Answered: A liquid has a molar volume of 0.2 L/mol at 400 K and 4 bar. This liquid is compressed with a pump that operates at steady-state and adiabatically. The liquid…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN: 9781259696527

Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher: McGraw-Hill Education

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A liquid has a molar volume of 0.2 L/mol at 400 K and 4 bar. This liquid is compressed with a pump that operates at steady-state and adiabatically. The liquid enters the pump at a rate of 3 mol/s, at a temperature of 400 K and pressureof 4 bar and exits at a pressure of 20 bar.

The liquid has the following properties which can be used (if needed) in the solution:
• Coefficient of thermal expansion, αV = 2 x 10-3 K-1
• Isothermal compressibility factor, κT = 4 x 10-5 bar-1
• Cp = Cv = 30 J/mol-K

Find the following:
A) Work done by the pump
B) Temperature of the liquid leaving the pump
C) Molar volume of the liquid leaving the pump

Expert Solution

Step 1: First Step:-

A. Work done by the pump

The work done by the pump is given by the following formula:

$W equals n R T left parenthesis 1 minus 1 divided by kappa T left parenthesis P 2 divided by P 1 right parenthesis right parenthesis$

where

n is the number of moles of liquid entering the pump per unit time ( $3 m o l divided by s$ )
R is the universal gas constant ( $8.314 space J divided by m o l minus K$ )
T is the temperature of the liquid entering the pump ( $400 K$ )
P1 is the pressure of the liquid entering the pump ( $4 b a r$ )
P2 is the pressure of the liquid leaving the pump ( $20 b a r$ )
κT is the isothermal compressibility factor of the liquid ( $4 space x 10 minus 5 space b a r minus 1$ )

Substituting the given values, we get the work done by the pump as follows:

$W equals left parenthesis 3 m o l divided by s right parenthesis left parenthesis 8.314 space J divided by m o l minus K right parenthesis left parenthesis 400 space K right parenthesis left parenthesis 1 minus 1 divided by left parenthesis 4 space x space 10 minus 5 space b a r minus 1 right parenthesis left parenthesis 20 space b a r divided by 4 space b a r right parenthesis right parenthesis equals 1263 J divided by s$

B. Temperature of the liquid leaving the pump

The temperature of the liquid leaving the pump can be found using the following formula:

$T 2 equals T 1 left parenthesis P 1 divided by P 2 right parenthesis to the power of 2 space end exponent left parenthesis 1 minus y right parenthesis$

where

T1 is the temperature of the liquid entering the pump ( $400 k$ )
P1 is the pressure of the liquid entering the pump ( $4 b a r$ )
P2 is the pressure of the liquid leaving the pump ( $20 b a r$ )
γ is the adiabatic index of the liquid ( $1.4$ )

Substituting the given values, we get the temperature of the liquid leaving the pump as follows:

$T 2 equals 400 K left parenthesis 4 b a r divided by 20 b a r right parenthesis squared left parenthesis 1 minus 1.4 right parenthesis equals 280 K$

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