Fluid Mechanics for MAP/Fluid Dynamics

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Introduction

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

Integral Approach: We focus on a control volume (CV), which is a finite region. 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 flux of 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.

Lagrangian versus Eulerian Approach: Substantial Derivative

File:Fluid Dynamics 3.png

Path of a fluid element

Let $\alpha$ be any flow variable (pressure, velocity, etc.). Eulerian approach deals with the description of $\alpha$ at each location $(x_i)$ and time (t). For example, measurement of pressure at all $x_i$ defines the pressure field: $P(x_1,x_2,x_3,t)$. Other field variables of the flow are:

Template:Center top$U_j(x_1,t),P(x_1,t),\rho (x_1,t),T(x_1),{\tau }_{jk}(x_1,t)\Rightarrow \alpha (x_i,t)$Template:Center bottom

Lagrangian approach tracks a fluid particle and determines its properties as it moves.

Template:Center top$x_i{|}_p(t+\mathrm{\Delta }t)=x_i{|}_p(t)+\underset{t}{\overset{t+\mathrm{\Delta }t}{\int }}{U}_i{|}_p(t^{\prime })d{t}^{\prime }$Template:Center bottom

Oceanographic measurements made with floating sensors delivering location, pressure and temperature data, is one example of this approach. X-ray opaque dyes, which are used to trace blood flow in arteries, is another example.

Let ${\alpha }_p$ be the variable of the particle (substance) P, this ${\alpha }_p$ is called "substantial variable".

For this variable:

${\alpha }_p(\overrightarrow{x_p},t)$ and $\overrightarrow{x_p}=\overrightarrow{x_p}(t)\to {\alpha }_p(\overrightarrow{x_p},t)={\alpha }_p(t)$

In other words, one observes the change of variable $\alpha$ for a selected amount of mass of fixed identity, such that for the fluid particle, every change is a function of time only.

In a fluid flow, due to excessive number of fluid particles, Lagrangian approach is not widely used.

Thus, for a particle P finding itself at point $x_i$ for a given time, we can write the equality with the field variable:

Template:Center top${\alpha }_p(t)=\alpha [(x_i{)}_p,t]$Template:Center bottom

Along the path of the particle:

Template:Center top${\alpha }_p(t+\mathrm{\Delta }t)=\alpha [(x_i+\mathrm{\Delta }{x}_i{)}_p,t+\mathrm{\Delta }t]$Template:Center bottom

Hence,

Template:Center top$d{\alpha }_p=\alpha [((x_i+\mathrm{\Delta }{x}_i{)}_p,t+\mathrm{\Delta }t)-\alpha ((x_i{)}_p,t)]$Template:Center bottom

Template:Center top$d{\alpha }_p=\frac{\partial \alpha }{\partial t}dt+\frac{\partial \alpha }{\partial x_i}d{x}_{ip}$Template:Center bottom

Template:Center top$\frac{d{\alpha }_p}{dt}=\frac{\partial \alpha }{\partial t}+\left(\frac{\partial \alpha }{\partial x_i}\right){\left(\frac{d{x}_i}{dt}\right)}_p=\underset{Localchangeintime}{\underbrace{\frac{\partial \alpha }{\partial t}}}+\underset{Changeinspace}{\underbrace{\frac{\partial \alpha }{\partial x_i}{U}_i}}$Template:Center bottom

The local change in time is the local time derivative (unsteadiness of the flow) and the change in space is the change along the path of the particle by means of the convective derivative.

Template:Center top$\frac{d{\alpha }_p}{dt}=\frac{D\alpha }{Dt}=(\frac{\partial }{\partial t}+U_i\frac{\partial }{\partial x_i})\alpha$Template:Center bottom

The substantial derivative connects the Lagrangian and Eulerian variables.

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.

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

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 do 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

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$F_i=\frac{d{P}_i}{dt}{|}_{system}$Template:Center bottom

Template:Center top$P_{i(system)}=\underset{mass(system)}{\int }{U}_{i}dm=\underset{V(system)}{\int }{U}_i\rho dV$Template:Center bottom

The first law of Thermodynamics

Template:Center top$\underset{heataddedonsystem}{\underbrace{dQ}}+\underset{workdoneonsystem}{\underbrace{dW}}=dE$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 $e=\underset{Intenralenergy}{\underbrace{u}}+\underset{Kineticenergy}{\underbrace{\frac{U_i{U}_i}{2}}}+\underset{Potentialenergy}{\underbrace{gz}}$

There are also other laws like the conservation of moment of momentum (angular momentum) and second law of thermodynamics, but they are not the subject of this course and will not be treated here.

Note that all basic laws are written for a system, i.e defined mass with fixed identity. We should rephrase these laws for a control volume.

Relation of a system derivative to the control volume derivative

Consider a fire extinguisher $\frac{dM}{dt}{|}_{system}=0$ whereas $\frac{dM}{dt}{|}_{cv}<0$ 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$B=M,b=1$Template:Center bottom Template:Center top$B=\overrightarrow{P},b=\overrightarrow{u}$Template:Center bottom Template:Center top$B=E,b=e$Template:Center bottom

File:Fire extinguisher final 01.svg Control Volume versus System

One dimensional Reynolds Transport Theorem

File:1D-RTT.svg

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

Consider a flow through a nozzle.

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$B_{II}(t+\mathrm{\Delta }t)={\rho }_2{b}_2\mathrm{\Delta }{V}_2$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$B_I(t+\mathrm{\Delta }t)={\rho }_1{b}_1\mathrm{\Delta }{V}_1={\rho }_1{b}_1{A}_1{U}_1\mathrm{\Delta }t=B_{in}$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 indicates the local unsteady effect. 2. The flux of B passing out of the CS. 3. The flux of B passing into the CS. There can be more than one inlet and outlet. 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 flux of $B$ across CS can be written as: Template:Center top${\dot{B}}_{out}=\underset{CSout}{\int }d{B}_{out}=\underset{CSout}{\int }\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA$Template:Center bottom Template:Center top${\dot{B}}_{in}=\underset{CSin}{\int }d{B}_{in}=-\underset{CSin}{\int }\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA$Template:Center bottom ${\dot{B}}_{in}$ 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$\frac{d{B}_{sys}}{dt}=\frac{\partial }{\partial t}\underset{cv}{\int }\rho bdV+\underset{cs}{\int }\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA$Template:Center bottom Since $\rho \overrightarrow{u}\cdot \overrightarrow{n}dA=dm$, RTT can be written as: Template:Center top$\frac{d{B}_{sys}}{dt}=\frac{\partial B_{cv}}{\partial t}+\underset{netfluxofBacrossCS}{\underbrace{\underset{CS}{\int }bdm}}$Template:Center bottom It is possible that CV can move with constant velocity or arbitrary acceleration. This form of RTT is valid if the CV has no acceleration with respect to a fixed (inertial) reference frame. RTT is then valid for a moving CV with constant velocity when: 1. All velocities are measured relative to the CV. 2. All time derivative measure relative to the CV. Thus for a CV moving with ${\overrightarrow{U}}_s$ ${\overrightarrow{U}}_r=\overrightarrow{U}-{\overrightarrow{U}}_s$ Template:Center top$\frac{d{B}_{sys}}{dt}=\frac{d}{dt}\left(\underset{cv}{\int }b\rho dV\right)+\underset{cs}{\int }b\rho \overrightarrow{U_r}\cdot \overrightarrow{n}dA$Template:Center bottom These issues will be covered again.

File:CV-multi-inlet-outlet.svg CV with multiple inlets and outlets File:Fluid Dynamics 13.png CV with arbitrary shape used to derive Reynolds transport theorem File:Flux-sign-through-CS.png Sign of the inflow and outflow fluxes to the CV File:Relative-absolute-velocity-vector-relation.png Relation between the absolute velocity vector with the velocity of the moving reference frame and the velocity w.r.t. to the moving reference frame

Conservation of mass

$B=M,b=1$

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

i.e

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

Assume $\rho$ = constant (incompressible)

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$0=\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA\to \text{Net mass flow rate is equal to zero}$Template:Center bottom

there is no-mass accumulation or deficit in the control volume.

Linear Momentum equation for inertial control volume

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

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

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

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

Examples

Example 1 Consider the mass balance in a stream tube by using the integral form of the conservatin of mass equation. 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 $A_2$. 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 The first term is zero and the second term can be analyzed by decomposing the integration area. Template:Center top$\underset{C{S}_I}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA+\underset{C{S}_{III}}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA+\underset{C{S}_{II}}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=0$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

File:Mass Balance for streamtube renew 02.svg Mass balance for stream tube inside a laminar flow

Example 2

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

File:Junction renew 01.svg mass balance for a connector device Hence, the velocity at section 2 can be calculated by $\underset{A_2}{\int }\rho {\overrightarrow{U}}_2\cdot {\overrightarrow{n}}_{2}dA=-[-\rho \left|U_1\right|A_1+{\dot{m}}_3-\rho Q_4]$ For ${\overrightarrow{n}}_2=(0,-1)\to U_{21}=0$ $-\rho U_{22}{A}_2=\rho \left|U_1\right|A_1-{\overrightarrow{m}}_3+\rho Q_4$ The term on the right side is positive if $U_2$ is negative (outflow) and it is negative if $U_2$ is positive (inflow).

Example 3

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. Template:Center top$F_i=F_{Si}+F_{Bi}=\frac{\partial }{\partial t}\underset{cv}{\int }{U}_i\rho dV+\underset{cs}{\int }{U}_i\rho U_j{n}_{j}dA$Template:Center bottom where the derivative with respect to time is zero due to steady state conditions. Template:Center top$F_i=\underset{C{S}_I}{\int }{U}_{1i}\rho U_{1j}{n}_{j}dA+\underset{C{S}_{II}}{\int }{U}_{2i}\rho U_{2j}{n}_{j}dA$Template:Center bottom Template:Center top$F_i=-U_{1i}\underset{{\dot{m}}_1}{\underbrace{\left|{\overrightarrow{U}}_1\right|A_1\rho }}+U_{2i}\underset{{\dot{m}}_2}{\underbrace{\left|{\overrightarrow{U}}_2\right|A_2{\rho }_2}}$Template:Center bottom ${\dot{m}}_1={\dot{m}}_2=\dot{m}$ Template:Center top$F_i=\dot{m}(U_{2i}-U_{1i})$Template:Center bottom

File:Force calculation fluid element renew 01.svg Force in a streamtube

Example 4

File:Nozzle test 2.svg

Experimental Setup for nozzle test

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

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 bottom

Template:Center top$F_{S1}=p_{a}A-p_{a}A+R_1$Template:Center bottom

Template:Center top$R_1=\underset{cs}{\int }{U}_1\rho U_j{n}_{j}dA=\underset{CS1}{\int }{U}_1^1\rho U_j^1{n}_{j}dA+\underset{0}{\underbrace{\underset{CS2}{\int }{U}_1^2\rho U_j^2{n}_{j}dA}}+\underset{0}{\underbrace{\underset{CS3}{\int }{U}_1^3\rho U_j^3{n}_{j}dA}}$Template:Center bottom

File:CV1andCV2 nozzle flow renew 01.svg

Control volume I and II and Free body diagram of the plate

$U_1=0$ at 2 and 3.

Template:Center top$R_1=-U_1\rho U_1{A}_{Jet}$Template:Center bottom

The force which acts on the plate (action-reaction) is $K_1=-R_1=U_1\rho U_1{A}_{Jet}$

It is also possible to solve the problem with $C{V}_2$

Template:Center top$F_{S1}=p_{a}A+R_1=-U_1^2{A}_{jet}\rho$Template:Center bottom

Template:Center top$R_1=-p_{a}A-U_1^2{A}_{Jet}\rho$Template:Center bottom

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

Template:Center top$K_1=-R_1$Template:Center bottom

Template:Center top$F_{net}=-R_1-p_{a}A$Template:Center bottom

Template:Center top$F_{net}=p_{a}A+U_1^{2}A\rho -p_{a}A$Template:Center bottom

Template:Center top$F_{net}=U_1^{2}A\rho$Template:Center bottom

Example 5

File:Fluid Dynamics 26.png

Velocity distribution of fluid over a 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.

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

Apply conservation of mass

Template:Center top$\rho \underset{cs}{\int }{U}_i{n}_{i}dA=0$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$U_{0}h=\underset{0}{\overset{\delta }{\int }}Ud{x}_2$Template:Center bottom

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

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$-D=-U_0\rho U_{0}hw+\int U_1\rho U_{1}wd{x}_2$Template:Center bottom

Template:Center top$D=U_0^2\rho hw-\rho w\underset{0}{\overset{\delta }{\int }}{U}_1^{2}d{x}_2=\rho U_0\underset{0}{\overset{\delta }{\int }}{U}_{1}d{x}_2-\rho w\underset{0}{\overset{\delta }{\int }}{U}_0{U}_{1}d{x}_2$Template:Center bottom

Insert the mass conservation result into the momentum equation.

Template:Center top$D=\rho w\underset{0}{\overset{\delta }{\int }}{U}_1(U_0-U_1)d{x}_2{|}_{x_1=L}$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

Example 6

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. 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. 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 bottom Template:Center top$F_{S_i}=R_1+p_{a}A-p_{a}A$Template:Center bottom Template:Center top$R_1=\underset{cs}{\int }{U}_1\rho U_j{n}_{j}dA$Template:Center bottom Template:Center top$R_1=-U_1^1\rho \left|{\overrightarrow{U}}^1\right|A_1+U_1^2\rho \left|{\overrightarrow{U}}^2\right|A_2$Template:Center bottom Template:Center top$=(U_1^2-U_1)\dot{m}((\left|{\overrightarrow{U}}_{jet}\right|)-\left|{\overrightarrow{U}}_v\right|(\cos \theta -1))$Template:Center bottom from continuity, Template:Center top$0=\underset{cs}{\int }\rho U\cdot ndA=-\left|{\overrightarrow{U}}^1\right|\rho A_1+\left|{\overrightarrow{U}}^2\right|\rho A_2$Template:Center bottom Template:Center top$\rho \left|{\overrightarrow{U}}^1\right|A_1=\rho \left|{\overrightarrow{U}}^2\right|A_2\to \dot{m}$Template:Center bottom Template:Center top$R_1=(U_1^2-U_1)\rho \left|U^1\right|A_1$Template:Center bottom Template:Center top$\left|U^1\right|=|{\overrightarrow{U}}_{jet}-{\overrightarrow{U}}_v|$Template:Center bottom Template:Center top$U_1^1=\left|{\overrightarrow{U}}_{jet}\right|-\left|{\overrightarrow{U}}_v\right|$Template:Center bottom Template:Center top$U_1^2=(\left|{\overrightarrow{U}}_{jet}\right|-\left|{\overrightarrow{U}}_v\right|)cos\theta$Template:Center bottom for $A_1=A_2$ Template:Center top$R_1=\rho {(\left|{\overrightarrow{U}}_{jet}\right|-\left|{\overrightarrow{U}}_v\right|)}^2(cos\theta -1)A_1$Template:Center bottom Template:Center top$=(U_1^2-U_i)\dot{m}((\left|{\overrightarrow{U}}_{jet}\right|)-(\left|{\overrightarrow{U}}_v\right|))$Template:Center bottom Template:Center top$R_2=\underset{cs}{\int }{U}_2\rho U_j{n}_{j}dA$Template:Center bottom Template: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, $U_2^2=\left|{\overrightarrow{U}}^2\right|sin\theta$. 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$R_2=U_2^2\rho \left|{\overrightarrow{U}}_2\right|A_2$Template:Center bottom Template:Center top$R_2={\left|{\overrightarrow{U}}^2\right|}^2\rho sin\theta A_2$Template:Center bottom Template:Center top$\left|{\overrightarrow{U}}^1\right|=\left|{\overrightarrow{U}}^2\right|=|{\overrightarrow{U}}_{jet}-{\overrightarrow{U}}_v|$Template:Center bottom Template:Center top$\overrightarrow{U}={\overrightarrow{U}}_{jet}-{\overrightarrow{U}}_v$Template:Center bottom

File:Water jet.png

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.

Denote an inertial reference frame with $X_1,X_2,X_3$ and another reference frame moving with the system $x_1,x_2,x_3$. Hence, $x_1,x_2,x_3$ becomes the non-inertial frame of reference. Let the system to move with a velocity and an acceleration ${\overrightarrow{U}}_{rf}$ and ${\overrightarrow{a}}_{rf}$ respectively. Here ${\overrightarrow{U}}_X={\overrightarrow{U}}_x+{\overrightarrow{U}}_{rf}$ and accordingly with time derivative dt, 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 These above relationship implies that the velocity and the acceleration is not same when considered from inertial and moving reference frame. The Newton's second law states that:

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

File:Momentum equation.png

Thus the following relation holds for the fluid velocity in the system

Template:Center top${\overrightarrow{U}}_x={\overrightarrow{U}}_X-{\overrightarrow{U}}_{rf}$Template:Center bottom

Where ${\overrightarrow{U}}_x$ is the velocity of the fluid in the system with respect to a non-inertial reference frame, ${\overrightarrow{U}}_X$ is the velocity of the fluid in the system with respect to the inertial reference frame. Accordingly, the acceleration reads:

Template:Center top$\frac{d{\overrightarrow{U}}_x}{dt}=\frac{d{\overrightarrow{U}}_X}{dt}-\frac{d{\overrightarrow{U}}_{rf}}{dt}(1)$Template:Center bottom

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

For a control volume moving with ${\overrightarrow{U}}_{rf}$ and ${\overrightarrow{a}}_{rf}$

Template:Center top$\frac{d{\overrightarrow{U}}_{CV}}{dt}=\frac{d{\overrightarrow{U}}_{rf}}{dt}$Template:Center bottom

Thus for cases, where $\frac{d{U}_{cv}}{dt}\ne 0$, the time derivative of ${\overrightarrow{P}}_X$ and ${\overrightarrow{P}}_x$ are not equal for a system accelerating relative to an inertial reference frame, i.e. RTT is not valid for an accelerating control volume.

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

Examples

Example 1

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. Neglecting drag on the rocket, find the relation for the velocity of the rocket U(t). Template:Center top$\underset{A}{\underbrace{F_2}}-\underset{B}{\underbrace{\underset{cv}{\int }{\overrightarrow{a}}_{rf2}\rho dV}}=\underset{C}{\underbrace{\frac{\partial }{\partial t}\underset{cv}{\int }{\overrightarrow{U}}_2\rho dV}}+\underset{D}{\underbrace{\underset{cs}{\int }{U}_2\rho {\overrightarrow{U}}_x\cdot \overrightarrow{n}dV}}$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$-\underset{cv}{\int }{a}_{rf2}\rho dV=-a_{rf2}\underset{cv}{\int }\rho dV$Template:Center bottom Template:Center top$-\underset{cv}{\int }{a}_{rf2}\rho dV=-a_{rf2}(M_0-\dot{m}t)$Template:Center bottom

File:Rocket real.png

is the time rate of change at $x_2$-momentum of the fluid in CV. One can treat the rocket CV as if it is composed of two CV's, i.e. the solid propellant section (CVI) and nozzle section (CVII): 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 velocity in CVI, $U_2=-U_e$ does not change in time at the nozzle and the mass in CVII does not change in time, this term can be neglected completely: Template:Center top$\frac{\partial }{\partial t}[\underset{U_2=0\to =0}{\underbrace{\underset{CVI}{\int }{U}_2\rho dV}}+\underset{CVII}{\int }{U}_2\rho dV]=\frac{\partial }{\partial t}\underset{CVII}{\int }{U}_2\rho dV\approx 0$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 bottom Template: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 $ln$ . 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.

File:Rocket CV 1 and 2 renew 01.svg Control Volume I and II

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

${\overrightarrow{\tau }}_s$ is the stress in the plane of dA.

${\overrightarrow{\tau }}_n$ is the normal stress normal to dA. In many cases Template:Center top${\overrightarrow{\tau }}_n=-p\overrightarrow{n}dA$Template:Center bottom since Template:Center top$\dot{W}=\overrightarrow{F}\cdot \overrightarrow{U}$Template:Center bottom Template:Center top$d\dot{W}=d\overrightarrow{F}\cdot \overrightarrow{U}$Template:Center bottom Template:Center top$d{\dot{W}}_{normal}={\overrightarrow{\tau }}_{n}dA\cdot \overrightarrow{U}\to {\dot{W}}_{normal}=\underset{cs}{\int }{\overrightarrow{\tau }}_n\cdot \overrightarrow{U}dA=-\underset{cs}{\int }p\overrightarrow{n}\cdot \overrightarrow{U}dA$Template:Center bottom Template:Center top$d{\dot{W}}_{shear}=\overrightarrow{\tau }dA\cdot \overrightarrow{U}\to {\dot{W}}_{shear}=\underset{cs}{\int }\overrightarrow{\tau }\cdot \overrightarrow{U}dA$Template:Center bottom Inserting those into the main energy equation: 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

File:Normal and shear stress renew.svg Shear and Normal stress component on the surface element dA /

Examples

Example 1

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. 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

File:Compressor renew.svg Energy balance for a compressor

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

For ${\dot{W}}_{shaft}=0$, ${\dot{W}}_{shear}=0$ and incompressible flow:

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

Thus with this equation I can calculate the loss of energy through a device.

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 work done by a pump or a turbine.

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 bottom

Differential Control Volume Analysis:Bernoulli Equation

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 Component of Momentum Equation 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$F_{S_s}=\underset{inlet}{\underbrace{pA}}-\underset{outlet}{\underbrace{(p+dp)(A+dA)}}+\underset{...}{\underbrace{(p+\frac{dp}{2})dA}}$Template:Center bottom Template:Center top$F_{S_s}=-Adp-\frac{1}{2}dpdA$Template:Center bottom

File:Differential control volume renew.svg Differential control volume for Bernoulli's Equation

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

File:Pressure on fluid element renew.svg Pressure Distribution on fluid element

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 theres 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 end for a flow around a car b) heat energy input in the heat engine c) addition of mechanical energy inside the flow of a ventilator

Examples

Example 1

Consider steady flow of water through a horizontal nozzle. Find $P_1$ as a function of flow rate $\dot{Q}$. 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 bottom From 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$P_1=P_{atm}+\rho \frac{8\rho {\dot{Q}}^2}{\pi D_1^4}({\left(\frac{D_1}{D_2}\right)}^4-1)$Template:Center bottom

Flow through horizontal nozzle Flow through horizontal nozzle

Example 2

Find a relation between the nozzle discharge velocity and the tank free surface height. Assume steady frictionless flow and uniform flow at 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, Template:Center top$U_2\cong \sqrt{2gh}$Template:Center bottom

File:Nozzle discharge velocity renew.svg Nozzle discharge velocity at the bottom of the tank

Example 3

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

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$(U_2^2-U_1^2)=\frac{2}{g}(p_1-p_2)$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.

File:Venturimeter 2.svg Application of Bernoulli Equation:Venturimeter