Chapter 2, Problem 2.48P

(a)

Interpretation Introduction

Interpretation:

To determine an expression for mass flow rate of water $\stackrel{•}{m}$ flowing through the nozzle with inlet pressure $P_1$, inlet diameter $D_1$ and outlet ambient pressure $P_2$, outlet diameter $D_2$ with conditions steady state flow, incompressible fluid, isothermal process and adiabatic process

Concept Introduction:

Since water is flowing through nozzle and mass of water is coming and going, so it must be an open system. For an open system steady state flow, the energy balance is therefore,

  $\stackrel{•}{Q}+\stackrel{•}{W}=\Delta [(H+\frac{1}{2}{u}^2+gz)\stackrel{•}{m}]+\frac{d(nU)}{dt}$

Where

$\stackrel{•}{Q}$ is rate of heat transfer throughout the system, $\stackrel{•}{W}$ is work rate and all the terms in the above equation have the dimension energy/mass.

And the relation between mass flow rate $\stackrel{•}{m}$ and velocity $u$ is

  $u=\frac{\stackrel{•}{m}}{A\rho }=\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D^2}$

Here, Area $A$ is cross sectional area of nozzle inlet and outlet with diameter $D_1$ and $D_2$

Answer to Problem 2.48P

The mass flow rate of water $\stackrel{•}{m}$ flowing through the nozzle is

  $\begin{array}{l}\stackrel{•}{m}={[}^2\\ \end{array}$

Explanation of Solution

Given information:

It is given that water is flowing with conditions steady state flows, incompressible fluid, isothermal process and adiabatic process and for water at constant temperature.

  $H_2-H_1=\raisebox{1ex}{$(P_2-P_1)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.$

The inlet pressure $P_1$, inlet diameter $D_1$ and outlet ambient pressure $P_2$, outlet diameter $D_2$ is also given.

For steady state flow;

  $\frac{d(nU)}{dt}=0$

So, the resultant energy balance will be

  $\stackrel{•}{Q}+\stackrel{•}{W}=\Delta [(H+\frac{1}{2}{u}^2+gz)\stackrel{•}{m}]$

After rearranging,

$Q+W=\Delta H+\frac{\Delta u^2}{2}+g\Delta z$............(1)

Where $Q$ is heat transfer through the system and $W$ is workdone by shaft and all the terms have dimension equal to energy.

For adiabatic and no work shaft process and incompressible flow, the heat transfer, work shaft and potential energy term in equation (1) shall be zero.

Hence

$\Delta H+\frac{\Delta u^2}{2}=0$........(2)

Put $H_2-H_1=\raisebox{1ex}{$(P_2-P_1)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.=\Delta H$ which is given and $u=\frac{\stackrel{•}{m}}{A\rho }=\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D^2}$ in equation (2)

  $\raisebox{1ex}{$(P_2-P_1)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.+\frac{{(\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D_2{}^2})}^2-\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D_1{}^2}{)}^2}{2}=0$

Rearranging

  $\begin{array}{l}\raisebox{1ex}{$(P_2-P_1)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.+\frac{{(\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D_2{}^2})}^2-{(\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D_1{}^2})}^2}{2}=0\\ {\stackrel{•}{m}}^2{(}^{\frac{4}{\rho \pi }}(\frac{1}{D_2{}^4}-\frac{1}{D_1{}^4})=\raisebox{1ex}{$2(P_1-P_2)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.\\ {\stackrel{•}{m}}^2{(}^{\frac{4}{\rho \pi }}(\frac{1}{D_2{}^4}-\frac{1}{D_1{}^4})=\raisebox{1ex}{$2(P_1-P_2)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.\\ {\stackrel{•}{m}}^2(\frac{1}{D_2{}^4}-\frac{1}{D_1{}^4})=\raisebox{1ex}{$2(P_1-P_2)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.{(}^{\frac{\pi }{4}}{\rho }^2\end{array}$

Rearranging and Hence the final solution will be

  $\begin{array}{l}\stackrel{•}{m}={[}^2\\ \end{array}$

(b)

Interpretation Introduction

Interpretation:

To determine the effect of temperature change on an expression for mass flow rate of water $\stackrel{•}{m}$ flowing through the nozzle with inlet pressure $P_1$, inlet diameter $D_1$ and outlet ambient pressure $P_2$, outlet diameter $D_2$ with conditions steady state flow, incompressible fluid, non isothermal process with $T_2>T_1$ owing to fluid friction and adiabatic process

Concept Introduction:

Since water is flowing through nozzle and mass of water is coming and going, so it must be an open system. For an open system steady state flow, the energy balance is therefore,

  $\stackrel{•}{Q}+\stackrel{•}{W}=\Delta [(H+\frac{1}{2}{u}^2+gz)\stackrel{•}{m}]+\frac{d(nU)}{dt}$

Where

$\stackrel{•}{Q}$ is rate of heat transfer throughout the system, $\stackrel{•}{W}$ is work rate and all the terms in the above equation have the dimension energy/mass.

And the relation between mass flow rate $\stackrel{•}{m}$ and velocity $u$ is

  $u=\frac{\stackrel{•}{m}}{A\rho }=\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D^2}$

Here, Area $A$ is cross sectional area of nozzle inlet and outlet with diameter $D_1$ and $D_2$

Answer to Problem 2.48P

The mass flow rate of water $\stackrel{•}{m}$ flowing through the nozzle is

  $\begin{array}{l}\stackrel{•}{m}={[}^2\\ \end{array}$

Explanation of Solution

Given information:

It is given that water is flowing with conditions steady state flows, incompressible fluid, non thermal process with $T_2>T_1$ owing to fluid friction and adiabatic process:

  $H_2-H_1=C(T_2-T_1)+\raisebox{1ex}{$(P_2-P_1)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.$

Here $C$ is the specific heat of water.

The inlet pressure $P_1$, inlet diameter $D_1$ and outlet ambient pressure $P_2$, outlet diameter $D_2$ is also given.

For steady state flow;

  $\frac{d(nU)}{dt}=0$

So, the resultant energy balance will be

  $\stackrel{•}{Q}+\stackrel{•}{W}=\Delta [(H+\frac{1}{2}{u}^2+gz)\stackrel{•}{m}]$

After rearranging,

$Q+W=\Delta H+\frac{\Delta u^2}{2}+g\Delta z$............(1)

Where, $Q$ is heat transfer through the system and $W$ is work done by shaft and all the terms have dimension equal to energy.

For adiabatic and no work shaft process and incompressible flow, the heat transfer, work shaft and potential energy term in equation (1) shall be zero.

Hence

$\Delta H+\frac{\Delta u^2}{2}=0$........(2)

Put $H_2-H_1=C(T_2-T_1)+\raisebox{1ex}{$(P_2-P_1)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.$ which is given and $u=\frac{\stackrel{•}{m}}{A\rho }=\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D^2}$ in equation (2)

  $C(T_2-T_1)+\raisebox{1ex}{$(P_2-P_1)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.+\frac{{(\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D_2{}^2})}^2-\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D_1{}^2}{)}^2}{2}=0$

Rearranging

  $\begin{array}{l}C(T_2-T_1)+\raisebox{1ex}{$(P_2-P_1)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.+\frac{{(\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D_2{}^2})}^2-{(\frac{4}{\pi }\frac{\stackrel{•}{m}}{\rho D_1{}^2})}^2}{2}=0\\ {\stackrel{•}{m}}^2{(}^{\frac{4}{\rho \pi }}(\frac{1}{D_2{}^4}-\frac{1}{D_1{}^4})=-2[C(T_2-T_1)+\raisebox{1ex}{$(P_2-P_1)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.]\\ {\stackrel{•}{m}}^2(\frac{1}{D_2{}^4}-\frac{1}{D_1{}^4})=-2[C(T_2-T_1)+\raisebox{1ex}{$(P_2-P_1)$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.]{(}^{\frac{\pi }{4}}{\rho }^2\end{array}$

Rearranging and Hence the final solution will be

  $\begin{array}{l}\stackrel{•}{m}={[}^2\\ \end{array}$