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

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 – gkˆ
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
Newton’s 2nd
!
!
!
!
! !
!
Dv
ra = r (¶ t v + (v × Ñ)v ) = r
= -Ñp – gkˆ
Dt
Euler’s 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) Euler’s 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
öˆ
R
ˆ
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 sˆ + an nˆ
Dt
and also write
!
v = vs sˆ + vn nˆ
(moves in s direction)
Chap. 4.2 Streamline coordinate system
!
Dvs
Dv D
Dsˆ
=
(vs sˆ) =
sˆ + vs
Dt Dt
Dt
Dt
Dvs ¶vs
¶vs
¶vs
=
+ vs
+ vn
Dt
¶t
¶s
¶n
Dsˆ ¶sˆ
¶sˆ
¶sˆ
= + 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 ¶sˆ
ˆ
=ç
+ vs
s
+
v
+
v
nˆ ÷
÷
sç
s
Dt è ¶t
¶s ø
 ø
è ¶t
(if steady) = vs
¶vs
1
sˆ + vs2 nˆ
¶s
Â
¶vs
a s = vs
¶s
2
vs
an =
Â
Chap. 3.2 Euler’s eqn. along a streamline
¶vs
¶v
¶p ˆ
ras = rvs
= rv = – – gk × sˆ
¶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 Euler’s 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
Bernoulli’s equation
Along a streamline
Chap. 3.2 Euler’s 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 ˆ
dp
dz
ran = – – gk × nˆ = – – 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  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
nˆ = -rˆ
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 Euler’s eqn. along a streamline Truck http://www.youtube.com/watch?v=1tjJmArxVBM Chap. 3.2 Euler’s 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)

Purchase answer to see full
attachment