Chapter 6.6, Problem 10P

A)

Interpretation Introduction

Interpretation:

The molar volume of the liquid leaving the valve has to be determined.

Concept introduction:

Write the expression for the energy balance for the valve.

$\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{v^2}{2}+gh\right)\right\}={\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{v_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{v_{out}^2}{2}+g{h}_{out}\right)+{\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}$

Here, time is t, total mass of the system is M, specific internal energy of the system is $\widehat{U}$, velocity of the system is v, height of the system is h, acceleration due to gravity is g, individual quantities of mass added to and removed from the process is ${\dot{m}}_{in}$ and ${\dot{m}}_{out}$, specific enthalpies of streams inlet and outlet is ${\widehat{H}}_{in}$ and ${\widehat{H}}_{out}$, heights at which streams enters and leave the system is $h_{in}$ and $h_{out}$, rate at which work is added to the system through expansion or contraction of the system is ${\dot{W}}_{EC}$, rate at which shaft work is added to the system is ${\dot{W}}_S$, and the rate at which heat is added to the system is $\dot{Q}$.

Write the expression for the coefficient of thermal expansion.

${\alpha }_V=\frac{1}{\underset{\_}{V}}{\left(\frac{\partial \underset{\_}{V}}{\partial T}\right)}_P$

Here, molar volume is $\underset{\_}{V}$, change in temperature at constant pressure is $\partial T$, and change in molar volume at constant pressure is $\partial \underset{\_}{V}$.

Write the expression for the isothermal compressibility.

${\kappa }_T=-\frac{1}{\underset{\_}{V}}{\left(\frac{\partial \underset{\_}{V}}{\partial P}\right)}_T$

Here, change in pressure at constant temperature is $\partial P$, and change in molar volume at constant temperature is $\partial \underset{\_}{V}$.

B)

Interpretation Introduction

Interpretation:

Estimate temperature at which the liquid would have a molar volume of $0.10002\frac{\text{L}}{\text{mol}}$.

Concept introduction:

The total derivative for molar volume from part A

$\frac{1}{\underset{\_}{V}}d\underset{\_}{V}=\left(\frac{0.001}{°\text{R}}\right)dT$