Energy Conservation (Bernoulli’s Equation)

Integration of Euler’s equation

Bernoulli’s equation2

222

1

211

22gz

Vpgz

Vp

Flow work + kinetic energy + potential energy = constant

p

A

x Under the action of the pressure, the fluid elementmoves a distance x within time tThe work done per unit time W/t (flow power) is

rate flow massunit per donework 1

,

t

W

AV

p

PAV

t

xA

p

t

xpA

t

W

02

1

2

1

2

1 gdzVdV

dp

Energy Conservation (cont.)

t)unit weighper (energy g where,22 2

222

1

211

z

g

Vpz

g

Vp

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

Examples: 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

Energy Equation(cont.)

Example: 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.

From Bernoulli' s equation, V = 2gh

From mass conservation, dm

dt

since

dh

hintegrate

h(t) = H

A V

m A hdh

dt

A

AV gh

dh

dth dt

t h t

hole

khole

k

drain

tantan

,( . )

. , . ,

. , , sec.

0 1

12

0 0443 0 0443

0 0215 0 93

2

2

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 t

Energy conservation (cont.)

Generalized energy concept:

1

211

2z

g

Vp

2

222

2z

g

Vp

Energy added, hA

(ex. pump, compressor)

Energy extracted, hE

(ex. turbine, windmill)

Energy loss, 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 equation, sBernoulli' Extended z

g

Vphhhz

g

VpLEA

Examples: 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.

pump

Z=15 in

1 2

Mercury (m=844.9 lb/ft3)water (w=62.4 lb/ft3)

1 hp=550 lb-ft/s

hE=hL=0, z1=z26-in dia. pipe 4-in dia.pipe

Q=300 gal/min=0.667 ft3/s=AVV1=Q/A1=3.33 ft/sV2=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)* . . /

zo

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

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, :

.

Energy Conservation (cont.)

Energy: E=U(internal thermal energy)+Emech (mechanical energy) =U+KE(kinetic energy)+PE(potential energy)Work: W=Wext(external work)+Wflow(flow work)Heat: Q heat transfer via conduction, convection & radiation

dE=dQ-dW, dQ>0 net heat transfer in dE>0 energy increase and vice versa dW>0, does positive work at the expense of decreasing energy, dE<0

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

Their difference is due to external heat transfer and work done on flow

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

Energy Conservation (cont.)

( )m up V

gzin in

2

2 ( )m u

p Vgzin out

2

2

Heat in q=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

Energy Conservation(cont.)Example: Superheated water vapor is entering the steam turbine with a mass flow rate of 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 power output of the turbine.

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 table: hin=3149.5 kJ/kg

From saturated steam table: 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