Energy Conservation (Bernoulli’s Equation)

If one integrates Euler’s eqn. along a streamline, between two points , &

Which gives us the Bernoulli’s Equation

Constant 22 2

222

1

211 gz

Vpgz

Vp

Flow work + kinetic energy + potential energy = constant

02

1

2

1

2

1 gdzVdV

dp

0 gdzVdVdp

Recall Euler’s equation:

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

We get :

Bernoulli’s Equation (Continued)

p

A

x

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 pressure p acting on the control surface as shown.

Due to the fluid pressure, the fluid element moves a distance x within time t. Hence, the work done per unit time W/t (flow power) is:

Flow work per unit mass

Flow work or Power

1/mass flow ratepv

p

Flow work is often also referred to as flow energy

t)unit weighper (energy g where,22 2

222

1

211

z

g

Vpz

g

Vp

Very Important: Bernoulli’s equation is only valid for :

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

Application of Bernoulli’s equation - Example 1:

Determine the velocity and mass flow rate of efflux from the circular hole (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

Bernoulli’s Equation (Cont)

Bernoulli’s 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 enclosedby the dotted line. Specify h=h(t) as the waterlevel as a function of time.

0 20 40 60 80 1000

1

2

3

4

time (sec.)

wat

er h

eigh

t (m

)

4

2.5e-007

h( )t

1000 tsec 90.3 t

0.0443t- 20

4

h

Energy exchange (conservation) in a thermal system

1

211

2z

g

Vp

2

222

2z

g

Vp

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 lossthrough pipes hL

loss throughelbows

hL

loss throughvalves

Energy conservation(cont.)

2

222

1

211

22 z

g

Vphhhz

g

VpLEA

Example: Determine the efficiency of the pump if the power input of the motoris 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=AVV1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/s

kinetic energy head gain

V V

gft2

212 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 1 25 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 Bernoulli’s Equation

Looking at the pressure term:

Energy conservation (cont.)

Example (cont.)

Pressure head gain:

pump work

p pft

hp p V V

gft

w

Aw

2 1

2 1 22

12

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

681 7

1 550

1 24

1 24

1 50 827 82 7%

Frictional losses in piping system

loss head frictional

22 equation, sBernoulli' Extended

21

2

222

1

211

L

LEA

hppp

zg

Vphhhz

g

Vp

P1

P2

Consider a laminar, fully developed circular pipe flow

p P+dp

w

Darcy’s Equation:

R: radius, D: diameterL: pipe lengthw: wall shear stress

[ ( )]( ) ( ) ,

,

p p dp R R dx

dpR

dx

p p ph

g

L

Df

L

D

V

g

w

w

Lw

FHIK FHIKFHGIKJ

2

1 22

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

w

f VFHIK

FHGIKJ4 2

2

24

2Vfw

g

V

D

Lf

D

L

gh w

L 2

4 2

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 diameter D

Such that D

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 roughness D

fVD

VD

f

f F

f F

FH IK

FHIK

64

64

, Re ,

Re,

(Re)

.

Re, :

.

DFf

Re,

D

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) inpipe flows also contribute to the total head loss of the system. Their contributionsare 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 vena contracta 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 Bernoulli’s equation to take intoconsideration this effect as described in the next page.

Kh

V g

pp K V

L

L

LV

2 2

12

2

2 12/

,

so that

12 0

3 7

2 51

f fD FHG

IKJ. log

.

.

Re,

valid for nonlaminar range

Minor Loss through flow entrance

V2 V3

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(

11,0,

2,

22:Equation sBernoulli' Extended

313131

23

3

233

1

211

gzV

p

2

2

Energy Conservation (cont.)

We will now also account for energy tranfer via Heat Transfer

The general 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 system where W = Wext(external work) + Wflow(flow work)

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

Convention: dQ > 0 net heat transfer into the system (Convention Q,q..) dW > 0, positive work done by the system

Q: What is Internal Energy ?

Energy Conservation (cont.)

U = mu, u(internal energy per unit mass), KE = (1/2)mV2 and PE = mgzFlow work Wflow= m (p/)

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

)

( ) , ( )

FHGIKJ

gz

m up V

gz m up V

gzin in out out

2 2

2 2

p

Energy Conservation (cont.)

( )m up V

gzin in

2

2 ( )m u

p Vgzin out

2

2

Heat in, =dQ/dt

Work out dW/dt

From 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 up

in out

in out

in out

2 2

2 2

2 2

2 2

Q

Conservation of Energy – ApplicationExample: Superheated water vapor enters a steam turbine at a mass flow rate1 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 MpaT=350 CV=80 m/sz=10 m

P=0.5 Mpa100% saturated steamV=50 m/sz=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 hV

gz m hV

gzdW

dtdW

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 1 95 0 049

392 8

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 the molecules • Due to the vibrational, translational and rotational energies of the moelcules.•Proprtional to the temperature of the gas.

Q – total 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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