chapter 5 fluid mechanics

Lecture 5: Control Volume Analysis

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Outline

• Introduction

• Basic Laws for a System

• Conservation of Mass

• Linear Momentum• Linear Momentum

• Energy Equation

• Examples

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Basic Laws for a System

• Conservation of Mass

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• Momentum Equation forInertial Control Volume

4


• The Angular Momentum Principle

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• The First Law of Thermodynamics

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• The Second Law of Thermodynamics

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Control Volume Analysis: Introduction

• Practical problems involve finite regions

• We call these regions control volumes

• Physical laws govern these regions

• We Apply Conservation Laws• We Apply Conservation Laws

• We look at Mass, Momentum, and Energy of the Region

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Conservation of Mass: Fixed Control Volume

Apply the Reynold’s Transport Theorem to the System of Mass:

With B = Mass, and b = 1, for a fixed non-deforming control volume:

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Time rate of change of the mass of the coincident system

Time rate of change of the mass of the contents of the coincident control volume

Net rate of flow of mass through the control surface

volume

Recall:“Coincident Condition”

Time = tTime = t - Time = t + 10

Conservation of Mass: Fixed Control VolumeRecall,

Then, Conservation of Mass in Control Volume Form:

If the flow is steady:

And, we sum up all the differential elements for mass flow through the surface:

= 0

where the control surface has the area A, is the density of the fluid, and Q is the volumetric flow rate.

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“outflow across the surface”

“inflow across the surface”

“no flow across the surface”

Mass flow rate:

“no flow across the surface”

The Average Velocity:

If the velocity, is uniformly distributed:

Control Volume12


If the flow is steady and incompressible, then:

Q is the volumetric flow rate.

If the flow is unsteady:

is important.is important.

(+) means mass is being added to the C.V.( - ) means mass is being subtracted from the C.V.

If the flow is one dimensional (uniform flow):

If the flow is not uniform:

For steady flow with one stream in and out:

For steady and incompressible flow with one stream:13


For steady flow, involving more than one stream:

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There are cases where it is convenient to have the control volume move. The most convenient is when the control volume moves with a constant velocity.

Conservation of Mass: Moving Control Volume

Reynolds Transport Theorem for a Moving Control Volume

With B = Mass, and b = 1, for a moving, non-deforming control volume:

Recall,

Then,

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Example

• An airplane moves forward at speed of 971 km/hr as shown inFigure below. The frontal intake area of the jet engine is0.80m2 and the entering air density is 0.736 kg/m3. Astationary observer determines that relative to the earth, the jetengine exhaust gases move away from the engine with a speedof 1050 km/hr. The engine exhaust area is 0.558 m2, and theexhaust gas density is 0.515 kg/m3. Estimate the mass flowrateexhaust gas density is 0.515 kg/m3. Estimate the mass flowrateof fuel into the engine in kg/hr.

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Solution

hrkgm

kmmhrkmmmkg

kmmhrkmmmkgm

fuelin

fuelin

/9050

)/1000)(/971)(80.0)(/736.0(

)/1000)(/2021)(558.0)(/515.0(

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W2 = V2- Vplane = 1050km/hr+971km/hr = 2021km/hr.

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Example

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Solution

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Conservation of Mass: Deforming Control Volume

The equation for the moving control volume can be used for a deforming control volume.

is non-zero.is non-zero.

W will vary as the velocity of the control surface varies.

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Conservation of Mass: Example Control Volumes

One inlet an one outlet:

Air in a Pipe:Steady Flow

Non-uniform velocity, V2 is an average velocity

Air Density varies at each location

Calculate: Calculate:

If we choose a control volume that excludes the fan and the condenser coils:

Dehumidifier:

Three inlet/outlet combinations, steady state:

If we choose the a second control volume:

Gives the same answer!

Five inlet/outlet combinations:

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Linear Momentum (Newtons 2nd Law): Fixed Control Volume

For “coincidence” of the system with the control volume:For “coincidence” of the system with the control volume:

Using Reynolds Transport Theorem with b = V, and B = Momentum:


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Linear Momentum: Fixed Control Volume

Time rate of change of the linear momentum of the coincident system

Time rate of change of the linear momentum of the contents of the coincident control volume

Net rate of flow of linear momentum through the control surface

volume

Recall:

“Coincident Condition”

Time = t 23


Then,

•The forces that act on the control volume are body forces and surface forces•The equation is a vector equation—linear momentum has direction.•Uniform (1-D) flows are easiest to work with in these equations•Momentum flow can be positive or negative out of the control volume•The time rate of change of momentum is zero for steady flow.

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•If the control surface is perpendicular to the flow where fluid enter or leaves the control volume, the surface force exerted by the fluid at the control surface will be due to pressure.•At an open exit, the surface pressure is atmospheric pressure.•Gage pressures may be used in certain situations.•The external forces have an algebraic sign, either positive or negative.•Only external forces acting on the control volume are considered.•Only external forces acting on the control volume are considered.•If the fluid alone is considered in the control volume, the reaction forces dues to any surfaces will need to be considered.•If the fluid and the surface are in the control volume are in the control volume, no reaction forces do not appear between the surface and the fluid.•Anchoring forces are considered external forces•Anchoring forces will generally exist in response surface stresses (shear and pressure acting on the control surface.

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Example

• Determine the anchoring force required to hold in place aconical nozzle attached to the end of a laboratory sink faucetwhen the water flow rate is 0.6 liter/s. The nozzle mass is0.1kg. The nozzle inlet and exit diameters are 16mm and 5mm,respectively. The nozzle axis is vertical and the axial distancerespectively. The nozzle axis is vertical and the axial distancebetween section (1) and (2) is 30mm. The pressure at section(1) is 464 kPa. to hold the vane stationary. Neglect gravity andviscous effects.

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Solution

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Solution

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Solution

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Example• A static thrust as sketched in Figure below is to be designed

for testing a jet engine. The following conditions are knownfor a typical test: Intake air velocity = 200 m/s; exhaust gasvelocity=500 m/s; intake cross-section area = 1m2; intakestatic pressure = -22.5 kPa=78.5 kPa (abs); intake statictemperature = 268K; exhaust static pressure = 0 kPa=101 kPatemperature = 268K; exhaust static pressure = 0 kPa=101 kPa(abs). Estimate the normal trust for which to design.

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Solution

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Linear Momentum: Moving Control Volume

Reynolds Transport Theorem for a Moving Control Volume

With B = Momentum, and b = V, for a fixed non-deforming control volume:

Then, substituting the above equation:

Substitute for V:

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Linear Momentum: Moving Control Volume

For steady flow in the control volume reference frame and VCV is constant:

And, then for an inertial frame, VCV is constant :

For steady flow (on a time average basis), “Mass conservation”:

Then,

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The Energy Equation: Fixed Control Volume

Heat Transfer Rate Work RateEnergy

Rewriting,

Also, noting that energy, e, can be rewritten (all per unit mass):

Internal Energy

Kinetic EnergyPotential Energy

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The Energy Equation: Fixed Control VolumeNow, invoking “coincidence” of the control volume and the system:

Using Reynolds Transport Theorem with b = e, and B = Total Energy:


Using Reynolds Transport Theorem with b = e, and B = Total Energy:

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The Energy Equation: Fixed Control Volume

Noting and Substituting,

=

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The Energy Equation: Work and Heat

represents heat transfer, conduction, convection, and radiation.

Heat transfer into the control volume is positive, heat transfer out is negative.

If the process is adiabatic, there is no heat transfer.

If the heat transfer in equals the heat transfer out, the net is zero:

Heat:

Work:Work transfer rate, power, is positive when the work is done on the contents of the control volume, by the surroundings.

Work includes shaft work such as turbines, fans, propellers, and other rotating equipment.

Other types of work are due to normal stresses and tangential stresses acting on fluid particles.

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The Energy Equation: Work and Heat

Work (continued):

Shaft Work:

Normal Stress:

Shear Stress:

Only non-zero at the control surface.

The tangential stress exists at the boundary, but due to “no-slip” condition, zero velocity, it is not transferred typically, and we consider it negligible if the appropriate control volume is chosen.

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The Energy Equation: Fixed Control VolumeNow, the Energy Equation take the following form:

+ =

Rearranging, and Substituting,

Then,

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The Energy Equation: Applications

(1) (2) (3)

(1) Assume Steady Sate then, = 0

(2)

Assume properties are uniformly distributed over the flow cross-section,

Assume one inlet and one outlet:

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However, the previous assumption of uniform 1D flow is often an oversimplification for control volumes, but its ease of use justifies it’s application to these situations.

The previous assumption is fairly good for a fluid particle following a stream tube in a steady state flow.

application to these situations.

Now, we can introduce shaft work. We note that shaft work is unsteady locally, but its effects downstream are steady.

One Dimensional Energy Equation for Steady-in-the-Mean Flow:

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Now, Introduce Enthalpy:

Then the 1D energy equation becomes the following:

With no shaft work—the fluid stream is constant throughout:

Or, the steady flow energy equation:

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Example• Steam enters a turbine with a velocity of 30m/s and enthalpy,

h1, of 3348 kJ/kg. The steam leaves the turbine as a mixture of vapor and liquid having a velocity of 60 m/s and an enthalpy of 2550 kJ/kg. If the flow through the turbine is adiabatic and changes in elevation are negligible, determine the work output involved per unit mass of steam through-flow.involved per unit mass of steam through-flow.

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Solution

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The Energy Equation: Compare to Bernoulli’s

If the flow is incompressible, in addition to being 1D and steady,

Divide the mass flow rate out:

Where,

If the flow is inviscid (frictionless), we obtain Bernoulli’s equation:

or, per unit mass,

Thus, the friction terms are the following:

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The Energy Equation: 1D, Steady, Incompressible, Friction Flow

For steady, incompressible, frictional flow:

Useful or available energy:

Loss terms:

Then we can rewrite the energy equation for 1D, Steady, incompressible Then we can rewrite the energy equation for 1D, Steady, incompressible Frictional flow:

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The Energy Equation: 1D, Steady-in-Mean Flow, Incompressible, Friction Flow

For Steady-in-Mean Flow, we introduce shaft work again:

Divide the mass flow rate out:

Where,

Then,

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The Energy Equation: In Terms of Heads

Multiply by density:

Then, divide by specific weight:

Where,

Turbine:

Pump:

is all other losses not associated with pumps or turbines

can be due to a turbine or pump

If we only have a pump or turbine, the terms on the R.H.S become these.

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Example• An axial-flow ventilating fan driven by a motor that delivers

0.4 kW of power to the fan blades produces a 0.6-m-diameteraxial stream of air having a speed of 12 m/s. The flowupstream of the fan involves negligible speed. Determine howmuch of the work to the air actually produces a useful effects,that is, a rise in available energy and estimate the fluidthat is, a rise in available energy and estimate the fluidmechanical efficiency of this fan.

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Solution

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752.08.95

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chapter 5 fluid mechanics

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