Chapter 2, Problem 2.21P

(a)

Interpretation Introduction

Interpretation:

A simplified form of the general steady-state energy balance equation for single-pipe heat exchanger should be developed.

Concept Introduction:

The general steady-state energy balance equation is as follows:

  $\left[\stackrel{•}{m}\left(\Delta H+\frac{\Delta u^2}{2}+g\Delta Z\right)\right]=Q+W$

Where

  $\begin{array}{c}\Delta H=\text{Enthalpy}\text{change}\\ \frac{\Delta u^2}{2}=\text{Kinectic}\text{Energy}\text{change}\\ g\Delta Z=\text{potential}\text{enegy}\text{change}\\ Q=\text{heat transfer}\\ W=\text{work done}\end{array}$

Answer to Problem 2.21P

Steady-state energy equation becomes $\stackrel{•}{m}\left(\Delta H\right)=Q$

Explanation of Solution

Single-pipe heat exchanger:

Assumptions:

1. Single stream is flowing within the pipe.

2. Neglect the change in potential and kinetic energies.

There is no any shaft work in heat exchanger.

So, $W=0$

Therefore steady-state energy equation becomes $\stackrel{•}{m}\left(\Delta H\right)=Q$

(b)

Interpretation Introduction

Interpretation: :

A simplified form of the general steady-state energy balance equation for double-pipe heat exchanger should be developed.

Concept Introduction:

The general steady-state energy balance equation is as follows:

  $\left[\stackrel{•}{m}\left(\Delta H+\frac{\Delta u^2}{2}+g\Delta Z\right)\right]=Q+W$

Where

  $\begin{array}{c}\Delta H=\text{Enthalpy}\text{change}\\ \frac{\Delta u^2}{2}=\text{Kinectic}\text{Energy}\text{change}\\ g\Delta Z=\text{potential}\text{enegy}\text{change}\\ Q=\text{heat transfer}\\ W=\text{work done}\end{array}$

Answer to Problem 2.21P

The steady-state energy balance becomes, $\stackrel{•}{m}\left(H_1-H_2\right)=0$

Explanation of Solution

Double-pipe heat exchanger:

Assumptions:

1. Two streams are flowing in two pipes.

2. Neglect the change in potential and kinetic energies.

There is no any shaft work in heat exchanger and heat transfer is internal between the two streams.

Therefore, the steady-state energy balance becomes,

  $\stackrel{•}{m}\left(H_1-H_2\right)=0$

(c)

Interpretation Introduction

Interpretation:

A simplified form of the general steady-state energy balance equation for pump should be developed.

Concept Introduction:

The general steady-state energy balance equation is as follows:

  $\left[\stackrel{•}{m}\left(\Delta H+\frac{\Delta u^2}{2}+g\Delta Z\right)\right]=Q+W$

Where

  $\begin{array}{c}\Delta H=\text{Enthalpy}\text{change}\\ \frac{\Delta u^2}{2}=\text{Kinectic}\text{Energy}\text{change}\\ g\Delta Z=\text{potential}\text{enegy}\text{change}\\ Q=\text{heat transfer}\\ W=\text{work done}\end{array}$

Answer to Problem 2.21P

The steady-state energy balance becomes, $\stackrel{•}{m}\left(\Delta {\rm H}\right)=W$

Explanation of Solution

Pump operating on a single liquid stream:

Assumptions:

1. Neglect the change in potential and kinetic energies.

2. Negligible heat transfer to the surroundings.

Therefore, the steady-state energy balance becomes,

  $\stackrel{•}{m}\left(\Delta {\rm H}\right)=W$

(d)

Interpretation Introduction

Interpretation:

A simplified form of the general steady-state energy balance equation for gas compressor should be developed.

Concept Introduction:

The general steady-state energy balance equation is as follows:

  $\left[\stackrel{•}{m}\left(\Delta H+\frac{\Delta u^2}{2}+g\Delta Z\right)\right]=Q+W$

Where

  $\begin{array}{c}\Delta H=\text{Enthalpy}\text{change}\\ \frac{\Delta u^2}{2}=\text{Kinectic}\text{Energy}\text{change}\\ g\Delta Z=\text{potential}\text{enegy}\text{change}\\ Q=\text{heat transfer}\\ W=\text{work done}\end{array}$

Answer to Problem 2.21P

The steady-state energy balance becomes, $\stackrel{•}{m}\left(\Delta {\rm H}\right)=W$

Explanation of Solution

Gas compressor:

Assumptions:

1. Neglect the change in potential and kinetic energies.

2. Negligible heat transfer to the surroundings.

Therefore, the steady-state energy balance becomes,

  $\stackrel{•}{m}\left(\Delta {\rm H}\right)=W$

(e)

Interpretation Introduction

Interpretation:

A simplified form of the general steady-state energy balance equation for gas turbine should be developed.

Concept Introduction:

The general steady-state energy balance equation is as follows:

  $\left[\stackrel{•}{m}\left(\Delta H+\frac{\Delta u^2}{2}+g\Delta Z\right)\right]=Q+W$

Where

  $\begin{array}{c}\Delta H=\text{Enthalpy}\text{change}\\ \frac{\Delta u^2}{2}=\text{Kinectic}\text{Energy}\text{change}\\ g\Delta Z=\text{potential}\text{enegy}\text{change}\\ Q=\text{heat transfer}\\ W=\text{work done}\end{array}$

Answer to Problem 2.21P

A simplified form of the general steady-state energy balance for Gas Turbine

  $\stackrel{·}{m}\left(\Delta H\right)=\stackrel{·}{W}$

Explanation of Solution

Gas turbine:

Assumptions:

1. Neglect the change in potential and kinetic energies.

2. Negligible heat transfer to the surroundings.

Therefore, the steady-state energy balance becomes,

  $\stackrel{·}{m}\left(\Delta H\right)=\stackrel{·}{W}$

(f)

Interpretation Introduction

Interpretation:

A simplified form of the general steady-state energy balance equation for throttle valve should be developed.

Concept Introduction:

The general steady-state energy balance equation is as follows:

  $\left[\stackrel{•}{m}\left(\Delta H+\frac{\Delta u^2}{2}+g\Delta Z\right)\right]=Q+W$

Where

  $\begin{array}{c}\Delta H=\text{Enthalpy}\text{change}\\ \frac{\Delta u^2}{2}=\text{Kinectic}\text{Energy}\text{change}\\ g\Delta Z=\text{potential}\text{enegy}\text{change}\\ Q=\text{heat transfer}\\ W=\text{work done}\end{array}$

Answer to Problem 2.21P

The steady-state energy balance becomes for throttle valve,

  $\Delta H=0$

Explanation of Solution

The purpose of the throttle valve is to reduce on a flowing stream.

Assumptions:

1. Neglect the change in potential and kinetic energies.

2. Adiabatic process.

There is no any shaft work in throttling process. Thus, $W=0$

Therefore, the steady-state energy balance becomes for throttle valve,

  $\Delta H=0$

(g)

Interpretation Introduction

Interpretation:

A simplified form of the general steady-state energy balance equation for nozzle should be developed.

Concept Introduction:

The general steady-state energy balance equation is as follows:

  $\left[\stackrel{•}{m}\left(\Delta H+\frac{\Delta u^2}{2}+g\Delta Z\right)\right]=Q+W$

Where

  $\begin{array}{c}\Delta H=\text{Enthalpy}\text{change}\\ \frac{\Delta u^2}{2}=\text{Kinectic}\text{Energy}\text{change}\\ g\Delta Z=\text{potential}\text{enegy}\text{change}\\ Q=\text{heat transfer}\\ W=\text{work done}\end{array}$

Answer to Problem 2.21P

The steady-state energy balance for Nozzle becomes

  $\left[\stackrel{•}{m}\left(\Delta H+\frac{\Delta u^2}{2}\right)\right]=0$

Explanation of Solution

The purpose of the nozzle is to product high velocity stream, for this change kinetic energy change must. Therefore it has to be taken into account.

Assumptions:

1. Neglect the change in potential energy.

2. Negligible heat transfer to the surroundings.

There is no any shaft work. Thus, . $W=0$

Therefore, the steady-state energy balance becomes,

  $\left[\stackrel{•}{m}\left(\Delta H+\frac{\Delta u^2}{2}\right)\right]=0$