# Fluid Mechanics for Mechanical Engineers/Integral Analysis of Fluid Flow

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## Introduction

Differential Approach seek solution at every point $\displaystyle (x_{1},x_{2},x_{3})$, 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.

## 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.

## Basic laws for a system
### Conservation of mass

The mass of a system does not change:
$\displaystyle {\frac {dM}{dt}}{\bigg |}_{system}=0$
where, $\displaystyle \displaystyle M=\int _{mass(system)}{dm}=\int _{V(system)}{\rho dV}$

### 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 ($\displaystyle {\vec {P}}$) of the system:
$\displaystyle {\vec {F}}_{on\,system}={\frac {d{\vec {P}}}{dt}}{\bigg |}_{system}$ or in tensor form $\displaystyle F_{i}={\frac {dP_{i}}{dt}}{\bigg |}_{system}$
where
$\displaystyle \displaystyle P_{i(system)}=\int _{mass(system)}{U_{i}dm}=\int _{V(system)}{U_{i}\rho dV}$
and
$\displaystyle {\vec {U}}=U_{x}{\vec {i}}+U_{y}{\vec {j}}+U_{z}{\vec {k}}=U_{1}{\vec {e}}_{1}+U_{2}{\vec {e}}_{2}+U_{3}{\vec {e}}_{3}$
where ${\vec {e}}_{1},{\vec {e}}_{2}\,{\text{and}}\,{\vec {e}}_{3}$ 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 $\displaystyle {\vec {\omega }}$, 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 ($\displaystyle {\vec {H}}$) of the system:
$\displaystyle {\vec {T}}_{on\,system}={\frac {d{\vec {H}}}{dt}}{\bigg |}_{system}$ or in tensor form $\displaystyle T_{i\,on\,system}={\frac {dH_{i}}{dt}}{\bigg |}_{system}$
where
$\displaystyle {\vec {H}}=\int _{mass(system)}{\vec {r}}\times {\vec {U}}dm$
$\displaystyle {\vec {T}}={\vec {r}}\times {\vec {F}}+{\vec {T}}_{shaft}$

### The First law of Thermodynamics

$\displaystyle \underbrace {dQ} _{heat\ added\ on\ system}+\underbrace {dW} _{work\ done\ on\ system}=dE_{system}$
in the rate form:
$\displaystyle {\dot {Q}}+{\dot {W}}={\frac {dE}{dt}}{\bigg |}_{system}$
where $\displaystyle \displaystyle E_{system}=\int _{mass(system)}{edm}=\int _{V(system)}{e\rho dV}$ and $\displaystyle \displaystyle e=\underbrace {u} _{Intenral\ energy}+\underbrace {\frac {U_{i}U_{i}}{2}} _{Kinetic\ energy}+\underbrace {gz} _{Potential\ energy}$

### The Second law of Thermodynamics

The second law states that if heat is added to a system at a temperature T{\textstyle T}, the entropy of the system rises more than the heat added per unit temperature:
$\displaystyle dS_{system}\geq {\frac {dQ}{T}}{\bigg |}_{system}$
The rate form of this law is:
$\displaystyle {\frac {dS}{dt}}{\bigg |}_{system}\geq {\frac {1}{T}}{\dot {Q}}_{system}$
where $S_{system}=\int _{mass(system)}sdm=\int _{V(system)}s\rho dV$

## Relation of a system derivative to the control volume derivative

Consider a fire extinguisher
$\displaystyle {\frac {dM}{dt}}{\bigg |}_{system}=0$ whereas $\displaystyle {\frac {dM}{dt}}{\bigg |}_{cv}<0$
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
$\displaystyle {\frac {dB}{dt}}{\bigg |}_{system}$ to $\displaystyle {\frac {dB}{dt}}{\bigg |}_{cv}$
The variables appear in the physical laws (balance laws) of a system are:
* Mass ($\displaystyle M$),
* Momentum ($\displaystyle P_{i}$),
* Energy ($\displaystyle E$),
* Moment of momentum ($\displaystyle H_{i}$),
* Entropy ($\displaystyle S$).
They are called extensive properties. Let $\displaystyle B$ be any arbitrary extensive property. The corresponding intensive property $\displaystyle b$ is the extensive property per unit mass:
$\displaystyle B_{system}=\int _{mass(system)}{b\ dm}=\int _{V(system)}{b\rho dV}$
Hence,
$\displaystyle B=M,\,b=1$
$\displaystyle B={\vec {P}},\,b={\vec {U}}$
$\displaystyle B={\vec {H}},\ b={\vec {r}}\times {\vec {U}}$
$\displaystyle B=E,\ b=e$
$\displaystyle B=S,\ b=s$

### One dimensional Reynolds Transport Theorem

Consider a flow through a nozzle.
If $\displaystyle B$ is an extensive variable of the system.
$\displaystyle B_{sys}(t)=B_{cv}(t)$
$\displaystyle \displaystyle B_{sys}(t+\Delta t)=B_{cv}(t+\Delta t)-B_{I}(t+\Delta t)+B_{II}(t+\Delta t)$
$\displaystyle {\frac {\Delta B_{sys}}{\Delta t}}={\frac {B_{sys}(t+\Delta t)-B_{sys}(t)}{\Delta t}}={\frac {B_{cv}(t+\Delta t)-B_{cv}(t)}{\Delta t}}\ -\ {\frac {B_{I}(t+\Delta t)}{\Delta t}}\ +\ {\frac {B_{II}(t+\Delta t)}{\Delta t}}$
The first term for $\displaystyle \Delta t\rightarrow 0$
$\displaystyle \lim _{\Delta t\rightarrow 0}{\frac {B_{cv}(t+\Delta t)-B_{cv}(t)}{\Delta t}}={\frac {\partial B_{cv}}{\partial t}}$
$\displaystyle B_{II}(t+\Delta t)$ for $\displaystyle \Delta t\rightarrow 0$
$\displaystyle B_{II}(t+\Delta t)=\rho _{2}b_{2}V_{II}$
$\displaystyle \displaystyle B_{II}(t+\Delta t)=\rho _{2}b_{2}A_{2}l_{2}=\rho _{2}b_{2}A_{2}U_{2}\Delta t$
$\displaystyle \displaystyle B_{out}=\rho _{2}b_{2}A_{2}U_{2}\Delta t$
Similarly
$\displaystyle \displaystyle B_{I}(t+\Delta t)=\rho _{1}b_{1}V_{I}=\rho _{1}b_{1}A_{1}U_{1}\Delta t=B_{in}$
Thus, for $\displaystyle \Delta t\rightarrow 0$, the terms in the equality for the time derivative of the system are
$\displaystyle {\frac {\Delta B_{sys}}{\Delta t}}\rightarrow {\frac {dB_{sys}}{dt}}$
so that,
$\displaystyle {\frac {dB_{sys}}{dt}}={\frac {\partial B_{cv}}{\partial t}}+{\dot {B}}_{out}-{\dot {B}}_{in}$
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, 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.

### 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 $({\vec {n}})$ defined at each point on the CS. The outflow and inflow flow rate of $B$ across CS can be written as:
$\displaystyle {\dot {B}}_{out}=\int _{CSout}{d{\dot {B}}_{out}}=\int _{CSout}{\rho b{\vec {U}}\cdot {\vec {n}}dA}$
$\displaystyle {\dot {B}}_{in}=\int _{CSin}{d{\dot {B}}_{in}}=-\int _{CSin}{\rho b{\vec {U}}\cdot {\vec {n}}dA}$
$\displaystyle {\dot {B}}_{in}$ and $\displaystyle {\dot {B}}_{out}$ are positive quantities. Therefore, the negative sign is introduced into $\displaystyle {\dot {B}}_{in}$, to compensate the negative value of $\displaystyle {\vec {U}}\cdot {\vec {n}}$.
$\displaystyle {\frac {dB_{sys}}{dt}}={\frac {\partial B_{cv}}{\partial t}}+\int _{CSout}{\rho b{\vec {U}}\cdot {\vec {n}}dA}+\int _{CSin}{\rho b{\vec {U}}\cdot {\vec {n}}dA}$
$\displaystyle {\frac {dB_{sys}}{dt}}={\frac {\partial }{\partial t}}\int _{cv}{\rho bdV}+\int _{cs}{\underbrace {\rho b{\vec {U}}\cdot {\vec {n}}} _{flux\ of\ B}dA}$
Since $\displaystyle \rho {\vec {U}}\cdot {\vec {n}}dA=d{\dot {m}}$, RTT can be written as:
$\displaystyle {\frac {dB_{sys}}{dt}}={\frac {\partial B_{cv}}{\partial t}}+\underbrace {\int _{CS}{bd{\dot {m}}}} _{net\ flow\ rate\ of\ B\ across\ CS}$

#### Conservation of mass

$\displaystyle B=M,b=1$
$\displaystyle {\frac {dM_{sys}}{dt}}={\frac {\partial }{\partial t}}\int _{cv}{1\rho dV}+\int _{cs}{1\rho {\vec {U}}\cdot {\vec {n}}dA}=0$
i.e
$\displaystyle \underbrace {{\frac {\partial }{\partial t}}\int _{cv}{\rho dV}} _{rate\ of\ change\ of\ mass\ in\ CV}+\underbrace {\int _{cs}{\underbrace {\rho {\vec {U}}\cdot {\vec {n}}} _{mass\ flux}dA}} _{net\ mass\ flow\ rate\ through\ the\ CS}=0$
Assume $\displaystyle \rho$ = constant (incompressible) and uniform in the CV
$\displaystyle 0=\rho {\frac {\partial }{\partial t}}\int _{cv}{dV}+\rho \int _{cs}{{\vec {U}}\cdot {\vec {n}}dA}$
As $\displaystyle V$ of CV is also constant, the time derivative drops out:
$\displaystyle 0=\rho \int _{cs}{{\vec {U}}\cdot {\vec {n}}dA}\rightarrow \underbrace {\int _{cs}{{\vec {U}}\cdot {\vec {n}}dA}} _{volume\ flow\ rate}=0$
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,
$\displaystyle 0=\int _{cs}{\rho {\vec {U}}\cdot {\vec {n}}dA}$
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)

$\displaystyle B={\vec {P}}$, $\displaystyle b={\vec {U}}$
$\displaystyle {\frac {d{\vec {P}}_{sys}}{dt}}={\vec {F}}_{on\ system}$
Using RTT, the rate of change of momentum of the system can be related to that of CV as follows:
$\displaystyle {\frac {d{\vec {P}}_{sys}}{dt}}={\frac {\partial }{\partial t}}\int _{cv}{{\vec {U}}\rho dV}+\underbrace {\int _{cs}{\underbrace {{\vec {U}}\rho {\vec {U}}\cdot {\vec {n}}} _{momentum\ flux}dA}} _{net\ momentum\ flow\ rate\ across\ CS}={\vec {F}}_{on\ system}$
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.
$\displaystyle {\vec {F}}_{on\ system}={\vec {F_{S}}}+{\vec {F_{B}}}$
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.
$\displaystyle {\vec {F}}_{pressure}=-\int _{A}{P{\vec {n}}dA}$
$\displaystyle {\vec {F}}_{v.stresses}=\int _{A}{\tau dA}$
The body forces can be due to gravity or magnetic field.
At the initial moment $\displaystyle t$
$\displaystyle {\vec {F}}_{on\ system}={\vec {F}}_{on\ CV}$
i.e. in the component form:
$\displaystyle F_{Si}+F_{Bi}={\frac {\partial }{\partial t}}\int _{cv}{U_{i}\rho dV}+\int _{cs}{U_{i}\rho U_{j}n_{j}dA}$

### Moment-of-momentum Equation

$\displaystyle B={\vec {H}}$, $\displaystyle b={\vec {r}}\times {\vec {U}}$
Net torque exerted on the system about the rotaional axis is equal to the rate of change of angular momentum of the system.
$\displaystyle \sum {T_{i}}={\frac {dH_{i}}{dt}}{\bigg |}_{system}$
expanding both sides by using the RTT
$\displaystyle {\vec {T}}_{shaft}+\sum {\vec {r}}\times {\vec {F}}_{s}+\int _{CV}{\vec {r}}\times {\vec {g}}\rho dV={\frac {\partial }{\partial t}}\int _{cv}{\vec {r}}\times {\vec {U}}\rho dV+\int _{cs}{\vec {r}}\times {\vec {U}}\rho {\vec {U}}\cdot {\vec {n}}\,dA$
where $\displaystyle {\vec {F}}_{s}$ is the surface force acting on the surfaces of the CV.

### First Law of Thermodynamics

$\displaystyle B=E,b=e$
$\displaystyle {\frac {dE}{dt}}_{system}={\frac {\partial }{\partial t}}\int _{cv}{e\rho dV}+\int _{cs}{e\rho {\vec {U}}\cdot {\vec {n}}dA}$
at the initial moment $\displaystyle t$, the following equality is valid:
$\displaystyle {\frac {dE}{dt}}_{system}=[{\dot {Q}}+{\dot {W}}]_{system}=[{\dot {Q}}+{\dot {W}}]_{cv}$
thus, the integral form of energy equation is:
$\displaystyle [{\dot {Q}}+{\dot {W}}]_{cv}={\frac {\partial }{\partial t}}\int _{cv}{e\rho dV}+\int _{cs}{e\rho {\vec {U}}\cdot {\vec {n}}dA}$

### Second Law of Thermodynamics

$\displaystyle B=S,b=s$
$\displaystyle {\frac {dS}{dt}}{\bigg |}_{system}\geq {\frac {1}{T}}{\dot {Q}}_{system}$
The left hand side of the inequality reads
$\displaystyle {\frac {dS}{dt}}{\bigg |}_{system}={\frac {\partial }{\partial t}}\int _{cv}{s\rho dV}+\int _{cs}{s\rho {\vec {U}}\cdot {\vec {n}}dA}$
The right hand side of the above inequality reads
$\displaystyle {\frac {1}{T}}{\dot {Q}}_{system}={\frac {1}{T}}{\dot {Q}}_{CV}=\int _{CS}{\frac {1}{T}}\left({\frac {\dot {Q}}{A}}\right)dA$
Hence the second law becomes:
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.
Let $\displaystyle A_{1}$ and $\displaystyle A_{2}$ be too small such that the velocities $\displaystyle {\vec {U}}$ at position 1 and 2 are uniform across $\displaystyle A_{1}$ and $\displaystyle A_{2}$.
When we write the conservation of mass equation in integral form:
$\displaystyle {\frac {\partial }{\partial t}}\int _{cv}{\rho dV}+\int _{cs}{\rho {\vec {U}}\cdot {\vec {n}}dA}=0$
We assume that;
1) In a steady flow volume of the stream tube, i.e. CV is constant, ( $\displaystyle {\frac {\partial }{\partial t}}\int _{cv}{dV}$ = 0)
2) Density is constant and uniform in the CV ($\rho _{1}=\rho _{2}=\rho :constant$).
Hence
$\displaystyle {\frac {\partial }{\partial t}}\int _{cv}{\rho dV}=0$
When we rewrite the conservation of mass equation:
$\int _{cs}{\rho {\vec {U}}\cdot {\vec {n}}dA}=0$
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 :
$CS_{I}$ ,$CS_{II}$ and $CS_{III}$.
* $CS_{I}$ : the surface area that the mass enters across.
$U_{1}=\left\vert {\vec {U}}\right\vert \left\vert {\vec {n}}\right\vert \cos \theta$
${\vec {n}}$ and ${\vec {U}}$ are in the opposite directions. $90^{\circ }\leq \theta \leq 180^{\circ }$and $\cos \theta$ is always negative. That is, the sign of mass input is always negative.
* $CS_{II}$ : the surface area from which the mass leaves.
$U_{2}=\left\vert {\vec {U}}\right\vert \left\vert {\vec {n}}\right\vert \cos \theta$
${\vec {n}}$ and ${\vec {U}}$ are in the same directions. $\theta \leq 90^{\circ }$and $\cos \theta$ is always positive. That is, the sign of mass output is always positive.
* $CS_{III}$ : the lateral surfaces of the CV.
$U_{3}=\left\vert {\vec {U}}\right\vert \left\vert {\vec {n}}\right\vert \cos \theta$
${\vec {n}}$ and ${\vec {U}}$ are perpendicular to each other . $\theta =90^{\circ }$and $\cos \theta$ is zero. That is, there is no flows.
$\displaystyle \int _{CS_{I}}{\rho {\vec {U}}\cdot {\vec {n}}dA_{1}}+\int _{CS_{III}}{\rho {\vec {U}}\cdot {\vec {n}}}dA_{3}+\int _{CS_{II}}{\rho {\vec {U}}\cdot {\vec {n}}dA_{2}}=0$
where the integration over $\displaystyle CS_{III}$ is zero, because there is no flow across the streamtube. Thus,
$\displaystyle \displaystyle -\rho U_{1}A_{1}+0+\rho U_{2}A_{2}=0$
$\displaystyle \rho U_{1}A_{1}=\rho U_{2}A_{2}\rightarrow U_{2}={\frac {U_{1}A_{1}}{A_{2}}}$
$\displaystyle {\dot {m}}_{1}={\dot {m}}_{2}$

#### 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 $\displaystyle A_{1}$, $\displaystyle A_{2}$ and $\displaystyle A_{3}$ = $\displaystyle A_{4}$.
The following parameters are known:
Mass flow out at 3 ($\displaystyle {\dot {m}}_{3}$).
Volume flow rate in through 4 ($\displaystyle {Q}_{4}$).
Velocity at 1 along $\displaystyle x_{1}$-direction $\displaystyle {\vec {U}}_{11}$, $\displaystyle {\vec {U}}_{1i}=(U_{11},0,0)$ so that $\displaystyle U_{11}>0$.
Find the flow velocity at section 2.Assume that the properties are uniform across the sections.
$\displaystyle {\frac {\partial }{\partial t}}\int _{cv}{\rho dV}+\int _{cs}{\rho {\vec {U}}\cdot {\vec {n}}dA}=0$
Where the first term is zero due to steady state conditions. At section 1:
$\displaystyle \int _{A_{1}}{\rho {\vec {U}}_{1}\cdot {\vec {n}}_{1}dA}=-\rho \left|U_{11}\right|A_{1}$ At section 3:
$\displaystyle \int _{A_{3}}{\rho {\vec {U}}_{3}}\cdot {\vec {n}}_{3}dA=\rho \left|U_{3}\right|A_{3}={\dot {m}}_{3}$
At section 4:
$\displaystyle \int _{A_{4}}{\rho {\vec {U}}_{4}\cdot {\vec {n}}_{4}dA}=-\rho \left|U_{4}\right|A_{4}=-\rho Q_{4}$
$\displaystyle \int _{A_{1}}{\rho {\vec {U}}_{1}\cdot {\vec {n}}_{1}dA}+\int _{A_{2}}{\rho {\vec {U}}_{2}\cdot {\vec {n}}_{2}}dA+\int _{A_{3}}{\rho {\vec {U}}_{3}\cdot {\vec {n}}_{3}dA}+\int _{A_{4}}{\rho {\vec {U}}_{4}\cdot {\vec {n}}_{4}dA}=0$
Hence, the velocity at section 2 can be calculated by
$\displaystyle \int _{A_{2}}{\rho {\vec {U}}_{2}\cdot {\vec {n}}_{2}dA}=-\left[-\rho \left|U_{11}\right|A_{1}+{\dot {m}}_{3}-\rho Q_{4}\right]$
For $\displaystyle {\vec {n}}_{2}=(0,-1)\rightarrow U_{21}=0$
$\displaystyle -\rho U_{22}A_{2}=\rho \left|U_{11}\right|A_{1}-{\dot {m}}_{3}+\rho Q_{4}$
If the term on the right side is positive, $\displaystyle U_{2}$ should be negative (outflow) and if it is negative, $\displaystyle U_{2}$ 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.
$\displaystyle F_{i}=F_{Si}+F_{Bi}={\frac {\partial }{\partial t}}\int _{cv}{U_{i}\rho _{1}dV}+\int _{cs}{U_{i}\rho _{2}U_{j}n_{j}dA}$
where the derivative with respect to time is zero due to steady state conditions. (${\frac {\partial }{\partial t}}\int _{cv}{U_{i}\rho dV}$=0 ).
There are 3 types of the control surfaces :
$CS_{I}$ ,$CS_{II}$ and $CS_{III}$.
* $CS_{I}$ : the surface area that the mass enters across.
$U_{1}=\left\vert {\vec {U}}\right\vert \left\vert {\vec {n}}\right\vert \cos \theta$ (general formula)
${\vec {n}}$ and ${\vec {U}}$ are in the opposite directions. $90^{\circ }\leq \theta \leq 180^{\circ }$and $\cos \theta$ is always negative. That is, the sign of mass input is always negative.
* $CS_{II}$ : the surface area from which the mass leaves.
$U_{2}=\left\vert {\vec {U}}\right\vert \left\vert {\vec {n}}\right\vert \cos \theta$ (general formula)
${\vec {n}}$ and ${\vec {U}}$ are in the same directions. $\theta \leq 90^{\circ }$and $\cos \theta$ is always positive. That is, the sign of mass output is always positive.
* $CS_{III}$ : the lateral surfaces of the CV.
$U_{3}=\left\vert {\vec {U}}\right\vert \left\vert {\vec {n}}\right\vert \cos \theta$ (general formula)
${\vec {n}}$ and ${\vec {U}}$ are perpendicular to each other . $\theta =90^{\circ }$and $\cos \theta$ is zero. That is, there is no flows.($\int _{CS_{III}}{U_{3i}\rho _{3}U_{3j}n_{j}dA_{3,lateral}}$ =0 )
$\displaystyle \displaystyle F_{i}=\int _{CS_{I}}{U_{1i}\rho _{1}U_{1j}n_{j}dA_{1}}+\int _{CS_{II}}{U_{2i}\rho _{2}U_{2j}n_{j}dA_{2}}+\int _{CS_{III}}{U_{3i}\rho _{3}U_{3j}n_{j}dA_{3,lateral}}$
$\displaystyle \displaystyle F_{i}=\int _{CS_{I}}{U_{1i}\rho _{1}U_{1j}n_{j}dA_{1}}+\int _{CS_{II}}{U_{2i}\rho _{2}U_{2j}n_{j}dA_{2}}$
Density is constant ($\rho _{1}=\rho _{2}=\rho _{3}=\rho =constant$)
$\displaystyle \displaystyle F_{i}=\rho \int _{CS_{I}}{U_{1i}U_{1j}n_{j}dA_{1}}+\rho \int _{CS_{II}}{U_{2i}U_{2j}n_{j}dA_{2}}$
When we integrate
$\displaystyle F_{i}=-U_{1i}\underbrace {\left|{\vec {U}}_{1}\right|A_{1}\rho } _{{\dot {m}}_{1}}\ +\ U_{2i}\underbrace {\left|{\vec {U}}_{2}\right|A_{2}\rho } _{{\dot {m}}_{2}}$
${\displaystyle {dm \over dt}-{\dot {m}}_{1}+{\dot {m}}_{2}=0}$ (steady state :$\displaystyle {dm \over dt}=0$)
${\displaystyle {\dot {m}}_{1}={\dot {m}}_{2}}={\dot {m}}$
$\displaystyle F_{i}={\dot {m}}(U_{2i}-U_{1i})$

#### 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 $\displaystyle x_{1}$ direction.
Independent from the selected CV.
$\displaystyle 0=\int _{cs}{\rho U_{i}n_{i}dA}\ (mass)$
$\displaystyle F_{i}=F_{Si}+F_{Bi}=\int _{cs}{U_{i}\rho U_{j}n_{j}dA}\ (momentum)$
No body force in $\displaystyle x_{1}$ direction.
$\displaystyle \displaystyle F_{1}=F_{S1}=\int _{cs}{U_{1}\rho U_{j}n_{j}dA}$
For the $CV_{I}$
$\displaystyle \displaystyle F_{S1}=p_{a}A-p_{a}A+R_{s1}$
$\displaystyle R_{s1}=\int _{cs}{U_{1}\rho U_{j}n_{j}dA}=\int _{CS1}{U_{1}^{1}\rho U_{j}^{1}n_{j}dA}+\underbrace {\int _{CS2}{U_{1}^{2}\rho U_{j}^{2}n_{j}dA}} _{0}+\underbrace {\int _{CS3}{U_{1}^{3}\rho U_{j}^{3}n_{j}dA}} _{0}$
$\displaystyle U_{1}=0$ at 2 and 3 and $\displaystyle U_{1}^{1}=U_{jet}$ at 1.
$\displaystyle \displaystyle R_{s1}=-U_{jet}\rho U_{jet}A_{Jet}$
The force which acts on the plate (action-reaction) is
$\displaystyle K_{1}=-R_{s1}=U_{1}\rho U_{1}A_{Jet}$
It is also possible to solve the problem with $\displaystyle CV_{II}$
$\displaystyle F_{S1}=p_{a}A+R_{p1}=-U_{jet}^{2}A_{jet}\rho$
$\displaystyle R_{p1}=-p_{a}A-U_{jet}^{2}A_{Jet}\rho$
Hence, the force exerted on the plate by the CV is
$\displaystyle K_{1}=-R_{p1}$
In 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:
$\displaystyle \displaystyle F_{net}=0=K_{1}-p_{a}A+Rs_{1}$
$\displaystyle \displaystyle F_{net}=0=p_{a}A+U_{jet}^{2}A\rho -p_{a}A+Rs_{1}$
$\displaystyle \displaystyle Rs_{1}=-U_{jet}^{2}A\rho$
It 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 $\displaystyle U_{1}$ can be approximated at L.
$\displaystyle U_{1}(L,x_{2})=U_{0}\left(2{\frac {x_{2}}{\delta }}-\left({\frac {x_{2}}{\delta }}\right)^{2}\right)$
Apply conservation of mass
$\displaystyle {\frac {\partial }{\partial t}}\int _{cv}{\rho dV}+\int _{cs}{\rho {\vec {U}}\cdot {\vec {n}}dA}=0$
Assumptions:
* The flow is steady; ( $\displaystyle {\frac {\partial }{\partial t}}$ = 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:
$\displaystyle \displaystyle \rho \int _{cs}{U_{i}n_{i}dA}=0$
As you know
${dm \over dt}-{\dot {m}}_{in}+{\dot {m}}_{out}=0$
$w$: the thickness of the plate.
${\dot {m}}=\rho UA$
$dA=wdx_{2}$
$\displaystyle \rho \int _{0}^{h}{-U_{0}\ w\ dx_{2}}+\rho \int _{0}^{\delta }{U\ w\ dx_{2}}=0$
$\displaystyle U_{0}h=\int _{0}^{\delta }{U_{1}dx_{2}}$
When we use RTT for ${\vec {F}}=d{\vec {P}}/dt$ where ${\vec {P}}$ is the momentum vector;
$\displaystyle F_{i}=\underbrace {{\frac {\partial }{\partial t}}\int {U_{i}\rho dV}} _{=0\ steady\ state}+\int _{cs}{U_{i}\rho U_{j}n_{j}dA}$
There are 4 control surfaces: $CS_{I}$ ,$CS_{II}$, $CS_{III}$ and $CS_{IV}$.
$CS_{I}$: the surface that the mass enters across.
$CS_{II}$: Lateral surfaces of the control volume, there is no flows.
$CS_{III}$: the surface from which mass leaves.
$CS_{IV}$: the wall of the plate. Because of no slip condition $U=0$ at the wall.
$\displaystyle F_{1}=-D=\int _{0}^{h}{U_{1}\rho U_{j}n_{j}dA}+\underbrace {\int _{CS2}{U_{1}\rho U_{j}n_{j}dA}} _{=0\ streamline}+\int _{0}^{\delta }{U_{1}\rho U_{j}n_{j}dA}+\underbrace {\int {U_{1}\rho U_{j}n_{j}dA}} _{=0\ wall}$
$\displaystyle -D=-U_{0}\rho wU_{0}h+\int _{0}^{\delta }{U_{1}\rho U_{1}\ w\ dx_{2}}$
$\displaystyle D=U_{0}\rho wU_{0}h-\rho \ w\int _{0}^{\delta }{U_{1}^{2}dx_{2}}$
Insert the mass conservation result $U_{0}h=\int _{0}^{\delta }{U_{1}dx_{2}}$ into the momentum equation.
$\displaystyle D=\rho wU_{0}\int _{0}^{\delta }{U_{1}dx_{2}}-\rho \ w\int _{0}^{\delta }{U_{1}U_{1}dx_{2}}$
Hence at $\displaystyle {x_{1}=L}$
$\displaystyle D=\rho w\int _{0}^{\delta }{U_{1}(U_{0}-U_{1})dx_{2}}$
$\displaystyle U_{1}$ is known. Here, using $\displaystyle {\frac {x_{2}}{\delta }}=\eta$
$\displaystyle D=\rho \ w\ U_{0}^{2}\delta \int _{0}^{1}{(2\eta -\eta ^{2})(1-2\eta +\eta ^{2})d\eta }={\frac {2}{15}}\rho U_{0}^{2}w\ \delta$

## RTT for a CV moving with constant speed

It is possible that CV can move with constant velocity or arbitrary acceleration.
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 $\displaystyle {\vec {U}}_{CV}$, the relative velocity w.r.t. to the reference frame attached to the moving CV is:
$\displaystyle {\vec {U}}_{r}={\vec {U}}-{\vec {U}}_{CV}$
$\displaystyle {\frac {dB_{sys}}{dt}}={\frac {\partial }{\partial t}}\left(\int _{cv}{b\rho dV}\right)+\int _{cs}{b\rho {\vec {U_{r}}}\cdot {\vec {n}}dA}$

#### 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 $\displaystyle {\vec {U}}_{v}$ in $\displaystyle x_{1}$ 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.
$\displaystyle \left|{\vec {U}}^{1}\right|=\left|{\vec {U}}_{jet}-{\vec {U}}_{v}\right|$
from the continuity equation
$\displaystyle 0=\int _{cs}{\rho U\cdot ndA}=-\left|{\vec {U}}^{1}\right|\rho A_{1}+\left|{\vec {U}}^{2}\right|\rho A_{2}$
$\displaystyle \rho \left|{\vec {U}}^{1}\right|A_{1}=\rho \left|{\vec {U}}^{2}\right|A_{2}={\dot {m}}$
for $A_{1}=A_{2}$, $\displaystyle \left|{\vec {U}}^{2}\right|=\left|{\vec {U}}^{1}\right|$
Thus, the component of ${\vec {U}}^{2}$ along $\displaystyle x_{1}$ direction is:
$\displaystyle U_{1}^{2}=\left|U_{1}^{1}\right|cos\theta =\left(\left|{\vec {U}}_{jet}\right|-\left|{\vec {U}}_{v}\right|\right)cos\theta$
The momentum equation reads
$\displaystyle F_{i}=F_{S_{i}}+\underbrace {F_{B_{i}}} _{=0}=\underbrace {{\frac {\partial }{\partial t}}\int _{cv}{U_{i}\rho dV}} _{=0{\textrm {(steady)}}}+\int _{cs}{U_{i}\rho U_{j}n_{j}dA}$
Specifically in $\displaystyle x_{1}$ direction is
$\displaystyle F_{S_{1}}=R_{1}+p_{a}A-p_{a}A$
$\displaystyle \displaystyle R_{1}=\int _{cs}{U_{1}\rho U_{j}n_{j}dA}$
$\displaystyle \displaystyle R_{1}=-U_{1}^{1}\rho \left|{\vec {U}}^{1}\right|A_{1}+U_{1}^{2}\rho \left|{\vec {U}}^{2}\right|A_{2}=-U_{1}^{1}{\dot {m}}+U_{1}^{2}{\dot {m}}$
since $\displaystyle U_{1}^{2}=\left|U_{1}^{1}\right|cos\theta =\left(\left|{\vec {U}}_{jet}\right|-\left|{\vec {U}}_{v}\right|\right)cos\theta$
$\displaystyle R_{1}={\dot {m}}U_{1}^{1}(cos\theta -1)={\dot {m}}\left(\left|{\vec {U}}_{jet}\right|-\left|{\vec {U}}_{v}\right|\right)(cos\theta -1)$
The force in the vertical direction (in $\displaystyle x_{2}$ direction) is:
$\displaystyle \displaystyle R_{2}=\int _{cs}{U_{2}\rho U_{j}n_{j}dA}$
$\displaystyle R_{2}=\int _{A_{1}}{U_{2}^{1}\rho U_{j}^{2}n_{j}^{2}dA}+\int _{A_{2}}{U_{2}^{2}\rho U_{j}^{2}n_{j}^{2}dA}$
at 1, $\displaystyle U_{2}^{1}=0$ and at 2, $\displaystyle U_{2}^{2}=\left|{\vec {U}}^{2}\right|sin\theta =\left|{\vec {U}}^{1}\right|sin\theta =\left|{\vec {U}}_{jet}-{\vec {U}}_{v}\right|sin\theta$.
$\displaystyle R_{2}=\int _{A_{2}}{U_{2}^{2}\rho U_{j}^{2}n_{j}^{2}dA}$
$\displaystyle R_{2}=U_{2}^{2}\rho \left|{\vec {U}}_{2}\right|A_{2}=U_{2}^{2}{\dot {m}}=\left|{\vec {U}}_{jet}-{\vec {U}}_{v}\right|{\dot {m}}\,sin\theta$

## Momentum Equation for CV with rectilinear acceleration

For an inertial CV the following transport equation for momentum holds:
$\displaystyle {\vec {F}}={\frac {\partial }{\partial t}}\int _{cv}{{\vec {U}}\rho \ dV}+\int _{cs}{{\vec {U}}\rho {\vec {U}}\cdot {\vec {n}}dA}$
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.
Denote an inertial reference frame with $\displaystyle X_{1},X_{2},X_{3}$ and another reference frame moving with the CV, $\displaystyle x_{1},x_{2},x_{3}$. Hence, $\displaystyle x_{1},x_{2},x_{3}$ becomes the non-inertial frame of reference. Let $\displaystyle x_{1},x_{2},x_{3}$reference frame move with a velocity and an acceleration $\displaystyle {\vec {U}}_{rf}$ and $\displaystyle {\vec {a}}_{rf}$, respectively. The velocity vector of a fluid particle w.r.t. to the inertial reference frame (w.r.t. to a ground observer) is$\displaystyle {\overrightarrow {U}}_{X}={\overrightarrow {U}}_{x}+{\overrightarrow {U}}_{rf}$ where ${\overrightarrow {U}}_{x}$ is the velocity of the fluid particle w.r.t. non-inertial reference frame $\displaystyle x_{1},x_{2},x_{3}$. Hence the time derivative of it is,
$\displaystyle {\frac {d{\overrightarrow {U}}_{X}}{dt}}={\frac {d{\overrightarrow {U}}_{x}}{dt}}+{\frac {d{\overrightarrow {U}}_{rf}}{dt}}\Rightarrow {\frac {d{\overrightarrow {U}}_{X}}{dt}}\neq {\frac {d{\overrightarrow {U}}_{x}}{dt}}$
or,
$\displaystyle {\vec {a}}_{X}={\vec {a}}_{x}+{\vec {a}}_{rf}\Rightarrow {\frac {d{\overrightarrow {P}}_{X}}{dt}}\neq {\frac {d{\overrightarrow {P}}_{x}}{dt}}$
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:
$\displaystyle \left.{\vec {F}}\right|_{system}=\left.{\frac {d{\vec {P}}_{X}}{dt}}\right|_{system}$
For a control volume moving with $\displaystyle {\vec {U}}_{rf}$ and $\displaystyle {\vec {a}}_{rf}$, i.e. $\displaystyle {\frac {d{\vec {U}}_{CV}}{dt}}={\frac {d{\vec {U}}_{rf}}{dt}}$, i.e. $\displaystyle {\frac {dU_{cv}}{dt}}\neq 0$, the time derivative of $\displaystyle {\vec {P}}_{X}$ and $\displaystyle {\vec {P}}_{x}$ 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 $\displaystyle {\vec {P}}_{X}$ to $\displaystyle {\vec {P}}_{x}$.
Previously we have seen that in a non-inertial reference frame having rectilinear acceleration, i.e. (translational acceleration).
$\displaystyle {\vec {a}}_{X}={\vec {a}}_{x}+{\vec {a}}_{rf}$
and also
$\displaystyle {\vec {F}}=\int _{mass\ (system)}{{\vec {a}}_{X}dm}$
$\displaystyle =\int _{mass\ (system)}{({\vec {a}}_{x}+{\vec {a}}_{rf})dm}$
$\displaystyle {\vec {F}}-\int _{mass\ (system)}{{\vec {a}}_{rf}dm}=\int _{mass\ (system)}{{\vec {a}}_{x}dm}$
$\displaystyle {\vec {F}}-\int _{mass\ (system)}{{\vec {a}}_{rf}dm}=\int _{mass\ (system)}{{\frac {d{\vec {U}}_{x}}{dt}}dm}={\frac {d}{dt}}\int _{mass\ (system)}{{\vec {U}}_{x}dm}$
$\displaystyle {\vec {F}}-\int _{V\ (system)}{{\vec {a}}_{rf}\rho \ dV}={\frac {d{\vec {P}}_{x}}{dt}}|_{system}$
For a moving CV we know that
$\displaystyle {\frac {d{\vec {P}}_{x}}{dt}}|_{system}={\frac {\partial }{\partial t}}\int _{cv}{{\vec {U}}_{x}\rho \ dV}+\int _{cs}{{\vec {U}}_{x}\rho {\vec {U}}_{x}\cdot {\vec {n}}dA}$
Let system and CV coincides at an instant $\displaystyle t_{0}$:
$\displaystyle {\vec {F}}_{on\ system}-\int _{V\ system}{{\vec {a}}_{rf}\rho dV}={\vec {F}}_{on\ cv}-\int _{cv}{{\vec {a}}_{rf}\rho dV}$
$\displaystyle {\vec {F}}_{on\ cv}-\int _{cv}{{\vec {a}}_{rf}\rho dV}={\frac {\partial }{\partial t}}\int _{cv}{{\vec {U}}_{x}\rho \ dV}+\int _{cs}{{\vec {U}}_{x}\rho {\vec {U}}_{x}\cdot {\vec {n}}dA}$

#### Example 7: Momentum equation for an accelerating control volume

A small rocket, with an inertial mass of $\displaystyle M_{0}$, is to be launched vertically. Assume a steady exhaust mass flow rate $\displaystyle {\dot {m}}$ and velocity $\displaystyle U_{e}$ relative to the rocket.
Neglecting drag on the rocket, find the relation for the velocity of the rocket U(t).
$\displaystyle \underbrace {F_{2}} _{A}-\underbrace {\int _{cv}{a_{rf2}\rho dV}} _{B}=\underbrace {{\frac {\partial }{\partial t}}\int _{cv}{U_{2}\rho dV}} _{C}+\underbrace {\int _{cs}{U_{2}\rho {\vec {U}}_{x}\cdot {\vec {n}}dV}} _{D}$
A:
$\displaystyle F_{2}=-g\int _{cv}{\rho dV}=-g\ M_{cv}$
$\displaystyle M_{cv}=M_{0}-{\dot {m}}t$
$\displaystyle F_{2}=-g(M_{0}-{\dot {m}}t)$
B: Since $\displaystyle {\vec {a}}_{rf}$ is not a function of the coordinates:
$\displaystyle \displaystyle -\int _{cv}{a_{rf2}\rho dV}=-a_{rf2}\int _{cv}{\rho dV}=-a_{rf2}(M_{0}-{\dot {m}}t)$
is the time rate of change at $\displaystyle x_{2}$-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):
$\displaystyle {\frac {\partial }{\partial t}}\int _{cv}{U_{2}\rho dV}={\frac {\partial }{\partial t}}\int _{cvI}{U_{2}\rho dV}+{\frac {\partial }{\partial t}}\int _{cvII}{U_{2}\rho dV}$
As solid propellant has no relative velocity in CV I and $\displaystyle U_{2}=-U_{e}$ 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:
$\displaystyle {\frac {\partial }{\partial t}}\left[\underbrace {\int _{CVI}{U_{2}\rho dV}} _{U_{2}=0\rightarrow =0}+\int _{CVII}{U_{2}\rho dV}\right]=\underbrace {{\frac {\partial }{\partial t}}\int _{CVII}{U_{2}\rho dV}} _{=0(U_{2}=-U_{e}=cst)}\approx 0$
D:
$\displaystyle \int _{cs}{U_{2}\rho {\vec {U}}_{x}\cdot {\vec {n}}dA}=U_{2}\int _{cs}{\rho {\vec {U}}_{x}\cdot {\vec {n}}dA}=U_{2}{\dot {m}}=-U_{e}{\dot {m}}$
Substitution of all the terms gives:
$\displaystyle -g(M_{0}-{\dot {m}}t)-a_{rf2}(M_{0}-{\dot {m}}t)=-U_{e}{\dot {m}}$
$\displaystyle a_{rf2}={\frac {U_{e}{\dot {m}}}{M_{0}-{\dot {m}}t}}-\ g$
$\displaystyle {\frac {dV_{cv}}{dt}}=a_{rf2}={\frac {U_{e}{\dot {m}}}{M_{0}-{\dot {m}}t}}-\ g$
$\displaystyle V_{cv}(t)=\int _{0}^{t}{\frac {U_{e}{\dot {m}}}{M_{0}-{\dot {m}}t}}dt-\int _{0}^{t}gdt$
$\displaystyle V_{cv}(t)=-U_{e}ln\left(1-{\frac {{\dot {m}}t}{M_{0}}}\right)-gt$
The first term is always positive due to the natural logarithm $\displaystyle ln$ . 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

$\displaystyle \left[{\dot {Q}}+{\dot {W}}\right]_{on\ system}=\left[{\dot {Q}}+{\dot {W}}\right]_{on\ cv}={\frac {\partial }{\partial t}}\int {e\rho \ dV}+\int _{cs}{e\rho {\vec {U}}\cdot {\vec {n}}dA}$
where $\displaystyle e=u+{\frac {\left|{\vec {U}}\right|^{2}}{2}}+g\ z$
$\displaystyle {\dot {W}}_{on\ cv}={\dot {W}}_{body}+{\dot {W}}_{surface}$
here $\displaystyle z=x_{2}$
$\displaystyle {\dot {W}}_{body}={\dot {W}}_{shaft}+{\dot {W}}_{elec}+{\dot {W}}_{other}$
$\displaystyle {\dot {W}}_{surface}={\dot {W}}_{normal}+{\dot {W}}_{shear}$
Since
$\displaystyle {\dot {W}}={\vec {F}}\cdot {\vec {U}}$
The differential work done per unit time on a differential surface is:
$\displaystyle d{\dot {W}}=d{\vec {F}}\cdot {\vec {U}}={\vec {\sigma }}dA\cdot {\vec {U}}$
where ${\vec {\sigma }}$ is the sum of normal stress $\displaystyle {\vec {\sigma }}_{n}$ and shearing stress $\displaystyle {\vec {\sigma }}_{s}$ on that surface, which is tangent to $dA$:
$\displaystyle {\vec {\sigma }}={\vec {\sigma }}_{n}+{\vec {\sigma }}_{s}$
Note that, although normal stress can be sum of the pressure and normal viscous stress, i.e. ${\vec {\sigma }}_{n}=-p{\vec {n}}+{\vec {\tau }}_{n}$, the shear stress is only due to viscous stress,i.e. ${\vec {\sigma }}_{s}={\vec {\tau }}_{s}$.
The differential work done per unit time by the normal forces on a differential surface area over CS is
$\displaystyle d{\dot {W}}_{n}=d{\vec {F}}_{n}\cdot {\vec {U}}={\vec {\sigma }}_{n}dA\cdot {\vec {U}}$
In many cases normal viscous stresses can be neglected and the normal stress exist only due to pressure:
$\displaystyle {\vec {\sigma }}_{n}=-p{\vec {n}}$
therefore the work done by normal forces per unit time over the whole CS becomes
$\displaystyle {\dot {W}}_{normal}=\int _{cs}{{\vec {\sigma }}_{n}\cdot {\vec {U}}dA}=-\int _{cs}{p{\vec {n}}\cdot {\vec {U}}dA}$
Hence, the work done by the shear stress per unit time can be for the differential area and written as
$\displaystyle d{\dot {W}}_{shear}={\vec {\sigma }}_{s}dA\cdot {\vec {U}}\rightarrow {\dot {W}}_{shear}=\int _{cs}{{\vec {\sigma }}_{s}\cdot {\vec {U}}dA}=\int _{cs}{{\vec {\tau }}_{s}\cdot {\vec {U}}dA}$
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:
$\displaystyle {\dot {Q}}\ +\ {\dot {W}}_{shaft}\ +\ {\dot {W}}_{shear}\ +\ {\dot {W}}_{other}={\frac {\partial }{\partial t}}\int {e\rho dV}+\int _{cs}{\left(\underbrace {u+{\frac {p}{\rho }}} _{h:\ enthalpy}+{\frac {\left|{\vec {U}}^{2}\right|}{2}}+gx_{2}\right)\rho {\vec {U}}\cdot {\vec {n}}dA}$

### Example 8

Air enters compressor at inlet 1 with negligible velocity and leaves at outlet 2. The power input to the machine is $\displaystyle P_{input}$ and the volume flow rate is $\displaystyle {\dot {V}}$. Find a relation for the rate of heat transfer in terms of the power, temperature, pressure, etc.
1:
$\displaystyle {\dot {Q}}\ +\ {\dot {W}}_{shaft}+\underbrace {{\dot {W}}_{shear}} _{=\ 0\ =\tau \cdot {\vec {U}}}+\underbrace {{\dot {W}}_{other}} _{=\ 0}=\underbrace {{\frac {\partial }{\partial t}}\inte\rho dV} _{=\ 0\ steady\ state}+\int _{cs}{\left(u+{\frac {p}{\rho }}+{\frac {U^{2}}{2}}+gx_{2}\right)\rho {\vec {U}}\cdot {\vec {n}}dA}$
$\displaystyle 0=\underbrace {{\frac {\partial }{\partial t}}\int _{cv}{\rho dV}} _{=\ 0\ steady\ state}+\int _{cs}{\rho {\vec {U}}\cdot {\vec {n}}dA}\rightarrow \left|\rho _{1}U_{1}A_{1}\right|=\left|\rho _{2}U_{2}A_{2}\right|={\dot {m}}$
2:
$\displaystyle {\dot {Q}}=-{\dot {W}}_{shaft}+\int _{cs}{\left(u+{\frac {p}{\rho }}+{\frac {U^{2}}{2}}+gz\right)\rho {\vec {U}}\cdot {\vec {n}}dA}$
For uniform properties at 1 and 2 and inserting the inserting the relation for the enthalpy $\displaystyle h=u+{\frac {p}{\rho }}$.
$\displaystyle {\dot {Q}}=-{\dot {W}}_{shaft}-\left(h_{1}+\underbrace {\frac {U_{1}^{2}}{2}} _{=\ 0}+gz_{1}\right)\left|\rho _{1}U_{1}A_{1}\right|+\left(h_{2}+{\frac {U_{2}^{2}}{2}}+gz_{2}\right)\left|\rho _{2}A_{2}U_{2}\right|$
$\displaystyle {\dot {Q}}=-{\dot {W}}_{shaft}+{\dot {m}}\left[h_{2}+{\frac {U_{2}^{2}}{2}}-h_{1}+\underbrace {g(z_{2}-z_{1})} _{=\ 0}\right]$
Assuming that air behaves like an ideal gas with a constant $\displaystyle c_{p}$.
$\displaystyle \displaystyle h_{2}-h_{1}=c_{p}(T_{2}-T_{1})$
$\displaystyle {\dot {Q}}=-{\dot {W}}_{shaft}+{\dot {m}}\left[c_{p}(T_{2}-T_{1})+{\frac {U_{2}^{2}}{2}}\right]$

## Special form of the Energy equation

For a CV with one inlet 1, one outlet 2 and steady uniform flow through it.
$\displaystyle \underbrace {{\dot {Q}}+{\dot {W}}_{shaft}+{\dot {W}}_{shear}} _{K}=\int _{cs}{\left(u+{\frac {p}{\rho }}+{\frac {\left|{\vec {U}}^{2}\right|}{2}}+gz\right)\rho {\vec {U}}\cdot {\vec {n}}dA}$
For uniform flow properties at the inlet and outlet.
$\displaystyle K=\left(u_{1}+{\frac {p_{1}}{\rho _{1}}}+{\frac {\left|{\vec {U}}_{1}^{2}\right|}{2}}+gz_{1}\right)\underbrace {\int _{1}{\rho _{1}{\vec {U}}_{1}}\cdot {\vec {n}}dA} _{-{\dot {m}}}+\left(u_{2}+{\frac {p_{2}}{\rho _{2}}}+{\frac {\left|{\vec {U}}_{2}^{2}\right|}{2}}+gz_{2}\right)\underbrace {\int _{2}{\rho _{2}{\vec {U}}_{2}}\cdot {\vec {n}}dA} _{\dot {m}}$
$\displaystyle K={\dot {m}}\left[(u_{2}-u_{1})+\left({\frac {p_{2}}{\rho _{2}}}-{\frac {p_{1}}{\rho _{1}}}\right)+\left({\frac {{\vec {U}}_{2}^{2}}{2}}-{\frac {{\vec {U}}_{1}^{2}}{2}}\right)+g(z_{2}-z_{1})\right]$
Reform:
$\displaystyle \underbrace {{\frac {p_{1}}{\rho _{1}}}+{\frac {\left|{\vec {U}}_{1}^{2}\right|}{2}}+gz_{1}} _{mechanical\ evergy\ per\ unit\ mass\ at\ flow\ cross\ section}={\frac {p_{2}}{\rho _{2}}}+{\frac {\left|{\vec {U}}_{2}^{2}\right|}{2}}+gz_{2}+(u_{2}-u_{1})-{\frac {\dot {Q}}{\dot {m}}}-{\frac {{\dot {W}}_{shaft}}{\dot {m}}}-{\frac {{\dot {W}}_{shear}}{\dot {m}}}$
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.
$\displaystyle {\frac {p_{1}}{g\rho _{1}}}+{\frac {\left|{\vec {U}}_{1}^{2}\right|}{2g}}+z_{1}={\frac {p_{2}}{g\rho _{2}}}+{\frac {\left|{\vec {U}}_{2}^{2}\right|}{g2}}+z_{2}+{\frac {1}{g}}(u_{2}-u_{1})-{\frac {\dot {Q}}{g{\dot {m}}}}-{\frac {{\dot {W}}_{shaft}}{g{\dot {m}}}}-{\frac {{\dot {W}}_{shear}}{g{\dot {m}}}}$
For $\displaystyle {\dot {W}}_{shaft}=0$, $\displaystyle {\dot {W}}_{shear}=0$ and incompressible flow through a passive device which involves no mechanical work and no heat addition, the energy equation per unit mass reads :
$\displaystyle {\frac {p_{1}}{\rho }}+{\frac {\left|{\vec {U}}_{1}^{2}\right|}{2}}+gz_{1}={\frac {p_{2}}{\rho }}+{\frac {\left|{\vec {U}}_{2}^{2}\right|}{2}}+gz_{2}+(u_{2}-u_{1})-{\frac {\dot {Q}}{\dot {m}}}$
$\displaystyle {\frac {p}{\rho }}+{\frac {U^{2}}{2}}+gz$: Mechanical energy per unit mass.
$\displaystyle u_{2}-u_{1}-{\frac {\dot {Q}}{\dot {m}}}$: Irreversible conversion of mechanical energy to unwanted thermal energy $\displaystyle (u_{2}-u_{1})$ and loss of energy via heat transfer $\displaystyle \left(-{\frac {\dot {Q}}{\dot {m}}}\right)$. 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.
$\displaystyle h_{loss}=\left[u_{2}-u_{1}-{\frac {\dot {Q}}{\dot {m}}}\right]{\frac {1}{g}}$
i.e.
$\displaystyle {\frac {p_{1}}{\rho g}}+{\frac {U_{1}^{2}}{2g}}+z_{1}={\frac {p_{2}}{\rho g}}+{\frac {U_{2}^{2}}{2g}}+z_{2}+h_{loss}$
One can add the shaft work done by a pump or a turbine, in to that equation
$\displaystyle {\frac {p_{1}}{\rho g}}+{\frac {U_{1}^{2}}{2g}}+z_{1}={\frac {p_{2}}{\rho g}}+{\frac {U_{2}^{2}}{2g}}+z_{2}+h_{loss}-h_{pump}+h_{turbine}$
Note 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

Consider the steady, incompressible, frictionless flow through the differential CV for a stream tube.
Continuity
$\displaystyle 0={\frac {\partial }{\partial t}}\underbrace {\int _{cv}{\rho dV}} _{=\ 0}+\int _{cs}{\rho {\vec {U}}\cdot {\vec {n}}dA}$
$\displaystyle 0=-{\dot {m}}_{in}+{\dot {m}}_{out}\ \leftrightarrow \ {\dot {m}}_{in}={\dot {m}}_{out}={\dot {m}}$
Streamwise-Component of Momentum Equation
The momentum equation along the streamline (streamwise-component) can be written as:
$\displaystyle F_{S_{s}}+F_{B_{s}}={\frac {\partial }{\partial t}}\underbrace {\int {U_{s}\rho dV}} _{=\ 0}+\int _{cs}{U_{s}\rho {\vec {U}}\cdot {\vec {n}}dA}$
$\displaystyle F_{S_{s}}=\underbrace {pA} _{inlet}-\underbrace {(p+dp)(A+dA)} _{outlet}+\underbrace {\left(p+{\frac {dp}{2}}\right)dA} _{periphery}$
$\displaystyle F_{S_{s}}=-Adp-{\frac {1}{2}}dpdA$
$\displaystyle F_{B_{s}}=\rho g_{s}dV=\rho (-gsin\theta )\underbrace {\left(A+{\frac {dA}{2}}\right)ds} _{approx.\,volume\,of\,CV}$
with:
$\displaystyle \displaystyle sin\theta ds=dz$
Then,
$\displaystyle F_{B_{s}}=-\rho g\left(A+{\frac {dA}{2}}\right)dz$
$\displaystyle \int _{cs}{U_{s}\rho {\vec {U}}\cdot {\vec {n}}dA}=-U_{s}{\dot {m}}+(U_{s}+dU_{s}){\dot {m}}={\dot {m}}dU_{s}$
$\displaystyle -Adp-\underbrace {{\frac {1}{2}}dpdA} _{\approx \ 0}-\rho g\left(A+{\frac {dA}{2}}\right)dz={\dot {m}}dU_{s}$
$\displaystyle -Adp-\rho g\ A\ dz={\dot {m}}dU_{s}$
where $\displaystyle {\dot {m}}=\rho U_{s}A\$
$\displaystyle -Adp-\rho g\ A\ dz=\rho U_{s}\ A\ dU_{s}$
$\displaystyle {\frac {dp}{\rho }}+U_{s}dU_{s}+gdz=0$
with:
$\displaystyle U_{s}dU_{s}=d\left({\frac {U_{s}^{2}}{2}}\right)$
Then,
$\displaystyle {\frac {dp}{\rho }}+d\left({\frac {U_{s}^{2}}{2}}\right)+gdz=0$
Integrate between 1 and 2 along a streamline:
$\displaystyle {\frac {p_{1}}{\rho }}+{\frac {U_{1}^{2}}{2}}+gz_{1}={\frac {p_{2}}{\rho }}+{\frac {U_{2}^{2}}{2}}+gz_{2}=constant$
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 there is also an unsteady Bernoulli equation.
2. Incompressible flow. For example, in aerodynamics, flow can be accepted to be incompressible for Mach number $\displaystyle (M={\frac {\text{speed of flow}}{\text{speed of sound}}})$ 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.

#### Example 9: Bernoulli equation for a flow through nozzle

Consider steady flow of water through a horizontal nozzle. Find $\displaystyle P_{1}$ as a function of flow rate $\displaystyle {\dot {Q}}$.
Assumptions: steady, incompressible, frictionless, flow along a streamline, $\displaystyle z_{2}=z_{1}$, uniform flow.
$\displaystyle {\frac {p_{1}}{\rho }}+{\frac {U_{1}^{2}}{2}}={\frac {p_{2}}{\rho }}+{\frac {U_{2}^{2}}{2}}$
$\displaystyle p_{1}=p_{atm}+\rho \left({\frac {U_{2}^{2}}{2}}-{\frac {U_{1}^{2}}{2}}\right)=p_{atm}+\rho {\frac {U_{1}^{2}}{2}}\left({\frac {U_{2}^{2}}{U_{1}^{2}}}-1\right)$
From continuity:
$\displaystyle -\left|\rho U_{1}A_{1}\right|+\left|\rho U_{2}A_{2}\right|=0$
$\displaystyle U_{1}={\frac {\dot {Q}}{A_{1}}}\ {\textrm {and}}\ U_{2}={\frac {\dot {Q}}{A_{2}}}$
$\displaystyle P_{1}=P_{atm}+\rho {\frac {U_{1}^{2}}{2}}\left(\left({\frac {A_{1}}{A_{2}}}\right)^{2}-1\right)$
$\displaystyle P_{1}=P_{atm}+\rho {\frac {{\dot {Q}}^{2}}{2A_{1}^{2}}}\left(\left({\frac {A_{1}}{A_{2}}}\right)^{2}-1\right)$
$\displaystyle P_{1}=P_{atm}+\rho {\frac {8{\dot {Q}}^{2}}{\pi D_{1}^{4}}}\left(\left({\frac {D_{1}}{D_{2}}}\right)^{4}-1\right)$

#### 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.
$\displaystyle {\frac {p_{1}}{\rho }}+{\frac {U_{1}^{2}}{2}}+gz_{1}={\frac {p_{2}}{\rho }}+{\frac {U_{2}^{2}}{2}}+gz_{2}$
$\displaystyle \displaystyle p_{1}=p_{2}=p_{atm}$
$\displaystyle \displaystyle U_{1}A_{1}=U_{2}A_{2}$
$\displaystyle U_{2}^{2}-U_{1}^{2}=2g(z_{2}-z_{1})$
$\displaystyle U_{2}^{2}\left[1-\left({\frac {A_{2}}{A_{1}}}\right)^{2}\right]=2gh$
$\displaystyle U_{2}^{2}={\frac {2gh}{\left[1-\left({\frac {A_{2}}{A_{1}}}\right)^{2}\right]}}$
for steadiness $\displaystyle A_{1}>>A_{2}$, thus $\displaystyle U_{2}\cong {\sqrt {2gh}}$

#### 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
$\displaystyle p_{1}+\rho {\frac {U_{1}^{2}}{2}}=p_{2}+\rho {\frac {U_{2}^{2}}{2}}$
$\displaystyle (U_{2}^{2}-U_{1}^{2})={\frac {2}{\rho }}(p_{1}-p_{2})$
$\displaystyle U_{1}A_{1}=U_{2}A_{2}={\dot {Q}}$
$\displaystyle U_{1}={\frac {\dot {Q}}{A_{1}}}\ {\textrm {and}}\ U_{2}={\frac {\dot {Q}}{A_{2}}}$
$\displaystyle \left({\frac {{\dot {Q}}^{2}}{A_{2}^{2}}}-{\frac {{\dot {Q}}^{2}}{A_{1}^{2}}}\right)={\frac {2}{\rho }}\Delta p$
$\displaystyle {\dot {Q}}^{2}\left({\frac {1}{A_{2}^{2}}}-{\frac {1}{A_{1}^{2}}}\right)={\frac {2}{\rho }}\Delta p$
$\displaystyle {\dot {Q}}^{2}{\frac {1}{A_{2}^{2}}}\left(1-{\frac {A_{2}^{2}}{A_{1}^{2}}}\right)={\frac {2}{\rho }}\Delta p$
If $\displaystyle \beta ={\frac {D_{2}}{D_{1}}}$, then:
$\displaystyle {\dot {Q}}={\sqrt {\frac {A_{2}^{2}\ 2\underbrace {\Delta p} _{measured}}{\rho (1-\beta ^{4})}}}$
That is the method for measuring flow rate.