Bernoulli’s Equation

Start with a force balance on a fluid element, i.e.,

the differential form of the steady momentum equation

use the vector identity

1

2

and substitute

1

2

V V p g z

V V V V V V

V V V V p g

rearranging and dividing by

1

2

z

pV V g z V V

2

take the dot product with a length element in the streamwise direction

1

2

Note that

and 0

10

2

Next, integrate along a str

ds

pV V g z ds V V ds

ds dx dy dz dx y z

V V ds

dpd V gdz

2

2

eamline from a ref. point to another point

in the flow

10 .

2

0 .2

dpd V gdz ds const

dp Vgz ds const

See next slide for

proof

The curl of

ˆˆ ˆ

ˆˆ ˆ

ˆˆ ˆ

ˆˆ ˆ

0

V V

w v w u v uV i j k

y z x z x y

V ui vj wk

V V v w i u w j u v k

V ui vj wkds

V V

u v w v u w w u vV V ds

V

Proof

2

2

1.

2

Can't integrate the first term unless we know

how relates to

Assuming ., i.e., incompressible

1.

2

dpV gz const

p

const

pV gz const

Case: Incompressible flow along a streamline

2

2

2

1

2

if 0 irrotational

then for any direction of integration

1.

2

.2

and for incompressible flow

.2

pV V g z V V

V

dpd V gdz const

dp Vgz const

p Vgz const

Case: Irrotational, Incompressible Flow

This is true throughout the flow field. Since the flow is irrotational, the integration may be carried out in any direction.

Solution Strategy

• Strategy for solving incompressible, inviscid flow – Obtain the velocity field from the governing

equations

– Obtain the pressure field from Bernoulli’s eqn using the known velocity field

2

2

- pressure (gage or absolute)

1 - dynamic pressure

2

- total pressure2

p

V

Vp

BERNOULLI’S EQN FROM AN ENERGY PERSPECTIVE

2 2

,

,

2 2

For 1d, steady flow with one inlet and one outlet

ˆ ˆ2

if there is no shaft work, 0

2

out inout in out in in sh out

out in

sh out

out in

out in

V Vp pm u u g z z Q W

W

V Vp pg z

2 2

ˆ ˆ

recall that for inviscid flow, Bernoulli's eqn was

02

Comparing the two, they are the same when

ˆ ˆ 0

inout in out in

out inout in

out in

inout in

Qz u u

m

V Vp pg z z

Qu u

m

They are the same when the internal energy change is equivalent to the heat addition per unit mass

2

Consider to be the useful or available energy2

ˆ ˆThen represents the loss of useful energy in an

incompressible fluid due to friction between the inlet and outlet or

inout in

in

p Vgz

Qu u

m

Qloss

m

ˆ ˆ empirical evidence shows that this is >0out inu u

2

2

- pressure (gage or absolute)

1 - dynamic pressure

2

- total pressure2

p

V

Vp

2

2 2 2

Now, let's assume the loss term is proportional to the kinetic

ˆ ˆenergy per unit mass, i.e., 2

02 2 2

neglecting changes in

out inL out in

out in outout in L

out in

V Qloss K u u

m

V V Vp pgz gz K

0,0,

2 2 2

ˆ ˆ

2

0, 0,

potential energy yields

2 2 2

ˆ ˆ 02

out inin out in

out in outout in L

p Qp u um

out inout in L out in

V V Vp p K

V Qp p K u u

m

outVoutV0inV

0inV

Compare volume flow rate for two different vent configurations each

having a hole diameter of 120 mm

1 a cylindrical hole with 0.5

2 a well-rounded cylindrical hole with 0.05

Assume incompressible,

L

L

K

K

steady flow. The room pressure is held constant

at 1 kPa. The outlet pressure is atmospheric.

2 2

2 2

out in

out in

V Vp p

out ing z z

2

2 22

2

2

2

solve for

2 22 2

2

1

The volume flow rate is 4

Substituting numbers, we get

4

outL

out

out in out outout L L

in out

in out

out

L

out h out h

out h

VK

V

V p p Vp pV K K

p pV

K

Q V A V d

V d

3

cyl. hole

32

rounded hole

0.372

0.445 4

out h

ms

mV ds

We get much more flow out of the hole having a well-rounded inlet.

cyl. hole

rounded hole

2

2

1

2 2

1

1

0.81

0.916

in outh d

in out

h

L

d

in out in outh h

d

L

d

d

p pQ A C

p pA

KQC

p p p pA A

CK

C

C

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

Loss coeff, KL

Dis

charg

e c

oeff

, C

d