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Fluid Mechanics for Mechanical Engineers/Integral Analysis of Fluid Flow

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Introduction

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Differential Approach seek solution at every point , 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.

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

System versus Control volume

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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

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Conservation of mass

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The mass of a system does not change:

where,

Newton's second law

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Linear Momentum Equation

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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 () of the system:


or in tensor form

where

and

where are the corresponding unit vectors along positive axis, i.e. 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

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For system rotating about an axis with an angular speed of , 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 () of the system:

or in tensor form

where

and

The First law of Thermodynamics

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in the rate form:

where and

The Second law of Thermodynamics

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

The rate form of this law is:

where

Relation of a system derivative to the control volume derivative

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Consider a fire extinguisher

Control Volume versus System

whereas

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

to

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

  • Mass (),
  • Momentum (),
  • Energy (),
  • Moment of momentum (),
  • Entropy ().

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

Hence,


One dimensional Reynolds Transport Theorem

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Consider a flow through a nozzle.

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

If is an extensive variable of the system.

The first term for

for

Similarly

Thus, for , the terms in the equality for the time derivative of the system are



so that,

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.

CV with multiple inlets and outlets

Three Dimensional Reynolds Transport Theorem

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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 defined at each point on the CS. The outflow and inflow flow rate of across CS can be written as:

CV with arbitrary shape used to derive Reynolds transport theorem

Inflow and outflow to the CV.

and are positive quantities. Therefore, the negative sign is introduced into , to compensate the negative value of .

Since , RTT can be written as:

Conservation of mass

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i.e

Assume = constant (incompressible) and uniform in the CV

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

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,

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)

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,

Using RTT, the rate of change of momentum of the system can be related to that of CV as follows:

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.

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.

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

At the initial moment

i.e. in the component form:

Moment-of-momentum Equation

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,

Net torque exerted on the system about the rotaional axis is equal to the rate of change of angular momentum of the system.

expanding both sides by using the RTT

where is the surface force acting on the surfaces of the CV.

First Law of Thermodynamics

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at the initial moment , the following equality is valid:

thus, the integral form of energy equation is:

Second Law of Thermodynamics

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The left hand side of the inequality reads

The right hand side of the above inequality reads

Hence the second law becomes:


Note that represents the local heat flux through the surfaces of the control volume.

Examples

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Example 1: Conservation of mass for a stream tube

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

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

Let and be too small such that the velocities at position 1 and 2 are uniform across and .

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

We assume that;

1) In a steady flow volume of the stream tube, i.e. CV is constant, ( = 0)

2) Density is constant and uniform in the CV ().

Hence

When we rewrite the conservation of mass equation:

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 :

, and .

  •  : the surface area that the mass enters across.

and are in the opposite directions. and is always negative. That is, the sign of mass input is always negative.

  •  : the surface area from which the mass leaves.

and are in the same directions. and is always positive. That is, the sign of mass output is always positive.

  •  : the lateral surfaces of the CV.

and are perpendicular to each other . and is zero. That is, there is no flows.

where the integration over is zero, because there is no flow across the streamtube. Thus,

Example 2: Conservation of mass with multiple inlets and outlets

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Consider the steady flow of water through the device. The inlet and outlet areas are , and = .

mass balance for a connector device (example 2)

The following parameters are known:

Mass flow out at 3 ().

Volume flow rate in through 4 ().

Velocity at 1 along -direction , so that .

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

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

At section 3:

At section 4:

Hence, the velocity at section 2 can be calculated by

For

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

Example 3: Momentum equation for a stream tube

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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.

Force in a streamtube (example 3)

where the derivative with respect to time is zero due to steady state conditions. (=0 ).

There are 3 types of the control surfaces :

, and .

  •  : the surface area that the mass enters across.

(general formula)

and are in the opposite directions. and is always negative. That is, the sign of mass input is always negative.

  •  : the surface area from which the mass leaves.

(general formula)

and are in the same directions. and is always positive. That is, the sign of mass output is always positive.

  •  : the lateral surfaces of the CV.

(general formula)

and are perpendicular to each other . and is zero. That is, there is no flows.( =0 )

Density is constant ()

When we integrate

(steady state :)

Example 4: Momentum equation

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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 direction.

Jet striking to a plate (example 4)
Control volume I and II and Free body diagram of the plate (example 4)

Independent from the selected CV.

No body force in direction.

For the

at 2 and 3 and at 1.

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

It is also possible to solve the problem with

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

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:

It means, the force of the support should be in the negative direction (to the left).

Example 5: Finding drag on a flat plate

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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 can be approximated at L.

Velocity distribution of fluid over a plate (example 5)

Apply conservation of mass

Assumptions:

  • The flow is steady; ( = 0).
  • Density is constant ( is constant).
  • The flow is parallel(There is no flow in direction.

When we rewrite conservation of mass:

As you know


: the thickness of the plate.

When we use RTT for  ;

There are 4 control surfaces: ,, and .

: the surface that the mass enters across.

: Lateral surfaces of the control volume, there is no flows.

: the surface from which mass leaves.

: the wall of the plate. Because of no slip condition at the wall.

Insert the mass conservation result into the momentum equation.

Hence at

is known. Here, using

RTT for a CV moving with constant speed

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

Relation between the absolute velocity vector with the velocity of the moving reference frame attached to moving CV and the relative velocity w.r.t. to the moving reference frame

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 , the relative velocity w.r.t. to the reference frame attached to the moving CV is:

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

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Consider the jet and the vane. Determine the force to be applied such that vane moves with a constant speed in direction.

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

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

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

from the continuity equation

for ,

Thus, the component of along direction is:

The momentum equation reads

Specifically in direction is

since

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

at 1, and at 2, .

Momentum Equation for CV with rectilinear acceleration

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

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.

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

Denote an inertial reference frame with and another reference frame moving with the CV, . Hence, becomes the non-inertial frame of reference. Let reference frame move with a velocity and an acceleration and , respectively. The velocity vector of a fluid particle w.r.t. to the inertial reference frame (w.r.t. to a ground observer) is where is the velocity of the fluid particle w.r.t. non-inertial reference frame . Hence the time derivative of it is,

or,

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:

For a control volume moving with and , i.e. , i.e. , the time derivative of and 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 to .

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

and also

For a moving CV we know that

Let system and CV coincides at an instant :

Example 7: Momentum equation for an accelerating control volume

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A small rocket, with an inertial mass of , is to be launched vertically. Assume a steady exhaust mass flow rate and velocity relative to the rocket.

Launching rocket (example 1)
Control Volume I and II (example 1)

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

A:

B: Since is not a function of the coordinates:

is the time rate of change at -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):

As solid propellant has no relative velocity in CV I and 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:

D:

Substitution of all the terms gives:

The first term is always positive due to the natural logarithm . 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

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where

here


Normal and shear stress components on a differential surface area

Since

The differential work done per unit time on a differential surface is:

where is the sum of normal stress and shearing stress on that surface, which is tangent to :

Note that, although normal stress can be sum of the pressure and normal viscous stress, i.e. , the shear stress is only due to viscous stress,i.e. .

The differential work done per unit time by the normal forces on a differential surface area over CS is

In many cases normal viscous stresses can be neglected and the normal stress exist only due to pressure:

therefore the work done by normal forces per unit time over the whole CS becomes

Hence, the work done by the shear stress per unit time can be for the differential area and written as

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:

Example 8

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

Energy balance for a compressor (example 1)

1:

2:

For uniform properties at 1 and 2 and inserting the inserting the relation for the enthalpy .

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

Special form of the Energy equation

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

For uniform flow properties at the inlet and outlet.

Reform:

If this equation is divided by , the head form of the energy equation can be obtained. This form of the energy equation has the unit in meters.

For , and incompressible flow through a passive device which involves no mechanical work and no heat addition, the energy equation per unit mass reads :

: Mechanical energy per unit mass.

: Irreversible conversion of mechanical energy to unwanted thermal energy and loss of energy via heat transfer . Note that, since the heat leaves the CV, it should be negative, i.e. . Hence this term is always positive.

In head form, the loss of energy through a device can be written as.

i.e.

One can add the shaft work done by a pump or a turbine, in to that equation

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

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Differential control on a streamline

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

Continuity

Streamwise-Component of Momentum Equation

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

Forces created by pressure on a fluid element

with:

Then,

where

with:

Then,

Integrate between 1 and 2 along a streamline:

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 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 wake of the car b) heat energy input in the heat engine c) addition of mechanical energy inside the flow of a ventilator

Example 9: Bernoulli equation for a flow through nozzle

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Consider steady flow of water through a horizontal nozzle. Find as a function of flow rate .

Flow through horizontal nozzle (example 1)

Assumptions: steady, incompressible, frictionless, flow along a streamline, , uniform flow.

From continuity:

Example 10: Bernoulli equation

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

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

for steadiness , thus

Example 11: Bernoulli equation and Venturi tube

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

Application of Bernoulli Equation:Venturimeter (example 3)