Chapter 3.9, Problem 20P

(A)

Interpretation Introduction

Interpretation:

The velocity of the steam entering the nozzle

Concept Introduction:

Write the expression for velocity from the continuity equation.

$v=\frac{\dot{m}\widehat{V}}{A}$

Here, mass flow rate is $\dot{m}$, specific volume is $\widehat{V}$, and area of a nozzle is A.

Use above equation, equate the mass flow rate for inlet and outlet in nozzle.

$\begin{array}{l}\frac{v_1{A}_1}{{\widehat{V}}_1}=\frac{v_2{A}_2}{{\widehat{V}}_2}\\ v_1=\left(\frac{v_2{A}_2}{{\widehat{V}}_2}\right)\frac{{\widehat{V}}_1}{A_1}\\ v_1=\left(\frac{v_2{D}_2^2}{{\widehat{V}}_2}\right)\frac{{\widehat{V}}_1}{D_1^2}\end{array}$

Here, velocity of the steam entering and exiting the nozzle are $v_1$ and $v_2$, area of the nozzle for inlet and outlet of nozzle are $A_1$ and $A_2$, diameter of the nozzle at inlet and outlet are $D_1$ and $D_2$, specific volume at inlet and outlet of the steam are ${\widehat{V}}_1$ and ${\widehat{V}}_2$ respectively.

(B)

Interpretation Introduction

Interpretation:

The rate at which this process loses heat.

Concept Introduction:

Write the steady state energy balance equation for nozzle.

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

Here, mass flow rate for inlet and outlet is ${\dot{m}}_{in}$ and ${\dot{m}}_{out}$, specific enthalpies of inlet and outlet is ${\widehat{H}}_i$ and ${\widehat{H}}_{out}$, heights at which streams enters and leave the system is $h_{in}$ 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 $\dot{Q}$.

At steady state left hand side is zero, assume the process is adiabatic and have specific enthalpy could no changed.

$0={\dot{m}}_{in}\left({\widehat{H}}_{in,1}+\frac{v_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{v_{out}^2}{2}+g{h}_{out}\right)+{\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}$

Here, the datum is same, so consider as zero.

Rewrite steady state equation.

$\begin{array}{l}0={\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{v_{in}^2}{2}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{v_{out}^2}{2}\right)+\dot{Q}\\ -\left({\widehat{H}}_{in}+\frac{v_{in}^2}{2}\right)+\left({\widehat{H}}_{out}+\frac{v_{out}^2}{2}\right)=\frac{\dot{Q}}{\dot{m}}\end{array}$

(C)

Interpretation Introduction

Interpretation:

The common assumption of the adiabatic nozzle which is the inlet velocity is negligible. Whether this assumption is consistent or not for the calculated answer in this case.