4.1: PROBLEM DEFINITION
Situation:
Path of a ‡uid particle.
Find:
If a light was attached to a ‡uid particle and take a time exposure, would the image
you photographed be a pathline or streakline?
SOLUTION
The pathline is de…ned as the path taken by a ‡uid particle moving through a …eld.
The photograph would yield a pathline.
1
4.2: PROBLEM DEFINITION
Situation:
Smoke rising from a chimney.
Find:
The pattern produced by smoke rising from a chimney on a windy day is analogous
to a pathline or streakline?
SOLUTION
The streakline is de…ned as a line generated by a tracer injected into ‡ow at starting
point. The tracer is the smoke and the starting point is the chimney so the smoke’s
pattern is analogous to a streakline. The di¤usion of the smoke prevents achieving
a …ne line. The wind also causes disperson, which is another reason that keeps the
line from being …ne.
2
4.3: PROBLEM DEFINITION
Situation:
A windsock is a sock-shaped device attached to a swivel on top of a pole. Windsocks
at airports are used by pilots to see instantaneous shifts in the direction of the wind.
If one drew a line co-linear with a windsock’s orientation at any instant, the line
would be best approximate a (a) pathline (b) streakline (c) streamline.
SOLUTION
Answer is (c) streamline. A windsock shows you the wind direction (‡ow …eld) at
a given location at one instant in time - it is not a parade of a series of particles
(streakline), or a path recorded on some medium (pathline). So the windsock reveals
the streamline.
3
4.4: PROBLEM DEFINITION
Situation:
For pathlines, streaklines, and streamlines to all be co-linear, the ‡ow must be
a) dividing
b) stagnant
c) steady
d) a tracer
SOLUTION
The answer is (c) steady.
4
4.5: PROBLEM DEFINITION
Situation:
Dye is injected into a ‡ow …eld and produces a streakline.
Pathline starts at t = 4 s, ends at t = 10 s: Flow speed is constant.
Find:
Draw a pathline of the particle.
SOLUTION
The streakline shows that the velocity …eld was originally in the horizontal direction
to the right and then the ‡ow …eld changed upward to the left. The pathline starts
o¤ to the right and then continues upward to the left.
5
4.6: PROBLEM DEFINITION
Situation:
A dye streak is produced in a ‡ow that has a constant speed.
Find:
Sketch a streamline at t = 8 s.
Sketch a particle pathline at t = 10 s for a particle that was released from point A
at time t = 2 s.
Sketch:
SOLUTION
At 8 seconds (near 10 sec) the streamlines of the ‡ow are horizontal to the right.
Streamlines at t = 8 s
Initially the ‡ow is downward to the right and then switches to the horizontal direction
to the right. Thus one has the following pathline.
Particle pathline for a particle released at t = 2 s
6
4.7: PROBLEM DEFINITION
Situation:
A velocity …eld is given mathematically as V = 2x + 3yj. The velocity …eld is:
a. 1D in x
b. 1D in y
c. 2D in x and y
SOLUTION
The answer is (b). Because there is only a j component, the vector is representing a
…eld that varies in 1 dimension .
7
4.8: PROBLEM DEFINITION
Situation:
Gasoline spill in a major river.
The mayor of a large downstream city demands an estimate of hours for the spill
to get to reach supply plant intake.
Emergency responders measure the speed of the leading edge of the spill, e¤ectively
focusing on one particle of ‡uid.
Environmental engineers employ a computer model which simulates the velocity
…eld for any stage of the river, and for all locations (including steep narrow canyon
sections with fast velocities, and an extremely wide reach with slow velocities).
To compare these two mathematical approaches, which statement is most correct?
a. The responders have an Eulerian approach, and the engineers have a Lagrangian
one
b. The responders have a Lagrangian approach, and the engineers have an Eulerian
one.
SOLUTION
Answer is (b). The responders have a Lagrangian approach, because they are making
a measurement based on following one particle. The engineers are using a Eulerian
approach, because they are de…ning velocity …elds for many locations in space.
8
4.9: PROBLEM DEFINITION
Situation:
Unsteady ‡ow.
Find:
Identify …ve examples of an unsteady ‡ow, and explain what features classify them
as unsteady.
SOLUTION
Answers will vary, but typical student answers might include:
1. Gust of wind blowing past a pole - velocity direction and magnitude vary with
time; the word gust means sudden burst of speed, rather than constant speed.
2. Flow next to a rock in a natural river - local velocities (direction and magnitude)
vary with time in the vicinity of the rock, even if the average river ‡owrate is
unchanging.
3. Flow past the lips due to inhaling and exhaling - direction keeps alternating,
speed goes to zero each time the direction changes.
4. The motion of water at the center of a boiling pot - in convection cells, ‡uid
is rising generally, but at any location there is some turbulent mixing as new
water ‡ows in from the side to be heated and then rise.
5. At the outlet hose of a manual tire pump - with each stroke of the pump handle
‡owrate and velocity are …rst increasing, then decreasing to zero.
9
4.10: PROBLEM DEFINITION
Situation:
Pouring a heavy syrup on pancakes.
Find:
Would the thin …lm of syrup be a laminar or turbulent ‡ow?
SOLUTION
The velocity is very low, the viscosity is high and the thickness of the layer is thin.
These conditions favor laminar ‡ow.
10
4.11: PROBLEM DEFINITION
Situation:
A velocity …eld is given by V = 10xyt i . It is
a. 1D and steady
b. 1D and unsteady
c. 2D and steady
d. 2D and unsteady
SOLUTION
Answer is (b), 1D and unsteady. It is unsteady because it varies as a function of t.
It is 1D because there is only one unit operator: i.
11
4.12: PROBLEM DEFINITION
Situation:
Which is the most correct way to characterize turbulent ‡ow?
a. 1D
b. 2D
c. 3D
SOLUTION
There are 2 possible answers.
The most general answer is (c) 3D; because it is generally applicable to water ‡ow in
a river, or air moving in the atmosphere.
However, if one is modeling turbulent ‡ow in a pipe, it is safe to represent it as (a) 1D,
because there is very little variation of velocity across the cross-section of turbulent
‡ow in a pipe, see Fig 4.13b, EFM 11e.
Turbulent ‡ow in a pipe is referred to as plug ‡ow, because it has a blunt-ended
velocity pro…le instead of laminar (bullet-shaped) pro…le.
12
4.13: PROBLEM DEFINITION
Situation:
The valve in a system is gradually opened to have a constant rate of increase in
discharge.
Find:
Describe the ‡ow at points A and B.
SOLUTION
A: Unsteady, uniform.
B: Non-uniform, unsteady.
13
4.14: PROBLEM DEFINITION
Situation:
Water ‡ows in a widening passage with ‡ow rate decreasing with time.
Find:
Describe the ‡ow.
PLAN
Steady or unsteady has to do with a time rate of change. Uniform or non-uniform
has to do with whether the streamlines are parallel or not.
SOLUTION
(b) Unsteady and (d) non-uniform.
14
4.15: PROBLEM DEFINITION
Situation:
A ‡ow pattern has converging streamlines.
Find:
Classify the ‡ow.
SOLUTION
De…nately non-uniform; could be either steady or unsteady.
15
4.16: PROBLEM DEFINITION
Find:
Match the given ‡ow labels with the mathematical descriptions.
SOLUTION
Steady ‡ow corresponds to @Vs =@t = 0:
Unsteady ‡ow corresponds to @Vs =@t 6= 0:
Uniform ‡ow corresponds to @Vs =@s = 0:
Non-uniform ‡ow corresponds to @Vs =@s 6= 0:
16
4.17: PROBLEM DEFINITION
Situation:
A series of ‡ows are either one, two or three dimensional.
Find:
Classify the ‡ows as one, two or three dimensional.
(a) Water ‡ow over the crest of a long spillway of a dam.
(b) Flow in a straight horizontal pipe.
(c) Flow in a constant-diameter pipeline that follows the contour of the ground in
hilly country.
(d) Air‡ow from a slit in a plate at the end of a large rectangular duct.
(e) Air‡ow past an automobile.
(f) Air ‡ow past a house.
(g) Water ‡ow past a pipe that is laid normal to the ‡ow across the bottom of a
wide rectangular channel.
SOLUTION
Note to Instructor: Student answers will vary. You may want to incorporate this
into a class discussion.
a. Two dimensional
b. One dimensional
c. One dimensional
d. Two dimensional
e.
f.
g.
17
Three dimensional
Three dimensional
Two dimensional
4.18: PROBLEM DEFINITION
Situation:
Acceleration.
Find:
Is the acceleration vector always aligned with the velocity vector?
SOLUTION
No. For ‡ow along a curved path, there is a centripetal acceleration which is normal
to the velocity vector.
18
4.19: PROBLEM DEFINITION
Situation:
Rotating bodies.
Find:
Is the acceleration toward the center of rotation a centripetal or centrifugal acceleration?
SOLUTION
The acceleration toward the center of rotation is centripetal acceleration. "Petal"
comes from Latin word"petere" which means to move toward so "centripetal" means
moving toward center. "Fugal" comes from Latin "fugere" which means to ‡ee so
"centrifugal" means moving from center.
19
4.20: PROBLEM DEFINITION
Situation:
In a ‡owing ‡uid, acceleration means that a ‡uid particle is
a. changing direction
b. changing speed
c. changing both speed and direction
d. any of the above
SOLUTION
The correct answer is d.
20
4.21: PROBLEM DEFINITION
Situation:
Flow through a nozzle is steady.
V increases between the entrance and the exit of the nozzle.
The acceleration halfway between the entrance and the nozzle is:
a. convective
b. local
c. both
SOLUTION
If the ‡ow is steady, there is no local acceleration.
Therefore the answer is (a) convective acceleration.
21
4.22: PROBLEM DEFINITION
Situation:
Local acceleration
a. is close to the origin
b. occurs in unsteady ‡ow
c. is always nonuniform
SOLUTION
The answer is b. Review the de…nition of local acceleration.
22
4.23: PROBLEM DEFINITION
Situation:
A path line is given with velocity as a function of distance and time.
V = s2 t1=2 , r = 0:5 m:
s = 3 m, t = 0:5 s.
Find:
Acceleration along and normal to pathline (m/s2 ):
PLAN
Apply Eq. 4.11 of EFM11e for acceleration along pathline.
SOLUTION
Equation 4.5
a = (V
@V
@V
+
)ut +
@s
@t
V2
r
un
Evaluation of velocity and derivatives at s = 2 m and t = 0:5 sec.
V
@V
@s
@V
@t
= s2 t1=2 = 32
0:51=2 = 6:364 m/s
= 2st1=2 = 2
3
=
1 2 1=2 1
st
=
2
2
0:51=2 = 4:243 1/s
32
0:5 1=2 = 6:364 m/s2
Evaluation of the acceleration
a = (6:36
4:24 + 6:36)ut +
6:362
0:5
a = 33:4ut + 81:0un (m/s2 )
23
un
4.24: PROBLEM DEFINITION
Situation:
Air is ‡owing around a sphere in a wind tunnel.
u = Uo (1 ro3 =x3 ).
Find:
An expression for the acceleration of a ‡uid particle on the x-axis. The form of the
answer should be ax = ax (x; ro ; Uo ).
PLAN
Use Eq. 4.7 (EFM11e) along x-axis which is a pathline. Replace V with u and s with
x:
SOLUTION
ax = u
@u @u
+
@x
@t
=
U0 1
=
U02 1
r03
x3
r03
x3
ax = (3 U02
@
@x
r03
)(1
x4
U0 1
r3
3 04
x
+0
r03
)
x3
24
r03
x3
@
+
@t
U0 1
r03
x3
4.25: PROBLEM DEFINITION
Situation:
Flow occurs in a tapered passage.
V = 4 m= s 2:25 tto m= s, to = 0:6 s
@V =@s = +2:1 s 1 at t = 0:5 s.
Find:
(a) local acceleration at section AA ( m= s2 ):
(b) Convective acceleration at section AA ( m= s2 ):
SOLUTION
a) Local acceleration
@V
2:25
=
@t
t0
2:25
=
0:6
al = 3:75 m/s2
al =
b) Convective acceleration
ac = V
@V
@s
0:5
) m/s
0:6
ac = 4:46 m/s2
= (4
2:25
25
2:2 s 1
4.26: PROBLEM DEFINITION
Situation:
One-dimensional ‡ow occurs in a nozzle.
Vtip = 5 ft= s, Vbase = 2 ft= s, L = 23 in = 1:917 ft.
Find:
Convective acceleration ( ft= s2 ).
SOLUTION
Velocity gradient.
dV
ds
=
Vtip
Vbase
L
2) ft= s
=
1:917 ft
= 1:565 s 1
(5
Acceleration at mid-point
V
ac
(2 + 5) ft= s
2
= 3:5 ft/s
dV
= V
ds
= 3:5 ft= s 1:565
=
ac = 5:48 ft= s2
26
4.27: PROBLEM DEFINITION
Situation:
One-dimensional ‡ow occurs in a nozzle and the velocity varies linearly with distance along the nozzle.
Vtip = 4t ft= s, Vbase = 1t ft= s, t = 2 s.
Find:
Local acceleration midway in the nozzle ( ft= s2 ).
SOLUTION
@V
@t
1t + 4t
=
2
= 2:5t (ft/s)
a` =
V
Then
a` =
@
(2:5t)
@t
a` = 2:5 ft/s2
27
4.28: PROBLEM DEFINITION
Situation:
Water ‡ow in a nozzle with
V =
(1
2t
0:5x=L)2
L = 4 ft, x = 0:5L; t = 3 s.
Find:
Local acceleration ( ft= s2 ):
Convective acceleration ( ft= s2 ).
SOLUTION
a` =
=
=
=
@V =@t
@=@t[2t=(1 0:5x=L)2 ]
2=(1 0:5x=L)2
2=(1 0:5 0:5L=L)2
a` = 3:56 ft/s2
ac = V (@V =@x)
= [2t=(1 0:5x=L)2 ]@=@x[2t=(1
4t2
0:5
=
( 2)
5
L(1 0:5x=L)
L
2
4 3
1
=
5
4 (1 0:5 0:5L=L) L
ac = 9:48 ft/s2
28
0:5x=L)2 ]
4.29: PROBLEM DEFINITION
Situation:
State Newton’s second law of motion.
Find:
Are there any limitations on the use of Newton’s second law?
SOLUTION
Newtons second law states
F~ = m~a
where m is the mass of the system. The velocity (and acceleration) must be measured
with respect to an inertial reference frame and the mass must be constant.
29
4.30: PROBLEM DEFINITION
Situation:
Force weight and force pressure.
Find:
What is the di¤erence between a force due to weight and a force due to pressure?
SOLUTION
The force due to weight is the gravitational attraction on the mass and the magnitude
of the force depends on the mass. The force due to pressure is the force acting on a
surface and depends on the magnitude of the pressure and the area of the surface.
30
4.31: PROBLEM DEFINITION
Situation:
Flow through an inclined pipe at 30o from horizontal.
a` = 0:3g.
Find:
Pressure gradient in ‡ow direction.
PLAN
Apply Euler’s equation.
SOLUTION
l
30
o
Euler’s equation
@
(p + z) =
@`
@p
@z
+
=
@`
@`
@p
=
@`
=
=
a`
a`
a`
g
(0:30
@p
=
@`
31
@z
@`
( 0:30g)
0:50)
0:20
sin 30
4.32: PROBLEM DEFINITION
Situation:
Kerosene is accelerated upward in vertical pipe.
SG = 0:81, az = 0:4g:
Find:
Pressure gradient required to accelerate ‡ow ( lbf= ft3 ).
Properties:
= 62:4 lbf= ft3 .
PLAN Apply Euler’s equation.
SOLUTION Applying Euler’s equation in the z direction.
@(p + z)
=
az =
0:40g
@z
g
@p
+
=
0:40
@z
@p
= SG
( 1 0:40)
@z
= 0:81 62:4 lbf= ft3 ( 1:40)
@p
=
@z
32
70:8 lbf/ft3
4.33: PROBLEM DEFINITION
Situation:
A hypothetical liquid ‡ows through a vertical tube.
pB pA = 8 kPa
inviscid ‡uid
Find:
Direction of acceleration.
Properties:
= 10 kN= m3 , pB
pA = 8 kPa.
PLAN Apply Euler’s equation.
SOLUTION Euler’s equation
@
(p + z)
@`
1
@z
@p
=
@`
@`
a` =
a`
Let ` be positive upward. Then @z=@` = +1 and @p=@` = (pA
Pa/m. Thus
a` =
g
a` = g
(8; 000
8; 000
pB )=1 =
8; 000
)
1
a` = g(0:8 1:0) m/s2
a` =
0:2g m/s2
a` has a negative value; therefore, acceleration is downward. Correct answer is (b).
33
4.34: PROBLEM DEFINITION
Situation:
Water stands with depth of 6 ft in a vertical pipe open at top and supported by
piston at the bottom.
z = 0 ft, z2 = 6 ft.
Find:
Acceleration of piston ( ft= s2 ).
Properties:
= 62:4 lbf= ft3 , = 1:94 slug= ft3 :
p1 = 8 psig; p2 = 0 psig:
PLAN
Apply Euler’s equation.
SOLUTION
Euler’s equation
@
(p + z) =
as
@s
Take s as vertically upward with point 1 at piston surface and point 2 at water surface.
(0
8 psig
(p + z) = as s
(p2 p1 )
(z2 z1 ) = as s
144 in2 = ft2 ) 62:4 lbf= ft3 6 ft = 1:94 slug= ft3 6as
(8 psig 144 in2 = ft2 62:4 lbf= ft3
as =
1:94 slug= ft3 6 ft
as = 66:8 ft/s2
34
6 ft)
4.35: PROBLEM DEFINITION
Situation:
Water accelerates in a horizontal pipe.
as = 7:7 m= s2 , = 1000 kg= m3 .
Find:
Pressure gradient ( N= m3 ).
PLAN
Apply Euler’s equation.
SOLUTION
Euler’s equation with no change in elevation
@p
=
@s
=
as
1; 000 kg= m3
@p
=
@s
7:7 m= s2
7; 700 N/m3
35
4.36: PROBLEM DEFINITION
Situation:
Water accelerated from rest in horizontal pipe.
L = 80 m, D = 30 cm, as = 5 m= s2 :
Find:
Pressure at upstream end ( kPa).
Properties:
= 1000 kg= m3 , pdownstream = 90 kPa.
PLAN
Apply Euler’s equation.
SOLUTION
Euler’s equation with no change in elevation
@p
=
@s
=
=
pdownstream
pupstream
pupstream
as
1; 000 kg= m3 5 m= s2
5; 000 N/m3
@p
=
s
@s
= 90; 000 Pa + 5; 000 N= m3 (80 m)
= 490; 000 Pa gage
pupstream = 490 kPa gage
36
4.37: PROBLEM DEFINITION
Situation:
A liquid ‡ows through a conduit.
Find:
Which statements can be discerned with certainty:
(a) The velocity is in the positive ` direction.
(b) The velocity is in the negative ` direction.
(c) The acceleration is in the positive ` direction.
(d) The acceleration is in the negative ` direction.
Assumptions:
Viscosity is zero.
Properties:
pA = 170 psf, pB = 100 psf;
= 100 lbf= ft3 :
PLAN
Apply Euler’s equation.
SOLUTION
Euler’s equation
where @p=@` = (pB
Then
@
(p + z) = a`
@`
@p
@z
= a`
@`
@`
pA )=` = (100 170)=2 = 35 lb/ft3 and @z=@` = sin 30 = 0:5:
1
a` = (35 lb= ft3
(100)(0:5))
1
= ( 15 ) lbf/ft3
Because a` has a negative value we conclude that Answer ) (d) the acceleration is in the negative ` direction .
Answer ) The ‡ow direction cannot be established; so answer (d) is the only
answer that can be discerned with certainty.
37
4.38: PROBLEM DEFINITION
Situation:
Velocity varies linearly with distance in water nozzle.
L = 1 ft, V1 = 30 ft= s; V2 = 80 ft= s.
Find: Pressure gradient midway in the nozzle (psf/ft).
Properties:
= 62:4 lbm= ft3 = 1:94 slug= ft3 :
PLAN
Apply Euler’s equation.
SOLUTION
Euler’s equation
@
(p + z) =
@x
ax
but z =const.; therefore
@p
=
ax
@x
@V
ax = aconvective = V
@x
@V
= (80 30)=1 = 50 s 1
@x
Vmid = (80 ft= s + 30 ft= s)=2 = 55 ft/s
ax = (55 ft/s)(50 ft/s/ft) = 2; 750 ft/s2
Finally
@p
= ( 1:94 slug/ft3 )(2; 750 ft/s2 )
@x
@p
= 5; 330 psf/ft
@x
38
4.39: PROBLEM DEFINITION
Situation:
Closed tank is full of liquid.
L = 3 ft, H = 4 ft, ax = 0:9g.
a` = 1:5g, SG = 1:2:
Find:
(a) pC
(b) pB
pA (psf):
pA (psf):
Properties:
= 1:94 slug= ft3 .
PLAN
Apply Euler’s equation.
SOLUTION
Euler’s equation. Take ` in the z-direction.
dp
d`
pB
d`
= al
d`
dp
=
(g + a` )
d`
=
1:2 1:94 slug= ft3 (32:2 ft= s2
= 37:5 psf/ft
pA =
37:5 psf/ft 4 ft
pB
pA =
1:5 32:2 ft= s2 )
150 psf
Take ` in the x-direction. Euler’s equation becomes
pC
pC
pC
dp
=
dx
pB =
=
=
pA =
pA =
ax
ax L
1:2 1:94 slug= ft3 0:9g
202:4 psf
pC pB + (pB pA )
202:4 150
pC pA = 52:4 psf
39
3 ft
4.40: PROBLEM DEFINITION
Situation:
Closed tank is full of liquid.
L = 2:5 m, H = 3 m, a` = 2=3g, ax = 1:0g, SG = 1:3:
Find:
(a) pC
(b) pB
pA ( kPa).
pA ( kPa).
Properties:
= 1000 kg= m3 .
PLAN
Apply Euler’s equation.
SOLUTION
Euler’s equation in z direction
dp
+
=
az
dz
dp
=
(g + az )
dz
dp
=
1:3 1; 000 kg= m3 (9:81 m= s2
dz
=
4; 251 N/m3
pB pA = 4; 251 N= m3 (3 m)
= 12; 753 Pa
pB pA = 12:7 kPa
6:54 m= s2 )
Euler’s equation in x-direction
pC
pC
pC
dp
=
dx
pB =
=
=
pA =
pA =
=
ax
ax L
1:3 1; 000 9:81 2:5
31; 882 Pa
pC pB + (pB pA )
31; 882 + 12; 753
44; 635 Pa
pC pA = 44:6 kPa
40
4.41: PROBLEM DEFINITION
Situation:
Aspirators.
Find:
How does an aspirator work?
SOLUTION
Air is forced through a constriction in a duct There is a port at the smallest area
connected to a reservoir of ‡uid to be aspirated. The Bernoulli equation predicts
a minimum pressure at the contraction which pulls ‡uid into the air ‡ow from the
reservoir and breaks it up into droplets that emerge from the aspirator.
41
4.42: PROBLEM DEFINITION
Situation:
When the Bernoulli Equation applies to a venturi, such as in Fig. 4.27 (EFM11e)
in §4.6, which of the following are true? (Select all that apply.)
a. if the velocity head and elevation head increase, then the pressure head must
decrease
b. pressure always decreases in the direction of ‡ow along a streamline
c. the total head of the ‡owing ‡uid is constant along a streamline
SOLUTION
The correct answers are a and c.
REVIEW
If you selected b, you were probably thinking of a pipe of constant diameter.
Because of the diameter changes in a venturi, at the location where the diameter is
small, velocity must increase (due to continuity), and thus pressure must drop at the
narrow location according to the Bernoulli Equation.
42
4.43: PROBLEM DEFINITION
Situation:
A water jet …res vertically from a nozzle.
V = 26 m= s.
Find:
Height jet will rise.
PLAN
Apply the Bernoulli equation from the nozzle to the top of the jet. Let point 1 be
in the jet at the nozzle and point 2 at the top.
SOLUTION
Bernoulli equation
p1
+
V12
p2 V22
+ z1 =
+
+ z2
2g
2g
where p1 = p2 = 0 gage
V1 = 26 m/s
V2 = 0
0+
(26 m= s)2
+ z1 = 0 + 0 + z2
2g
676 m2 = s2
z2 z1 = h =
2 9:81 m= s2
h = 34:5 m
43
4.44: PROBLEM DEFINITION
Situation:
Water discharges from a pressurized tank.
z1 = 0:8 m, z2 = 0 m; V1 = 0 m= s.
p1 = 15 kPa gage = 15; 000 Pa gage
Find:
Velocity of water at outlet (m/s).
Properties:
Water (20 C, 10 kPa), Table A.5:
= 998 kg= m3 ,
= 9790 N= m3 .
SOLUTION
Apply the Bernoulli equation between the water surface in the tank (1) and the outlet
(2)
V2
V2
p1 + z 1 + 1 = p2 + z 2 + 2
2
2
Neglect V1 (V1
V2 ):Also p2 = 0 gage. The Bernoulli equation reduces to
V22
= p1 + (z1 z2 )
2
s
2 (p1 + (z1 z2 ))
V2 =
Elevation di¤erence z1
N/m3 :Therefore
z2 = 0:8 m. For water at 20o C,
V2 =
s
= 998kg/m3 and
2(15; 000 Pa + 9790 N= m3 (0:8 m))
998 kg= m3
V2 = 6:76 m= s
44
= 9790
4.45: PROBLEM DEFINITION
Situation:
Water ‡ows through a vertical venturi con…guration.
V1 = 5 ft= s, z = 0:5 ft.
Find:
Velocity at minimum area (ft/s).
Properties:
T = 68 F.
SOLUTION
Apply the Bernoulli equation between the pipe (1) and the minimum area (2)
p1 + z 1 +
V2
V12
= p2 + z 2 + 2
2
2
From problem statement, V1 = 5 ft/s. Rewriting equation
V22
V2
= 1 + (p1 + z1 )
2
2
(p2 + z2 )
The di¤erence in the elevation in piezometers gives the change in piezometric pressure,
(p1 + z1 ) (p2 + z2 ) =
h so
s
q
2 h
2
= V12 + 2g h
V1 +
V2 =
q
=
52 (ft/s)2 + 2 32:2 ft/s2 (0:5 ft)
V2 = 7:56 ft= s
45
4.46: PROBLEM DEFINITION
Situation:
Kerosene ‡ows through a contraction section and a pressure is measured between
pipe and contraction section.
V2 = 8:7 m= s:
Find:
Velocity in upstream pipe (m/s).
Properties:
Table A.4:
T = 20 C,
= 814 kg= m3 .
p = 25 kPa.
SOLUTION
Apply the Bernoulli equation between pipe (1) and contraction section (2)
V12
V2
= p2 + z 2 + 2
2
2
V12
V22
pz1 +
= pz2 +
2
2
p1 + z 1 +
The pressure gage measures the di¤erence in piezometric pressure, pz1
Rewrite the Bernoulli equation for V1
V12
V22
=
2
s2
V1 =
V22
(pz1
2(pz1
pz2 )
pz2 )
The density of kerosene at 20o C is 814 kg/m3 :Solving for V1
s
2(25; 000 kPa)
V1 =
(8:7 m= s)2
(814 kg= m3 )
V1 = 3:78 m= s
46
pz2 = 20 kPa.
4.47: PROBLEM DEFINITION
Situation:
A stagnation tube placed in a river (select all that apply)
a. can be used to determine air pressure
b. can be used to determine water velocity
c. measures kinetic pressure + static pressure
SOLUTION
Correct answers are b and c.
47
4.48: PROBLEM DEFINITION
Situation:
A Pitot tube on an airplane is used to measure airspeed
z2 = 10000 ft, hH2 O = 8 in.
T = 23 F; p = 9 psia.
Find:
Airspeed ( ft= s).
Properties:
Water (23 F), Table A.5: = 62:4 lbf= ft3 .
Air. Table A.2: R = 1716 J= kg K.
PLAN
Since the airspeed can be found by applying the Pitot-static tube equation, the steps
to reach the goal are:
1. Find pz by using the hydrostatic equation.
2. Find density by applying the ideal gas law.
3. Substitute values into the Pitot-static tube equation.
SOLUTION
1. Hydrostatic equation.
pz =
H 2 O hH 2 O
= 62:4 lbf= ft3
8
ft
12
= 41:6 psf
2. Ideal gas law
p
RT
=
(9psi)(144psi=psf)
((1; 716 lbf ft= slug R)(483 R))
= 0:00156 slugs/ft3
=
3. Pitot-Static Tube equation.
V
V
=
=
s
s
2 pz
2 41:6 lbf/ft2
(0:00156 slugs/ft3 )
V = 231 ft/s
48
4.49: PROBLEM DEFINITION
Situation:
A Pitot tube is inserted into a stream of water.
l = 15 cm = 0:15 m.
Find:
Velocity at point A.
PLAN
1. Apply hydrostatic equation to relate stagnation point to water column in tube
2. Apply the Bernoulli equation between the free stream (location A) and the
stagnation point
SOLUTION
Hydrostatic equation (between stagnation point and water surface in tube)
ps
=l+d
where d is depth below surface and l is distance above water surface.
Bernoulli equation (between A and the stagnation point)
V2
2g
V2
l+d = d+
2g
2
V
= l
2g
ps
V
V
= d+
p
2gl
p
=
2 9:81 m= s2
=
V = 1:72 m=s
49
0:15 m
4.50: PROBLEM DEFINITION
Situation:
A glass tube with 90o bend inserted into a stream of water.
V = 6:6 m= s.
Find:
Rise in vertical leg above water surface (m).
PLAN
Apply the Bernoulli equation.
SOLUTION
Hydrostatic equation (between stagnation point and water surface in tube)
ps
=h+d
where d is depth below surface and h is distance above water surface.
Bernoulli equation (between free stream and stagnation point)
V2
2g
V2
h+d = d+
2g
2
V
h =
2g
ps
= d+
(6:6 m= s)2
h =
2 (9:81 m= s2 )
h = 2:22 m
50
4.51: PROBLEM DEFINITION
Situation:
An air-water manometer is connected to a Pitot-static tube to measure air velocity.
T = 160 F, h = 4 in.
Find:
Velocity ( ft= s).
Properties:
Air, p = 14 psia
Table A.2: Air R = 1716 J= kg K.
Water (140 F, 14 psia), Table A.5:
= 61:0 lbf= ft3 .
PLAN Apply the Pitot tube equation calculate velocity. Apply the ideal gas law
to solve for density.
SOLUTION Ideal gas law
p
RT
=
14 psia 144 in2 = ft2
(1; 716 J= kg K)(160 + 460) K
= 0:00190 slugs/ft
=
Pitot tube equation
2 pz
V =
1=2
From the manometer equation
pz =
w
h 1
a
w
but
a= w
1 so
V
2 w h
=
=
"
1=2
2 61:0 lbf= ft3 (4:0=12) ft
0:00190 slug= ft
V = 146 ft= s
51
#1=2
4.52: PROBLEM DEFINITION
Situation:
Two Pitot tubes are connected to air-water manometers to measure air and water
velocities.
Find:
The relationship between VA and VW .
p
V = 2g h =
s
SOLUTION
The
pz is the same for both; however,
w >>
a
Therefore VA > VW . The correct choice is b).
52
2 pz
4.53: PROBLEM DEFINITION
Situation:
A Pitot tube measures the velocity of kerosene at center of a pipe.
D = 12 in, h = 5:5 in,
Find:
Velocity ( ft= s).
Properties:
From Table A.4: ker = 1:58 slugs/ft3 :
T = 68 F, ker = 51 lbf/ft3 , HG = 847 lbf/ft3 .
PLAN
Apply the Pitot tube equation and the hydrostatic equation.
SOLUTION
Hydrostatic equation
pz =
h( HG
5:5
ft(847
=
12
= 364:8 psf
ker )
51) lbf= ft3
Pitot tube equation
V
=
=
2 pz
1=2
2 (364:8psf)
1:58 slug= ft3
V = 21:5 ft= s
53
1=2
4.54: PROBLEM DEFINITION
Situation:
A Pitot tube for measuring velocity of air.
Find:
Air velocity ( m= s).
Properties:
Air (10o C), Table A.3:
pz = 3 kPa.
= 1:25 kg/m3 .
PLAN
Apply the Pitot tube equation.
SOLUTION
Pitot tube equation
V
=
=
2 pz
1=2
2 (3; 000 Pa)
1:25 kg= m3
V = 69:3 m= s
54
1=2
4.55: PROBLEM DEFINITION
Situation:
A Pitot tube is used to measure the velocity of air.
pz = 15 psf, T = 200 F.
Find:
Air velocity ( ft= s).
Properties:
Air (200o F), Table A.3:
= 0:00187 slug= ft3 .
PLAN
Apply the Pitot tube equation.
SOLUTION
Pitot tube equation
V
V
=
s
=
2 (15psf)
0:00187 slug= ft3
2 pz
V = 127 ft= s
55
1=2
4.56: PROBLEM DEFINITION
Situation:
A Pitot tube measures gas velocity in a duct.
Find:
Gas velocity in duct ( ft= s).
Properties:
pz = 3 psi,
= 0:019 lbm= ft3 .
PLAN
Apply the Pitot tube equation.
SOLUTION
Pitot tube equation The density is
V
0:019 lbm
ft3
1 slug
32:2 lbm
=
s
=
2 (3psi) (144psf=psi)
0:000590 slug= ft3
= 0:000590 slug/ft3
2 pz
V = 1210 ft/s
56
1=2
4.57: PROBLEM DEFINITION
Situation:
A ‡ow-metering device is described in the problem.
V2 = 1:5V1 , h = 10 cm:
Find:
Velocity at station 2 ( m= s).
Properties:
= 1:2 kg= m3 .
PLAN
Apply the Bernoulli equation and the manometer equation.
SOLUTION
Bernoulli equation
p1
+
V12
p2 V22
pt
=
+
=
2g
2g
Manometer equation
p1 + 0:1
9810
z
0:1
neglect
}|
1:2
pt
{
9:81 = pt
p1 = 981 N/m2 =
V12
2
2 (981 N= m2 )
1:2 kg= m3
= 40:4 m= s
= 1:5V1
V12 =
V1
V2
V2 = 60:7 m= s
57
4.58: PROBLEM DEFINITION
Situation:
A spherical Pitot tube is used to measure the ‡ow velocity in water.
V2 = 1:5V0
Find:
Free stream velocity ( m= s).
Properties:
= 965 kg= m3 ,
p = 3 kPa.
PLAN
Apply the Bernoulli equation between the two points. Let point 1 be the stagnation
point and point 2 at 90 around the sphere.
SOLUTION
Bernoulli equation
V2
V12
= pz2 + 2
2
2
(1:5V0 )2
pz1 + 0 = pz2 +
2
pz1 pz2 = 1:125 V02
3; 000 Pa
V02 =
= 2:755 m2 =s2
1:125 (965 kg= m3 )
pz1 +
V0 = 1:66 m= s
58
4.59: PROBLEM DEFINITION
Situation:
A device for measuring the water velocity in a pipe consists of a cylinder with
pressure taps at forward stagnation point and at the back on the cylinder.
= 1000 kg= m3 , p = 500 Pa, Pressure Coe¢ cient is -0.3.
Find:
Water velocity ( m= s).
PLAN
Apply the Bernoulli equation between the location of the two pressure taps. Let point
1 be the forward stagnation point and point 2 in the wake of the cylinder.
SOLUTION
The piezometric pressure at the forward pressure tap (stagnation point, Cp = 1) is
pz1 = pz0 +
At the rearward pressure tap
pz2
pz0
V02
V2
2
=
0:3
0:3
V02
2
2
or
pz2 = pz0
The pressure di¤erence is
V02
2
The pressure gage records the di¤erence in piezometric pressure so
V0 =
pz1
pz2 = 1:3
2
1:3
pz
1=2
2
=
(500 Pa)
1:3 (1000 kg= m3 )
= 0:88 m= s
V0 = 0:88 m= s
59
1=2
4.60: PROBLEM DEFINITION
Situation:
A body moving horizontally through still water.
VA = 14 m= s, VB = 8 m= s; VC = 1 m= s.
Find:
pB
pC ( kPa).
Properties:
= 1000 kg/m3
SOLUTION
Apply the Bernoulli equation.
pB
pC = (VC2
2
VB2 )
(1)
Reference all velocities to an observer situated on the sphere. From this reference
frame, the ‡ow is steady and the Bernoulli equation is applicable.
VC = 14 m= s
VB = 14 m= s
1 m= s = 13 m= s
8 m= s = 6 m= s
Combine Eqs. (1) to (3)
pB
pC =
pB
pC
(VC2
VB2 )
2
1; 000 kg= m3
)[(13 m= s)2
= (
2
= 66; 500 Pa
pB pC = 66:5 kPa
60
(6 m= s)2 ]
(2)
(3)
4.61: PROBLEM DEFINITION
Situation:
Water is in a ‡ume with a pressure gage along the bottom.
Da = Db , Va = 0 m= s, Vb = 3 m= s.
Find:
If gage A will read greater or less than gage B.
SOLUTION
Both gage A and B will read the same, due to hydrostatic pressure distribution in
the vertical in both cases. There is no acceleration in the vertical direction.
61
4.62: PROBLEM DEFINITION
Situation:
An instrument used to …nd gas velocity in smoke stacks.
CpA = 1, CpB = 0:2, h = 5 mm.
Find:
Velocity of stack gases ( m= s).
Properties:
T = 20 C, R = 200 J= kg K.
Tgas = 250 C, pgas = 101 kPa abs.
SOLUTION
Ideal gas law
=
p
RT
101; 000 Pa
(200 J= kg K) (250 + 273) K
= 0:966 kg/m3
=
Manometer equation
pz = ( w
but
w
a)
h
a so
pz = w h
= 9790 N= m3 (0:005 m)
= 48:95 Pa
(pA
(pA
pB )z = (CpA
pB )z
V02
CpB )
V02
2
V02
= 1:2
2
2 (48:95 Pa)
=
1:2 (0:966 kg= m3 )
V0 = 9:19 m/s
62
4.63: PROBLEM DEFINITION
Situation:
An airplane uses a Pitot-static tube to measure airspeed.
z2 = 3000 m, Vind = 56 m= s.
Find:
True air-speed (m/s).
Properties:
TSL = 17 C, T = 6:3 C.
pSL = 101 kPa; p = 70 kPa:
PLAN
Apply the Pitot-tube equation and correct for density change.
SOLUTION
The Pitot-static tube equation is
2 p
V =
1=2
Multiplying and dividing by the sea level density
V =
2 p
1=2
1=2
SL
SL
The factor
2 p
SL
1=2
is the indicated airspeed so
1=2
Vtrue = Vind
SL
From the ideal gas law
SL
=
pSL T
101 kPa (273 6:3) K
=
= 1:327
TSL p
70 kPa (273 + 17) K
True air speed
Vtrue = 56 m/s
p
1:327
Vtrue = 64:5 m= s
63
4.64: PROBLEM DEFINITION
Situation:
Check equations for pitot tube velocity measurement provided by instrument company.
p
V = 1096:7 hv =d, d = 1:325Pa =T .
Find:
Validity of Pitot tube equations provided.
PLAN
Apply the Bernoulli equation
SOLUTION
Applying the Bernoulli equation to the Pitot tube, the velocity is related to the change
in piezometric pressure by
V2
pz =
2
3
where pz is in psf, is in slugs/ft and V is in ft/s. The piezometric pressure
di¤erence is related to the "velocity pressure" by
3
w (lbf/ft )hv (in)
2
pz (lbf/ft ) =
12(in/ft)
62:4 hv
=
12
= 5:2hv
The density in slugs/ft3 is given by
d (lbm/ft3 )
(slug/ft ) =
gc (lbm/slug)
d
=
32:2
= 0:03106d
3
The velocity in ft/min is obtained by multiplying the velocity in ft/s by 60. Thus
r
2 5:2hv
V = 60
0:03106d
r
hv
= 1098
d
This di¤ers by less than 0.1% from the manufacturer’s recommendations. This could
be due to the value used for gc but the di¤erence is probably not signi…cant compared
to accuracy of "velocity pressure" measurement.
64
From the ideal gas law, the density is given by
=
p
RT
where is in slugs/ft3 ; p in psfa and T in o R. The gas constant for air is 1716 ftlbf/slug-o R. The pressure in psfg is given by
p (psfg) =
13:6 62:4 (lbf/ft3 )
12(in/ft)
Pa (in-Hg)
= 70:72Pa
where 13.6 is the speci…c gravity of mercury. The density in lbm/ft3 is
d = gc
70:72Pa
1716 T
= 32:2
= 1:327
Pa
T
which is within 0.2% of the manufacturer’s recommendation.
65
4.65: PROBLEM DEFINITION
Situation:
The ‡ow of water over di¤erent surfaces.
Find:
Relationship of pressures.
(a) pC > pB > pA.
(b) pB > pC > pA.
(c) pC = pB = pA.
(d) pB < pC < pA.
(e) pA < pB < pC.
SOLUTION
The ‡ow curvature requires that pB > pD + d where d is the liquid depth. Also,
because of hydrostatics pC = pD + d: Therefore pB > pC . Also pA < pD + d so
pA < pC : So pB > pC > pA .
The valid statement is (b).
66
4.66: PROBLEM DEFINITION
Situation:
Fluid element rotation.
Find:
What is meant by rotation of a ‡uid element?
SOLUTION
An arbitrary cubical element is selected in a ‡ow. One side lies along the x-axis. As
the element moves through the ‡ow it will be deformed. If the angle between the
bisectors of the sides and the x-axis does not change, there is no rotation.
67
4.67: PROBLEM DEFINITION
Situation:
A spherical ‡uid element in an inviscid ‡uid.
Find:
If pressure and gravitational forces are the only forces acting on the element, can
they cause the element to rotate?
SOLUTION
The result force due to pressure passes through the center of the sphere so no moment
arm to create rotation. The resultant forces due to gravity also pass though the center
so cannot cause rotation.
68
4.68: PROBLEM DEFINITION
Situation:
A two-dimensional velocity …eld is represented by the vector V = 10xi
10yj.
Find:
Is the ‡ow irrotational?
SOLUTION
In a two dimensional ‡ow in the x
y plane, the ‡ow is irrotational if (Eq. 4.39a)
@v
@u
=
@x
@y
The velocity components and derivatives are
u = 10x
v =
10y
Therefore the ‡ow is irrotational
69
@u
=0
@y
@v
=0
@x
4.69: PROBLEM DEFINITION
Situation:
A ‡ow …eld has velocity components described by u =
!y and v = !x.
Find:
Vorticity:
Rate of rotation:
SOLUTION
Rate of rotation
! z = (1=2)(
@v
@x
@u
)
@y
1
(! ( !))
2
1
=
(2!)
2
!z = !
=
Vorticity is twice the average rate of rotation; therefore, the vorticity = 2!
70
4.70: PROBLEM DEFINITION
Situation:
A two-dimensional velocity …eld is given by:
, v = (x2Cy
.
u = (x2Cx
+y 2 )
+y 2 )
Find:
Check if ‡ow is irrotational.
SOLUTION
Apply equations for ‡ow rotation in x
@v
@x
y plane.
@u
2xCy
2yCx
=
2
@y
(x2 + y 2 )
(x2 + y 2 )2
= 0
The ‡ow is irrotational
71
4.71: PROBLEM DEFINITION
Situation:
A two-dimensional ‡ow …eld is de…ned by:
u = x2 y 2 ; v = 2xy:
Find:
If the ‡ow is rotational or irrotational.
SOLUTION
Rate of ‡ow rotation about the z-axis,
z
1 @u @v
2 @y @x
1
=
( 2y + 2y) = 0
2
=
Therefore, the ‡ow is irrotational.
72
4.72: PROBLEM DEFINITION
Situation:
Incompressible and inviscid liquid ‡ows around a bend.
V = 1r m= s, ri = 1 m, ro = 3 m.
Find:
Depth of liquid from inside to outside radius (m).
PLAN
Flow …eld is irotational so apply the Bernoulli equation across streamlines between
the outside of the bend at the surface (point 2) and the inside of the bend at the
surface (point 1).
SOLUTION
Bernoulli equation
V22
p1 V12
+ z2 =
+
+ z1
2g
2g
V2
V2
0 + 2 + z2 = 0 + 1 + z1
2g
2g
2
V1
V22
z2 z1 =
2g
2g
p2
+
where V2 = (1=3) m/s; V1 = (1=1) m/s. Then
z2
z1 =
1
((1 m= s)2 (0:33 m= s)2 )
2g
z2 z1 = 0:045 m
73
4.73: PROBLEM DEFINITION
Situation:
An outlet pipe from a reservoir.
V = 30 ft= s, h = 18 ft:
Find:
Pressure at point A (psig):
PLAN
Apply the Bernoulli equation.
SOLUTION
Bernoulli equation. Let point 1 be at surface in reservoir.
p1
V12
pA VA2
+ z1 =
+
+ zA
2g
2g
pA
(30 ft= s)2
+0
0 + 0 + 18 =
+
62:4 lbf= ft3 2 32:2 ft= s2
pA = (18 ft 13:98 ft) 62:4 lbf= ft3
pA = 251 psfg
pA = 1:74 psig
+
74
4.74: PROBLEM DEFINITION
Situation:
An outlet pipe from a reservoir.
V = 8 m= s, h = 19 m:
Find:
Pressure at point A (kPa):
Assumptions:
Flow is irrotational.
PLAN
Apply the Bernoulli equation.
SOLUTION
Bernoulli equation. Let point 1 be at reservoir surface.
p1
V12
pA VA2
+ z1 =
+
+ zA
2g
2g
pA
(8 m= s)2
0 + 0 + 19 =
+
+0
9810 N= m3 2 9:81 m= s2
pA = (19 m 3:26 m) 9810 N= m3
pA = 154; 390 Pa, gage
pA = 154 kPa, gage
+
75
4.75: PROBLEM DEFINITION
Situation:
Air ‡ows past a cylinder: Highest velocity at the maximum width of sphere is twice
the free stream velocity.
V0 = 40 m= s, Vmax = 2V0 :
Find:
Pressure di¤erence between highest and lowest pressure (kPa).
Assumptions:
Hydrostatic e¤ects are negligible and the wind has density of 1.2 kg/m3 .
PLAN
Apply the Bernoulli equation between points of highest and lowest pressure.
SOLUTION
The maximum pressure will occur at the stagnation point where V = 0 and the point
of lowest pressure will be where the velocity is highest (Vmax = 80 m/s).
Bernoulli equation
Vh2
V`2
ph +
= p` +
2
2
2
)
ph + 0 = p` + (Vmax
2
1:2 kg= m3
ph p` =
(80 m= s)2
2
= 3; 840 Pa
ph p` = 3:84 kPa
76
4.76: PROBLEM DEFINITION
Situation:
Velocity and pressure given at two points in a duct.
V1 = 1 m= s, V2 = 2 m= s:
Find:
Determine which is true:
(a) Flow in contration in nonuniform and irrotational.
(b) Flow in contration is uniform and irrotational.
(c) Flow in contration is nonuniform and rotational.
(d) Flow in contration is uniform and rotational.
Assumptions:
Elevations are equal.
Properties:
p1 = 10 kPa, p2 = 7 kPa:
= 1000 kg= m3 :
PLAN
Check to see if it is irrotational by seeing if it satis…es Bernoulli’s equation.
SOLUTION
The ‡ow is non-uniform.
Bernoulli equation
V12
p2 V22
+ z1 =
+
+ z2
2g
2g
10; 000 Pa
(1 m= s)2
7; 000 Pa
(2 m= s)2
+0 =
+0
+
+
9; 810 N= m3 2 (9:81 m= s2 )
9; 810 N= m3 2 (9:81 m= s2 )
1:070 6= 0:917
p1
+
Flow is rotational. The correct choice is c.
77
4.77: PROBLEM DEFINITION
Situation:
Water ‡owing from a large ori…ce in bottom of tank.
VA = 4 ft= s, VB = 12 ft= s:
zA = 1 ft, zB = 0 ft:
Find:
pA
pB (psf):
Properties:
= 62:4 lb= ft3 :
PLAN
Apply the Bernoulli equation.
SOLUTION
Bernoulli equation
pA
+ zA +
pA
VA2
pB
V2
=
+ zB + B
2g
2g
2
2
(VB VA )
zA
pB =
2g
(144 16) ft2 = s2
= 62:4 lb= ft3
2 (32:2 ft= s2 )
pA
pB = 61:6 psf
78
1 ft
4.78: PROBLEM DEFINITION
Situation:
A ‡ow pattern past an airfoil.
V0 = 80 m= s, V1 = 85 m= s, V2 = 75 m= s.
Find:
Pressure di¤erence between bottom and top (kPa).
Assumptions:
The pressure due to elevation di¤erence between points is negligible.
Properties:
= 1:2 kg= m3 :
SOLUTION
The ‡ow is ideal and irrotational so the Bernoulli equation applies between any two
points in the ‡ow …eld
p1
+
V12
p2 V22
+ z1 =
+
+ z2
2g
2g
(V 2 V22 )
2 1
1:2 kg= m3
=
(852 752 ) m= s
2
= 960 Pa
p2 p1 = 0:960 kPa
p2
p1 =
p2
p1
79
4.79: PROBLEM DEFINITION
Situation:
Flow of water between parallel plates.
Find:
Is the Bernoulli equation valid between plates?
SOLUTION
The ‡ow between the two plates is rotational. The Bernoulli equation cannot be
applied across streamlines in rotational ‡ows.
80
4.80: PROBLEM DEFINITION
Situation:
A ‡uid is ‡owing around a cylinder as shown in Fig 4.37 in §4.10 (EFM11e). A
favorable pressure gradient can be found:
a. upstream of the stagnation point
b. at the stagnation point
c. between the stagnation point and separation point
SOLUTION
A favorable pressure gradient is de…ned as where the ‡ow is accelerating.
Therefore the correct answer is (c); ‡ow is accelerating from v = 0 at the
stagnation point to a positive velocity.
Upstream of the stagnation point the ‡ow is decelerating.
At the stagnation point the ‡ow has stopped, so it is neither decelerating or accelerating.
81
4.81: PROBLEM DEFINITION
Situation:
The wake of a sphere which separates at 120o :
V0 = 100 m= s.
V = 1:5V0 , = 120 .
Find:
(a) Gage pressure ( kPa).
(b) Pressure coe¢ cient.
Properties:
= 1:2 kg= m3 .
PLAN
Apply the Bernoulli equation from the free stream to the point of separation and the
pressure coe¢ cient equation.
SOLUTION
Pressure coe¢ cient
Cp =
p p0
V 2 =2
Bernoulli equation
u2
U2
= p+
2
2
2
p p0 =
(U
u2 )
2
p0 +
or
p p0
= (1
U 2 =2
u
U
2
)
but
u = 1:5U sin
u = 1:5U sin 120
u = 1:5U 0:866
82
At the separation point
u
U
u
U
= 1:299
2
= 1:687
Cp = 1
1:687
Cp =
0:687
U2
2
= ( 0:687)(1:2 kg= m3 =2)(100 m= s)2
=
4; 122 Pa
pgage = 4:12 kPa gage
pgage = Cp
83
4.82: PROBLEM DEFINITION
Situation:
Irrotational ‡ow past a circular cylinder with constant approach velocity.
Find:
Describe the ‡ow as:
(a) Steady or unsteady.
(b) One dimensional, two dimensional, or three dimensional.
(c) Locally accelerating or not, and is so, where.
(d) Convectively accelerating or not, and if so, where.
SOLUTION
(a) Steady.
(b) Two-dimensional; velocity is changing in the x and y direction (but we assume
it is not changing in the axis that comes out of the page)
(c) No; the ‡ow is steady, so by de…nition there is no local acceleration.
(d) Convective acceleration is present at each where a ‡uid particles changes speed as
it moves along the streamline. Centripetal acceleration, which is a form of convective
acceleration. occurs where there is streamline curvature.
84
4.83: PROBLEM DEFINITION
Situation:
Application of the Bernoulli equation between a point upstream and in the wake
of a sphere.
Find:
Is the Bernoulli equation valid between these two points?
SOLUTION
The ‡ow in the wake is irrotational so the Bernoulii equation cannot be applied
between two arbitrary points
85
4.84: PROBLEM DEFINITION
Situation:
A closed tank …lled with water is rotated about a vertical axis.
D = 4 ft, ! = 10 rad= s:
Find:
Pressure at bottom center of tank (psig):
Properties:
= 62:4 lbm= ft3 = 1:94 slug= ft3 ,
= 62:4 lbf= ft3 :
PLAN
Apply the equation for pressure variation equation- rotating ‡ow.
SOLUTION
Pressure variation equation- rotating ‡ow
r2 ! 2
= pp + z p
2
where pp = 0; rp = 3 ft and r = 0; then
p+ z
p =
=
=
(rp !)2 + (zp
rp2 ! 2
2
z)
2
1:94 slug= ft3
(3 ft 10)2 + 62:4 lbf= ft3 (2:5 ft)
2
717 psfg = 4:98 psig
p=
4:98 psig
86
4.85: PROBLEM DEFINITION
Situation:
A tank of liquid is rotated on an arm.
SG = 0:80, D = 1 ft:
h = 1 ft, r = 2 ft.
VA = 20 ft= s, pA = 25 psf.
Find:
Pressure at B (psf).
Properties:
= 62:4 lbm= ft3 = 1:94 slug= ft3 ,
= 62:4 lbf= ft3 :
PLAN
Apply the pressure variation equation- rotating ‡ow from point A to point B:
SOLUTION
Pressure variation equation- rotating ‡ow
pA + z A
2 2
2 2
rB
!
rA
!
= pB + z B
2
2
2
2
2
pB = pA + (! )(rB rA
) + (zA
2
where ! = VA =rA = 20=1:5 = 13:333 rad/s and
= 0:8
1:94 slugs/ft3 : Then
pB = 25psf + 1:94 slug= ft3 (0:80=2) (13:33 rad= s)2 (2:5 ft)2
= 25 + 551:5 49:9
pB = 527 psf
87
zB )
(1:5 ft)2 + 62:4 lbf= ft3 (0:8) ( 1)
4.86: PROBLEM DEFINITION
Situation:
A cream separator is in operation.
D = 20 cm, f = 9000 rpm.
Find:
Centripetal acceleration (m/s2 ).
RCF.
SOLUTION
The centripetal acceleration is
ar =
V2
= !2r
r
The rotational rate of the separator is
!=2
9000rpm
60 s= min
= 942:5 rad/s
The radius of the separator is 10 cm or 0.1 m. The acceleration is
ar = (942:5 rad= s)2 (0:1 m)
ar = 88800 m/s2
The RCF is
RCF = 88831 m/s2 =9:81 m/s2
RCF = 9060
88
4.87: PROBLEM DEFINITION
Situation:
A closed tank with liquid is rotated about the vertical axis.
! = 10 rad/s, rB = 0:5 m, az = 4 m= s2 .
Find:
Di¤erence in pressure between points A and B (kPa).
Properties:
= 1000 kg= m3 ; SG = 1:2.
PLAN
Apply the pressure variation equation for rotating ‡ow between points B & C. Let
point C be at the center bottom of the tank.
SOLUTION
Pressure variation equation- rotating ‡ow
2 2
!
rC2 ! 2
rB
= pC
2
2
where rB = 0:5 m, rC = 0 and ! = 10 rad/s. Then
pB
pB
pC =
=
pC
pA
=
=
=
=
(! 2 )(r2 )
2
1200 kg= m3
(100 rad2 = s2 )(0:25 m2 )
2
15; 000 Pa
2 + az `
2 11; 772 N= m3 + 1; 200 kg= m3 4 m= s2 (2)
33:1 kPa
Then
pB
pA = pB pC + (pC pA )
= 15; 000 Pa + 33; 144 Pa
= 48; 144 Pa
pB pA = 48:1 kPa
89
4.88: PROBLEM DEFINITION
Situation:
A U-tube rotating about the leg on the right side.
r1 = 0:5 m, z1 = 0:5 m:
z2 = 0 m; r2 = 0 m.
Find:
Maximum rotational speed so that no liquid escapes from the leg on the left side
(rad/s).
PLAN
Since the ‡uid is in rigid body rotation, apply the pressure variation equation for
rotating ‡ow. At the condition of imminent spilling, the liquid will be to the top of
the left leg and at the bottom of the right leg. Thus, locate point 1 be at top of the
left (outside) leg. Locate point 2 at the bottom of the right (inside) leg.
SOLUTION
Pressure variation equation- rotating ‡ow
p1 + z 1
r12 ! 2
= p2 + z 2
2
r22 ! 2
2
Term-by-term analysis
p1
z1
r1
z2
r2
=
=
=
=
=
p2 = 0 kPa-gage
0:5 m
0:5 m
0m
0m
Substitute values into Eq. 1.
r12 ! 2
r22 ! 2
= p2 + z 2
2
2
2
2
2
(0:5 m ) !
0 + g (0:5 m)
=0+0 0
2
(0:52 m2 ) ! 2
g (0:5 m)
=0
2
! 2 = 4g
p
!=2 g
p1 + z 1
! = 6:26 rad/s
90
(1)
4.89: PROBLEM DEFINITION
Situation:
A stagnation tube in a tank is rotated:
! = 100 rad/ s, r = 20 cm, = 10000 N= m3 .
Find:
Location of liquid surface in central tube.
PLAN
Pressure variation equation for rotating ‡ow from pt. 1 to pt. 2 where pt. 1 is at
liquid surface in vertical part of tube and pt. 2 is just inside the open end of the
Pitot tube.
SOLUTION
1
0
2
10 cm
Elevation view
Plan view
Pressure variation equation- rotating ‡ow
V12
p2
+ z1 =
2g
p2
0 + (0:10 + `) =
p1
0
V22
+ z2
2g
r2 ! 2
0
2g
(1)
where z1 = z2 : If we reference the velocity of the liquid to the tip of the Pitot tube
then we have steady ‡ow and Bernoulli’s equation will apply from pt. 0 (point ahead
of the Pitot tube) to point 2 (point at tip of Pitot tube).
V02
p2 V22
+ z0 =
+
+ z2
2g
2g
0:1
r2 ! 2
p2
+
=
+0
2g
p0
+
(2)
Solve Eqs. (1) & (2) for `
` = 0 liquid surface in the tube is the same as the elevation as outside liquid surface.
91
4.90: PROBLEM DEFINITION
Situation:
A manometer is rotated about one leg.
z = 20 cm, r = 10 cm, S = 0:8.
Find:
Acceleration in g’s in leg with greatest amount of oil.
PLAN
Apply the pressure variation equation for rotating ‡ow between the liquid surfaces of
1 & 2. Let leg 1 be the leg on the axis of rotation. Let leg 2 be the other leg of the
manometer.
SOLUTION
Pressure variation equation- rotating ‡ow
p1 + z 1
r12 ! 2
= p2 + z 2
2
0 + z1
0 =
z2
r22 ! 2
g 2
r22 ! 2
= z2 z1
2g
an = r! 2
(z2 z1 )2g
=
r2
(0:20)(2g)
=
0:1
an = 4g
92
r22 ! 2
2
4.91: PROBLEM DEFINITION
Situation:
A fuel tank rotated in zero-gravity environment.
f = 3 rpm, r1 = 1:5 m, zA = 1 m:
Find:
Pressure at exit (Pa).
Properties:
= 800 kg= m3 , p1 = 0:1 kPa.
PLAN
Apply the pressure variation equation for rotating ‡ow from liquid surface to point
A. Call the liquid surface point 1.
SOLUTION
Pressure variation equation- rotating ‡ow
2 2
rA
!
r12 ! 2
= pA + z A
2
2
2
! 2
pA = p1 +
(r
r12 ) + (z1 zA )
2 A
p1 + z 1
However (z1
zA ) = 0 in zero-g environment. Thus
800 kg/m3
6
2
60 rad/s
= 100 Pa + 49:3 Pa
pA = 149 Pa
pA = p1 +
93
2
((1:5 m)2
(1 m)2 )
4.92: PROBLEM DEFINITION
Situation:
Water …lls a tube that is closed at one end.
D = 1 cm, r = 40 cm, ! = 50 rad/s.
Find:
Force exerted on closed end (N).
Properties:
= 1000 kg= m3
PLAN
Apply the pressure variation equation for rotating ‡ow from the open end of the tube
to the closed end.
SOLUTION
Pressure variation equation- rotating ‡ow
p1 + z 1
r12 ! 2
= p2 + z 2
2
r22 ! 2
2
where z1 = z2 . Also let point 2 be at the closed end; therefore r1 = 0 and r2 = 0:40
m.
(0:4 m)2 (50 rad/s)2
2
= 500 kg= m3 0:16 m2 2500 rad2 = s2
= 200 kPa
p2 =
Then
F = p2 A = 200; 000 Pa( =4)(:01 m)2
F = 15:7 N
94
4.93: PROBLEM DEFINITION
Situation:
Water sits in a U-tube that is closed at one end.
D = 1 cm, ` = 2 cm.
Find:
Rotational speed when water will begin to spill from open tube (rad/s).
Properties:
= 1000 kg= m3 ,
= 9810 N= m3 .
PLAN
Apply the pressure variation equation for rotating ‡ow between water surface in leg
A-A to water surface in open leg after rotation.
SOLUTION
When the water is on the verge of spilling from the open tube, the air volume in the
closed part of the tube will have doubled. Therefore, we can get the pressure in the
air volume with this condition.
pi V i = p f V f
and i and f refer to initial and …nal conditions
Vi
= 101 kPa
Vf
= 50:5 kPa, abs =
pf = pi
pf
1
2
50:5 kPa, gage
Pressure variation equation- rotating ‡ow
2
2 2
!2
ropen
rA
!
pA + z A
= popen + zopen
2
2
(6`)2 ! 2
pA + 0 0 = 0 +
6`
2
50:5
3
3
10 Pa = 9810 N= m (6) (0:02 m)
1000 kg= m
! = 84:72 rad/s
95
3
(6
2
0:02 m)
!2
2
4.94: PROBLEM DEFINITION
Situation:
Water is pumped from a reservoir by a centrifugal pump consisting of a disk with
radial ports.
r = 5 cm, f = 3000 rpm, z1 = 0 m.
Find:
Maximum operational height (m).
PLAN
Apply the pressure variation equation for rotating ‡ow
Locate point 1 at the liquid surface where z = 0:
Locate point 2 at the outer edge of the rotating disk.
SOLUTION
Pressure variation equation
p1 + z 1
r12 ! 2
= p2 + z 2
2
0+0
0 = 0 + z2
z2 =
r22 ! 2
2
r22 ! 2
2
r22 ! 2
2g
Rotational Rate
! = (3000 rev/min)(1min=60 s)(2 rad/rev) = 314:1 rad/s
Find z2
z2 = =
r22 ! 2
(0:05 m)2 (314:1 rad= s)2
=
2g
2 (9:81 m= s2 )
z2 = 12:6 m
96
4.95: PROBLEM DEFINITION
Situation:
A tank rotated about the horizontal axis and water in tank rotates as a solid body.
V = r!, z = 1; 0; +1 m, ! = 5 rad/ s.
Find:
Pressure gradient each value of z (kPa/m).
Properties:
= 1000 kg= m3 .
PLAN
Apply the pressure variation equation for rotating ‡ow.
SOLUTION
Pressure variation equation- rotating ‡ow.
@p
@z
+
=
@z
@z
@p
=
@z
when z =
r! 2
r! 2
1m
@p
=
@z
!2
=
1+
=
!2
g
9; 810 N= m3 (1 +
@p
=
@z
25
)
9:81 m= s2
34:8 kPa/m
when z = +1 m
@p
=
@z
+ !2
=
1+
=
9810 N= m3
!2
g
1
@p
= 15:2 kPa/m
@z
97
25
9:81 m= s2
At z = 0
@p
=
@z
@p
=
@z
98
9:81 kPa/m