THE FIRST LAW OF THERMODYNAMICS(Energy Equation for a Control Volume)

We obtain the general energy equation by combining the first law ofthermodynamics and the transport theorem. For a system, conservationof energy can be written as,

system a ofenergy the

of change of Rate

system theenergy to

of transfer of rateNet

with mathematical terms,

systemdt

dEWQ

……………….(1)

Note: Heat and work are both energy. In general energy can be classifiedin two groups

1. Mechanical energy• Work - Chemical Energy• Kinetic energy - Nuclear Energy• Potential energy

2. Thermal energy• Heat• Internal Energy

Mechanical energies are associated with force and motion. Thermalenergies are associated with temperature, molecular structure and heattransfer.

1Flow Analysis 3

syssys

deedmEM

sys

and

gzV

ue 2

2

……………….(2)

The system and control volume formulations are related by Reynoldstransport theorem

CSCsystem

AdVdtdt

dN

To derive energy equation for a control volume, we set N=E and η=e, then

CSCsystem

AdVedetdt

dE ……………….(3)

Note that in deriving transport equation, the system and control volumecoincided at t = t0, hence we can write

volumecontrolsystem WQWQ ……………….(4)

Substituting Eqs. (3) and (4) into Eq. (1), we obtain

CSC

AdVedet

WQ

……………….(5)

?? WQ

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Rate of Work Done on a Control Volume

Control Volume

ττ

ττ

work done by pressure moving fluid

particles

work done against pressure

work transmitted by a rotating shaft

work done by shear stress

Note: Rate of work power

If we neglect electrical and other equivalent forms of work, three types of work mightbe done on or by the fluid inside the control volume as shown in the figure below:

1. Shaft work, Ws2. Work done on the control surface by normal stresses (Pressure Work)3. Shear Work (Wshear)

1. Shaft Work ( ): is transmitted by a rotating shaft such as pump drive shaft or aturbine output shaft that is “cut” by the control surface. This work is done by shearstresses in the “cut” shaft, so it is somewhat similar to shear work. Shaft work issometimes called ‘pump work’ or ‘turbine work’ if these devices are present.

2. Work Done on the Control Surface by Normal Stresses (Pressure Work):Pressure work is done by fluid pressure acting on the boundaries of the controlvolume.

The work done by force moved through distance ds is

sW

F

sdFW

Rate of the work VFt

sdF

t

WW

tt

00limlim

3Flow Analysis 3

The rate of work done on an element of area of the control surface bynormal stress is given by

Ad

VAdVFd nn

The total rate of work done on the entire surface by normal stresses isgiven by (Since the work out across the boundaries of the control volumeis the negative of the work done on the control volume)

AdVVAdW

CS

nn

CS

nnnormal

pnn Note:

Hence, AdVpW

CS

normal

………………. (6)

3. Shear Work ( ):Shear work is done by shear stresses in the fluid acting on boundaries ofthe control volume. Similar to normal work,

shearW

dAVWCS

shear

Shear force acting on an area element isdA

dAFd

: shear stress acting in the plane of dA

We often choose a control volume with control surfaces lying adjacent tosolid boundaries, and with control surfaces cutting through inlet andoutlet ports. Hence, the shear work can be expressed as two terms

AdVAdVW

portsASurface

SolidA

shear

)()

(

4Flow Analysis 3

At solid surfaces , so the first term is zero (for a fixed control volume)0V

The last term can be made zero by a proper choice of control surfaces. Ifwe choose a control surface that cuts across each port perpendicular tothe flow, then dA is parallel to V and hence, τ is perpendicular to V.Thus, for control surfaces perpendicular to V

0V

and 0shearW

Hence, the energy equation for a control volume becomes

CSC

othershear

CS

s AdVedet

WWAdVpWQ

where

gzV

ue 2

2

hpvu

5

Substituting

AdVpvAdVp

CSC

othershears AdVpvedet

WWWQ

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6

Note: is zero if there is no control surface that lies within amoving fluid

shearW

V

If the control surface is perpendicular to the flow, shear stress does no work,

Fluid is at rest at solid boundary. Shear stress does no work

V

Shear stress does work on moving fluid

Flow Analysis 3

Example: A compressor compresses 6 kg/s of air from inlet conditions T1 =300 K and P1 = 90 kPa to discharge conditions T2 = 390 K and P2 = 310 kPa.The air in the inlet pipe has a uniform velocity profile. The air in thedischarge pipe has a parabolic velocity profile given by

2

2

max 1R

ruu

where R2 is the inside radius of the of the discharge pipe. Elevation changesare negligible, and the internal energy change of the air is given by

1212 TTCuu v

with kgKjCv /720

Assuming steady flow and negligible heat transfer, find the power requiredto drive the compressor.

u

u1

.

m=6 kg/sn

.

1

2

Wshaft

T1=300 K

P1=90 kPa

R1=0.25 m

T2=390 K

P2=310 kPa

R2=0.25 m

.

Ws=?Assumptions:Elevation changes are negligibleSteady flowHeat transfer is negligible

To be completed in class

7Flow Analysis 3

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Example: Turbines convert the energy contained within a fluid intomechanical energy or shaft work. A turbine is installed in a dam as shownin the figure. Water is permitted to flow through a passage way to theturbine after which the water drains downstream. For the data given in thefigure, determine the power available to the turbine when the discharge atthe outlet is 30 m3/s.

Assumptions:• Steady flow• Incompressible flow• No heat transfer• Internal energy change can be neglected

To be completed in class

10Flow Analysis 3

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Average Properties and Velocities

Usually, the uniform flow assumption is only an approximation and weuse average velocities and property values to calculate flow of energy atvarious inlet and outlet planes. For accurate calculations, we mustcarefully define the averages so that they truly represent associatedenergy flows. In most cases, appropriate average values of u, p, and zare readily apparent because these properties are often closely uniformacross the section. However, determination of average velocity andhence kinetic energy flux requires a careful attention.

Average Velocity is defined as

A

AdV

V A

Average Kinetic Energy

Since velocity is usually non-uniform, representing the kinetic energy fluxin terms of uniform velocity slightly more complicated.

AdVV

E

A

k

2

2

Since velocity is not uniform, AVV

Ek2

2

hence, a kinetic energy correction factor is defined by

AV

dAV

3

3

2

12

1

The true kinetic energy flux across a plane is

AVEk

3

2

1

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For flow in a circular pipe, ranges from 2 for fully developed laminarflow to about 1.05 for fully developed turbulent flow.

Laminar flow

2.0

Turbulent flow

1.0

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INTERNAL INCOMPRESSIBLE VISCOUS FLOW

Energy Equation for a Flow in a Pipe

Assumptions1. No shaft work2. Incompressible flow3. Steady flow4. Internal energy and pressure are uniform at cross sections 1 and 2

CSC

othershears AdVpedet

WWWQ

0000

gzV

ue 2

2

11

21

22

22

1212

12

2222

dAVV

dAVV

zzgmpp

muumQ

AA

Note: At cross-sections 1 and 2, velocity profiles are non-uniform.However, integrals in the above equation can be expressed in terms of

average velocity and kinetic energy correction factor, i.e.

mV

AVV

dAVV

A

222

22

2

2

Therefore, energy equation becomes,

22

21

1

22

21212

12

VVmzzgm

ppmuumQ

14

Considering, u, p and are uniform over inlet and outlet cross-sections, we can write

Flow Analysis 3

dividing by mass flow rate and rearranging, we get

mg

Q

g

uuz

g

V

g

pz

g

V

g

p

12

2

2

22

21

2

11

1

22

heat transfer per unit mass of moving fluid.m

Q

For incompressible flow (combining the first and the second law ofthermodynamics),

quugh f 12

Here hf represents the loss of potential to perform useful work. It shows usthat the internal energy (and hence temperature) of an incompressiblefluid can be increased by two ways: heat transfer to the fluid and friction.Only one effect can cause an internal energy decrease; namely heattransfer from fluid, as ghf cannot be negative.

fhzV

g

pz

g

V

g

p

2

2

22

21

2

11

1

22

EXTENDED BERNOULLI EQUATION

hf is called head loss

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An incompressible viscousfluid flows between twohorizontal parallel plates asshown. The plates arespaced 0.5 cm apart and arevery wide perpendicular topage. Flow is laminar andvelocity profile at any crosssection is given by,

Example:

= 850 kg/m3

22

12 Y

y

L

PYu

where P is the pressure change that occurs in length L. Calculateaverage mechanical energy loss hf between the pressure gages. Thenshow that mechanical energy loss also satisfies the equations

m

dy

u

gh CVf

2

2Re

2V

Y

Lgh f

YVRe

ad

here

and

where

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CLASSIFYING THE FLOW IN A PIPE OR DUCT

Laminar and Turbulent Flow

If the flow in a pipe is laminar, the fluid moves along smooth streamlines.

If the flow is turbulent, a rather violent mixing of the fluid occurs, and thefluid velocity at a point varies randomly with time.

The difference between laminar and turbulent flows were classified byOsborne Reynolds in 1883. Reynolds performed a series of experiments.

Pipe-flow transition experiment. (a) laminar flow. (b) High ReD, turbulentflow. (c) Spark photograph of turbulent flow condition. (After O.Reynolds, an experimental investigation of the circumstances whichdetermine whether the motion of water shall be direct or sinuous and ofthe law of resistance in parallel channels, Phil. Trans. Roy. Soc., London,A174:935–982, 1883)

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Reynolds’ experiments showed that the nature of the pipe flow dependson the Reynolds number,

dVRe

Developing and Fully Developed Flow

The flow in a constant area duct or pipe is said to be fully developed if theshape of the velocity profile is the same at all cross sections.

The length Le is called entrance length or the developing length

flowturbulent Re4.4

flowlaminar Re06.0

(Re)

6/1

d

L

d

L

d

L

e

e

e f

18Flow Analysis 3