Chapter 7

ENERGY PRINCIPLE

Fluid Mechanics, Spring Term 2011

1st Law of Thermodynamics:

This is a statement of conservation of energy of a system.

!E is the change in energy of the system.

Q is the heat input.

W is the work output (work done by the system)

The sign convention comes from the old days of steamengines: You put heat into the engine, and you getmechanical work out of the engine.

Energy conservation can be written interms of time changes:

The total energy E may take different forms, such as1) Kinetic energy Ek2) Potential energy Ep3) Internal energy Eu

We will now use Reynolds Transport Theorem towrite the energy equation for a Eulerian setup, i.e.,for a control volume cv.

We go through the derivation only to help youunderstand what the terms are and where they comefrom. You do not need to memorize this derivation.

The integral equations are quite complicated, but inthe end weʼll get a simpler algebraic form which isthe form that we will apply to examples andproblems.

Reynolds Transport Theorem Applied to Energy

Recall Reynolds Transport Theorem (Chapter 5):

Now let E = B, and e = b (e = energy per unit mass):

or

a) b) c)

One more time, letʼs look at the terms:

b) The total amount of energy inside the cv.

a) Net heat input and work input into the system.

c) Correction between system and cv: Energycarried across the cs by fluid flow.

Note that heat carried by fluid particles is part of

, not of

The same way we wrote

We write for the energies per unit mass:

The energy equation becomes:

We do not measure energies directly, so the present formof our energy equation is not very useful.

We therefore replace the energy terms with more easilyapplied quantities:

where we assume that all potential energy is gravitationaland that it is relative to some point with z = 0.

where V is the velocity relative to inertial reference frame.

Flow Work versus Shaft Work

Flow work is the work done by pressure forces; itʼs thesame work as weʼve seen in the Bernoulli equation.

It doesnʼt matter whether thiswork is done by a piston or by adjacent fluid. For a fluid,

and hence

This is Flow Work.

Shaft work Ws is by definition all other work.

It is usually work done through a shaft, such as apump or turbine shaft.

A pump does work on the flow, thereby increasingits energy. By the definition

this means negative work.

Work done by the flow on a turbine is positivework.

Throwing all of this into the energy equation:

Notice that we may combine the area integrals into one:

Sometimes we combine the last 2 terms in the 2nd integral:

where h is the specific enthalpy.

What we have so far is the general energy equation:

We now simplify this for specific applications.

For example, for steady flow the energy accumulation term(the volume integral) is zero.Furthermore, if inflow and outflow is via pipes with constantproperties across, then

(still messy, but easier to apply…)

Example 7.1:Steam Turbine

Given:

Inflow at p1 = 1.4 MPa andT1 = 400oC;This corresponds tospecific enthalpyh1 = 3121 kJ/kg

Outflow at p2 = 101 kPa and 100oC; h2 = 2676 kJ/kgSteam enters at V1 = 15 m/s and exits at V2 = 60 m/sHeat lost through turbine wall is 7600 kJ/hMass flow through turbine is 0.5 kg/s

What is the power output of the turbine?

Example 7.1: Solution

Simplifying the equation for steadyflow we just derived:

For constant elevation (zo = zi) and steady flow

…all parameters on right-hand side are given.

Another specific application:Steady flow of incompressible fluid in pipe

Notice that we now allow velocity variations acrossthe pipe.

Start again with general (integral) equation:

Again, flow is steady (no energy accumulation in cv), so thatthe volume integral is zero.

Between points 1 and 2:

At points 1 and 2, there is no acceleration normal to thestreamlines, and hence

is constant across the sections.

Also, u is usually constant across a section. These termscan thus be taken out of the integral.

Recall so that…

To get rid of that last integral, we define

where " is a kinetic energy correction factor

Example 7.2: What is this “magic” correction factor afor laminar flow in a pipe?

The velocity profile isgiven as

Example 7.2: Solution

The mean velocity is given by

Example 7.2 (continued)

From our definition of "

we see

For laminar flow in a pipe (parabolic velocity profile):

For turbulent flow in a pipe with constant velocity:

In general, from

we see that

Now going back to the energy equation for flow in pipes:

Introduce

(Shaft power is turbine power minus pump power)

And divide the entire equation by g (to get units of length):

It is common practice to associate pressures with theirequivalent “head” (height of hydrostatic fluid column):

Head supplied by pump:

Head given up to turbine:

Notice that all thermal terms are those in brackets on theright, while all other terms are mechanical.

Thermal energy is generated from mechanical energy byviscous action between fluid particles. This action is notreversible, and the thermal energy is not recoverable asmechanical energy.

We group all thermal terms into one parameter:

Head loss (due toviscous dissipation)

Recall that this is still the energy equation.

It is a lot simpler than the general equation we startedout with, but thatʼs because it is now tailored towardsvery specific applications.

The reason for the red box around the equation is thatthis is the one we will use (not because it is particularlyfundamental)!

Notice from our definition of the head for the pumpand the turbine that

Power supplied to flow by pump.

Power delivered by turbine (takenfrom the flow)

Example 7.3:Flow in pipewith head loss.

Given:"= 1 (turbulent flow)L = length of pipeQ = discharge through pipeD = diameter of pipez1 = elevation of water level in tankz2 = elevation of pipe

Head loss in pipe is given by

What is the pressure at point 2?

Example 7.3: Solution

Write energy equation between water surface in tank andsection at point 2 in pipe:

With and without pumps or turbines simplifies to:

Example 7.3 (continued)

Velocity in pipe in terms of discharge:

Head loss in terms of Q:

Everything in the energy equation is now known;

Just solve for p2 and substitute the values given in book…

Example 7.4:Turbulent flow in apipeGiven:p1 = pressure at 1p2 = pressure at 2Q = constant discharge"1 = "2 = 1

hL = head lossz1 = elevation at 1z2 = elevation at 2A1 = A2 = pipe crosssectional area

What power has to besupplied by the pump inorder to keep the pressure atpoint 2 at the indicated levelp2?

Example 7.4: Solution

Start with basic energy equation for pipes:

With and we get

Everything is given except hp. Solve for hp and substituteparameters given in book…

Example 7.5: Powergeneration byhydroelectric powerplant.

Given:

hL = head lossQ = discharge atmaximum rate of powergeneration

What is the maximum rate of power generation?

(The book lists some additional parameters, but we donʼtneed them.)

Example 7.5: Solution

Again, start with basic energy equation for pipes:

The book sets .Thatʼs perhaps too simple, but andgets rid of the velocity terms,anyway.

Also:

Example 7.5 (continued)

The energy equation simplifies to:

Once we have found the head used up by the turbine,we turn this into the power delivered:

where Iʼve used:

A word about the Bernoulli Equation and theEnergy Equation

Energy equation for incompressible flow in pipes:

If velocity is constant at given section (" = 1), and there areno pumps or turbines and no head loss:

The Bernoulli equation is just a special case of the energyequation, where the “pipe” is just a narrow imaginary tubeenclosing the streamline.

And if we apply the Bernoulli equation to a problem inwhich the velocities are zero:

becomes

This is just the hydrostatic pressure equation fromchapter 3. The hydrostatic equation is thus also just aspecial form of the energy equation.

Putting it all together:An example that uses the energy, momentum andcontinuity principles in combination (Section 7.3)

Head loss due to abrupt expansion in a pipe.

For turbulent conditions,

The energy equation becomes

The momentum equation

becomes (neglecting shear stresses against the pipe walls)

The momentum equation can also be written as

From continuity for a steady flow, we have

Only when we combine the information from all 3equations can we find

The head loss is due to turbulent flow in the larger pipe.

Note that the equation is also valid for flow into areservoir with V2 = 0.

One more basic concept: Hydraulic and Energy Grade Lines

At any point along thepipe, we can insert amanometer and astagnation tube (recallthat the velocity at theentrance to thestagnation tube is zero).

From energy eqn:

(since p = 0 at top of stagnation tube)

The stagnation tube measures the same plus the velocity-term (since v = 0 at entrance to stagnation tube):

(Stagnation tube)

The height of the fluid column in the manometer is thehydraulic grade line;that in the stagnation tube is the energy grade line.

Energy eqn. between top of reservoir and top of tube:

(Manometer)

EGL = energy gradeline

HGL = hydraulicgrade line

The linear slope is due to viscous head loss, which isa constant per unit length of pipe.

A pump gives anabrupt rise in EGLand HGL.

A turbine gives anabrupt drop.