Chapter 3.9, Problem 29P

(A)

Interpretation Introduction

Interpretation:

The flow rate of existing steam $\left({\dot{m}}_{steam}\right)$ of the heat exchanger.

Concept Introduction:

The steady state energy balance equation for heat exchanger.

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

Here, mass flow rate for inlet and outlet are ${\dot{m}}_{in}$ and ${\dot{m}}_{out}$, specific enthalpies of inlet and outlet are ${\widehat{H}}_i$ and ${\widehat{H}}_{out}$, heights at which streams enters and leave the system are $h_{in}$ and $h_{out}$, time taken is t, mass of the system is M, specific internal energy is $\widehat{U}$, velocity is v, height is h, acceleration due to gravity is g, rate at which work is added to the system is ${\dot{W}}_{EC}$, rate of shaft work is ${\dot{W}}_S$, and the rate of heat addition to the system is $\dot{Q}$.

The formula to calculate the net specific enthalpy change of compound.

${\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)}_{compound}={\left({\displaystyle {\int }_{T_1}^{T_2}{C}_{P,liquid}dT+\Delta {\widehat{H}}_{\text{vap}}+{\displaystyle {\int }_{T_2}^{T_3}{C}_{P,vapor}dT}}\right)}_{compound}$

Here, initial temperature of compound is $T_1$, normal boiling of compound is $T_2$, constant heat capacity in the liquid phase is $C_{P,liquid}$, enthalpy of vaporization is $\Delta {\widehat{H}}_{\text{vap}}$, constant heat capacity in the vapor phase is $C_{P,vapor}$.

(B)

Interpretation Introduction

Interpretation:

The flow rate of existing steam $\left({\dot{m}}_{steam}\right)$ of the heat exchanger

Concept Introduction:

The mass flow rate of the existing steam at heat exchanger using the steady state balance equation.

$\begin{array}{l}0=\left[\begin{array}{l}{\dot{m}}_{out,steam}\left({\widehat{H}}_{out,steam}\right)+{\dot{m}}_{out,compound}\left({\widehat{H}}_{out,compound}\right)-{\dot{m}}_{in,steam}\left({\widehat{H}}_{in,steam}\right)\\ -{\dot{m}}_{in,compound}\left({\widehat{H}}_{in,compound}\right)\end{array}\right]\\ 0={\dot{m}}_{steam}\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)+{\dot{m}}_{compound}\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)\\ {\dot{m}}_{steam}\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)=-{\dot{m}}_{compound}\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)\end{array}$

${\dot{m}}_{steam}=\frac{-{\dot{m}}_{compound}{\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)}_{compound}}{{\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)}_{steam}}$

Here, rate of mass flow rate of leaving steam is ${\dot{m}}_{out,steam}$, specific enthalpy of leaving steam is ${\widehat{H}}_{out,steam}$, rate of mass flow rate of leaving compound is ${\dot{m}}_{out,compound}$, specific enthalpy of leaving compound is ${\widehat{H}}_{out,compound}$, rate of mass flow rate of initial steam is ${\dot{m}}_{in,steam}$, specific enthalpy of initial steam is ${\widehat{H}}_{in,steam}$, rate of mass flow rate of initial compound is ${\dot{m}}_{in,compound}$, and specific enthalpy of initial compound is ${\widehat{H}}_{in,compound}$.