7.3-Energy Equation_ Pipe Flow.pdf

Energy Equation: Pipe Flow

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7.3 Energy Equation: Pipe Flow

Section 7.2 showed how to derive the general form of the energy equation for a control volume. Now, this

section will simplify this general form to give an equation that is much easier to use yet still applicable to most

situations. The first step is to develop a way to account for the kinetic energy distribution in the flowing fluid.

Kinetic Energy Correction Factor

Figure 7.4 shows fluid that is pumped through a pipe. At sections 7.1 and 7.2, kinetic energy is transported

across the control surface by the flowing fluid. To derive an equation for this kinetic energy, start with the mass

flow rate equation.

This integral can be conceptualized as adding up the mass of each fluid particle that is crossing the section area

and then dividing by the time interval associated with this crossing. To convert this integral to kinetic energy

(KE), multiply the mass of each fluid particle by ( V 2 /2).

Figure 7.4 Flow carries kinetic energy into and out of a control surface.

The kinetic energy correction factor is defined as

For a constant density fluid, this equation simplifies to

(7.21)

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Energy Equation: Pipe Flow

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In most cases, α takes on a value of 1 or 2. When the velocity profile in a pipe is uniformly distributed, then

α = 1. When flow is laminar, the velocity distribution is parabolic and α = 2. When flow is turbulent, the velocity

profile is pluglike and α ≈ 1.05. For turbulent flow it is common practice to let α = 1.

To establish a value for α, integrate the velocity profile using Eq. 7.21. This approach is illustrated in Example

7.1.

Derivation of a Simplified Form of the Energy Equation

Now that the KE correction factor is available for representing the distribution of kinetic energy, the derivation

may by completed. Begin by applying Eq. 7.18 to the control volume shown in Fig. 7.4. Assuming steady flow,

Eq. 7.18. simplifies to:

(7.22)

As explained in Chapter 4, (see p. 87), piezometric head p /ρ + gz is constant across sections 7.1 and 7.2 because

the streamlines are straight and parallel. If temperature is also assumed constant across each section, then

p /ρ + gz + u can be taken outside the integral to yield

(7.23)

EXAMPLE 7.1 KIETIC EERGY CORRECTIO FACTOR

FOR LAMIAR FLOW

The velocity distribution for laminar flow in a pipe is given by the equation

Here r 0 is the radius of the pipe and r is the radial distance from the center. Find the kineticenergy

correction factor α.

PROBLEM DEFINITION

Situation: Laminar flow in a round pipe. Velocity profile is given.

Find: Kineticenergy correction factor α (no units).

Sketch:

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Energy Equation: Pipe Flow

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PLAN

1. Find dA using the above sketch.

2. Find the mean velocity using the flow rate equation (5.8).

3. Find α using Eq. 7.21.

SOLUTION

1. Differential area

· From the sketch, dA can be visualized as a rectangular strip of length 2π r and width dr .

Thus, dA = 2π r dr .

2. Mean velocity

Interpretation: For laminar flow in a round pipe, the mean velocity is onehalf of the maximum

(centerline) velocity.

3. Kineticenergy correction factor(α).

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Energy Equation: Pipe Flow

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To evaluate the integral, make a change of variable by letting

. The integral

becomes

Next, factor out

from each term in Eq. 7.23. Since does not appear as a factor of ∫(ρ

V 3 /2) dA , express ∫(ρ V 3 /2) dA as

, where α is the kinetic energy correction factor:

(7.24)

Divide through by :

(7.25)

Introduce Eq. 7.10 into Eq. 7.25:

(7.26)

Introduce pump head and turbine head :

(7.27)

Equation 7.26 becomes

(7.28)

Equation 7.28 is separated into terms that represent mechanical energy (nonbracketed terms) and terms that

represent thermal energy (the bracketed term). This bracketed term is always positive because of the second law

of thermodynamics. This term is called head loss and is represented by h L . Head loss is the conversion of useful

mechanical energy to waste thermal energy through viscous action between fluid particles. Head loss is

analogous to thermal energy (heat) that is produced by Coulomb friction. When the bracketed term is replace by

head loss h L , Eq. 7.28 becomes the energy equation.

(7.29)

Equation 7.29 is based on three main assumptions: (a) the flow is steady; (b) the control volume has one inlet

port and one exit port; and (c) the density of the flow is constant.

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Energy Equation: Pipe Flow

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In words, Eq. 7.29 can be stated as

Terms in Eq. 7.29 use the concept of head. A head term has a primary dimension of length, and it represents an

energy or work concept. Head is related to energy or work:

Thus, another way to interpret Eq. 7.29 is that work and energy flows are balanced as shown in Fig. 7.5. Energy

can enter the control volume in two ways: energy can be transported across the control surface by the flowing

fluid * or a pump can do work on the fluid and thereby add energy to the fluid. Energy can leave the cv in three

ways. Energy within the flow can be used to do work on a turbine, energy can be transported across the control

surface by the flowing fluid, or mechanical energy can be converted to waste heat via head loss.

Figure 7.5 Equation 7.29 involves a balance of work and energy terms.

The recommended way to apply the energy equation is to start with Eq. 7.29 and then analyze each term in a

stepwise fashion as shown in Example 7.2.

EXAMPLE 7.2 PRESSURE I A PIPE

A horizontal pipe carries cooling water at 10°C for a thermal power plant from a reservoir as shown.

The head loss in the pipe is

where L is the length of the pipe from the reservoir to the point in question, V is the mean velocity in

the pipe, and D is the diameter of the pipe. If the pipe diameter is 20 cm and the rate of flow is

0.06 m 3 /s, what is the pressure in the pipe at L = 2000 m. Assume α 2 = 1.

PROBLEM DEFINITION

Situation: Cooling water for a power plant is flowing in a horizontal pipe.

Find: Pressure (kPa) in the pipe at section 2.

Sketch:

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