The energy balance equation for a turbine is: W ˙ S n ˙ = H _ o u t − H _ i n Here, molar flow rate is n ˙ , rate of shaft work is W ˙ S , exit reversible molar enthalpy is H _ o u t , and inlet molar enthalpy is H _ i n . The residual enthalpy formula for ideal gas taken from Appendix D-1 is: H _ 2 − H _ 1 = ( H _ 2 − H _ 2 i g ) + ( H _ 2 i g − H _ 1 i g ) − ( H _ 1 − H _ 1 i g ) Here, inert gas molar enthalpy at state 1 and 2 is H _ 1 i g and H _ 2 i g , gas molar enthalpy at state 1 and 2 is H _ 1 and H _ 2 respectively. The general expression for the residual enthalpy using the van der Waal equation is: H _ R = R T ( Z − 1 ) + ∫ T = T , V _ = ∞ T = T , V _ = V _ [ T ( ∂ P ∂ T ) V _ − P ] d V _ (2) Here, compressibility factor is Z , and change in pressure with respect to change in temperature at constant molar volume is ( ∂ P ∂ T ) V _ . The change in pressure with respect to change in temperature at constant molar volume can be written as: ( ∂ P ∂ T ) V _ = R V _ − b Substitute R V _ − b for ( ∂ P ∂ T ) V _ and R T V _ − b − a V _ 2 for P in Equation (2). H _ R = R T ( Z − 1 ) + ∫ V _ = ∞ V _ = V _ [ T ( R V _ − b ) − R T V _ − b − a V _ 2 ] d V _ = R T ( Z − 1 ) + ∫ V _ = ∞ V _ = V _ [ − a V _ 2 ] d V _ = R T ( Z − 1 ) − a V _ (3) Using the given Equation of state, substitute for Z in Equation (3): H _ R = P V _ − R T − a V _ Substitute R T V _ − b − a V _ 2 for P . H _ R = ( R T V _ − b − a V _ 2 ) V _ − R T − a V _ = R T ( V _ V _ − b − 1 ) − 2 a V _ The adjustment in enthalpy can be assessed if T and V are both known for both the conditions. P and T are both given so V _ at the channel conditions can be found from the condition of state itself. Substitute 8.314 × 10 3 m 3 ⋅ Pa mol ⋅ K 10 bar for P , 325 ° C for T , 0.5542 Pa ⋅ m 6 mol 2 for a , and 3.051 × 10 − 5 m 3 / mol for b in Equation (1) 10 bar = [ 8.314 × 10 3 m 3 ⋅ Pa mol ⋅ K ] [ 325 ° C ] V _ i n − [ 3.051 × 10 − 5 m 3 / mol ] − [ 0.5542 Pa ⋅ m 6 mol 2 ] V _ i n 2 10 bar = [ 8.314 × 10 3 m 3 × ( 100 cm m ) 3 ⋅ Pa mol ⋅ K ] [ 325 ° C ] V _ i n − [ 3.051 × 10 − 5 m 3 / mol ] − [ 0.5542 Pa ⋅ m 6 mol 2 ] V _ i n 2 V _ i n = 4890 cm 3 / mol At the outlet conditions, just P is known. Hence we require at the least two conditions to find V _ and T . One is the condition of state and the other being the entropy parity from the reversible turbine. Thus we require the expressions for the change in entropy and the remaining entropy. The ideal gas change in entropy is. d S _ = C V * d T T + R d V _ V _ Here, ideal gas constant volume heat capacity is C V * . The ideal gas heat capacity for water from Appendix D is obtained as: S _ o u t i g − S _ i n i g = R ∫ T i n T o u t ( A − 1 ) + B T + C T 2 + D T 3 + E T 4 T d T + R d V _ V _ = R [ ( A − 1 ) ln ( T o u t T i n ) + B ( T o u t − T i n ) + C ( T o u t 2 − T i n 2 ) 2 + D ( T o u t 3 − T i n 3 ) 3 + E ( T o u t 4 − T i n 4 ) 4 ] + R ln ( V _ o u t V _ i n ) (4) Here, constants are A , B , C , D , E , molar volume of outlet state is V _ o u t , and molar volume of an inlet state is V _ i n . Referring to Appendix D.2, “Liquid Heat Capacities”, the constants A, B, C, D , and E for water are obtained: A = 4.395 B = − 4.186 × 10 − 3 C = 1.405 × 10 − 5 D = − 1.564 × 10 − 8 E = 0.632 × 10 − 11 Substitute 4.395 for A , − 4.186 × 10 − 3 for B , 1.405 × 10 − 5 for C , − 1.564 × 10 − 8 for D , 0.632 × 10 − 11 for E , 10 bar for P i n , 598 K for T i n , 4890 cm 3 mol for V _ i n , 8.314 J mol ⋅ K for R , 1 bar for P o u t , and 3.051 × 10 − 5 m 3 / mol for b in Equation (4). S _ o u t i g − S _ i n i g = { [ 8.314 J mol ⋅ K ] [ ( 4.395 − 1 ) ln ( T o u t 598 K ) + [ − 4.186 × 10 − 3 ] ( T o u t − ( 598 K ) ) + ( 1.405 × 10 − 5 ) ( T o u t 2 − ( 598 K ) 2 ) 2 + ( − 1.564 × 10 − 8 ) ( T o u t 3 − ( 598 K ) 3 ) 3 + ( 0.632 × 10 − 11 ) ( T o u t 4 − ( 598 K ) 4 ) 4 ] + [ 8.314 J mol ⋅ K ] ln ( V _ o u t 4890 cm 3 mol ) } (5) There are two unknowns in the entropy balance equation. The outlet pressure using EOS can be written as: P o u t = R T o u t V _ o u t − b − a V _ o u t 2 Substitute 83.14 bar ⋅ cm 3 mol ⋅ K for R , 1 bar for P o u t , 0.5542 Pa ⋅ m 6 mol 2 for a , and 3.051 × 10 − 5 m 3 / mol for b . 1 bar = 83.14 bar ⋅ cm 3 mol ⋅ K ( T o u t ) V _ o u t − 3.051 × 10 − 5 m 3 / mol − 0.5542 Pa ⋅ m 6 mol 2 V _ o u t 2 (6) The residual molar entropy is. S _ R = R ln ( Z ) + ∫ T = T , V _ = ∞ T = T , V _ = V _ [ ( ∂ P ∂ T ) V _ − R V _ ] d V _ = R ln ( Z ) + ∫ T = T , V _ = ∞ T = T , V _ = V _ [ R V _ − b − R V _ ] d V _ S _ R = R ln ( Z ) + R [ ln ( V _ − b ) − ln ( V _ ) ] V _ = ∞ V _ = V _ = R ln ( Z ) + R ln ( V _ − b V _ ) (7) Substitute P V _ R T for Z in Equation (7). S _ R = R ln ( P V _ R T ) + R ln ( V _ − b V _ ) Apply the residuals to the entropy balance for a reversible turbine. 0 = S _ o u t − S _ i n = S _ o u t R + ( S _ o u t i g − S _ i n i g ) − S _ i n R Substitute R ln ( P V _ R T ) + R ln ( V _ − b V _ ) for S _ i n R and S _ o u t R . 0 = R ln ( P o u t V _ o u t R T o u t ) + R ln ( V _ o u t − b V _ o u t ) + ( S _ o u t i g − S _ i n i g ) − R ln ( P i n V _ i n R T i n ) + R ln ( V _ i n − b V _ i n ) (8) Substitute 10 bar for P i n , 598 K for T i n , 4890 cm 3 mol for V _ i n , 83

To redo problem 4-9 using van der Waals equation instead of steam tables and comment on the accuracy of van der Waals equation in this case. Concept introduction: The ideal gas equation treats the molecules as perfectly elastic and this works well in many cases also. But the molecules are not point masses and have a determinable effect on the properties. Hence van der Waals modified the ideal gas equation taking into consideration the molecule size and their interaction forces and proposed the van der Waals equation given by [ P + a ( n V ) 2 ] ( V n − b ) = R T Here, gas constant is R, temperature is T , constants are a, b, and volume is V In this problem, the properties obtained from steam tables for problem 4-9 will be calculated using the van der Waals equation and the same procedure will be continued.

Fundamentals of Chemical Engineering Thermodynamics (MindTap Course List)

1st Edition

ISBN: 9781111580704

Author: Kevin D. Dahm, Donald P. Visco

Publisher: Cengage Learning

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Question

Chapter 6.6, Problem 22P

Interpretation Introduction

Interpretation:

To redo problem 4-9 using van der Waals equation instead of steam tables and comment on the accuracy of van der Waals equation in this case.

Concept introduction:

The ideal gas equation treats the molecules as perfectly elastic and this works well in many cases also. But the molecules are not point masses and have a determinable effect on the properties. Hence van der Waals modified the ideal gas equation taking into consideration the molecule size and their interaction forces and proposed the van der Waals equation given by

[P+a(nV)2](Vn−b)=RT

Here, gas constant is R, temperature is T, constants are a, b, and volume is V

In this problem, the properties obtained from steam tables for problem 4-9 will be calculated using the van der Waals equation and the same procedure will be continued.

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