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.
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.
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:
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:
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
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, 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:
and are positive quantities. Therefore,
the negative sign is introduced into , to compensate the negative value of .
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.
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.
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 )
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.
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).
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.
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.
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 , 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
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.
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
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.
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.
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:
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.
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 .
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
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:
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.
Example 9: Bernoulli equation for a flow through nozzle