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.

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

Lagrangian versus Eulerian Approach: Substantial Derivative

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:

$U_j(x_1,t),P(x_1,t),\rho (x_1,t),T(x_1),{\tau }_{jk}(x_1,t)\Rightarrow \alpha (x_i,t)$

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

$x_i{|}_p(t+\Delta t)=x_i{|}_p(t)+{\int }_t^{t+\Delta t}{U}_i{|}_p(t^{\prime })d{t}^{\prime }$

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:

${\alpha }_p(t)=\alpha [(x_i{)}_p,t]$

Along the path of the particle:

${\alpha }_p(t+\Delta t)=\alpha [(x_i+\Delta x_i{)}_p,t+\Delta t]$

Hence,

$d{\alpha }_p=\alpha [((x_i+\Delta x_i{)}_p,t+\Delta t)-\alpha ((x_i{)}_p,t)]$

$d{\alpha }_p=\frac{\partial \alpha }{\partial t}dt+\frac{\partial \alpha }{\partial x_i}d{x}_{ip}$

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

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.

$\frac{d{\alpha }_p}{dt}=\frac{D\alpha }{Dt}=(\frac{\partial }{\partial t}+U_i\frac{\partial }{\partial x_i})\alpha$

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

Basic laws for a system

Conservation of mass

The mass of a system do not change:

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

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

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:

$F_i=\frac{d{P}_i}{dt}{|}_{system}$

$P_{i(system)}={\int }_{mass(system)}{U}_{i}dm={\int }_{V(system)}{U}_i\rho dV$

The first law of Thermodynamics

$\underset{heataddedonsystem}{\underbrace{dQ}}+\underset{workdoneonsystem}{\underbrace{dW}}=dE$

in the rate form:

$\dot{Q}+\dot{W}=\frac{dE}{dt}{|}_{system}$

where $E_{system}={\int }_{mass(system)}edm={\int }_{V(system)}e\rho dV$

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: $B_{system}={\int }_{mass(system)}bdm={\int }_{V(system)}b\rho dV$ Hence, $B=M,b=1$ $B=\overrightarrow{P},b=\overrightarrow{u}$ $B=E,b=e$

Control Volume versus System

One dimensional Reynolds Transport Theorem

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.

$B_{sys}(t)=B_{cv}(t)$

$B_{sys}(t+\Delta t)=B_{cv}(t+\Delta t)-B_I(t+\Delta t)+B_{II}(t+\Delta t)$

$\frac{\Delta B_{sys}}{\Delta t}=\frac{B_{sys}(t+\Delta t)-B_{sys}(t)}{\Delta t}=\frac{B_{cv}(t+\Delta t)-B_{cv}(t)}{\Delta t}-\frac{B_I(t+\Delta t)}{\Delta t}+\frac{B_{II}(t+\Delta t)}{\Delta t}$

The first term for $\Delta t\to 0$

$\underset{\Delta t\to 0}{lim}\frac{B_{cv}(t+\Delta t)-B_{cv}(t)}{\Delta t}=\frac{\partial B_{cv}}{\partial t}$

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

$B_{II}(t+\Delta t)={\rho }_2{b}_2\Delta V_2$

$B_{II}(t+\Delta t)={\rho }_2{b}_2{A}_2{l}_2={\rho }_2{b}_2{A}_2{U}_2\Delta t$

$B_{out}={\rho }_2{b}_2{A}_2{U}_2\Delta t$

Similarly

$B_I(t+\Delta t)={\rho }_1{b}_1\Delta V_1={\rho }_1{b}_1{A}_1{U}_1\Delta t=B_{in}$

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

$\frac{\Delta B_{sys}}{\Delta t}\to \frac{d{B}_{sys}}{dt}$

$\frac{B_{in}}{\Delta t}={\dot{B}}_{in}$

$\frac{B_{out}}{\Delta t}={\dot{B}}_{out}$

so that,

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

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: ${\dot{B}}_{out}={\int }_{CSout}d{B}_{out}={\int }_{CSout}\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA$ ${\dot{B}}_{in}={\int }_{CSin}d{B}_{in}=-{\int }_{CSin}\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA$ ${\dot{B}}_{in}$ and ${\dot{B}}_{out}$ are positive quantities. Therefore, the negative sign is introduced into ${\dot{B}}_{in}$, to compensate the negative value of $\overrightarrow{U}\cdot \overrightarrow{n}$. $\frac{d{B}_{sys}}{dt}=\frac{\partial B_{cv}}{\partial t}+{\int }_{CSout}\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA+{\int }_{CSin}\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA$ $\frac{d{B}_{sys}}{dt}=\frac{\partial }{\partial t}{\int }_{cv}\rho bdV+{\int }_{cs}\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA$ Since $\rho \overrightarrow{u}\cdot \overrightarrow{n}dA=dm$, RTT can be written as: $\frac{d{B}_{sys}}{dt}=\frac{\partial B_{cv}}{\partial t}+\underset{netfluxofBacrossCS}{\underbrace{{\int }_{CS}bdm}}$ 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 ${\overrightarrow{U}}_s$ ${\overrightarrow{U}}_r=\overrightarrow{U}-{\overrightarrow{U}}_s$ $\frac{d{B}_{sys}}{dt}=\frac{d}{dt}\left({\int }_{cv}b\rho dV\right)+{\int }_{cs}b\rho \overrightarrow{U_r}\cdot \overrightarrow{n}dA$ These issues will be covered again.

CV with multiple inlets and outlets CV with arbitrary shape used to derive Reynolds transport theorem Sign of the inflow and outflow fluxes to the CV 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

$B=M,b=1$

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

i.e

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

Assume $\rho$ = constant (incompressible)

$0=\rho \frac{\partial }{\partial t}{\int }_{cv}dV+\rho {\int }_{cs}\overrightarrow{U}\cdot \overrightarrow{n}dA$

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

$0=\rho {\int }_{cs}\overrightarrow{U}\cdot \overrightarrow{n}dA\to \underset{volumeflowrate}{\underbrace{{\int }_{cs}\overrightarrow{U}\cdot \overrightarrow{n}dA}}=0$

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,

$0={\int }_{cs}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA\to \text{Net mass flow rate is equal to zero}$

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

Linear Momentum equation for inertial control volume

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

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

${\frac{d\overrightarrow{P}}{dt}}_{system}={\overrightarrow{F}}_{onsystem}$

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.

${\overrightarrow{F}}_{onsystem}=\overrightarrow{F_S}+\overrightarrow{F_B}$

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.

${\overrightarrow{F}}_{pressure}=-{\int }_{A}P\overrightarrow{n}dA$

${\overrightarrow{F}}_{v.stresses}={\int }_A\tau dA$

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

at the initial moment $t$

${\overrightarrow{F}}_{onsystem}={\overrightarrow{F}}_{onCV}$

i.e. in component form:

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

First law of Thermodynamics

$B=E,b=e$

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

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

${\frac{dE}{dt}}_{system}=[\dot{Q}+\dot{W}{]}_{system}=[\dot{Q}+\dot{W}{]}_{cv}$

thus, the integral form of energy equation is:

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

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 $A_2$. $\frac{\partial }{\partial t}{\int }_{cv}\rho dV+{\int }_{cs}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=0$ The first term is zero and the second term can be analyzed by decomposing the integration area. ${\int }_{C{S}_I}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA+{\int }_{C{S}_{III}}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA+{\int }_{C{S}_{II}}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=0$ where the integration over $C{S}_{III}$ is zero, because there is no flow across the streamtube. Thus, $-\rho U_1{A}_1+0+\rho U_2{A}_2=0$ $\rho U_1{A}_1=\rho U_2{A}_2\to U_2=\frac{U_1{A}_1}{A_2}$ ${\dot{m}}_1={\dot{m}}_2$

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. $\frac{\partial }{\partial t}{\int }_{cv}\rho dV+{\int }_{cs}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=0$ Where the first term is zero due to steady state conditions. At section 1: ${\int }_{A_1}\rho {\overrightarrow{U}}_1\cdot {\overrightarrow{n}}_{1}dA=-\rho \left|U_{11}\right|A_1$ At section 3: ${\int }_{A_3}\rho {\overrightarrow{U}}_3\cdot {\overrightarrow{n}}_{3}dA=\rho \left|U_3\right|A_3={\dot{m}}_3$ At section 4: ${\int }_{A_4}\rho {\overrightarrow{U}}_4\cdot {\overrightarrow{n}}_{4}dA=-\rho \left|U_4\right|A_4=-\rho Q_4$ ${\int }_{A_1}\rho {\overrightarrow{U}}_1\cdot {\overrightarrow{n}}_{1}dA+{\int }_{A_2}\rho {\overrightarrow{U}}_2\cdot {\overrightarrow{n}}_{2}dA+{\int }_{A_3}\rho {\overrightarrow{U}}_3\cdot {\overrightarrow{n}}_{3}dA+{\int }_{A_4}\rho {\overrightarrow{U}}_4\cdot {\overrightarrow{n}}_{4}dA=0$

mass balance for a connector device Hence, the velocity at section 2 can be calculated by ${\int }_{A_2}\rho {\overrightarrow{U}}_2\cdot {\overrightarrow{n}}_{2}dA=-[-\rho \left|U_1\right|A_1+{\dot{m}}_3-\rho Q_4]$ 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. $F_i=F_{Si}+F_{Bi}=\frac{\partial }{\partial t}{\int }_{cv}{U}_i\rho dV+{\int }_{cs}{U}_i\rho U_j{n}_{j}dA$ where the derivative with respect to time is zero due to steady state conditions. $F_i={\int }_{C{S}_I}{U}_{1i}\rho U_{1j}{n}_{j}dA+{\int }_{C{S}_{II}}{U}_{2i}\rho U_{2j}{n}_{j}dA$ $F_i=-U_{1i}\underset{{\dot{m}}_1}{\underbrace{\left|{\overrightarrow{U}}_1\right|A_1\rho }}+U_{2i}\underset{{\dot{m}}_2}{\underbrace{\left|{\overrightarrow{U}}_2\right|A_2{\rho }_2}}$ ${\dot{m}}_1={\dot{m}}_2=\dot{m}$ $F_i=\dot{m}(U_{2i}-U_{1i})$

Force in a streamtube

Example 4

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.

$0={\int }_{cs}\rho U_i{n}_{i}dA(mass)$

$F_i=F_{Si}+F_{Bi}={\int }_{cs}{U}_i\rho U_j{n}_{j}dA(momentum)$

No body force in $x_1$ direction.

$F_1=F_{S1}={\int }_{cs}{U}_1\rho U_j{n}_{j}dA$

$F_{S1}=p_{a}A-p_{a}A+R_1$

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

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

$U_1=0$ at 2 and 3.

$R_1=-U_1\rho U_1{A}_{Jet}$

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 $C{V}_2$

$F_{S1}=p_{a}A+R_1=-U_1^2{A}_{jet}\rho$

$R_1=-p_{a}A-U_1^2{A}_{Jet}\rho$

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

$K_1=-R_1$

$F_{net}=-R_1-p_{a}A$

$F_{net}=p_{a}A+U_1^{2}A\rho -p_{a}A$

$F_{net}=U_1^{2}A\rho$

Example 5

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.

$U_1(L,x_2)=U(2\frac{x_2}{\delta }-{\left(\frac{x_2}{\delta }\right)}^2)$

Apply conservation of mass

$\rho {\int }_{cs}{U}_i{n}_{i}dA=0$

$\rho {\int }_0^h-U_{0}wd{x}_2+\rho {\int }_0^{\delta }Uwd{x}_2=0$

$U_{0}h={\int }_0^{\delta }Ud{x}_2$

$F_i=\underset{=0steadystate}{\underbrace{\frac{\partial }{\partial t}\int U_i\rho dV}}+{\int }_{cs}{U}_i\rho U_j{n}_{j}dA$

$F_1=-D={\int }_0^h{U}_1\rho U_j{n}_{j}dA+\underset{=0streamline}{\underbrace{{\int }_{CS2}{U}_1\rho U_j{n}_{j}dA}}+{\int }_0^{\delta }{U}_1\rho U_j{n}_{j}dA+\underset{=0wall}{\underbrace{\int U_1\rho U_j{n}_{j}dA}}$

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

$D=U_0^2\rho hw-\rho w{\int }_0^{\delta }{U}_1^{2}d{x}_2=\rho U_0{\int }_0^{\delta }{U}_{1}d{x}_2-\rho w{\int }_0^{\delta }{U}_0{U}_{1}d{x}_2$

Insert the mass conservation result into the momentum equation.

$D=\rho w{\int }_0^{\delta }{U}_1(U_0-U_1)d{x}_2{|}_{x_1=L}$

$U_1$ is known.Here,using $\frac{x_2}{\delta }=\eta$

$D=\rho w{U}_0^2\delta {\int }_0^1(2\eta -{\eta }^2)(1-2\eta +{\eta }^2)d\eta =\frac{2}{15}\rho U_0^{2}w\delta$

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. $F_i=F_{S_i}+\underset{=0}{\underbrace{F_{B_i}}}=\underset{=0\text{(steady)}}{\underbrace{\frac{\partial }{\partial t}{\int }_{cv}{U}_i\rho dV}}+{\int }_{cs}{U}_i\rho U_j{n}_{j}dA$ $F_{S_i}=R_1+p_{a}A-p_{a}A$ $R_1={\int }_{cs}{U}_1\rho U_j{n}_{j}dA$ $R_1=-U_1^1\rho \left|{\overrightarrow{U}}^1\right|A_1+U_1^2\rho \left|{\overrightarrow{U}}^2\right|A_2$ $=(U_1^2-U_1)\dot{m}((\left|{\overrightarrow{U}}_{jet}\right|)-\left|{\overrightarrow{U}}_v\right|(\cos \theta -1))$ from continuity, $0={\int }_{cs}\rho U\cdot ndA=-\left|{\overrightarrow{U}}^1\right|\rho A_1+\left|{\overrightarrow{U}}^2\right|\rho A_2$ $\rho \left|{\overrightarrow{U}}^1\right|A_1=\rho \left|{\overrightarrow{U}}^2\right|A_2\to \dot{m}$ $R_1=(U_1^2-U_1)\rho \left|U^1\right|A_1$ $\left|U^1\right|=|{\overrightarrow{U}}_{jet}-{\overrightarrow{U}}_v|$ $U_1^1=\left|{\overrightarrow{U}}_{jet}\right|-\left|{\overrightarrow{U}}_v\right|$ $U_1^2=(\left|{\overrightarrow{U}}_{jet}\right|-\left|{\overrightarrow{U}}_v\right|)cos\theta$ for $A_1=A_2$ $R_1=\rho {(\left|{\overrightarrow{U}}_{jet}\right|-\left|{\overrightarrow{U}}_v\right|)}^2(cos\theta -1)A_1$ $=(U_1^2-U_i)\dot{m}((\left|{\overrightarrow{U}}_{jet}\right|)-(\left|{\overrightarrow{U}}_v\right|))$ $R_2={\int }_{cs}{U}_2\rho U_j{n}_{j}dA$ $R_2={\int }_{A_1}{U}_2^1\rho U_j^2{n}_j^{2}dA+{\int }_{A_2}{U}_2^2\rho U_j^2{n}_j^{2}dA$ at 1, $U_2^1=0$ and at 2, $U_2^2=\left|{\overrightarrow{U}}^2\right|sin\theta$. $R_2={\int }_{A_2}{U}_2^2\rho U_j^2{n}_j^{2}dA$ $R_2=U_2^2\rho \left|{\overrightarrow{U}}_2\right|A_2$ $R_2={\left|{\overrightarrow{U}}^2\right|}^2\rho sin\theta A_2$ $\left|{\overrightarrow{U}}^1\right|=\left|{\overrightarrow{U}}^2\right|=|{\overrightarrow{U}}_{jet}-{\overrightarrow{U}}_v|$ $\overrightarrow{U}={\overrightarrow{U}}_{jet}-{\overrightarrow{U}}_v$

Momentum Equation for CV with rectilinear acceleration

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

$\overrightarrow{F}=\frac{\partial }{\partial t}{\int }_{cv}\overrightarrow{U}\rho dV+{\int }_{cs}\overrightarrow{U}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$

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,

$\frac{d{\overrightarrow{U}}_X}{dt}=\frac{d{\overrightarrow{U}}_x}{dt}+\frac{d{\overrightarrow{U}}_{rf}}{dt}\Rightarrow \frac{d{\overrightarrow{U}}_X}{dt}\ne \frac{d{\overrightarrow{U}}_x}{dt}$

or,

${\overrightarrow{a}}_X={\overrightarrow{a}}_x+{\overrightarrow{a}}_{rf}\Rightarrow \frac{d{\overrightarrow{P}}_X}{dt}\ne \frac{d{\overrightarrow{P}}_x}{dt}$

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:

${\overrightarrow{F}|}_{system}={\frac{d{\overrightarrow{P}}_X}{dt}|}_{system}$

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

${\overrightarrow{U}}_x={\overrightarrow{U}}_X-{\overrightarrow{U}}_{rf}$

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:

$\frac{d{\overrightarrow{U}}_x}{dt}=\frac{d{\overrightarrow{U}}_X}{dt}-\frac{d{\overrightarrow{U}}_{rf}}{dt}(1)$

${\overrightarrow{a}}_x={\overrightarrow{a}}_X-{\overrightarrow{a}}_{rf}$

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

$\frac{d{\overrightarrow{U}}_{CV}}{dt}=\frac{d{\overrightarrow{U}}_{rf}}{dt}$

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).

${\overrightarrow{a}}_X={\overrightarrow{a}}_x+{\overrightarrow{a}}_{rf}$

and also

$\overrightarrow{F}={\int }_{mass(system)}{\overrightarrow{a}}_{X}dm$

$={\int }_{mass(system)}({\overrightarrow{a}}_x+{\overrightarrow{a}}_{rf})dm$

$\overrightarrow{F}-{\int }_{mass(system)}{\overrightarrow{a}}_{rf}dm={\int }_{mass(system)}{\overrightarrow{a}}_{x}dm$

$\overrightarrow{F}-{\int }_{mass(system)}{\overrightarrow{a}}_{rf}dm={\int }_{mass(system)}\frac{d{\overrightarrow{U}}_x}{dt}dm=\frac{d}{dt}{\int }_{mass(system)}{\overrightarrow{U}}_{x}dm$

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

For a moving CV we know that

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

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

${\overrightarrow{F}}_{onsystem}-{\int }_{Vsystem}{\overrightarrow{a}}_{rf}\rho dV={\overrightarrow{F}}_{oncv}-{\int }_{cv}{\overrightarrow{a}}_{rf}\rho dV$

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

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). $\underset{A}{\underbrace{F_2}}-\underset{B}{\underbrace{{\int }_{cv}{\overrightarrow{a}}_{rf2}\rho dV}}=\underset{C}{\underbrace{\frac{\partial }{\partial t}{\int }_{cv}{\overrightarrow{U}}_2\rho dV}}+\underset{D}{\underbrace{{\int }_{cs}{U}_2\rho {\overrightarrow{U}}_x\cdot \overrightarrow{n}dV}}$ A: $F_2=-g{\int }_{cv}\rho dV=-g{M}_{cv}$ $M_{cv}=M_0-\dot{m}t$ $F_2=-g(M_0-\dot{m}t)$ B: Since ${\overrightarrow{a}}_{rf}$ is not a function of the coordinates: $-{\int }_{cv}{a}_{rf2}\rho dV=-a_{rf2}{\int }_{cv}\rho dV$ $-{\int }_{cv}{a}_{rf2}\rho dV=-a_{rf2}(M_0-\dot{m}t)$

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): $\frac{\partial }{\partial t}{\int }_{cv}{U}_2\rho dV=\frac{\partial }{\partial t}{\int }_{cvI}{U}_2\rho dV+\frac{\partial }{\partial t}{\int }_{cvII}{U}_2\rho dV$ 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: $\frac{\partial }{\partial t}[\underset{U_2=0\to =0}{\underbrace{{\int }_{CVI}{U}_2\rho dV}}+{\int }_{CVII}{U}_2\rho dV]=\frac{\partial }{\partial t}{\int }_{CVII}{U}_2\rho dV\approx 0$ D: ${\int }_{cs}{U}_2\rho {\overrightarrow{U}}_x\cdot \overrightarrow{n}dA=U_2{\int }_{cs}\rho {\overrightarrow{U}}_x\cdot \overrightarrow{n}dA=U_2\dot{m}=-U_e\dot{m}$ Substitution of all the terms gives: $-g(M_0-\dot{m}t)-a_{rf2}(M_0-\dot{m}t)=-U_e\dot{m}$ $a_{rf2}=\frac{U_e\dot{m}}{M_0-\dot{m}t}-g$ $\frac{d{V}_{cv}}{dt}=a_{rf2}=\frac{U_e\dot{m}}{M_0-\dot{m}t}-g$ $V_{cv}(t)={\int }_0^t\frac{U_e\dot{m}}{M_0-\dot{m}t}dt-{\int }_0^{t}gdt$ $V_{cv}(t)=-U_{e}ln(1-\frac{\dot{m}t}{M_0})-gt$ 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.

Control Volume I and II

Extension of Energy equation for CV

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

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

${\dot{W}}_{oncv}={\dot{W}}_{body}+{\dot{W}}_{surface}$

here $z=x_2$

${\dot{W}}_{body}={\dot{W}}_{shaft}+{\dot{W}}_{elec}+{\dot{W}}_{other}$

${\dot{W}}_{surface}={\dot{W}}_{normal}+{\dot{W}}_{shear}$

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

${\overrightarrow{\tau }}_n$ is the normal stress normal to dA. In many cases ${\overrightarrow{\tau }}_n=-p\overrightarrow{n}dA$ since $\dot{W}=\overrightarrow{F}\cdot \overrightarrow{U}$ $d\dot{W}=d\overrightarrow{F}\cdot \overrightarrow{U}$ $d{\dot{W}}_{normal}={\overrightarrow{\tau }}_{n}dA\cdot \overrightarrow{U}\to {\dot{W}}_{normal}={\int }_{cs}{\overrightarrow{\tau }}_n\cdot \overrightarrow{U}dA=-{\int }_{cs}p\overrightarrow{n}\cdot \overrightarrow{U}dA$ $d{\dot{W}}_{shear}=\overrightarrow{\tau }dA\cdot \overrightarrow{U}\to {\dot{W}}_{shear}={\int }_{cs}\overrightarrow{\tau }\cdot \overrightarrow{U}dA$ Inserting those into the main energy equation: $\dot{Q}+{\dot{W}}_{shaft}+{\dot{W}}_{shear}+{\dot{W}}_{other}=\frac{\partial }{\partial t}\int e\rho dV+{\int }_{cs}(\underset{h:enthalpy}{\underbrace{u+\frac{p}{\rho }}}+\frac{\left|{\overrightarrow{U}}^2\right|}{2}+g{x}_2)\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$

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: $\dot{Q}+{\dot{W}}_{shaft}+\underset{=0=\tau \cdot \overrightarrow{U}}{\underbrace{{\dot{W}}_{shear}}}+\underset{=0}{\underbrace{{\dot{W}}_{other}}}=\underset{=0steadystate}{\underbrace{\frac{\partial }{\partial t}\int e\rho dV}}+{\int }_{cs}(u+\frac{p}{\rho }+\frac{U^2}{2}+g{x}_2)\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ $0=\underset{=0steadystate}{\underbrace{\frac{\partial }{\partial t}{\int }_{cv}\rho dV}}+{\int }_{cs}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA\to \left|{\rho }_1{U}_1{A}_1\right|=\left|{\rho }_2{U}_2{A}_2\right|=\dot{m}$ 2: $\dot{Q}=-{\dot{W}}_{shaft}+{\int }_{cs}(u+\frac{p}{\rho }+\frac{U^2}{2}+gz)\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ For uniform properties at 1 and 2 and inserting the inserting the relation for the enthalpy $h=u+\frac{p}{\rho }$. $\dot{Q}=-{\dot{W}}_{shaft}-(h_1+\underset{=0}{\underbrace{\frac{U_1^2}{2}}}+g{z}_1)\left|{\rho }_1{U}_1{A}_1\right|+(h_2+\frac{U_2^2}{2}+g{z}_2)\left|{\rho }_2{A}_2{U}_2\right|$ $\dot{Q}=-{\dot{W}}_{shaft}+\dot{m}[h_2+\frac{U_2^2}{2}-h_1+\underset{=0}{\underbrace{g(z_2-z_1)}}]$ Assuming that air behaves like an ideal gas with a constant $c_p$. $h_2-h_1=c_p(T_2-T_1)$ $\dot{Q}=-{\dot{W}}_{shaft}+\dot{m}[c_p(T_2-T_1)+\frac{U_2^2}{2}]$

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.

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

For uniform flow properties at the inlet and outlet.

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

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

Reform:

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

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

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

$\frac{p}{\rho }+\frac{U^2}{2}+gz$: 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.

$h_{loss}=[u_2-u_1-\frac{\dot{Q}}{\dot{m}}]\frac{1}{g}$

i.e.

$\frac{p_1}{\rho g}+\frac{U_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{U_2^2}{2g}+z_2+h_{loss}$

One can add the work done by a pump or a turbine.

$\frac{p_1}{\rho g}+\frac{U_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{U_2^2}{2g}+z_2+h_{loss}-h_{pump}+h_{turbine}$

Differential Control Volume Analysis:Bernoulli Equation

Consider the steady, incompressible, frictionless flow through the differential CV for a stream tube. Continuity $0=\frac{\partial }{\partial t}\underset{=0}{\underbrace{{\int }_{cv}\rho dV}}+{\int }_{cs}\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ $0=-{\dot{m}}_{in}+{\dot{m}}_{out}\leftrightarrow {\dot{m}}_{in}={\dot{m}}_{out}=\dot{m}$ Component of Momentum Equation $F_{S_s}+F_{B_s}=\frac{\partial }{\partial t}\underset{=0}{\underbrace{\int U_s\rho dV}}+{\int }_{cs}{U}_s\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$ $F_{S_s}=\underset{inlet}{\underbrace{pA}}-\underset{outlet}{\underbrace{(p+dp)(A+dA)}}+\underset{\mathrm{...}}{\underbrace{(p+\frac{dp}{2})dA}}$ $F_{S_s}=-Adp-\frac{1}{2}dpdA$

Differential control volume for Bernoulli's Equation

$F_{B_s}=\rho g_{s}dV=\rho (-gsin\theta )(A+\frac{dA}{2})ds$ with: $sin\theta ds=dz$ Then, $F_{B_s}=-\rho g(A+\frac{dA}{2})dz$ ${\int }_{cs}{U}_s\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=-U_s\dot{m}+(U_s+d{U}_s)\dot{m}=\dot{m}d{U}_s$ $-Adp-\underset{\approx 0}{\underbrace{\frac{1}{2}dpdA}}-\rho g(A+\frac{dA}{2})dz=\dot{m}d{U}_s$ $-Adp-\rho gAdz=\dot{m}d{U}_s$ where $\dot{m}=\rho U_{s}A$ $-Adp-\rho gAdz=\rho U_{s}Ad{U}_s$ $\frac{dp}{\rho }+U_{s}d{U}_s+gdz=0$ with: $U_{s}d{U}_s=d\left(\frac{U_s^2}{2}\right)$ Then, $\frac{dp}{\rho }+d\left(\frac{U_s^2}{2}\right)+gdz=0$

Pressure Distribution on fluid element

Integrate between 1 and 2 along a streamline:

$\frac{p_1}{\rho }+\frac{U_1^2}{2}+g{z}_1=\frac{p_2}{\rho }+\frac{U_2^2}{2}+g{z}_2=constant$

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.

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. $\frac{p_1}{\rho }+\frac{U_1^2}{2}=\frac{p_2}{\rho }+\frac{U_2^2}{2}$ $p_1=p_{atm}+\rho (\frac{U_2^2}{2}-\frac{U_1^2}{2})=p_{atm}+\rho \frac{U_1^2}{2}(\frac{U_2^2}{U_1^2}-1)$ From continuity: $-\left|\rho U_1{A}_1\right|+\left|\rho U_2{A}_2\right|=0$ $U_1=\frac{\dot{Q}}{A_1}\text{and}{U}_2=\frac{\dot{Q}}{A_2}$ $P_1=P_{atm}+\rho \frac{U_1^2}{2}({\left(\frac{A_1}{A_2}\right)}^2-1)$ $P_1=P_{atm}+\rho \frac{{\dot{Q}}^2}{2A_1^2}({\left(\frac{A_1}{A_2}\right)}^2-1)$ $P_1=P_{atm}+\rho \frac{8\rho {\dot{Q}}^2}{\pi D_1^4}({\left(\frac{D_1}{D_2}\right)}^4-1)$

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. $\frac{p_1}{\rho }+\frac{U_1^2}{2}+g{z}_1=\frac{p_2}{\rho }+\frac{U_2^2}{2}+g{z}_2$ $p_1=p_2=p_{atm}$ $U_1{A}_1=U_2{A}_2$ $U_2^2-U_1^2=2g(z_2-z_1)$ $U_2^2[1-{\left(\frac{A_2}{A_1}\right)}^2]=2gh$ $U_2^2=\frac{2gh}{[1-{\left(\frac{A_2}{A_1}\right)}^2]}$ for steadiness $A_1>>A_2$, thus, $U_2\cong \sqrt{2gh}$

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

$p_1+\rho \frac{U_1^2}{2}=p_2+\rho \frac{U_2^2}{2}$ $(U_2^2-U_1^2)=\frac{2}{g}(p_1-p_2)$ $U_1{A}_1=U_2{A}_2=\dot{Q}$ $U_1=\frac{\dot{Q}}{A_1}\text{and}{U}_2=\frac{\dot{Q}}{A_2}$ $(\frac{{\dot{Q}}^2}{A_2^2}-\frac{{\dot{Q}}^2}{A_1^2})=\frac{2}{\rho }\Delta p$ ${\dot{Q}}^2(\frac{1}{A_2^2}-\frac{1}{A_1^2})=\frac{2}{\rho }\Delta p$ ${\dot{Q}}^2\frac{1}{A_2^2}(1-\frac{A_2^2}{A_1^2})=\frac{2}{\rho }\Delta p$ If $\beta =\frac{D_2}{D_1}$, then: $\dot{Q}=\sqrt{\frac{A_2^{2}2\underset{measured}{\underbrace{\Delta p}}}{\rho (1-{\beta }^4)}}$ That is the method for measuring flow rate.

Application of Bernoulli Equation:Venturimeter