Fluid Mechanics for Mechanical Engineers/Integral Analysis of Fluid Flow

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Introduction

Differential Approach seek solution at every point $(x_1,x_2,x_3)$, i.e describe the detailed flow pattern at all points.

Integral approach for a Control Volume (CV) is interested in a finite region and it determines gross flow effects such as force or torque on a body or the total energy exchange. For this purpose, balances of incoming and outgoing flow rate mass, momentum and energy are made through this finite region. It gives very fast engineering answers, sometimes crude but useful.

File:Fluid Dynamics 1.png

Flow over an airfoil: Lagrangian vs Eulerian approach and differential vs integral approach.

System versus Control volume

In mechanics, system is a collection of matter of fixed identity (always the same atoms or fluid particles) which may move, flow and interact with its surroundings.

Hence, the mass is constant for a system, although it may continually change size and shape. This approach is very useful in statics and dynamics, in which the system can be isolated from its surrounding and its interaction with the surrounding can be analysed by using a free-body diagram.

In fluid dynamics, it is very hard to identify and follow a specific quantity of the fluid. Imagine a river and you have to follow a specific mass of water along the river.

Mostly, we are rather interested in determining forces on surfaces, for example on the surfaces of airplanes and cars. Hence, instead of system approach, we identify a specific volume in space (associated with our geometry of interest) and analyse the flow within, through or around this volume. This specific volume is called "Control Volume". This control volume can be fixed, moving or even deforming.

The control volume is a specific geometric entity independent of the flowing fluid. The matter within a control volume may change with time, and the mass may not remain constant.

File:Different type control volume 03.svg

Example of different types of control volume. A)Fixed CV: Flow through a pipe. B)Moving CV: Flow through a jet engine of a flying aircraft C)Deforming CV: Flow from a deflating balloon

Basic laws for a system

Conservation of mass

The mass of a system does not change:

Template:Center top $\frac{dM}{dt}{|}_{system}=0$ Template:Center bottom

where, $M=\underset{mass(system)}{\int }dm=\underset{V(system)}{\int }\rho dV$

Newton's second law

Linear Momentum Equation

For a system moving relative to a inertial reference frame, the sum of all external forces acting on the system is equal to the time rate of change of linear momentum ($\overrightarrow{P}$) of the system:

Template:Center top ${\overrightarrow{F}}_{onsystem}=\frac{d\overrightarrow{P}}{dt}{|}_{system}$ or in tensor form $F_i=\frac{d{P}_i}{dt}{|}_{system}$ Template:Center bottom where Template:Center top $P_{i(system)}=\underset{mass(system)}{\int }{U}_{i}dm=\underset{V(system)}{\int }{U}_i\rho dV$ Template:Center bottomandTemplate:Center top $\overrightarrow{U}=U_x\overrightarrow{i}+U_y\overrightarrow{j}+U_z\overrightarrow{k}=U_1{\overrightarrow{e}}_1+U_2{\overrightarrow{e}}_2+U_3{\overrightarrow{e}}_3$ Template:Center bottomwhere ${\overrightarrow{e}}_1,{\overrightarrow{e}}_2\text{and}{\overrightarrow{e}}_3$ are the corresponding unit vectors along positive $x_1,x_2\text{and}{x}_3$ axis, i.e. $x,y\text{and}z$ axis. This kind of subscript notation is a part of tensor notation which is explained at Fluid Mechanics for Mechanical Engineers/Scalar, Vectors and Tensors chapter.

Moment-of-Momentum Equation

For system rotating about an axis with an angular speed of $\overrightarrow{\omega }$, the sum of torque created by all external forces about the axis of rotation, is equal to the time rate of change of angular momentum ($\overrightarrow{H}$) of the system:

Template:Center top ${\overrightarrow{T}}_{onsystem}=\frac{d\overrightarrow{H}}{dt}{|}_{system}$ or in tensor form $T_{ionsystem}=\frac{d{H}_i}{dt}{|}_{system}$ Template:Center bottom where Template:Center top $\overrightarrow{H}=\underset{mass(system)}{\int }\overrightarrow{r}\times \overrightarrow{U}dm$ and $\overrightarrow{T}=\overrightarrow{r}\times \overrightarrow{F}+{\overrightarrow{T}}_{shaft}$ Template:Center bottom

The First law of Thermodynamics

Template:Center top $\underset{heataddedonsystem}{\underbrace{dQ}}+\underset{workdoneonsystem}{\underbrace{dW}}=d{E}_{system}$ Template:Center bottom

in the rate form:

Template:Center top $\dot{Q}+\dot{W}=\frac{dE}{dt}{|}_{system}$ Template:Center bottom

where $E_{system}=\underset{mass(system)}{\int }edm=\underset{V(system)}{\int }e\rho dV$ and $e=\underset{Intenralenergy}{\underbrace{u}}+\underset{Kineticenergy}{\underbrace{\frac{U_i{U}_i}{2}}}+\underset{Potentialenergy}{\underbrace{gz}}$

The Second law of Thermodynamics

The second law states that if heat is added to a system at a temperature $T$, the entropy of the system rises more than the heat added per unit temperature:

Template:Center top $d{S}_{system}\ge \frac{dQ}{T}{|}_{system}$ Template:Center bottom

The rate form of this law is:

Template:Center top $\frac{dS}{dt}{|}_{system}\ge \frac{1}{T}{\dot{Q}}_{system}$ Template:Center bottom

where $S_{system}=\underset{mass(system)}{\int }sdm=\underset{V(system)}{\int }s\rho dV$

Relation of a system derivative to the control volume derivative

Consider a fire extinguisher

File:Fire extinguisher final 01.svg

Control Volume versus System

$\frac{dM}{dt}{|}_{system}=0$ whereas $\frac{dM}{dt}{|}_{cv}<0$

All basic laws are written for a system, i.e defined mass with fixed identity. We should rephrase these laws for a control volume. In other words, we would like to relate

$\frac{dB}{dt}{|}_{system}$ to $\frac{dB}{dt}{|}_{cv}$

The variables appear in the physical laws (balance laws) of a system are:

• Mass ($M$),
• Momentum ($P_i$),
• Energy ($E$),
• Moment of momentum ($H_i$),
• Entropy ($S$).

They are called extensive properties. Let $B$ be any arbitrary extensive property. The corresponding intensive property $b$ is the extensive property per unit mass:

Template:Center top $B_{system}=\underset{mass(system)}{\int }bdm=\underset{V(system)}{\int }b\rho dV$ Template:Center bottom

Hence,

Template:Center top $B=M,b=1$

$B=\overrightarrow{P},b=\overrightarrow{U}$

$B=\overrightarrow{H},b=\overrightarrow{r}\times \overrightarrow{U}$

$B=E,b=e$

$B=S,b=s$ Template:Center bottom

One dimensional Reynolds Transport Theorem

Consider a flow through a nozzle.

File:1D-RTT.svg

Flow through a nozzle used to derive the 1-D Reynolds transport theorem

If $B$ is an extensive variable of the system.

Template:Center top $B_{sys}(t)=B_{cv}(t)$ Template:Center bottom

Template:Center top $B_{sys}(t+\mathrm{\Delta }t)=B_{cv}(t+\mathrm{\Delta }t)-B_I(t+\mathrm{\Delta }t)+B_{II}(t+\mathrm{\Delta }t)$ Template:Center bottom

Template:Center top $\frac{\mathrm{\Delta }{B}_{sys}}{\mathrm{\Delta }t}=\frac{B_{sys}(t+\mathrm{\Delta }t)-B_{sys}(t)}{\mathrm{\Delta }t}=\frac{B_{cv}(t+\mathrm{\Delta }t)-B_{cv}(t)}{\mathrm{\Delta }t}-\frac{B_I(t+\mathrm{\Delta }t)}{\mathrm{\Delta }t}+\frac{B_{II}(t+\mathrm{\Delta }t)}{\mathrm{\Delta }t}$ Template:Center bottom

The first term for $\mathrm{\Delta }t\to 0$

Template:Center top $\underset{\mathrm{\Delta }t\to 0}{\lim }\frac{B_{cv}(t+\mathrm{\Delta }t)-B_{cv}(t)}{\mathrm{\Delta }t}=\frac{\partial B_{cv}}{\partial t}$ Template:Center bottom

$B_{II}(t+\mathrm{\Delta }t)$ for $\mathrm{\Delta }t\to 0$

Template:Center top $B_{II}(t+\mathrm{\Delta }t)={\rho }_2{b}_2{V}_{II}$ Template:Center bottom

Template:Center top $B_{II}(t+\mathrm{\Delta }t)={\rho }_2{b}_2{A}_2{l}_2={\rho }_2{b}_2{A}_2{U}_2\mathrm{\Delta }t$ Template:Center bottom

Template:Center top $B_{out}={\rho }_2{b}_2{A}_2{U}_2\mathrm{\Delta }t$ Template:Center bottom

Similarly

Template:Center top $B_I(t+\mathrm{\Delta }t)={\rho }_1{b}_1{V}_I={\rho }_1{b}_1{A}_1{U}_1\mathrm{\Delta }t=B_{in}$ Template:Center bottom

Thus, for $\mathrm{\Delta }t\to 0$, the terms in the equality for the time derivative of the system are

Template:Center top $\frac{\mathrm{\Delta }{B}_{sys}}{\mathrm{\Delta }t}\to \frac{d{B}_{sys}}{dt}$ Template:Center bottom

Template:Center top $\frac{B_{in}}{\mathrm{\Delta }t}={\dot{B}}_{in}$ Template:Center bottom

Template:Center top $\frac{B_{out}}{\mathrm{\Delta }t}={\dot{B}}_{out}$ Template:Center bottom

so that,

Template:Center top $\frac{d{B}_{sys}}{dt}=\frac{\partial B_{cv}}{\partial t}+{\dot{B}}_{out}-{\dot{B}}_{in}$ Template:Center bottom

This is the equation of 1 dimensional Reynolds transport theorem (RTT).

The three terms on the RHS of RTT are:

1. The rate of change of B within CV, which indicates also the local unsteady effect.

2. The flow rate of B flowing out of the CS.

3. The flow rate of B flowing into the CS.

There can be more than one inlet and outlet.

File:CV-multi-inlet-outlet.svg

CV with multiple inlets and outlets

Three Dimensional Reynolds Transport Theorem

Hence, for a quite complex, unsteady, three dimensional situation, we need a more general form of RTT. Consider an arbitrary 3-D CV and the outward unit normal vector $(\overrightarrow{n})$ defined at each point on the CS. The outflow and inflow flow rate of $B$ across CS can be written as:

File:Fluid Dynamics 13.png

CV with arbitrary shape used to derive Reynolds transport theorem

Template:Center top ${\dot{B}}_{out}=\underset{CSout}{\int }d{\dot{B}}_{out}=\underset{CSout}{\int }\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

Template:Center top ${\dot{B}}_{in}=\underset{CSin}{\int }d{\dot{B}}_{in}=-\underset{CSin}{\int }\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

File:Flux-sign-through-CS.png

Inflow and outflow to the CV.

${\dot{B}}_{in}$ and ${\dot{B}}_{out}$ are positive quantities. Therefore, the negative sign is introduced into ${\dot{B}}_{in}$, to compensate the negative value of $\overrightarrow{U}\cdot \overrightarrow{n}$.

Template:Center top $\frac{d{B}_{sys}}{dt}=\frac{\partial B_{cv}}{\partial t}+\underset{CSout}{\int }\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA+\underset{CSin}{\int }\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

Template:Center top $\frac{d{B}_{sys}}{dt}=\frac{\partial }{\partial t}\underset{cv}{\int }\rho bdV+\underset{cs}{\int }\underset{fluxofB}{\underbrace{\rho b\overrightarrow{U}\cdot \overrightarrow{n}}}dA$ Template:Center bottom

Since $\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=d\dot{m}$, RTT can be written as:

Template:Center top $\frac{d{B}_{sys}}{dt}=\frac{\partial B_{cv}}{\partial t}+\underset{netflowrateofBacrossCS}{\underbrace{\underset{CS}{\int }bd\dot{m}}}$ Template:Center bottom

Conservation of mass

$B=M,b=1$

Template:Center top $\frac{d{M}_{sys}}{dt}=\frac{\partial }{\partial t}\underset{cv}{\int }1\rho dV+\underset{cs}{\int }1\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=0$ Template:Center bottom

i.e

Template:Center top $\underset{rateofchangeofmassinCV}{\underbrace{\frac{\partial }{\partial t}\underset{cv}{\int }\rho dV}}+\underset{netmassflowratethroughtheCS}{\underbrace{\underset{cs}{\int }\underset{massflux}{\underbrace{\rho \overrightarrow{U}\cdot \overrightarrow{n}}}dA}}=0$ Template:Center bottom

Assume $\rho$ = constant (incompressible) and uniform in the CV

Template:Center top $0=\rho \frac{\partial }{\partial t}\underset{cv}{\int }dV+\rho \underset{cs}{\int }\overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

As $V$ of CV is also constant, the time derivative drops out:

Template:Center top $0=\rho \underset{cs}{\int }\overrightarrow{U}\cdot \overrightarrow{n}dA\to \underset{volumeflowrate}{\underbrace{\underset{cs}{\int }\overrightarrow{U}\cdot \overrightarrow{n}dA}}=0$ Template:Center bottom

The net volume flow rate should be zero through the control surfaces.

Note that we did not assume a steady flow. This equation is valid for both steady and unsteady flows.

If the flow is steady,

Template:Center top $0=\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

there is no-mass accumulation or deficit in the control volume, i.e. net  mass  flow rate is equal to zero.

Linear Momentum equation for inertial control volume (Newton's 2nd law of motion)

$B=\overrightarrow{P}$, $b=\overrightarrow{U}$Template:Center top $\frac{d{\overrightarrow{P}}_{sys}}{dt}={\overrightarrow{F}}_{onsystem}$ Template:Center bottomUsing RTT, the rate of change of momentum of the system can be related to that of CV as follows:Template:Center top $\frac{d{\overrightarrow{P}}_{sys}}{dt}=\frac{\partial }{\partial t}\underset{cv}{\int }\overrightarrow{U}\rho dV+\underset{netmomentumflowrateacrossCS}{\underbrace{\underset{cs}{\int }\underset{momentumflux}{\underbrace{\overrightarrow{U}\rho \overrightarrow{U}\cdot \overrightarrow{n}}}dA}}={\overrightarrow{F}}_{onsystem}$ Template:Center bottom

This equation states that the sum of all forces acting on a non-accelerating CV is equal to the sum of the net rate of change of momentum inside the CV and the net rate of momentum flux through the CS.

Force on the system is the sum of surface forces and body forces.

Template:Center top ${\overrightarrow{F}}_{onsystem}=\overrightarrow{F_S}+\overrightarrow{F_B}$ Template:Center bottom

The surface forces are mainly due to pressure, which is normal to the surface, and viscous stresses, which can be both normal or tangential to the surfaces.

Template:Center top ${\overrightarrow{F}}_{pressure}=-\underset{A}{\int }P\overrightarrow{n}dA$ Template:Center bottom

Template:Center top ${\overrightarrow{F}}_{v.stresses}=\underset{A}{\int }\tau dA$ Template:Center bottom

The body forces can be due to gravity or magnetic field.

At the initial moment $t$

Template:Center top ${\overrightarrow{F}}_{onsystem}={\overrightarrow{F}}_{onCV}$ Template:Center bottom

i.e. in the component form:

Template:Center top $F_{Si}+F_{Bi}=\frac{\partial }{\partial t}\underset{cv}{\int }{U}_i\rho dV+\underset{cs}{\int }{U}_i\rho U_j{n}_{j}dA$ Template:Center bottom

Moment-of-momentum Equation

$B=\overrightarrow{H}$, $b=\overrightarrow{r}\times \overrightarrow{U}$

Net torque exerted on the system about the rotaional axis is equal to the rate of change of angular momentum of the system. Template:Center top $\sum T_i=\frac{d{H}_i}{dt}{|}_{system}$ Template:Center bottom

expanding both sides by using the RTT

Template:Center top ${\overrightarrow{T}}_{shaft}+\sum \overrightarrow{r}\times {\overrightarrow{F}}_s+\underset{CV}{\int }\overrightarrow{r}\times \overrightarrow{g}\rho dV=\frac{\partial }{\partial t}\underset{cv}{\int }\overrightarrow{r}\times \overrightarrow{U}\rho dV+\underset{cs}{\int }\overrightarrow{r}\times \overrightarrow{U}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

where ${\overrightarrow{F}}_s$ is the surface force acting on the surfaces of the CV.

First Law of Thermodynamics

$B=E,b=e$

Template:Center top ${\frac{dE}{dt}}_{system}=\frac{\partial }{\partial t}\underset{cv}{\int }e\rho dV+\underset{cs}{\int }e\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

at the initial moment $t$, the following equality is valid:

Template:Center top ${\frac{dE}{dt}}_{system}=[\dot{Q}+\dot{W}{]}_{system}=[\dot{Q}+\dot{W}{]}_{cv}$ Template:Center bottom

thus, the integral form of energy equation is:

Template:Center top $[\dot{Q}+\dot{W}{]}_{cv}=\frac{\partial }{\partial t}\underset{cv}{\int }e\rho dV+\underset{cs}{\int }e\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

Second Law of Thermodynamics

$B=S,b=s$

Template:Center top $\frac{dS}{dt}{|}_{system}\ge \frac{1}{T}{\dot{Q}}_{system}$ Template:Center bottom The left hand side of the inequality reads Template:Center top $\frac{dS}{dt}{|}_{system}=\frac{\partial }{\partial t}\underset{cv}{\int }s\rho dV+\underset{cs}{\int }s\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

The right hand side of the above inequality reads

Template:Center top $\frac{1}{T}{\dot{Q}}_{system}=\frac{1}{T}{\dot{Q}}_{CV}=\underset{CS}{\int }\frac{1}{T}\left(\frac{\dot{Q}}{A}\right)dA$ Template:Center bottom

Hence the second law becomes:

Template:Center top $\frac{\partial }{\partial t}\underset{cv}{\int }s\rho dV+\underset{cs}{\int }s\rho \overrightarrow{U}\cdot \overrightarrow{n}dA\ge \underset{CS}{\int }\frac{1}{T}\left(\frac{\dot{Q}}{A}\right)dA$ Template:Center bottom

Note that $\left(\frac{\dot{Q}}{A}\right)$ represents the local heat flux through the surfaces of the control volume.

Examples

Example 1: Conservation of mass for a stream tube

Consider the mass balance in a stream tube in a steady incompressible flow, by using the integral form of the conservation of mass equation.

File:Mass Balance for streamtube renew 02.svg

Mass balance for stream tube inside a laminar flow (example 1)

Let $A_1$ and $A_2$ be too small such that the velocities $\overrightarrow{U}$ at position 1 and 2 are uniform across $A_1$ and $A_2$.

When we write the conservation of mass equation in integral form:

Template:Center top $\frac{\partial }{\partial t}\underset{cv}{\int }\rho dV+\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=0$ Template:Center bottom

We assume that;

1) In a steady flow volume of the stream tube, i.e. CV is constant, ( $\frac{\partial }{\partial t}\underset{cv}{\int }dV$ = 0)

2) Density is constant and uniform in the CV (${\rho }_1={\rho }_2=\rho :constant$).

Hence

$\frac{\partial }{\partial t}\underset{cv}{\int }\rho dV=0$

When we rewrite the conservation of mass equation:

$\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=0$

The first term is zero and the second term can be analyzed by decomposing the integration area. There are 3 types of the control surfaces :

$C{S}_I$ ,$C{S}_{II}$ and $C{S}_{III}$.

• $C{S}_I$ : the surface area that the mass enters across.

$U_1=\left|\overrightarrow{U}\right|\left|\overrightarrow{n}\right|\cos \theta$

$\overrightarrow{n}$ and $\overrightarrow{U}$ are in the opposite directions. $90^{\circ }\le \theta \le 180^{\circ }$and $\cos \theta$ is always negative. That is, the sign of mass input is always negative.

• $C{S}_{II}$ : the surface area from which the mass leaves.

$U_2=\left|\overrightarrow{U}\right|\left|\overrightarrow{n}\right|\cos \theta$

$\overrightarrow{n}$ and $\overrightarrow{U}$ are in the same directions. $\theta \le 90^{\circ }$and $\cos \theta$ is always positive. That is, the sign of mass output is always positive.

• $C{S}_{III}$ : the lateral surfaces of the CV.

$U_3=\left|\overrightarrow{U}\right|\left|\overrightarrow{n}\right|\cos \theta$

$\overrightarrow{n}$ and $\overrightarrow{U}$ are perpendicular to each other . $\theta =90^{\circ }$and $\cos \theta$ is zero. That is, there is no flows.

Template:Center top $\underset{C{S}_I}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}d{A}_1+\underset{C{S}_{III}}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}d{A}_3+\underset{C{S}_{II}}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}d{A}_2=0$ Template:Center bottom

where the integration over $C{S}_{III}$ is zero, because there is no flow across the streamtube. Thus,

Template:Center top $-\rho U_1{A}_1+0+\rho U_2{A}_2=0$ Template:Center bottom

Template:Center top $\rho U_1{A}_1=\rho U_2{A}_2\to U_2=\frac{U_1{A}_1}{A_2}$ Template:Center bottom

Template:Center top ${\dot{m}}_1={\dot{m}}_2$ Template:Center bottom

Example 2: Conservation of mass with multiple inlets and outlets

Consider the steady flow of water through the device. The inlet and outlet areas are $A_1$, $A_2$ and $A_3$ = $A_4$.

File:Junction renew 01.svg

mass balance for a connector device (example 2)

The following parameters are known:

Mass flow out at 3 (${\dot{m}}_3$).

Volume flow rate in through 4 ($Q_4$).

Velocity at 1 along $x_1$-direction ${\overrightarrow{U}}_{11}$, ${\overrightarrow{U}}_{1i}=(U_{11},0,0)$ so that $U_{11}>0$.

Find the flow velocity at section 2.Assume that the properties are uniform across the sections.

Template:Center top $\frac{\partial }{\partial t}\underset{cv}{\int }\rho dV+\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=0$ Template:Center bottom

Where the first term is zero due to steady state conditions. At section 1:

$\underset{A_1}{\int }\rho {\overrightarrow{U}}_1\cdot {\overrightarrow{n}}_{1}dA=-\rho \left|U_{11}\right|A_1$ At section 3:

$\underset{A_3}{\int }\rho {\overrightarrow{U}}_3\cdot {\overrightarrow{n}}_{3}dA=\rho \left|U_3\right|A_3={\dot{m}}_3$

At section 4:

$\underset{A_4}{\int }\rho {\overrightarrow{U}}_4\cdot {\overrightarrow{n}}_{4}dA=-\rho \left|U_4\right|A_4=-\rho Q_4$

Template:Center top $\underset{A_1}{\int }\rho {\overrightarrow{U}}_1\cdot {\overrightarrow{n}}_{1}dA+\underset{A_2}{\int }\rho {\overrightarrow{U}}_2\cdot {\overrightarrow{n}}_{2}dA+\underset{A_3}{\int }\rho {\overrightarrow{U}}_3\cdot {\overrightarrow{n}}_{3}dA+\underset{A_4}{\int }\rho {\overrightarrow{U}}_4\cdot {\overrightarrow{n}}_{4}dA=0$ Template:Center bottom

Hence, the velocity at section 2 can be calculated by

$\underset{A_2}{\int }\rho {\overrightarrow{U}}_2\cdot {\overrightarrow{n}}_{2}dA=-[-\rho \left|U_{11}\right|A_1+{\dot{m}}_3-\rho Q_4]$

For ${\overrightarrow{n}}_2=(0,-1)\to U_{21}=0$

$-\rho U_{22}{A}_2=\rho \left|U_{11}\right|A_1-{\dot{m}}_3+\rho Q_4$

If the term on the right side is positive, $U_2$ should be negative (outflow) and if it is negative, $U_2$ should be positive (inflow).

Example 3: Momentum equation for a stream tube

Consider the steady flow through a stream tube. The velocity and density are uniform at the inlet and outlet of the fixed CV. Find an expression for the net force on the control volume.

File:Force calculation fluid element renew 01.svg

Force in a streamtube (example 3)

Template:Center top $F_i=F_{Si}+F_{Bi}=\frac{\partial }{\partial t}\underset{cv}{\int }{U}_i{\rho }_{1}dV+\underset{cs}{\int }{U}_i{\rho }_2{U}_j{n}_{j}dA$ Template:Center bottom

where the derivative with respect to time is zero due to steady state conditions. ($\frac{\partial }{\partial t}\underset{cv}{\int }{U}_i\rho dV$=0 ).

There are 3 types of the control surfaces :

$C{S}_I$ ,$C{S}_{II}$ and $C{S}_{III}$.

• $C{S}_I$ : the surface area that the mass enters across.

$U_1=\left|\overrightarrow{U}\right|\left|\overrightarrow{n}\right|\cos \theta$ (general formula)

$\overrightarrow{n}$ and $\overrightarrow{U}$ are in the opposite directions. $90^{\circ }\le \theta \le 180^{\circ }$and $\cos \theta$ is always negative. That is, the sign of mass input is always negative.

• $C{S}_{II}$ : the surface area from which the mass leaves.

$U_2=\left|\overrightarrow{U}\right|\left|\overrightarrow{n}\right|\cos \theta$ (general formula)

$\overrightarrow{n}$ and $\overrightarrow{U}$ are in the same directions. $\theta \le 90^{\circ }$and $\cos \theta$ is always positive. That is, the sign of mass output is always positive.

• $C{S}_{III}$ : the lateral surfaces of the CV.

$U_3=\left|\overrightarrow{U}\right|\left|\overrightarrow{n}\right|\cos \theta$ (general formula)

$\overrightarrow{n}$ and $\overrightarrow{U}$ are perpendicular to each other . $\theta =90^{\circ }$and $\cos \theta$ is zero. That is, there is no flows.($\underset{C{S}_{III}}{\int }{U}_{3i}{\rho }_3{U}_{3j}{n}_{j}d{A}_{3,lateral}$ =0 )

Template:Center top $F_i=\underset{C{S}_I}{\int }{U}_{1i}{\rho }_1{U}_{1j}{n}_{j}d{A}_1+\underset{C{S}_{II}}{\int }{U}_{2i}{\rho }_2{U}_{2j}{n}_{j}d{A}_2+\underset{C{S}_{III}}{\int }{U}_{3i}{\rho }_3{U}_{3j}{n}_{j}d{A}_{3,lateral}$ Template:Center bottom

$F_i=\underset{C{S}_I}{\int }{U}_{1i}{\rho }_1{U}_{1j}{n}_{j}d{A}_1+\underset{C{S}_{II}}{\int }{U}_{2i}{\rho }_2{U}_{2j}{n}_{j}d{A}_2$

Density is constant (${\rho }_1={\rho }_2={\rho }_3=\rho =constant$)

$F_i=\rho \underset{C{S}_I}{\int }{U}_{1i}{U}_{1j}{n}_{j}d{A}_1+\rho \underset{C{S}_{II}}{\int }{U}_{2i}{U}_{2j}{n}_{j}d{A}_2$

When we integrate Template:Center top $F_i=-U_{1i}\underset{{\dot{m}}_1}{\underbrace{\left|{\overrightarrow{U}}_1\right|A_1\rho }}+U_{2i}\underset{{\dot{m}}_2}{\underbrace{\left|{\overrightarrow{U}}_2\right|A_2\rho }}$ Template:Center bottom

$\frac{dm}{dt}-{\dot{m}}_1+{\dot{m}}_2=0$ (steady state :$\frac{dm}{dt}=0$) Template:Center top ${\dot{m}}_1={\dot{m}}_2=\dot{m}$ Template:Center bottom

Template:Center top $F_i=\dot{m}(U_{2i}-U_{1i})$ Template:Center bottom

Example 4: Momentum equation

Water from a two dimensional stationary nozzle strikes to a plate. Assume that the flow is normal to the plate and in the jet velocity is steady and uniform. Determine the force on the plate in $x_1$ direction.

File:Nozzle test 2.svg

Jet striking to a plate (example 4)

Control volume I and II and Free body diagram of the plate (example 4)

Independent from the selected CV.

Template:Center top $0=\underset{cs}{\int }\rho U_i{n}_{i}dA(mass)$ Template:Center bottom

Template:Center top $F_i=F_{Si}+F_{Bi}=\underset{cs}{\int }{U}_i\rho U_j{n}_{j}dA(momentum)$ Template:Center bottom

No body force in $x_1$ direction.

Template:Center top $F_1=F_{S1}=\underset{cs}{\int }{U}_1\rho U_j{n}_{j}dA$ Template:Center bottomFor the $C{V}_I$Template:Center top $F_{S1}=p_{a}A-p_{a}A+R_{s1}$ Template:Center bottom

$R_{s1}=\underset{cs}{\int }{U}_1\rho U_j{n}_{j}dA=\underset{CS1}{\int }{U}_1^1\rho U_j^1{n}_{j}dA+\underset{0}{\underbrace{\underset{CS2}{\int }{U}_1^2\rho U_j^2{n}_{j}dA}}+\underset{0}{\underbrace{\underset{CS3}{\int }{U}_1^3\rho U_j^3{n}_{j}dA}}$

$U_1=0$ at 2 and 3 and $U_1^1=U_{jet}$ at 1.

Template:Center top $R_{s1}=-U_{jet}\rho U_{jet}{A}_{Jet}$ Template:Center bottom

The force which acts on the plate (action-reaction) is

$K_1=-R_{s1}=U_1\rho U_1{A}_{Jet}$

It is also possible to solve the problem with $C{V}_{II}$

Template:Center top $F_{S1}=p_{a}A+R_{p1}=-U_{jet}^2{A}_{jet}\rho$ Template:Center bottom

Template:Center top $R_{p1}=-p_{a}A-U_{jet}^2{A}_{Jet}\rho$ Template:Center bottom

Hence, the force exerted on the plate by the CV is

Template:Center top $K_1=-R_{p1}$ Template:Center bottomIn order to calculate the force exerted by the support on the plate to hold the plate, we need to look to the free body diagram of the plate:Template:Center top $F_{net}=0=K_1-p_{a}A+R{s}_1$ Template:Center bottom

Template:Center top $F_{net}=0=p_{a}A+U_{jet}^{2}A\rho -p_{a}A+R{s}_1$ Template:Center bottom

Template:Center top $R{s}_1=-U_{jet}^{2}A\rho$ Template:Center bottomIt means, the force of the support should be in the negative direction (to the left).

Example 5: Finding drag on a flat plate

Consider the plate exposed to uniform velocity. The flow is steady and incompressible. A boundary layer builds up on the plate. Determine the Drag force on the plate. Note that $U_1$ can be approximated at L.

File:Fluid Dynamics 26.png

Velocity distribution of fluid over a plate (example 5)

Template:Center top $U_1(L,x_2)=U_0(2\frac{x_2}{\delta }-{\left(\frac{x_2}{\delta }\right)}^2)$ Template:Center bottom

Apply conservation of mass

Template:Center top $\frac{\partial }{\partial t}\underset{cv}{\int }\rho dV+\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=0$ Template:Center bottom

Assumptions:

• The flow is steady; ( $\frac{\partial }{\partial t}$ = 0).

• Density is constant ($\rho$ is constant).

• The flow is parallel(There is no flow in $x_2$ direction.

When we rewrite conservation of mass: Template:Center top $\rho \underset{cs}{\int }{U}_i{n}_{i}dA=0$ Template:Center bottom

As you know

Template:Center top $\frac{dm}{dt}-{\dot{m}}_{in}+{\dot{m}}_{out}=0$

$w$: the thickness of the plate.

$\dot{m}=\rho UA$

$dA=wd{x}_2$ Template:Center bottom

Template:Center top $\rho \underset{0}{\overset{h}{\int }}-U_{0}wd{x}_2+\rho \underset{0}{\overset{\delta }{\int }}Uwd{x}_2=0$ Template:Center bottom

Template:Center top $U_{0}h=\underset{0}{\overset{\delta }{\int }}{U}_{1}d{x}_2$ Template:Center bottom

When we use RTT for $\overrightarrow{F}=d\overrightarrow{P}/dt$ where $\overrightarrow{P}$ is the momentum vector; Template:Center top $F_i=\underset{=0steadystate}{\underbrace{\frac{\partial }{\partial t}\int U_i\rho dV}}+\underset{cs}{\int }{U}_i\rho U_j{n}_{j}dA$ Template:Center bottom

There are 4 control surfaces: $C{S}_I$ ,$C{S}_{II}$, $C{S}_{III}$ and $C{S}_{IV}$.

$C{S}_I$: the surface that the mass enters across.

$C{S}_{II}$: Lateral surfaces of the control volume, there is no flows.

$C{S}_{III}$: the surface from which mass leaves.

$C{S}_{IV}$: the wall of the plate. Because of no slip condition $U=0$ at the wall.

Template:Center top $F_1=-D=\underset{0}{\overset{h}{\int }}{U}_1\rho U_j{n}_{j}dA+\underset{=0streamline}{\underbrace{\underset{CS2}{\int }{U}_1\rho U_j{n}_{j}dA}}+\underset{0}{\overset{\delta }{\int }}{U}_1\rho U_j{n}_{j}dA+\underset{=0wall}{\underbrace{\int U_1\rho U_j{n}_{j}dA}}$ Template:Center bottom

Template:Center top $-D=-U_0\rho w{U}_{0}h+\underset{0}{\overset{\delta }{\int }}{U}_1\rho U_{1}wd{x}_2$ Template:Center bottom

Template:Center top $D=U_0\rho w{U}_{0}h-\rho w\underset{0}{\overset{\delta }{\int }}{U}_1^{2}d{x}_2$ Template:Center bottom

Insert the mass conservation result $U_{0}h=\underset{0}{\overset{\delta }{\int }}{U}_{1}d{x}_2$ into the momentum equation. Template:Center top $D=\rho w{U}_0\underset{0}{\overset{\delta }{\int }}{U}_{1}d{x}_2-\rho w\underset{0}{\overset{\delta }{\int }}{U}_1{U}_{1}d{x}_2$ Template:Center bottomHence at $x_1=L$Template:Center top $D=\rho w\underset{0}{\overset{\delta }{\int }}{U}_1(U_0-U_1)d{x}_2$ Template:Center bottom

$U_1$ is known. Here, using $\frac{x_2}{\delta }=\eta$ Template:Center top $D=\rho w{U}_0^2\delta \underset{0}{\overset{1}{\int }}(2\eta -{\eta }^2)(1-2\eta +{\eta }^2)d\eta =\frac{2}{15}\rho U_0^{2}w\delta$ Template:Center bottom

RTT for a CV moving with constant speed

It is possible that CV can move with constant velocity or arbitrary acceleration.

File:Relative-absolute-velocity-vector-relation.png

Relation between the absolute velocity vector $\overrightarrow{U}$ with the velocity of the moving reference frame attached to moving CV ${\overrightarrow{U}}_{CV}$ and the relative velocity w.r.t. to the moving reference frame ${\overrightarrow{U}}_r$

RTT is valid if the CV has no acceleration with respect to a fixed (inertial) reference frame. In other words, the standard form of RTT is then valid for a moving CV with constant velocity when:

1. All velocities are measured relative to the CV.

2. All time derivatives are measured relative to the CV.

Thus for a CV moving with ${\overrightarrow{U}}_{CV}$, the relative velocity w.r.t. to the reference frame attached to the moving CV is:

${\overrightarrow{U}}_r=\overrightarrow{U}-{\overrightarrow{U}}_{CV}$

Template:Center top $\frac{d{B}_{sys}}{dt}=\frac{\partial }{\partial t}\left(\underset{cv}{\int }b\rho dV\right)+\underset{cs}{\int }b\rho \overrightarrow{U_r}\cdot \overrightarrow{n}dA$ Template:Center bottom

Example 6: Momentum equation for a CV moving with constant velocity

Consider the jet and the vane. Determine the force to be applied such that vane moves with a constant speed ${\overrightarrow{U}}_v$ in $x_1$ direction.

File:Water jet.png

Jet hitting to a vane moving with constant speed (example 6)

Assume: steady flow, properties are uniform at 1 and 2, nobody forces, incompressible flow.

Note that for an inertial CV (static or moving with constant speed) RTT is valid, but velocities should be written with respect to the moving CV.

$\left|{\overrightarrow{U}}^1\right|=|{\overrightarrow{U}}_{jet}-{\overrightarrow{U}}_v|$

from the continuity equation

$0=\underset{cs}{\int }\rho U\cdot ndA=-\left|{\overrightarrow{U}}^1\right|\rho A_1+\left|{\overrightarrow{U}}^2\right|\rho A_2$

$\rho \left|{\overrightarrow{U}}^1\right|A_1=\rho \left|{\overrightarrow{U}}^2\right|A_2=\dot{m}$

for $A_1=A_2$, $\left|{\overrightarrow{U}}^2\right|=\left|{\overrightarrow{U}}^1\right|$

Thus, the component of ${\overrightarrow{U}}^2$ along $x_1$ direction is:

$U_1^2=\left|U_1^1\right|cos\theta =(\left|{\overrightarrow{U}}_{jet}\right|-\left|{\overrightarrow{U}}_v\right|)cos\theta$

The momentum equation reads

Template:Center top $F_i=F_{S_i}+\underset{=0}{\underbrace{F_{B_i}}}=\underset{=0\text{(steady)}}{\underbrace{\frac{\partial }{\partial t}\underset{cv}{\int }{U}_i\rho dV}}+\underset{cs}{\int }{U}_i\rho U_j{n}_{j}dA$ Template:Center bottomSpecifically in $x_1$ direction isTemplate:Center top $F_{S_1}=R_1+p_{a}A-p_{a}A$ Template:Center bottomTemplate:Center top $R_1=\underset{cs}{\int }{U}_1\rho U_j{n}_{j}dA$ Template:Center bottomTemplate:Center top $R_1=-U_1^1\rho \left|{\overrightarrow{U}}^1\right|A_1+U_1^2\rho \left|{\overrightarrow{U}}^2\right|A_2=-U_1^1\dot{m}+U_1^2\dot{m}$

since $U_1^2=\left|U_1^1\right|cos\theta =(\left|{\overrightarrow{U}}_{jet}\right|-\left|{\overrightarrow{U}}_v\right|)cos\theta$

$R_1=\dot{m}{U}_1^1(cos\theta -1)=\dot{m}(\left|{\overrightarrow{U}}_{jet}\right|-\left|{\overrightarrow{U}}_v\right|)(cos\theta -1)$

The force in the vertical direction (in $x_2$ direction) is:

Template:Center top $R_2=\underset{cs}{\int }{U}_2\rho U_j{n}_{j}dA$ Template:Center bottomTemplate:Center top $R_2=\underset{A_1}{\int }{U}_2^1\rho U_j^2{n}_j^{2}dA+\underset{A_2}{\int }{U}_2^2\rho U_j^2{n}_j^{2}dA$ Template:Center bottom

at 1, $U_2^1=0$ and at 2, $U_2^2=\left|{\overrightarrow{U}}^2\right|sin\theta =\left|{\overrightarrow{U}}^1\right|sin\theta =|{\overrightarrow{U}}_{jet}-{\overrightarrow{U}}_v|sin\theta$.

Template:Center top $R_2=\underset{A_2}{\int }{U}_2^2\rho U_j^2{n}_j^{2}dA$ Template:Center bottom

Template:Center top $R_2=U_2^2\rho \left|{\overrightarrow{U}}_2\right|A_2=U_2^2\dot{m}=|{\overrightarrow{U}}_{jet}-{\overrightarrow{U}}_v|\dot{m}sin\theta$ Template:Center bottom

Momentum Equation for CV with rectilinear acceleration

For an inertial CV the following transport equation for momentum holds:

Template:Center top $\overrightarrow{F}=\frac{\partial }{\partial t}\underset{cv}{\int }\overrightarrow{U}\rho dV+\underset{cs}{\int }\overrightarrow{U}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

However, not all CV are inertial: for example a rocket must accelerate if it is to get off the ground. A CV attached to the rocket has to be used and it is non-inertial.

File:Momentum equation.png

A system and coordinate axis attached to it moving with the same rectilinear acceleration

Denote an inertial reference frame with $X_1,X_2,X_3$ and another reference frame moving with the CV, $x_1,x_2,x_3$. Hence, $x_1,x_2,x_3$ becomes the non-inertial frame of reference. Let $x_1,x_2,x_3$reference frame move with a velocity and an acceleration ${\overrightarrow{U}}_{rf}$ and ${\overrightarrow{a}}_{rf}$, respectively. The velocity vector of a fluid particle w.r.t. to the inertial reference frame (w.r.t. to a ground observer) is${\overrightarrow{U}}_X={\overrightarrow{U}}_x+{\overrightarrow{U}}_{rf}$ where ${\overrightarrow{U}}_x$ is the velocity of the fluid particle w.r.t. non-inertial reference frame $x_1,x_2,x_3$. Hence the time derivative of it is, Template:Center top $\frac{d{\overrightarrow{U}}_X}{dt}=\frac{d{\overrightarrow{U}}_x}{dt}+\frac{d{\overrightarrow{U}}_{rf}}{dt}\Rightarrow \frac{d{\overrightarrow{U}}_X}{dt}\ne \frac{d{\overrightarrow{U}}_x}{dt}$ Template:Center bottom

or,Template:Center top ${\overrightarrow{a}}_X={\overrightarrow{a}}_x+{\overrightarrow{a}}_{rf}\Rightarrow \frac{d{\overrightarrow{P}}_X}{dt}\ne \frac{d{\overrightarrow{P}}_x}{dt}$ Template:Center bottom Above relationships imply that the velocity and the acceleration are not the same when considered from inertial and moving reference frames.

However, the Newton's second law states that:

Template:Center top ${\overrightarrow{F}|}_{system}={\frac{d{\overrightarrow{P}}_X}{dt}|}_{system}$ Template:Center bottom

For a control volume moving with ${\overrightarrow{U}}_{rf}$ and ${\overrightarrow{a}}_{rf}$, i.e. $\frac{d{\overrightarrow{U}}_{CV}}{dt}=\frac{d{\overrightarrow{U}}_{rf}}{dt}$, i.e. $\frac{d{U}_{cv}}{dt}\ne 0$, the time derivative of ${\overrightarrow{P}}_X$ and ${\overrightarrow{P}}_x$ are not equal i.e. RTT is not valid for an accelerating control volume and has to be modified.

To develop momentum equation for an accelerating CV, it is necessary to relate ${\overrightarrow{P}}_X$ to ${\overrightarrow{P}}_x$.

Previously we have seen that in a non-inertial reference frame having rectilinear acceleration, i.e. (translational acceleration).

Template:Center top ${\overrightarrow{a}}_X={\overrightarrow{a}}_x+{\overrightarrow{a}}_{rf}$ Template:Center bottom and also

Template:Center top $\overrightarrow{F}=\underset{mass(system)}{\int }{\overrightarrow{a}}_{X}dm$ Template:Center bottom

Template:Center top $=\underset{mass(system)}{\int }({\overrightarrow{a}}_x+{\overrightarrow{a}}_{rf})dm$ Template:Center bottom

Template:Center top $\overrightarrow{F}-\underset{mass(system)}{\int }{\overrightarrow{a}}_{rf}dm=\underset{mass(system)}{\int }{\overrightarrow{a}}_{x}dm$ Template:Center bottom

Template:Center top $\overrightarrow{F}-\underset{mass(system)}{\int }{\overrightarrow{a}}_{rf}dm=\underset{mass(system)}{\int }\frac{d{\overrightarrow{U}}_x}{dt}dm=\frac{d}{dt}\underset{mass(system)}{\int }{\overrightarrow{U}}_{x}dm$ Template:Center bottom

Template:Center top $\overrightarrow{F}-\underset{V(system)}{\int }{\overrightarrow{a}}_{rf}\rho dV=\frac{d{\overrightarrow{P}}_x}{dt}{|}_{system}$ Template:Center bottom

For a moving CV we know that

Template:Center top $\frac{d{\overrightarrow{P}}_x}{dt}{|}_{system}=\frac{\partial }{\partial t}\underset{cv}{\int }{\overrightarrow{U}}_x\rho dV+\underset{cs}{\int }{\overrightarrow{U}}_x\rho {\overrightarrow{U}}_x\cdot \overrightarrow{n}dA$ Template:Center bottom

Let system and CV coincides at an instant $t_0$:

Template:Center top ${\overrightarrow{F}}_{onsystem}-\underset{Vsystem}{\int }{\overrightarrow{a}}_{rf}\rho dV={\overrightarrow{F}}_{oncv}-\underset{cv}{\int }{\overrightarrow{a}}_{rf}\rho dV$ Template:Center bottom

Template:Center top ${\overrightarrow{F}}_{oncv}-\underset{cv}{\int }{\overrightarrow{a}}_{rf}\rho dV=\frac{\partial }{\partial t}\underset{cv}{\int }{\overrightarrow{U}}_x\rho dV+\underset{cs}{\int }{\overrightarrow{U}}_x\rho {\overrightarrow{U}}_x\cdot \overrightarrow{n}dA$ Template:Center bottom

Example 7: Momentum equation for an accelerating control volume

A small rocket, with an inertial mass of $M_0$, is to be launched vertically. Assume a steady exhaust mass flow rate $\dot{m}$ and velocity $U_e$ relative to the rocket.

File:Rocket real.png

Launching rocket (example 1)

File:Rocket CV 1 and 2 renew 01.svg

Control Volume I and II (example 1)

Neglecting drag on the rocket, find the relation for the velocity of the rocket U(t).

Template:Center top $\underset{A}{\underbrace{F_2}}-\underset{B}{\underbrace{\underset{cv}{\int }{a}_{rf2}\rho dV}}=\underset{C}{\underbrace{\frac{\partial }{\partial t}\underset{cv}{\int }{U}_2\rho dV}}+\underset{D}{\underbrace{\underset{cs}{\int }{U}_2\rho {\overrightarrow{U}}_x\cdot \overrightarrow{n}dV}}$ Template:Center bottom

A:

Template:Center top $F_2=-g\underset{cv}{\int }\rho dV=-g{M}_{cv}$ Template:Center bottom

Template:Center top $M_{cv}=M_0-\dot{m}t$ Template:Center bottom

Template:Center top $F_2=-g(M_0-\dot{m}t)$ Template:Center bottom

B: Since ${\overrightarrow{a}}_{rf}$ is not a function of the coordinates:

Template:Center top $-\underset{cv}{\int }{a}_{rf2}\rho dV=-a_{rf2}\underset{cv}{\int }\rho dV=-a_{rf2}(M_0-\dot{m}t)$ Template:Center bottom

is the time rate of change at $x_2$-momentum of the mass content in CV. One can treat the rocket CV as if it is composed of two CV's, i.e. the solid propellant section (CV I) and nozzle section (CV II): Template:Center top $\frac{\partial }{\partial t}\underset{cv}{\int }{U}_2\rho dV=\frac{\partial }{\partial t}\underset{cvI}{\int }{U}_2\rho dV+\frac{\partial }{\partial t}\underset{cvII}{\int }{U}_2\rho dV$ Template:Center bottom

As solid propellant has no relative velocity in CV I and $U_2=-U_e$ does not change in time at the nozzle and the mass in CV II does not change in time, this term can be neglected completely:

Template:Center top $\frac{\partial }{\partial t}[\underset{U_2=0\to =0}{\underbrace{\underset{CVI}{\int }{U}_2\rho dV}}+\underset{CVII}{\int }{U}_2\rho dV]=\underset{=0(U_2=-U_e=cst)}{\underbrace{\frac{\partial }{\partial t}\underset{CVII}{\int }{U}_2\rho dV}}\approx 0$ Template:Center bottom

D: Template:Center top $\underset{cs}{\int }{U}_2\rho {\overrightarrow{U}}_x\cdot \overrightarrow{n}dA=U_2\underset{cs}{\int }\rho {\overrightarrow{U}}_x\cdot \overrightarrow{n}dA=U_2\dot{m}=-U_e\dot{m}$ Template:Center bottom

Substitution of all the terms gives:

Template:Center top $-g(M_0-\dot{m}t)-a_{rf2}(M_0-\dot{m}t)=-U_e\dot{m}$ Template:Center bottomTemplate:Center top $a_{rf2}=\frac{U_e\dot{m}}{M_0-\dot{m}t}-g$ Template:Center bottom

Template:Center top $\frac{d{V}_{cv}}{dt}=a_{rf2}=\frac{U_e\dot{m}}{M_0-\dot{m}t}-g$ Template:Center bottom

Template:Center top $V_{cv}(t)=\underset{0}{\overset{t}{\int }}\frac{U_e\dot{m}}{M_0-\dot{m}t}dt-\underset{0}{\overset{t}{\int }}gdt$ Template:Center bottom

Template:Center top $V_{cv}(t)=-U_{e}ln(1-\frac{\dot{m}t}{M_0})-gt$ Template:Center bottom

The first term is always positive due to the natural logarithm $ln$ . To overcome gravity one should have enough exit velocity, i.e. momentum. Moreover, it can be seen from this equation that if the fuel mass burned is a large fraction of the initial mass, the final rocket velocity can exceed the exit velocity of the fluid.

Extension of Energy equation for CV

Template:Center top ${[\dot{Q}+\dot{W}]}_{onsystem}={[\dot{Q}+\dot{W}]}_{oncv}=\frac{\partial }{\partial t}\int e\rho dV+\underset{cs}{\int }e\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

where $e=u+\frac{{\left|\overrightarrow{U}\right|}^2}{2}+gz$

Template:Center top ${\dot{W}}_{oncv}={\dot{W}}_{body}+{\dot{W}}_{surface}$ Template:Center bottom

here $z=x_2$

Template:Center top ${\dot{W}}_{body}={\dot{W}}_{shaft}+{\dot{W}}_{elec}+{\dot{W}}_{other}$ Template:Center bottom

Template:Center top ${\dot{W}}_{surface}={\dot{W}}_{normal}+{\dot{W}}_{shear}$ Template:Center bottom

Normal and shear stress components on a differential surface area

Since Template:Center top $\dot{W}=\overrightarrow{F}\cdot \overrightarrow{U}$ Template:Center bottom The differential work done per unit time on a differential surface is: Template:Center top $d\dot{W}=d\overrightarrow{F}\cdot \overrightarrow{U}=\overrightarrow{\sigma }dA\cdot \overrightarrow{U}$ Template:Center bottom where $\overrightarrow{\sigma }$ is the sum of normal stress ${\overrightarrow{\sigma }}_n$ and shearing stress ${\overrightarrow{\sigma }}_s$ on that surface, which is tangent to $dA$: Template:Center top $\overrightarrow{\sigma }={\overrightarrow{\sigma }}_n+{\overrightarrow{\sigma }}_s$ Template:Center bottom Note that, although normal stress can be sum of the pressure and normal viscous stress, i.e. ${\overrightarrow{\sigma }}_n=-p\overrightarrow{n}+{\overrightarrow{\tau }}_n$, the shear stress is only due to viscous stress,i.e. ${\overrightarrow{\sigma }}_s={\overrightarrow{\tau }}_s$.

The differential work done per unit time by the normal forces on a differential surface area over CS is Template:Center top $d{\dot{W}}_n=d{\overrightarrow{F}}_n\cdot \overrightarrow{U}={\overrightarrow{\sigma }}_{n}dA\cdot \overrightarrow{U}$ Template:Center bottom In many cases normal viscous stresses can be neglected and the normal stress exist only due to pressure: Template:Center top ${\overrightarrow{\sigma }}_n=-p\overrightarrow{n}$ Template:Center bottom

therefore the work done by normal forces per unit time over the whole CS becomes Template:Center top ${\dot{W}}_{normal}=\underset{cs}{\int }{\overrightarrow{\sigma }}_n\cdot \overrightarrow{U}dA=-\underset{cs}{\int }p\overrightarrow{n}\cdot \overrightarrow{U}dA$ Template:Center bottom Hence, the work done by the shear stress per unit time can be for the differential area and written as Template:Center top $d{\dot{W}}_{shear}={\overrightarrow{\sigma }}_{s}dA\cdot \overrightarrow{U}\to {\dot{W}}_{shear}=\underset{cs}{\int }{\overrightarrow{\sigma }}_s\cdot \overrightarrow{U}dA=\underset{cs}{\int }{\overrightarrow{\tau }}_s\cdot \overrightarrow{U}dA$ Template:Center bottom

Inserting those into the main energy equation gives the following form of the energy equation, in which the work done by the normal force on CS due to pressure is interpreted as the net flow rate of enthalpy through CS:

Template:Center top $\dot{Q}+{\dot{W}}_{shaft}+{\dot{W}}_{shear}+{\dot{W}}_{other}=\frac{\partial }{\partial t}\int e\rho dV+\underset{cs}{\int }(\underset{h:enthalpy}{\underbrace{u+\frac{p}{\rho }}}+\frac{\left|{\overrightarrow{U}}^2\right|}{2}+g{x}_2)\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

Example 8

Air enters compressor at inlet 1 with negligible velocity and leaves at outlet 2. The power input to the machine is $P_{input}$ and the volume flow rate is $\dot{V}$. Find a relation for the rate of heat transfer in terms of the power, temperature, pressure, etc.

File:Compressor renew.svg

Energy balance for a compressor (example 1)

1:

Template:Center top $\dot{Q}+{\dot{W}}_{shaft}+\underset{=0=\tau \cdot \overrightarrow{U}}{\underbrace{{\dot{W}}_{shear}}}+\underset{=0}{\underbrace{{\dot{W}}_{other}}}=\underset{=0steadystate}{\underbrace{\frac{\partial }{\partial t}\int e\rho dV}}+\underset{cs}{\int }(u+\frac{p}{\rho }+\frac{U^2}{2}+g{x}_2)\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

Template:Center top $0=\underset{=0steadystate}{\underbrace{\frac{\partial }{\partial t}\underset{cv}{\int }\rho dV}}+\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA\to \left|{\rho }_1{U}_1{A}_1\right|=\left|{\rho }_2{U}_2{A}_2\right|=\dot{m}$ Template:Center bottom

2:

Template:Center top $\dot{Q}=-{\dot{W}}_{shaft}+\underset{cs}{\int }(u+\frac{p}{\rho }+\frac{U^2}{2}+gz)\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

For uniform properties at 1 and 2 and inserting the inserting the relation for the enthalpy $h=u+\frac{p}{\rho }$.

Template:Center top $\dot{Q}=-{\dot{W}}_{shaft}-(h_1+\underset{=0}{\underbrace{\frac{U_1^2}{2}}}+g{z}_1)\left|{\rho }_1{U}_1{A}_1\right|+(h_2+\frac{U_2^2}{2}+g{z}_2)\left|{\rho }_2{A}_2{U}_2\right|$ Template:Center bottom

Template:Center top $\dot{Q}=-{\dot{W}}_{shaft}+\dot{m}[h_2+\frac{U_2^2}{2}-h_1+\underset{=0}{\underbrace{g(z_2-z_1)}}]$ Template:Center bottom

Assuming that air behaves like an ideal gas with a constant $c_p$.

Template:Center top $h_2-h_1=c_p(T_2-T_1)$ Template:Center bottom

Template:Center top $\dot{Q}=-{\dot{W}}_{shaft}+\dot{m}[c_p(T_2-T_1)+\frac{U_2^2}{2}]$ Template:Center bottom

Special form of the Energy equation

For a CV with one inlet 1, one outlet 2 and steady uniform flow through it.

Template:Center top $\underset{K}{\underbrace{\dot{Q}+{\dot{W}}_{shaft}+{\dot{W}}_{shear}}}=\underset{cs}{\int }(u+\frac{p}{\rho }+\frac{\left|{\overrightarrow{U}}^2\right|}{2}+gz)\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

For uniform flow properties at the inlet and outlet.

Template:Center top $K=(u_1+\frac{p_1}{{\rho }_1}+\frac{\left|{\overrightarrow{U}}_1^2\right|}{2}+g{z}_1)\underset{-\dot{m}}{\underbrace{\underset{1}{\int }{\rho }_1{\overrightarrow{U}}_1\cdot \overrightarrow{n}dA}}+(u_2+\frac{p_2}{{\rho }_2}+\frac{\left|{\overrightarrow{U}}_2^2\right|}{2}+g{z}_2)\underset{\dot{m}}{\underbrace{\underset{2}{\int }{\rho }_2{\overrightarrow{U}}_2\cdot \overrightarrow{n}dA}}$ Template:Center bottom

Template:Center top $K=\dot{m}[(u_2-u_1)+(\frac{p_2}{{\rho }_2}-\frac{p_1}{{\rho }_1})+(\frac{{\overrightarrow{U}}_2^2}{2}-\frac{{\overrightarrow{U}}_1^2}{2})+g(z_2-z_1)]$ Template:Center bottom

Reform: Template:Center top $\underset{mechanicalevergyperunitmassatflowcrosssection}{\underbrace{\frac{p_1}{{\rho }_1}+\frac{\left|{\overrightarrow{U}}_1^2\right|}{2}+g{z}_1}}=\frac{p_2}{{\rho }_2}+\frac{\left|{\overrightarrow{U}}_2^2\right|}{2}+g{z}_2+(u_2-u_1)-\frac{\dot{Q}}{\dot{m}}-\frac{{\dot{W}}_{shaft}}{\dot{m}}-\frac{{\dot{W}}_{shear}}{\dot{m}}$ Template:Center bottom

If this equation is divided by $g$, the head form of the energy equation can be obtained. This form of the energy equation has the unit in meters.Template:Center top $\frac{p_1}{g{\rho }_1}+\frac{\left|{\overrightarrow{U}}_1^2\right|}{2g}+z_1=\frac{p_2}{g{\rho }_2}+\frac{\left|{\overrightarrow{U}}_2^2\right|}{g2}+z_2+\frac{1}{g}(u_2-u_1)-\frac{\dot{Q}}{g\dot{m}}-\frac{{\dot{W}}_{shaft}}{g\dot{m}}-\frac{{\dot{W}}_{shear}}{g\dot{m}}$ Template:Center bottom

For ${\dot{W}}_{shaft}=0$, ${\dot{W}}_{shear}=0$ and incompressible flow through a passive device which involves no mechanical work and no heat addition, the energy equation per unit mass reads :

Template:Center top $\frac{p_1}{\rho }+\frac{\left|{\overrightarrow{U}}_1^2\right|}{2}+g{z}_1=\frac{p_2}{\rho }+\frac{\left|{\overrightarrow{U}}_2^2\right|}{2}+g{z}_2+(u_2-u_1)-\frac{\dot{Q}}{\dot{m}}$ Template:Center bottom

$\frac{p}{\rho }+\frac{U^2}{2}+gz$: Mechanical energy per unit mass.

$u_2-u_1-\frac{\dot{Q}}{\dot{m}}$: Irreversible conversion of mechanical energy to unwanted thermal energy $(u_2-u_1)$ and loss of energy via heat transfer $(-\frac{\dot{Q}}{\dot{m}})$. Note that, since the heat leaves the CV, it should be negative, i.e. $\dot{Q}<0$. Hence this term is always positive.

In head form, the loss of energy through a device can be written as. Template:Center top $h_{loss}=[u_2-u_1-\frac{\dot{Q}}{\dot{m}}]\frac{1}{g}$ Template:Center bottom

i.e.

Template:Center top $\frac{p_1}{\rho g}+\frac{U_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{U_2^2}{2g}+z_2+h_{loss}$ Template:Center bottom

One can add the shaft work done by a pump or a turbine, in to that equation Template:Center top $\frac{p_1}{\rho g}+\frac{U_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{U_2^2}{2g}+z_2+h_{loss}-h_{pump}+h_{turbine}$ Template:Center bottomNote that, pump work appears with a negative sign because it adds energy and turbine works appears with a positive sign because it extracts energy.

Differential Control Volume Analysis:Bernoulli Equation

Differential control on a streamline

Consider the steady, incompressible, frictionless flow through the differential CV for a stream tube.

Continuity

Template:Center top $0=\frac{\partial }{\partial t}\underset{=0}{\underbrace{\underset{cv}{\int }\rho dV}}+\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

Template:Center top $0=-{\dot{m}}_{in}+{\dot{m}}_{out}\leftrightarrow {\dot{m}}_{in}={\dot{m}}_{out}=\dot{m}$ Template:Center bottom

Streamwise-Component of Momentum Equation

The momentum equation along the streamline (streamwise-component) can be written as:

Template:Center top $F_{S_s}+F_{B_s}=\frac{\partial }{\partial t}\underset{=0}{\underbrace{\int U_s\rho dV}}+\underset{cs}{\int }{U}_s\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ Template:Center bottom

Template:Center top $F_{S_s}=\underset{inlet}{\underbrace{pA}}-\underset{outlet}{\underbrace{(p+dp)(A+dA)}}+\underset{periphery}{\underbrace{(p+\frac{dp}{2})dA}}$ Template:Center bottom

File:Pressure on fluid element renew.svg

Forces created by pressure on a fluid element

$F_{S_s}=-Adp-\frac{1}{2}dpdA$

Template:Center top $F_{B_s}=\rho g_{s}dV=\rho (-gsin\theta )\underset{approx.volumeofCV}{\underbrace{(A+\frac{dA}{2})ds}}$ Template:Center bottom

with:

Template:Center top $sin\theta ds=dz$ Template:Center bottom

Then,

Template:Center top $F_{B_s}=-\rho g(A+\frac{dA}{2})dz$ Template:Center bottom

Template:Center top $\underset{cs}{\int }{U}_s\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=-U_s\dot{m}+(U_s+d{U}_s)\dot{m}=\dot{m}d{U}_s$ Template:Center bottom

Template:Center top $-Adp-\underset{\approx 0}{\underbrace{\frac{1}{2}dpdA}}-\rho g(A+\frac{dA}{2})dz=\dot{m}d{U}_s$ Template:Center bottom

Template:Center top $-Adp-\rho gAdz=\dot{m}d{U}_s$ Template:Center bottom

where $\dot{m}=\rho U_{s}A$

Template:Center top $-Adp-\rho gAdz=\rho U_{s}Ad{U}_s$ Template:Center bottom

Template:Center top $\frac{dp}{\rho }+U_{s}d{U}_s+gdz=0$ Template:Center bottom

with:

Template:Center top $U_{s}d{U}_s=d\left(\frac{U_s^2}{2}\right)$ Template:Center bottom

Then,

Template:Center top $\frac{dp}{\rho }+d\left(\frac{U_s^2}{2}\right)+gdz=0$ Template:Center bottom

Integrate between 1 and 2 along a streamline:

Template:Center top $\frac{p_1}{\rho }+\frac{U_1^2}{2}+g{z}_1=\frac{p_2}{\rho }+\frac{U_2^2}{2}+g{z}_2=constant$ Template:Center bottom

Bernoulli equation is clearly related to the steady flow energy equation for a stream line. This form of the Bernoulli equation, when the following conditions are satisfied:

1. Steady flow. Note that there is also an unsteady Bernoulli equation.

2. Incompressible flow. For example, in aerodynamics, flow can be accepted to be incompressible for Mach number $(M=\frac{\text{speed of flow}}{\text{speed of sound}})$ less than 0.3.

3. Frictionless flow, e.g. in the absence of solid walls.

4. Flow along a single streamline. Different streamline has a different constant.

5. No shaft work between 1 and 2.

6. No heat transfer between 1 and 2.

File:Bernoullis law inapplicability 02.svg

Bernoulli's equation is inapplicable due to a) friction loss on the surface and the wake of the car b) heat energy input in the heat engine c) addition of mechanical energy inside the flow of a ventilator

Example 9: Bernoulli equation for a flow through nozzle

Consider steady flow of water through a horizontal nozzle. Find $P_1$ as a function of flow rate $\dot{Q}$.

File:Streamline in nozzle 01.svg

Flow through horizontal nozzle (example 1)

Assumptions: steady, incompressible, frictionless, flow along a streamline, $z_2=z_1$, uniform flow.

Template:Center top $\frac{p_1}{\rho }+\frac{U_1^2}{2}=\frac{p_2}{\rho }+\frac{U_2^2}{2}$ Template:Center bottom

Template:Center top $p_1=p_{atm}+\rho (\frac{U_2^2}{2}-\frac{U_1^2}{2})=p_{atm}+\rho \frac{U_1^2}{2}(\frac{U_2^2}{U_1^2}-1)$ Template:Center bottomFrom continuity:

Template:Center top $-\left|\rho U_1{A}_1\right|+\left|\rho U_2{A}_2\right|=0$ Template:Center bottom

Template:Center top $U_1=\frac{\dot{Q}}{A_1}\text{and}{U}_2=\frac{\dot{Q}}{A_2}$ Template:Center bottom

Template:Center top $P_1=P_{atm}+\rho \frac{U_1^2}{2}({\left(\frac{A_1}{A_2}\right)}^2-1)$ Template:Center bottom

Template:Center top $P_1=P_{atm}+\rho \frac{{\dot{Q}}^2}{2A_1^2}({\left(\frac{A_1}{A_2}\right)}^2-1)$ Template:Center bottom

Template:Center top $P_1=P_{atm}+\rho \frac{8{\dot{Q}}^2}{\pi D_1^4}({\left(\frac{D_1}{D_2}\right)}^4-1)$ Template:Center bottom

Example 10: Bernoulli equation

Find a relation between the nozzle discharge velocity and the tank free surface height. Assume steady frictionless flow and uniform flow at 2.

File:Nozzle discharge velocity renew.svg

Nozzle discharge velocity at the bottom of the tank (example 2)

Template:Center top $\frac{p_1}{\rho }+\frac{U_1^2}{2}+g{z}_1=\frac{p_2}{\rho }+\frac{U_2^2}{2}+g{z}_2$ Template:Center bottom

Template:Center top $p_1=p_2=p_{atm}$ Template:Center bottom

Template:Center top $U_1{A}_1=U_2{A}_2$ Template:Center bottom

Template:Center top $U_2^2-U_1^2=2g(z_2-z_1)$ Template:Center bottom

Template:Center top $U_2^2[1-{\left(\frac{A_2}{A_1}\right)}^2]=2gh$ Template:Center bottom

Template:Center top $U_2^2=\frac{2gh}{[1-{\left(\frac{A_2}{A_1}\right)}^2]}$ Template:Center bottom

for steadiness $A_1>>A_2$, thus $U_2\cong \sqrt{2gh}$

Example 11: Bernoulli equation and Venturi tube

Find a relationship between the flow rate and the pressure difference inside a pipe which could be measured by venturimeter

File:Venturimeter 2.svg

Application of Bernoulli Equation:Venturimeter (example 3)

Template:Center top $p_1+\rho \frac{U_1^2}{2}=p_2+\rho \frac{U_2^2}{2}$ Template:Center bottom

Template:Center top $(U_2^2-U_1^2)=\frac{2}{\rho }(p_1-p_2)$ Template:Center bottom

Template:Center top $U_1{A}_1=U_2{A}_2=\dot{Q}$ Template:Center bottom

Template:Center top $U_1=\frac{\dot{Q}}{A_1}\text{and}{U}_2=\frac{\dot{Q}}{A_2}$ Template:Center bottom

Template:Center top $(\frac{{\dot{Q}}^2}{A_2^2}-\frac{{\dot{Q}}^2}{A_1^2})=\frac{2}{\rho }\mathrm{\Delta }p$ Template:Center bottom

Template:Center top ${\dot{Q}}^2(\frac{1}{A_2^2}-\frac{1}{A_1^2})=\frac{2}{\rho }\mathrm{\Delta }p$ Template:Center bottom

Template:Center top ${\dot{Q}}^2\frac{1}{A_2^2}(1-\frac{A_2^2}{A_1^2})=\frac{2}{\rho }\mathrm{\Delta }p$ Template:Center bottom

If $\beta =\frac{D_2}{D_1}$, then:

Template:Center top $\dot{Q}=\sqrt{\frac{A_2^{2}2\underset{measured}{\underbrace{\mathrm{\Delta }p}}}{\rho (1-{\beta }^4)}}$ Template:Center bottom

That is the method for measuring flow rate.