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Energy Conservation (Bernoullis Equation)

If one integrates Eulers eqn. along a streamline, between two points , &

Which gives us theBernoullis Equation

Constant22

2

2

221

2

11 gzVpgzVp

Flow work + kinetic energy + potential energy = constant

02

1

2

1

2

1 gdzVdVdp

0 gdzVdVdp

Recall Eulers equation:

Also recall that viscous forces were neglected, i.e. f low is invisicd

We get :

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Bernoullis Equation (Continued)

p

A

Dx

D

D

D

D

D

D

D

D

t

W

AV

p

pAV

t

xA

p

t

xpA

t

W

1

,

Flow Work (p/):It is the work required to move fluid across the control volume boundaries.

Consider a fluid element of cross-sectional area

A with pressurep acting on the control surface

as shown.

Due to the fluid pressure, the fluid element moves a distance Dx within

time Dt. Hence, the work done per unit time DW/Dt (flow power) is:

F low workper unit mass

Flow work or Power

1/mass flow rate

pvp

F low workis often also referred to as f low energy

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t)unit weighper(energygwhere,22

2

2

221

2

11

z

g

Vpz

g

Vp

Very Important: Bernoullis equation is only valid for:

incompressible fluids,steady flow along a streamline, no energy loss dueto friction, no heat transfer.

Application of Bernoullis equation - Example 1:

Determine the velocity and mass flow rate of efflux from the circularhole (0.1 m dia.) at the bottom of the water tank (at this instant). The

tank is open to the atmosphere and H=4 m

H

1

2

p1 = p2, V1=0

)/(5.69

)85.8()1.0(4

*1000

)/(85.84*8.9*2

2)(2

2

212

skg

AVm

sm

gHzzgV

Bernoullis Equation (Cont)

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Bernoullis Eqn/Energy Conservation (cont.)

Example 2: If the tank has a cross-sectional area of 1 m2, estimate the time

required to drain the tank to level 2.

h(t)

1

2

First, choose the control volume as enclosed

by the dotted line. Specify h=h(t) as the water

level as a function of time.

From Bernoulli' s equation, V = 2gh

From mass conservation,dm

dt

since

dhh

integrate

A V

m A hdh

dt

A

AV gh

dhdt

h dt

hole

khole

k

tan

tan

,( . )

. , . ,

0 1

12

0 0443 0 0443

2

2

0 20 40 60 80 1000

1

2

3

4

time (sec.)

waterheight

(m)

4

2.5e-007

h ( )t

1000 t

sec90.3t

0.0443t-20

4

h

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Energy exchange (conservation) in a thermal system

1

211

2z

gVp

2

2

22

2z

gVp

Energy added, hA

(ex. pump, compressor)

Energy extracted, hE

(ex. turbine, windmill)

Energy lost, hL

(ex. friction, valve, expansion)

pump turbine

heat exchanger

condenser

hEhA

hL, friction loss

through pipes hL

loss through

elbows

hL

loss through

valves

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Energy conservation(cont.)

2

2

221

2

11

22

z

g

Vphhhz

g

VpLEA

Example: Determine the efficiency of the pump if the power input of the motor

is measured to be 1.5 hp. It is known that the pump delivers 300 gal/min of water.

Mercury (m=844.9 lb/ft3)

water (w=62.4 lb/ft3)

1 hp=550 lb-ft/s

No turbine work and frictional losses, hence: hE=hL=0. Also z1=z2

6-in dia. pipe 4-in dia.pipe Given: Q=300 gal/min=0.667 ft3/s=AV

V1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/skinetic energy head gain

V V

gft2

2

1

2 2 2

2

7 54 3 33

2 32 20 71

( . ) ( . )

* .. ,

p z z p z z

p p z

lb ft

w o m w o w

m w

1 2

2 1

29 62 125 97813

( )

(844. .4)* . . /

pump

Z=15 in

1 2

zo

If energy is added, removed or lost via pumps turbines, friction, etc.then we use

Extended Bernoullis Equation

Looking at the pressure term:

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Energy conservation (cont.)

Example (cont.)

Pressure head gain:

pump work

p pft

hp p V V

gft

w

A w

2 1

2 1 2

2

1

2

97813

6215 67

216 38

.

.4. ( )

. ( )

Flow power delivered by pump

P =

Efficiency =P

P

w

input

Qh

ft lb s

hp ft lb s

P hp

A

( .4)( . )( . )

. ( / )

/

.

.

.

. .

62 0 667 16 38

6817

1 550

1 24

124

15

0 827 82 7%

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Frictional losses in piping system

lossheadfrictional

22equation,sBernoulli'Extended

21

2

2

221

2

11

D

L

LEA

hppp

zg

Vphhhz

g

Vp

P1

P2

Consider a laminar, fully developed circular pipe flow

p P+dp

w

Darcys Equation:

R: radius, D: diameter

L: pipe length

w: wall shear stress

[ ( )]( ) ( ) ,

,

p p dp R R dx

dp

R

dx

p p ph

g

L

Df

L

D

V

g

w

w

Lw

FHIKFHIKFHGIKJ

2

1 2

2

2

2

4

2

Pressure force balances frictional force

integrate from 1 to 2

where f is defined as frictional factor characterizing

pressure loss due to pipe wall shear stress

D

w

f VHKH

G4 2

2

24

2Vf

w

g

V

D

Lf

D

L

gh wL

2

42

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When the pipe flow is laminar, it can be shown (not here) that

by recognizing that as Reynolds number

Therefore, frictional factor is a function of the Reynolds number

Similarly, for a turbulent flow, f = function of Reynolds number also

. Another parameter that influences the friction is the surface

roughness as relativeto the pipe diameterD

Such thatD

Pipe frictional factor is a function of pipe Reynolds

number and the relative roughness of pipe.This relation is sketched in the Moody diagram as shown in the following page.

The diagram shows f as a function of the Reynolds number (Re), with a series of

parametric curves related to the relative roughnessD

fVD

VD

f

f F

f F

FHIK

FHIK

64

64

, Re ,

Re,

(Re)

.

Re, :

.

DFf

Re,

D

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Losses in Pipe Flows

Major Losses: due to friction, significant head loss is associated with the straight

portions of pipe flows. This loss can be calculated using the Moody chart or

Colebrook equation.

Minor Losses: Additional components (valves, bends, tees, contractions, etc) in

pipe flows also contribute to the total head loss of the system. Their contributions

are generally termed minor losses.

The head losses and pressure drops can be characterized by using the loss coefficient,

KL, which is defined as

One of the example of minor losses is the entrance flow loss. A typical flow pattern

for flow entering a sharp-edged entrance is shown in the following page. A venacontracta region is formed at the inlet because the fluid can not turn a sharp corner.

Flow separation and associated viscous effects will tend to decrease the flow energy;

the phenomenon is fairly complicated. To simplify the analysis, a head loss and the

associated loss coefficient are used in the extended Bernoullis equation to take into

consideration this effect as described in the next page.

Kh

V g

pp K V

L

L

LV

2 2

12

2

2 12/,

DD

so that

12 0 3 7

2 51

f fD HG KJ. log .

.

Re ,

valid for nonlaminar range

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Minor Loss through flow entrance

V2 V

3

V1

(1/2)V22 (1/2)V3

2

KL(1/2)V32 pp

ghK

zzgK

VVppp

g

VKhz

g

Vphz

g

Vp

LL

LLL

1

2)(2(

1

1,0,

2,

22:EquationsBernoulli'Extended

313131

2

33

2

331

2

11

gzV

p

2

2

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Energy Conservation (cont.)Let us now also account for energy transfer via Heat Transfer, e.g. in

a heat exchanger

The most general form of conservation of energy for a system can be

written as: dE = dQ-dW where (Ch. 3, YAC)

dE Change in Total Energy, E

and E = U(internal energy)+Em(mechanical energy) (Ch. 1 YAC)

E = U + KE (kinetic energy) + PE(potential energy)

dW Work done by the systemwhere

W = Wext(external work) + Wflow(flow work)

dQ = Heat transferinto the system(via conduction, convection & radiation)

Convention: dQ > 0 net heat transfer into the system (Symbols Q,q..)

dW > 0, positive work done by the system

Q: What is Internal Energy ?

mechanical

energy

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Energy Conservation (cont.)

U = mu, u(internal energy per unit mass),

KE = (1/2)mV2 and PE = mgz

Flow work Wflow= m (p/)

It is common practice to combine the total energy with flow work.

Thus:

The difference between energy in and out is due to heat transfer (into or out)

and work done (by or on) the system.

Energy flow rate: m(u +

V

2 plus Flow work rate m

p

Flow energy in Energy out =

2

)

( ) , ( )

HGKJ

gz

m up V

gz m up V

gzin in out out

2 2

2 2

p

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Energy Conservation (cont.)

( )m up V

gzin in

2

2 ( )m u

p Vgzin out

2

2

Heat in, =dQ/dt

Work out dW/dtFrom mass conservation:

From the First law of Thermodynamics (Energy Conservation):

dQ

dt or

dQ

dt

where is defined as "enthaply"

( )

( ) ,

( ) ( )

m m m

m up V

gz m up V

gzdW

dt

m hV

gz m hV

gzdW

dt

h u

p

in out

in out

in out

2 2

2 2

2 2

2 2

Q

Hence, a system exchanges energy with the environment due to:

1) Flow in/out 2) Heat Transfer, Q and 3) Work, W

This energy exchange is governed by the F ir st Law of Thermodynamics

system

Enthalpy

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Conservation of EnergyApplication

Example: Superheated water vapor enters a steam turbine at a mass flow rate

1 kg/s and exhausting as saturated steam as shown. Heat loss from the turbine is

10 kW under the following operating condition. Determine the turbine power output.

P=1.4 Mpa

T=350 C

V=80 m/s

z=10 m

P=0.5 Mpa

100% saturated steam

V=50 m/s

z=5 m

10 kw

From superheated vapor tables:

hin=3149.5 kJ/kg

From saturated steam tables: hout=2748.7 kJ/kg

dQ

dt

( )

( )

( ) ( )[( . . )

( )

( . )( )

]

. . .

. ( )

m h

V

gz m h

V

gz

dW

dt

dW

dt

kW

in out

2 2

2 2

2 2

10 1 3149 5 2748 7

80 50

2 1000

9 8 10 5

1000

10 400 8 195 0 049

392 8

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Internal Energy ?

Internal energy, U (total) or u (per unit mass) is the sum of all

microscopic forms of energy.

It can be viewed as the sum of the kinetic and potential energies of themolecules

Due to the vibrational, translational and rotational energies of the moelcules.

Proportional to the temperature of the gas.

Qtotal heat transfer (J)

rate of total heat transfer (J/s, W)

q heat transfer per unit mass (J/kg)

Heat Flux, heat transfer per unit area (J/m2)

Q

q

Q, q ?!%

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