MECH 324 Marymount University Chapter 3 & 4 Bernoulli Equation Worksheet do the first question and the second question and question number 3 i will upload a lectures slide i want you to copy hand writing from page 11-15 on question 3 after you finish scan them and send it to me pdf file all together the 3 questions do it with a blue pin and papers Chap. 3-4 Bernoulli equation

Streamlines past an airfoil

Chap. 4.2 Material Derivative

Material Derivative

!

In ra = -Ñp – gk

What does

!

a

e.g

pump

Tapering Tube

–

Accelerate fluid element by

(1) increase flow rate on pump

¶v x ¶u

a1x =

=

¶t

¶t

Partial derivative

mean?

Chap. 4.2 Material Derivative

(2) keep pump constant, tapering makes fluid accelerate

dx

æ dx ö

uç x – ÷

2ø

è

!

x

æ dx ö

uç x + ÷

2ø

è

dx

dx

dx

dx

u( x + ) – u( x – )

u( x + ) – u( x – )

2

2 = u ( x)

2

2

a2 x =

dt

dx

¶u

dx é

¶u

dx ù

u ( x) + ( x) – êu ( x) + ( x)(- )ú + 0(dx 2 )

¶x

2 ë

¶x

2 û

= u ( x)

dx

¶u

= u ( x) ( x)

¶x

Chap. 4.2 Material Derivative

other ways to accelerate in the x-direction

z

e.g

linear shear flow,

w

u ( z ) = g!z

w = vz

x

rising fluid element

dz

æ dx ö

uç x + ÷

2ø

è

æ dx ö

uç x – ÷

2ø

è

Traverses

dz

in

dz

dt =

w

dz

dzx

) – u( z )

2

2 w

a3 x =

dz

¶u

Tayler => a3 x = w

¶z

u( z +

Chap. 4.2 Material Derivative

Can also happen in y-direction

Analogous for

a y az

a4 x = u

¶u

¶u

u = uy

é

¶u

¶u

¶u

¶u ù

ê a x = ¶t + u ¶x + v ¶y + w ¶z ú

ê

ú

ê a = ¶v + u ¶v + v ¶v + w ¶v ú

ê y ¶t

¶x

¶y

¶z ú

ê

¶w

¶w

¶w

¶w ú

êa z =

+u

+v

+w ú

¶t

¶x

¶y

¶z úû

êë

Chap. 4.2 Material Derivative

Shorter in vector notation

! ¶v ! ! !

! ! ! ! D !

a=

+ (v × Ñ )v = ¶ t v + (v × Ñ )v =

v

¶t

Dt

! !

D ¶

Material derivative :

= + (v × Ñ )

Dt ¶t

æ u ö æ ¶ / ¶x ö

÷

! ! ç ÷ ç

¶

¶

¶

v × Ñ = ç v ÷ × ç ¶ / ¶y ÷ = u + v + w

¶x

¶y

¶z

ç w ÷ ç ¶ / ¶z ÷

è ø è

ø

Newtons 2nd

!

!

!

!

! !

!

Dv

ra = r (¶ t v + (v × Ñ)v ) = r

= -Ñp – gk

Dt

Eulers equation

Chap. 4.2 Material Derivative

Remarks:

!

¶

v

(1) flows with t are called Steady

! ! !

(2) terms (v × Ñ)v are non-linear in

=> eqns. much harder to solve

(3) Material derivative not an issue for solids,

(4) Eulers eqn. does not contain viscosity

Chap. 4.2 Material Derivative

An incompressible, inviscid fluid flows steadily past a tennis ball of radius R, as shown.

According to a more analysis of the flow. The fluid velocity along streamline A-B is given by

3

æ

ö

R

V = u ( x)i = V0 çç1 + 3 ÷÷i

x ø

è

Where V0 is the upstream velocity far ahead of the sphere

Determine the acceleration experienced by fluid particles as they flow along this streamline

Chap. 4.2 Material Derivative

Chap. 4.2 Streamline coordinate system

Chap. 4.2 Streamline coordinate system

!

Streamlines : lines everywhere tangent to v

– normal lines ( ? to streamlines)

describe fluid motion along the s and n axis

Analogously,

! D !

a=

v = as s + an n

Dt

and also write

!

v = vs s + vn n

(moves in s direction)

Chap. 4.2 Streamline coordinate system

!

Dvs

Dv D

Ds

=

(vs s) =

s + vs

Dt Dt

Dt

Dt

Dvs ¶vs

¶vs

¶vs

=

+ vs

+ vn

Dt

¶t

¶s

¶n

Ds ¶s

¶s

¶s

= + v s + vn

Dt ¶t

¶s

¶n

1

n?

Inverse radius of curvature of path (streamline) =

Â

!

¶vs ö

Dv æ ¶vs

1 ö

æ ¶s

=ç

+ vs

s

+

v

+

v

n ÷

÷

sç

s

Dt è ¶t

¶s ø

Â ø

è ¶t

(if steady) = vs

¶vs

1

s + vs2 n

¶s

Â

¶vs

a s = vs

¶s

2

vs

an =

Â

Chap. 3.2 Eulers eqn. along a streamline

¶vs

¶v

¶p

ras = rvs

= rv = – – gk × s

¶s

¶s

¶s

dz

!

sin

q

=

steady

v = v = vs

ds

v ¹ vy

Along streamline : only functions of s =>

¶

d

=>

¶s

ds

Chap. 3.2 Eulers eqn. along a streamline

dv

d æ1 2ö

d

d 1

rv = r ç v ÷ = – p – g z

ds

ds è 2 ø

ds

ds r

4

ò ds

1 2

1

v = – ò dp – gz + const.

2

r

If

r =const. =>

=>

1 2

1

v = – p – gz + const.

2

r

1 2

rv + p + gz = const.

2

Bernoullis equation

Along a streamline

Chap. 3.2 Eulers eqn. along a streamline

Conditions for Bernoulli to hold:

(1) viscosity must be negligible (Euler) : inviscid flow

(2) Steady flow

(3) incompressible flow

(4) along a streamline

Chap. 3.4 Normal to streamlines

¶p

dp

dz

ran = – – gk × n = – – g

¶n

dn

dn

cos q =

dz

dn

Only along n?

For steady flow,

v2

dp

dz

ran = r = – – g

Â

dn

dn

=>

v2

ò r Â dn = – p – gz + const.

Chap. 3.4 Normal to streamlines

If

r = const.

=>

v 2 ( n)

p + gz + r ò

dn = const.

Â( n )

Across streamlines

Straight streamlines

(2)

Â- > ¥

(1)

p + gz = const.

p1 – p2 = g ( z 2 – z1 )

like hydrostatics

Chap. 3.4 Normal to streamlines

typically,

e.g. Swirling flow (sink)

r?

y

c

v(r ) = vq (r ) =

r

n = -r

r

q

x

r2

=>

c=const.

¶p

¶p

=¶n

¶r

v 2 (r )

dp

dp

ran = r

= – +0=

Â( r )

dn

dr

c2 1

=r 2×

r r

r

2

c2

1

1

1 2æ 1 1 ö

ö

2æ

p(r2 ) – p(r1 ) = ò r 3 dg = rc ç – 3 ÷ = rc çç 2 – 2 ÷÷

r

è 2 r ø r1 2

è r1 r2 ø

r1

Example Problems

Chap. 3.4 Energy conservation

1 2

p + rv + gz = const.

2

Along streamline

pdV dW

work

p=

=

=

dV

dV volume

1 2 1 mv 2 Kin.Energy

rv =

=

2

2 V

volume

mgz Grav.PotentialEnergy

gz = rgz =

=

V

volume

=> Inviscid flows conserve total energy per volume

Chap. 3.4 Energy conservation

Free jet:

If large reservoir v1 = 0

p1 = p0

(1) v1

z

h

v2

1

p0 + r 0 v12 + gh

2

1

= p0 + r 0 v22 + 0

2

(2)

p 2 = p0

v2 =

2gh

r

= 2 gh

Chap. 3.4 Energy conservation

Alternative :

If large reservoir v1 = 0

(1) Dv

E pot1 = mgh = rDVgh = gDVh

Ekin = 0

z

h

(2) E pot 2 = 0

E pot1 + Ekin1 = E pot 2 + Ekin 2

1

1

Ekin 2 = mv 2 = rDVv22

2

2

v2 = 2 gh

Example Problems

3.29 Water flows from a pressurized tank, through a 6-in diameter pipe, exists from a 2-in

diameter nozzle, and rises 20ft above the nozzle as shown in Fig. P3.36. Determine the

pressure in the tank if the flow is steady, frictionless, and incompressible.

Chap. 3.5 Static, Stagnation

Remark : nomenclature

1 2

p + rv + gz = const.

2

Static pressure

(Thermodynamic

pressure)

Hydrostatic pressure

Dynamic pressure

Chap. 3.5 Static, Stagnation

e.g. Two paper strips

(2)

v2 @ 0

z1 = z 2

p3 = p3′ = p0 > p1

Pushing force

(3)

(3)

v1

(1)

Blowing air

1 2

p1 + rv1 = p2 = p0

2

1 2

p1 = p0 – rv1 < p0
2
Chap. 3.5 Static, Stagnation
S=stagnation point
!
v =0
Stagnation streamline
1 2
ps = p + rv1
2
If hydrostatic pressures negligible
Þ
Þ
ps Highest pressure on stag. Streamline
p s highest pressure on body
Chap. 3.5 Static, Stagnation
Chap. 3.2 Eulers eqn. along a streamline
Truck
http://www.youtube.com/watch?v=1tjJmArxVBM
Chap. 3.2 Eulers eqn. along a streamline
F1
http://www.youtube.com/watch?v=jYaIXWNOa_A
Chap. 3.6 Bernoulli Equation
Chap. 3.6 Bernoulli Equation
Mass flowing through A1 per time
=Mass flowing through A2 per time
But volumes are only conserved for
incompressible flows
Chap. 3.6 Bernoulli Equation
Mass per time
dx = v1dt
m! = r1 A1v1 = const.
= r 2 A2 v2
v1
(Mass conservation)
If incompressible fluid, r1 = r 2 => A1v1 = A2 v2

Q = vA is the (volumetric) flow rate

Mass

m = r1dV = r1 A1dx

= r1 A1v1dt

2

circular cross section ( Ai = pRi )

2

2

=> v1 R1 = v2 R2

Chap. 3.6 Bernoulli Equation

Bernoulli

1 2

1 2

p1 + rv1 = p2 + rv2

2

2

v2 > v1

( A2 < A1 )
p2 > p1

2

2

4

æ

ö

æ

ö 1

æ

ö

æ

ö

æ

ö

1

1

A

1

A

R

p1 – p2 = r (v22 – v12 ) = r ç v12 çç 1 ÷÷ – v12 ÷ = rv12 ç çç 1 ÷÷ – 1÷ = rv12 çç 14 – 1÷÷

÷ 2

ç è A2 ø

÷ 2

2

2 ç è A2 ø

è R2 ø

è

ø

è

ø

1

=> measure velocity

or flow rate

é

ù2

ê

ú

2( p – p )

v1 = ê 1 4 2 ú

ê æ R1

öú

ê r çç 4 – 1÷÷ ú

êë è R2 ø úû

Q = v1 A1

[ Venturi flow meter ]

Chap. 3.6 Bernoulli Equation

Example

h very large

=> V very large at jet

far away v=0

1 2

large v => p < 0
p
=
p
rv
conduit :
0
2
in reality, P drops to pvap (vapor pressure of liquid)
=> boiling/ bubbles form

=> bubbles re-collapse when transported to regions of higher pressure

=> Cavitation damage

Example Problems

3.45 Water (assumed inviscid and incompressible) flows steadily in the vertical variable-area

pipe shown in Fig. P3.45. Determine the flowrate if the pressure in each of the gages reads 50kPa.

Chap. 3.6 Bernoulli for Measurements

Venturi flow meter

v2 @ 0 (2) is stagnation point

(highest pressure)

Bernoulli

h1

h2

(1)

(2)

1

p1 + rv12 = p2 + 0

2

1 2

p

p

=

rv1 = g (h2 – h1 ) = rg (h2 – h1 )

=> 2

1

2

=> v1 = 2 g (h2 – h1 )

measure speed by measuring dynamic pressure

more compact version : Pitot tube

v1 = 2 g (h2 – h1 )

Chap. 3.6 Bernoulli for Measurements

Chap. 3.6 Bernoulli for Measurements

Chap. 3.6 Bernoulli for Measurements

Example Problems

Example Problems

Example Problems

Example Problems

Example Problems

Example Problems

Problem 1(30pt)

A square gate of width w=1mhas its top edge at a depth h=10m below the water surface. It is

inclined at 45° angle and its bottom edge is hinged as shown in the figure below.

0

V

water

F

FR

12h

8=45

Find the force F needed to just open the gate.

The siphon method is used to steadily pull fluid from a large tank (water height ho = 3m)

over a height of elevation H above the tank surface (figure A). The siphon tube’s bottom

end is h below the tank floor. The tank contains water; viscous effects are negligible.

Assume also that the vapor pressure of water is negligible.

(a) If you want to apply Bernoulli’s equation, what conditions need to be fulfilled? (5 pts)

(b) A tube of constant cross section is used. What is the exit velocity of water at the

bottom (horizontal) end of the tube as a function of h? (15 pts)

(c) What is the maximum height H attainable before cavitation sets in, if the bottom end is

held at h= Imbelow the tank? Where does cavitation (absolute pressure ?0) occur?

(15 pts)

(d) In case (c), what is the flow rate through the tube, if De = 10cm (15pts)

A

B

3m

D

To.

3m

Problem 3 (20pt)

Please copy & write down Lecture slide Chap.3 page 11-15 with figures (HAND WRITING)

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