# ME 215.3 Fluid Mechanics I

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```							         ME 215.3
Fluid Mechanics I

Example Problems

c James D. Bugg

January 2009

Department of Mechanical Engineering
1. The tank shown holds oil of speciﬁc gravity 0.89. The top of the tank is closed and the
space in the tank above the oil contains air. The U-tube manometer contains water
and the displacements are as indicated. Atmospheric pressure is 101.3 kPa. What is
the pressure of the air in the tank? (Solution: page 49)

Open to atmosphere
Air

4 cm
Air

6 cm

15 cm
Water
Oil

2. A ﬂat, vertical gate holds back a pool of water as shown. Find the force that the water
exerts on the gate. (Solution: page 51)

Patm = 100 kPa
3m

ρ = 1000 kg/m3
hinge

gate (5 m wide)
3m

stop

1                                                               ME 215.3 Example Problems
3. A Tainter gate is constructed from a quarter-cylinder and is used to hold back a pool of
water. The radius of the gate is 1.22 m and it is 2.44 m long. Calculate the hydrostatic
force on the gate. (Solution: page 53)

m
22
length=2.44 m

1.
gate

=
R
ρ = 1000 kg/m3

4. A ﬂat, vertical gate holds back a pool of water as shown. Find the force that the stop
exerts on the gate. (Solution: page 56)

Patm = 100 kPa
3m

ρ = 1000 kg/m3
hinge

gate (5 m wide)
3m

stop

ME 215.3 Example Problems                                                                  2
5. A circular hatch on a submarine is hinged as shown. The radius of the hatch is 35 cm
above the centre of the hatch. Determine how hard a sailor has to push on the centre
of the hatch to open it. The seawater density is 1025 kg/m3 and the centre of the hatch
is 2 m below the surface of the ocean. The air pressure inside the submarine is equal
to atmospheric pressure.

hinge
A

g                                                                                   70 cm

opening
force
hatch
hatch
inside of    outside of                                      View A-A
submarine     submarine

A

6. A rectangular gate is hinged along the top edge as shown. The gate is 4 m long, 15 cm
thick, and is made of concrete (s.g.=2.3). Determine the water depth H if the gate is
just about to open. What force does the gate exert on the hinge? The water density
is 1000 kg/m3 .

g
m
2

H
hinge
gate
o
45

3                                                                      ME 215.3 Example Problems
7. A pool of ﬂuid has a density that varies linearly from 1000 kg/m3 at the surface to
1600 kg/m3 at a depth of 4 m. A 2 m by 2 m square gate is hinged along its bottom
edge and held in place by a force F at its top edge. Find the force F .

g

2m

F

2m
hinge

8. A long, square wooden block is pivoted along one edge. The block is in equilibrium
when immersed in water (ρ = 1000 kg/m3 ) to the depth shown. Evaluate the speciﬁc
gravity of the wood.

1.2 m

wood        1.2 m
0.6 m

water
hinge

ME 215.3 Example Problems                                                             4
√
9. A cylindrical gate with a radius of curvature of 2 m holds back a pool of water with
a layer of oil ﬂoating on the surface. The gate is 8 m long. What force does the pinned
connection at A exert on the gate?

2m
g                √                               1m                   Oil (s.g. = 0.8)
A

1m                   Water (ρ = 1000 kg/m3)

10. A parabolic gate 4 m wide (in the z direction) holds a pool of water as shown. Find
the tension in the cable AB required to hold the gate in the position shown. Find the
reaction force which the hinge at C exerts on the gate. Note that point A is directly
above point C.

g                                 A           Cable

1.5 m                    B
0.5 m
Parabolic gate
2m      y                                  y = x2
water
ρ = 1000 kg/m3                C
x
Hinge

5                                                                 ME 215.3 Example Problems
11. A partially submerged pipe rests against a frictionless wall at B as shown. The speciﬁc
gravity of the pipe material is 2.0 and the pipe has an outside diameter of 2 m. The
ﬂuid is water with ρ = 1000 kg/m3 . Find the inside diameter of the pipe and the force
which the wall exerts on the pipe. The ends of the pipe are closed.

s.g. = 2.0

0.5 m                              B                        g

12. A gate of mass 2000 kg is mounted on a frictionless hinge along its lower edge. The
width of the gate (perpendicular to the plane of view) is 8 m. For the equilibrium
position shown, calculate the length of the gate, b.

b

1m
Water                              30o

hinge

ME 215.3 Example Problems                                                                 6
13. A vertical, plane wall holds back a pool of water (ρ = 1000 kg/m3 ) which is 1.5 m
deep. The wall has a triangular gate in it that is hinged along the bottom edge and
held closed by a horizontal force F applied at the top corner. Calculate the force F
required to hold it closed.

A

F

gate
gate

2m
1.5 m

g

A

hinge                           1m
View A-A

14. A Tainter gate holds back a pool of water (ρ = 1000 kg/m3 ) which is 2 m deep. The
radius of curvature of the gate is 2 m and it is 6 m long. The mass of the gate is 1000 kg
and its centre of gravity is at the position indicated on the diagram. Assume that the
point where the gate contacts the bottom of the pool is frictionless. Calculate the
reaction force on the hinge.

gate c. of g.                             2m

45o

g                                       45o
2m           1.8 m
hinge

7                                                                        ME 215.3 Example Problems
15. Determine H when the L-shaped gate shown is just about to open. Neglect the weight
of the gate and let the density of the ﬂuid be ρ.

gate

H
g

hinge
L

16. What force F is needed to hold the 4 m wide gate closed? The ﬂuid is water and it
has a density of 1000 kg/m3 .

hinge
9m

3m
F

ME 215.3 Example Problems                                                            8
17. A semi-circular gate is hinged at the bottom as shown. The density of the ﬂuid varies
linearly from 1000 kg/m3 at the surface of the reservoir to 1500 kg/m3 at the bottom
of the reservoir. Find the force F required to hold the gate in place?

A

g
3m

F                                                    hinge
gate
hinge
1m

Section A-A                                                  A

18. A circular cylinder holding back a pool of water is held in place by a stop as shown.
The height of the stop is 0.5 m and the water depth is 1.5 m. If the water depth were
increased beyond 1.5 m the cylinder would roll over the stop. Assuming that the water
contacts the cylinder surface right up to point A, what is the speciﬁc gravity of the
cylinder.

Diameter = 2 m

1.5 m
A
0.5 m

9                                                               ME 215.3 Example Problems
19. A gate composed of a quarter circle portion and a straight portion holds back a pool
of water. The gate is hinged at A and held in place by a force F applied as shown.
Find the magnitude of the force F required to hold the gate in the position shown.
The width of the gate is 8 m.

F

0.5 m

g                                                Water (ρ = 1000 kg/m3 )
1.5 m

Width of Gate is 8 m
R = 1m

A

Hinge

ME 215.3 Example Problems                                                             10
20. A circular cylinder of radius R = 50 mm, length b = 100 mm, and density ρc =
800 kg/m3 blocks a slot in the bottom of a water tank as shown. The line joining
the centre of the cylinder and the point where the cylinder contacts the edge of the
slot subtends an angle of α = 30o with the vertical. If h = 150 mm what is the force
exerted on the cylinder by the tank?

h

g

R

α = 30o

11                                                          ME 215.3 Example Problems
21. Water at 20o C is retained in a pool by a triangular gate which is hinged along its top
edge and held in place by a stop at its lowest point. What force does the stop exert
on the gate? What force does the hinge exert on the gate?

A

g

hinge
1m

7m

3.5 m
stop

A                          Section A-A

22. A 35 kg, 10 cm cube of material is suspended from a wire in a ﬂuid of unknown den-
sity. The tension in the wire is 335.5 N. Determine the speciﬁc gravity of the ﬂuid.
(Solution: page 59)

23. A U-tube is used as a crude way to measure linear acceleration. Determine the mag-
nitude of the acceleration as a function of the geometry of the tube, the acceleration
due to gravity, and the displacement of the ﬂuid in the tube. (Solution: page 61)

ME 215.3 Example Problems                                                                12
24. An open-top cart half full of water (ρ = 1000 kg/m3) is shown at rest in the ﬁgure
below. The cart begins to accelerate to the right at a constant 3 m/s2 . After some
time, the ﬂuid reaches a hydrostatic state. Determine the net hydrostatic force on the
rear, vertical end of the cart. The cart is 0.8 m wide.

2m

3 m/s2           g
0.5 m
1m

25. A container of water (ρ = 1000 kg/m3 ) accelerates on a 30o slope. It is completely
closed except for a small hole in the position indicated. If the gauge pressure at point
A is 25 kPa, what is the acceleration a? If the width of the container (perpendicular
to the page) is 0.5 m, what is the net hydrostatic force on the top of the container?

hole
1m

a

g

0.75 m           A

30o

13                                                             ME 215.3 Example Problems
26. A partially full can with an open top spins around an axis as shown. The diameter of
the can is 50 cm and is spinning at 50 rpm. If the depth of the ﬂuid at the outer edge
of the can is 30 cm, what is the depth of the ﬂuid on the axis of rotation? (Solution:
page 63)

ω = 50 rpm

30 cm

50 cm

27. A cylindrical container is rotated about its axis. Derive a general relationship for the
shape of the free surface as a function of the rotation rate, container radius, and the
depth of the ﬂuid when it is not spinning. (Solution: page 65)

ω

Ho

R

ME 215.3 Example Problems                                                                 14
28. A cylindrical can of radius 4 cm and height 12 cm has an open top. It is initially at
rest and completely full of liquid. It is rotated about its axis at 250 rev/min until the
ﬂuid inside it achieves solid body rotation. The rotation is then stopped and the ﬂuid
within the container is allowed to come to rest. How deep will the ﬂuid in the container
be?

g
12 cm

4 cm

29. A U-tube manometer contains two ﬂuids with diﬀerent densities as shown. The ﬂuid
positions shown in the diagram are for the case when the tube is not spinning. It is
then spun around the axis shown until the liquid level in both legs is equal? Find ω.

1000 kg/m3

1200 kg/m3
10 cm

g

ω
20 cm

15                                                                 ME 215.3 Example Problems
30. The U-tube shown is rotated about the vertical axis indicated on the diagram at
60 rev/min. Determine the displacement of the water in each leg from its rest position.
Perform this calculation on the centreline of the tubes. What is the pressure at point
A?

Axis of rotation

Diameter, 5 mm
Diameter, 8 mm

Rest level
A

5 cm
g

3 cm                   10 cm

31. A 80 cm diameter cylindrical can has a closed top except for a small vent hole at the
centre. If the density of the ﬂuid is 1000 kg/m3 and the can spins at 60 rpm, what is
the force on the top of the can? (Solution: page 67)

60 rpm

vent

ρ = 1000 kg/m3

cylindrical can

ME 215.3 Example Problems                                                                  16
32. An upright, 10-cm diameter, cylindrical paint can 20 cm deep spins around its axis of
symmetry at 500 rpm. It is completely full of mineral oil (sg = 0.87) and there is a
very small hole in the lid at the rim of the can. Determine the net hydrostatic force
on the lid.

33. A quarter-circle, cylindrical gate 5 m long and with a 1.5 m radius is hinged at the
point indicated below. It holds back a pool of water (ρ = 998 kg/m3 ) with a horizontal
force F applied as shown. Determine the force F and the reaction at the hinge.

F

5m
g
hinge

1.

34. You are driving down the road at 100 km/hr with a cup of coﬀee in your drink holder.
The cup has no lid and is 8 cm in diameter. The coﬀee is initially 1 cm from the top
of the cup. You come to a curve in the highway which is not banked. Determine the
minimum radius of the curve for which the coﬀee will not spill. If you go around a
curve whose radius is 75% of the minimum, how much coﬀee will spill (in cm3 )?

35. A square gate (1.25 m x 1.25 m) is hinged along a line 0.25 m from its top edge. A force
F1 applied at the top of the gate holds it closed. The pool of water (ρ = 998 kg/m3 ) is
1.75 m deep. Calculate the force F1 required to hold the gate closed.

hinge
g                   F1
gate
1.75 m

1.25 m
1m

17                                                                   ME 215.3 Example Problems
36. A cylindrical container is 40 cm deep and 30 cm in diameter and is open at the top.
Water is put in the container to a depth of 20 cm. The container is then spun around
its axis at an angular speed of ω until the water reaches a hydrostatic state. The
spinning is stopped and the water is allowed to come to rest. After coming to rest, the
water is 10 cm deep. Determine ω in revolutions per minute (rpm).

initial water level

40 cm
g
20 cm

30 cm

37. The police are using a ﬁre hose to move a ﬂat barricade. What is the horizontal force
on the barricade due to the stream of water? (Solution: page 69)

V = 15 m/s

Aj = 0.01 m2

ME 215.3 Example Problems                                                                18
38. A jet of water from issues from a 0.01 m2 nozzle at 15 m/s. It impinges on a vane
mounted on a movable cart and is deﬂected through 45o . If the cart is moving away from
the nozzle at 5 m/s, what is the force which the water exerts on the cart? (Solution:
page 72)

45o

Aj = 0.01 m2

cart

39. A 50 mm diameter pipe carries water at 1 m/s. A nozzle with an exit diameter of
25 mm is attached at the end of the pipe. If the pressure just upstream of the nozzle
is 10 kPa, what is the force on the bolts at the ﬂange connection attaching the nozzle
to the pipe? (Solution: page 79)

40. Consider the entrance region of a circular pipe for laminar ﬂow. What is the frictional
drag on the ﬂuid between axial locations 1 and 2 in terms of the pressure at those
locations, the density of the ﬂuid, the mean velocity of the ﬂuid, and the pipe radius.
(Solution: page 81)

Uo                                  u(r) = Umax (1 − (r/R)2)

r
1                       z             2

circular pipe

19                                                               ME 215.3 Example Problems
41. A 50 mm diameter pipe carries water at 1 m/s. A nozzle with an exit diameter of
25 mm is attached at the end of the pipe. Assuming frictionless ﬂow, ﬁnd the force on
the ﬂange connection. (Solution: page 84)

42. If the nozzle from the previous problem has a pressure of 10 kPa at its entrance, what
is the loss coeﬃcient? (Solution: page 86)

43. A transition piece turns 45o and expands from 50 mm to 80 mm. The pressure at the
entrance is 20 kPa and the loss coeﬃcient is 0.50 based on the discharge velocity. Water
ﬂows through the transition at 4 kg/s. Find the force required to hold the transition
in place. (Solution: page 87)

D2 = 80 mm
20 kPa

45o

4 kg/s
transition piece
Km = 0.50 (based on
discharge velocity)
D1 = 50 mm

ME 215.3 Example Problems                                                                 20
44. Air (ρ = 1.2 kg/m3 ) blows parallel to a ﬂat plate upon which a boundary layer grows.
The plate is 10 m long in the streamwise direction and 4 m wide (normal to the page).
At the leading edge of the plate the air speed is uniform at U∞ = 10 m/s. The velocity
proﬁle is measured at the end of the plate and is found to vary with distance y from the
plate according to u = U∞ (y/δ)1/7 where δ = 170 mm is the boundary layer thickness.
Determine the viscous force on the plate. Consider only the ﬂow over the top surface
of the plate. The dashed line labeled “A” is a streamline in this ﬂow.

A

170 mm
10 m/s
ﬂat plate                             y

10 m

45. An object is placed in a 1 m diameter wind tunnel and the air velocity downstream
of the object is found to linearly vary from zero at the centreline of the wind tunnel
to a maximum at the wind tunnel wall. The air velocity upstream of the object is
a uniform 20 m/s. A mercury manometer indicates a 10 mmHg pressure diﬀerence as
shown. Assume the pressure is uniform across the wind tunnel at any axial location.
Neglect shear at the wind tunnel walls and let ρ = 1.2 kg/m3 . Determine the drag
force on the object.

20 m/s                                                     1 m diameter

mercury (s.g.=13.56)     g
10 mm

21                                                             ME 215.3 Example Problems
46. A tank on wheels contains water (ρ = 1000 kg/m3 ) and is held in place by a cable as
shown below. A pump mounted on top of the tank draws water from the tank and
discharges the water through a horizontal, 1 cm diameter nozzle at a mass ﬂow rate
of 5 kg/s. The cable breaks and the cart begins to move. At some instant in time
later, the total mass of the tank, pump assembly, and the water remaining in the tank
is 100 kg and the velocity of the system is 20 m/s. Determine the acceleration of the
system at this instant in time. The ﬂow rate supplied by the pump remains constant.

Pump
5 kg/s
g
1 cm dia.

Cable

47. A new type of lawn sprinkler is developed that has two arms of diﬀerent lengths. The
arms lie in a horizontal plane and rotate about a vertical axis. Both nozzles are aimed
upward at 20o from the horizontal plane and have an exit diameter of 2 mm. The ﬂow
to each arm is equal. One arm is 10 cm from the pivot and the other is 20 cm from the
pivot. If the pivot is assumed to be frictionless and the sprinkler delivers 2 L/min, ﬁnd
the angular speed of the sprinkler?

10 cm                 20 cm

D = 2 mm

ME 215.3 Example Problems                                                                  22
48. An aluminium block weighing 10 N is supported by a jet of water issuing from a 2.5 cm
diameter nozzle. What jet exit velocity is required to hold the block 10 cm above the
nozzle exit? The jet of water is deﬂected through 180o when it strikes the block and
the guides holding the block in place are frictionless.

guides

g                                   aluminum block

10 cm

2.5 cm

49. An elbow connected to a 5 cm diameter pipe discharges water (ρ = 1000 kg/m3) as
shown. The diﬀerence between the stagnation and static pressures ∆P measured by
the pitot-static tube is 500 Pa. The mass of the elbow and the water which it contains
is 1 kg. The elbow has a minor loss coeﬃcient of 0.75 based on the inlet velocity. What
force does the pipe exert on the elbow at the ﬂange connection?

∆P
D1 = 5 cm

Elbow
g
˙
m

Pipe                                                 D2 = 3 cm

Flange

23                                                            ME 215.3 Example Problems
50. A pipe 4 cm in diameter supplies water to a 2 cm diameter nozzle. The gauge pressure
just upstream of the nozzle is 60 kPa and the head loss in the nozzle is given by
hL = 0.2Ve2 /2g where Ve is the velocity at the nozzle exit. The jet of water (not
shown) hits a frictionless splitter plate which is inclined at 30o to the axis of the pipe.
Half of the water is deﬂected downwards along this plate. Find the force required to
hold the plate in this position.

60 kPa

water

ρ = 1000 kg/m3
30o

51. Water (ρ = 1000 kg/m3 ) is pumped at volume ﬂowrate Q through the nozzle shown.
If the ﬂowrate is high enough, a force F will be required to keep the cart stationary.
Assume that the depth of the water in the cart remains at 0.8 m. Derive an expression
for F as a function of Q and indicate the range of Q for which it is valid. Sketch the
function.

g

F                                              1m
0.8 m
D = 5 cm       45o

2m                            2m

Q

ME 215.3 Example Problems                                                                     24
52. An engineer is measuring the lift and drag on an aerofoil section mounted in a two–
dimensional wind tunnel. The wind tunnel is 0.5 m high and 0.5 m deep (into the
paper). The upstream air velocity is uniform at 10 m/s. The downstream velocity is
uniform at 12 m/s in the lower half of the wind tunnel. The downstream velocity in
the upper half is uniform. The vertical component of velocity is zero at the beginning
and end of the test section. The test section is 1 m long. The engineer measures the
pressure distribution in the tunnel along the upper and lower walls and ﬁnds

Pu = 100 − 10x − 20x(1 − x) [Pa, gauge]

Pl = 100 − 10x + 20x(1 − x) [Pa, gauge]
where x is the distance in metres measured from the beginning of the test section.
The air density is constant at 1.2 kg/m3 . Find the lift and drag forces acting on the
aerofoil. Neglect shear on the walls of the wind tunnel.

Beginning                              End
of                                  of
Test section                        Test section

g
0.25 m
10 m/s
0.25 m                     12 m/s

1m

x

53. A cart mounted on straight, level rails is used to test rocket engines. A 400 kg cart
has a rocket mounted on it which has an initial mass of 500 kg. Eighty percent of the
mass of the rocket is fuel. The products of combustion exhaust at a speed of 1000 m/s
relative to the rocket nozzle. Neglect friction in the wheels of the cart and aerodynamic
drag. Determine the speed of the rocket when the fuel is all used.

g

25                                                             ME 215.3 Example Problems
54. Water (ρ = 1000 kg/m3 ) exits from a circular pipe 10 cm in diameter with a mass ﬂow
rate of 4 kg/s and strikes a ﬂat plate at 90o . The velocity at the exit varies linearly
from a maximum at the pipe centreline to zero at the pipe wall as shown. Determine
the force exerted on the ﬂat plate. Compare this to the force that would be exerted at
the same mass ﬂow rate if the pipe exit velocity had been uniform.

10 cm diameter

˙
m = 4 kg/s

55. A two-arm sprinkler is constructed as shown below. The total mass ﬂow rate of water
(ρ = 1000 kg/m3 ) is 1 kg/s and it is divided equally between the two arms. At the end
of one arm a 12 mm diameter nozzle is oriented perpendicular to the arm and is in the
same horizontal plane as the arm. The other arm has a 1 mm wide slot that also emits
water in the horizontal plane. Find the rotational speed of the sprinkler in revolutions
per minute assuming that the pivot is frictionless.

15 cm                 15 cm
12 mm dia.    2 cm
A

1 mm

axis of rotation
A                        Section A-A
Plan View                                       (enlarged)

ME 215.3 Example Problems                                                                 26
56. Two circular coaxial jets of incompressible liquid with speed V collide as shown. The
interaction region is open to atmosphere. Liquid leaves the interaction region as a
conical sheet. Obtain an expression for the angle θ of the resulting ﬂow in terms of d1
and d2 .

V
θ

V                                         d2
d1
V

V

57. A tank of water sitting on a weigh scale is being ﬁlled with water (ρ = 1000 kg/m3 )
from a 2 cm diameter pipe at a rate of 5 kg/s. The tank is circular and has a diameter of
0.5 m. The empty tank has a mass of 2 kg. At the instant when the water is 0.5 m deep
in the tank, what force will the scale read? Carefully explain each of your assumptions.

5 kg/s

2 cm
0.5 m
0.5 m diameter            g

0.5 m

Scale

27                                                                 ME 215.3 Example Problems
58. Calculate the force of the water (ρ = 1000 kg/m3 ) on the frictionless vane if

(b) the blade moves to the right at 20 m/s, and
(c) the blade moves to the left at 20 m/s.

40 m/s

60o
5 cm
dia.

59. A jet of water (ρ = 998 kg/m3 ) strikes a frictionless splitter vane as shown. The
ﬂowrate is 1 kg/s while the diameter of the nozzle is 1 cm. The position of the splitter
vane is adjusted in the z direction so that there is no reaction in the z direction. What
is the reaction in the x direction?

z
o
1 kg/s                                                  45
x

splitter
vane
D = 1 cm

ME 215.3 Example Problems                                                                  28
60. Water at 20o C ﬂows from a nozzle of diameter D = 5 mm at speed V = 10 m/s and
strikes a vane which splits the ﬂow and deﬂects it as shown. Thirty-ﬁve percent of the
mass ﬂow deﬂects through 90o (to the left). What force does the ﬂuid exert on the
vane?

45o

D
V

61. Water at 20o C ﬂows through an elbow/nozzle arrangement and exits to atmosphere as
shown below. The inlet pipe diameter is 12 cm while the nozzle exit diameter is 3 cm.
The mass ﬂowrate is 1 kg/s. If the pressure at the ﬂange connection is 200 kPa(gauge),
what is the force and moment on the ﬂanged connection? Neglect the weight of the
elbow and water.

200 kPa

Water                                                          g

ﬂanged                                                   30 cm
connection

45o

29                                                           ME 215.3 Example Problems
62. A nozzle for a spray system is designed to produce a ﬂat radial sheet of water. The
sheet leaves the nozzle at V2 = 10 m/s, covers 180o of arc and has thickness t = 1.5 mm.
The nozzle discharge radius is R = 50 mm. The water supply pipe is 35 mm in diameter
and the inlet pressure P1 is 50 kPa above atmospheric. Calculate the force exerted by
the spray nozzle on the supply pipe through the ﬂanged connection.

P1                              V2
35 mm
supply pipe                 diameter

Water                                      R

thickness, t

ﬂanged            spray nozzle
connection

63. A ride at an amusement park consists of a wheeled cart that zooms down an inclined
plane onto a straight and level track where it decelerates to rest by means of a high-
speed jet of air projected directly forward through a nozzle. The jet is supplied by
a compressed air cylinder aboard the cart. The initial gross mass of the cart and its
occupants is 500 kg. Air escapes from the braking jet at a constant mass ﬂow rate of
20 kg/s and a constant velocity (relative to the nozzle) of 150 m/s. At the instant when
the braking jet is activated, the speed of the cart is 40 m/s. Determine how much air
must escape in order to stop the cart. What is the stopping distance?

ME 215.3 Example Problems                                                                 30
64. Fluid enters a 5 cm diameter pipe with a uniform velocity of Uo . The ﬂuid exits the
pipe at two locations. One is a 5 cm diameter exit with a turbulent velocity proﬁle
described by
1/7
r
uz = Uc1 1 −                 .
R1

where uz is the axial component of velocity, Uc1 is the centreline velocity, r is the radial
coordinate, and R1 is the radius of the pipe. The second exit is 1.5 cm in diameter and
has a laminar, parabolic velocity proﬁle described by
2
r
uz = Uc2   1−                    .
R2

If Uc1 is measured to be 2 m/s and Uc2 is measured to be 1 m/s, what is Uo ?

uniform velocity                                 5 cm diameter

turbulent velocity proﬁle

1.5 cm diameter

laminar velocity proﬁle

65. A 6 mm diameter angled nozzle is attached to the end of a 2 cm diameter pipe with
a ﬂanged connection as shown below. The angle between the nozzle and the supply
pipe is 45o . The pressure just upstream of the ﬂanged connection is 200 kPa (gauge).
Water (ρ = 998 kg/m3 ) exits the nozzle at 20 m/s. Determine the force and torque in
the ﬂanged connection.

200
kPa
ﬂanged connection

6 mm diameter

2 cm diameter                          4 cm

31                                                                        ME 215.3 Example Problems
66. A two-arm lawn sprinkler, as viewed from above, is shown in the sketch. Water is
delivered to the sprinkler at a volume ﬂow rate of 5 L/min and it can be assumed that
this ﬂow is split evenly between the two nozzles. The nozzles both have a diameter
of 2 mm and are both angled upwards from the horizontal plane at 30o . The density
of the water is 998 kg/m3 . Assume the pivot is frictionless. Calculate the rotational
speed of the sprinkler.

15 cm          15 cm

45o

67. Fluid (ρ = 850 kg/m3 ) enters a 5-cm diameter pipe with a uniform velocity of 3 m/s
at location A. At location B, the ﬂuid has a turbulent velocity proﬁle described by
r   1/5
uz = Uc 1 −
R
where uz is the axial component of velocity, Uc is the centreline velocity, r is the radial
coordinate, and R is the radius of the pipe. The pressure at location A is 4 kPa larger
than at location B. Determine the viscous force on the wall of the pipe between location
A and B.

uniform velocity                       5 cm diameter

A                                                    B

68. A 2”ID steel pipe 50 m long carries water at a rate of 0.04 m3 /s. There are two 90o
regular ﬂanged elbows and an open ﬂanged globe valve. The net elevation change is
30 m. Calculate the pressure diﬀerence between the ends of the pipe. (Solution: page
90)

ME 215.3 Example Problems                                                                    32
69. Two reservoirs are connected by a pipe as shown. The elevation change between the
two reservoirs is 20 m. Find the volume ﬂowrate between the reservoirs. (Solution:
page 92)

ρ = 1000 kg/m3
µ = 0.001 kg/(m · s)
20 m

b
a

D = 2 cm              D = 4 cm
L = 5m                L = 5m
ǫ = 0.05 mm

70. Calculate the frictional pressure drop in 100 m of 3-cm diameter smooth pipe with
water (ρ = 1000 kg/s, µ = 0.001 Pa · s) ﬂowing at 1 kg/s. What would the frictional
pressure loss be if the roughness were 0.5 mm?

71. The frictional pressure drop in 100 m of 3-cm diameter smooth pipe with water (ρ =
1000 kg/s, µ = 0.001 Pa · s) ﬂowing is 1 MPa. Determine the volume ﬂow rate.

72. A piping system contains a valve (Km = 6.5) and discharges water (ρ = 998 kg/m3 ,
ν = 10−6 m2 /s) to atmosphere. A mercury manometer 20 m from the end of the pipe
reads as shown. The right leg of the manometer is open to atmosphere. The pipe is
smooth and its diameter is 5 cm. Determine the volume ﬂowrate through the pipe in
L/min.

g
20 cm

mercury (s.g.=13.56)
20 cm

33                                                          ME 215.3 Example Problems
73. A pump has a characteristic curve that can be approximated by a parabola as shown.
It pumps water through 100 m of 20 cm diameter cast iron pipe. What is the ﬂowrate?
(Solution: page 94)

100

80

60

40

20

0
0.0   0.5         1.0        1.5   2.0         2.5
3
Flowrate(m /s)

74. Two large water (ν = 10−6 m2 /s) reservoirs are joined by two equal-length pipes as
shown. The 25 mm diameter pipe has a roughness of 0.5 mm while the 20 mm diameter
pipe is smooth. The ﬂow rates through the two pipes are equal. What is the total ﬂow
rate between the two reservoirs? Neglect minor losses.

20 mm, smooth

25 mm, ǫ = 0.5 mm                             g

ME 215.3 Example Problems                                                             34
75. Water (ρ = 1000 kg/m3 , µ = 0.001 Pa · s) ﬂows through a horizontal section of 4 cm
diameter pipe. The pipe has a roughness of 0.2 mm. A stagnation pressure tap and a
static pressure tap are mounted 1 m apart as shown. The pressure diﬀerence between
these two taps is measured with a mercury (s.g.=13.56) manometer which shows a
displacement of 5 cm. The manometer tubes are otherwise full of water. What is the
volume ﬂow rate of the water?

Stagnation tap
Static tap

Q
g

5 cm

1m                          mercury (s.g.=13.56)

35                                                         ME 215.3 Example Problems
76. A vertical section of 4 cm diameter pipe with ǫ = 0.2 mm has two pressure gauges
mounted 5 m apart which, at the current ﬂow rate, read the same (PA = PB ) . A ﬂuid
with a kinematic viscosity of 4 × 10−6 m2 /s ﬂows through the pipe. Does the ﬂuid ﬂow
up or down? What is the volume ﬂow rate?

PA

5m                                          g

PB

D = 4 cm

ME 215.3 Example Problems                                                              36
77. A pump draws water (ρ = 1000 kg/m3 , µ = 0.001 Pa · s) from a reservoir and discharges
it into 100 m of 10 cm diameter pipe which has a roughness of 0.1 mm. The discharge
of the pipe is 20 m lower than the surface of the reservoir. Neglect minor losses. If the
pump characteristics are represented by the pump curve shown below, estimate the
ﬂowrate.

60
55
50
45
40

35
30
25
20
15
10
5
0
0.0   0.5   1.0   1.5     2.0   2.5   3.0     3.5    4.0   4.5   5.0

Flow Rate (m3 /min)

37                                                                          ME 215.3 Example Problems
78. An engineer needs to measure the minor loss coeﬃcient of a reducing elbow which
system, the only available pressure taps are 1 m upstream and 0.75 m downstream of
the elbow. A mercury manometer is attached at these locations and reads 7.5 cmHg
when the mass ﬂowrate is 1 kg/s. What is the minor loss coeﬃcient for the elbow?
Assume that all pipes are smooth. The ﬂowing ﬂuid is water with ρ = 1000 kg/m3 and
µ = 0.001 Pa · s.

Reducing elbow                       0.75 m

2 cm diameter

not to scale

1m                                   g

4 cm diameter

7.5 cm
˙
m = 1 kg/s

mercury (s.g.=13.56)

ME 215.3 Example Problems                                                               38
79. A centrifugal pump has a pump curve that can be described by

hp = 4 − 10Q2

where hp is the pump head in metres of water and Q is the volume ﬂowrate in L/min.
This pump is used to supply water (ρ = 1000 kg/m3 , ν = 10−6 m2 /s) to the system
shown below. The sum of all minor loss coeﬃcients is 12 and the total length of tube
is 8 m. The tube has an inside diameter of 5 mm and roughness ǫ = 0.025 mm. What
is the ﬂowrate achieved?

Discharge to
Total length L = 8 m            atmosphere
2.0 m                    Diameter D = 5 mm
Roughness ǫ = 0.05 mm

80. A centrifugal pump draws water (ρ = 1000 kg/m3 , µ = 0.001 Pa · s) from a large
reservoir and pumps it through 1335 m of 30 cm inside diameter pipe with roughness
ǫ = 0.5 mm. Over the range of interest the pump head can be expressed by

hp = A − BQ2

where hp is the head produced by the pump, Q is the volume ﬂowrate, and A and B
are constants. When the pipe exit is at the same elevation as the reservoir surface
the ﬂowrate is 17.0 m3 /min. However, when the exit of the same pipe is raised to an
elevation of 50 m the ﬂow reduces to 13.3 m3 /min. How high can the exit be raised
before the ﬂow will be zero? For all conditions the pipe discharges to atmosphere.
Neglect minor losses in this problem.

39                                                          ME 215.3 Example Problems
81. A 6 mm internal diameter, thin-walled, smooth rubber hose 12 m long is used to siphon
water (ρ = 1000 kg/m3 , ν = 10−6 m2 /s) from a large tank. The outlet of the hose is
6 m below the water surface in the tank. What will the volume ﬂowrate be? Neglect
minor losses due to bends in the hose.

6m

6 mm I.D., 12 m long

82. The piping system shown is ﬁtted with a centrifugal pump whose characteristics can
be approximated by hp = 46 − 2.5Q2 where hp is the head produced by the pump in m
of water and Q is the volume ﬂowrate in m3 /min. What is the ﬂowrate through this
piping system?

All elbows shown are 90o , regular, ﬂanged.
Sudden          All pipes have ǫ = 0.15 mm
expansion

D = 100 mm
Pump                     L = 170 m
D = 200 mm
L = 340 m

Gate valve           Sudden
contraction
Fully open

ME 215.3 Example Problems                                                                40
83. You are going to measure the minor loss coeﬃcient of a new valve design using the
apparatus shown. Assume that the water tank is large and the pipe discharges to
atmosphere. The 2-cm diameter smooth pipe is 4 m long. The free surface in the tank
is 5 m above the pipe exit. If you measure the ﬂowrate to be 60 L/min what is the
minor loss coeﬃcient of the valve? Let ν = 10−6 m2 /s.

5m
L = 4m                         New valve
D = 2 cm

84. A centrifugal pump is connected to a piping system as shown below. The pipe is 5 cm
in diameter and has a roughness of ǫ = 0.5 mm. The two valves are identical and have
minor loss coeﬃcients of 1.0 when fully open and 20 when 50% open. With both valves
fully open, the pressure guage reads PA = 250 kPa. When both valves are 50% open,
the ﬂow is 355 L/min. The pump curve can be represented by a parabola. The free
surface in the tank is at the same elevation as the pipe exit. Let ν = 10−6 m2 /s and
ρ = 998 kg/m3 . Determine PA if the ﬁrst valve is fully open and the second valve is
fully closed. Neglect losses which occur on the suction side of the pump.

PA
Pump
Valve                           Valve

25 m            25 m           25 m               25 m

41                                                           ME 215.3 Example Problems
85. A portion of a piping system is shown below. At point B, the piping system discharges
to the atmosphere. The two elbows each have a minor loss coeﬃcient of 2.4. The valve
has a minor loss coeﬃcient of 6.9. The pipe is 25 mm in diameter and has a roughness
of 0.05 mm. There is a total of 15 m of pipe between point A and point B. Point B is
3 m above point A. The pressure PA is 200 kPa guage. Determine the volume ﬂow rate
(in L/min) if the ﬂuid is water with ρ = 998 kg/m3 and µ = 0.001 Pa · s.

Valve                                 B

PA

A

ME 215.3 Example Problems                                                              42
86. Flow around a certain bridge pier can be modelled by a freestream and a single source.
If the freestream velocity is 5 m/s and the pressure a long distance upstream is 50 kPa,
what is the pressure at point A on the pier surface? The stagnation point will be 1 m
upstream of the source. (Solution: page 96)

y

x

2m
A

43                                                             ME 215.3 Example Problems
87. A pair of doublets of equal strength are placed in a freestream as indicated below. The
streamline which passes through the origin also passes through the point (4 m, 1 m).
What is the doublet strength? Referenced to P∞ , what is the pressure at the origin?
(Solution: page 99)

2m             2m

doublets
2m                         y

x
o
30

U∞ = 10 m/s

88. The landing gear strut on a small aircraft has a cross–section as indicated below. This
shape can be modelled as a Rankine Oval with a source and a sink placed as shown.
The aircraft is ﬂying at 45 m/s and the air density is 1.2 kg/m3 . What is the diﬀerence
in static pressure between points A and B? (Solution: page 101)

B

A

source           sink
U∞
12 cm             12 cm

30 cm

ME 215.3 Example Problems                                                                 44
89. Consider the irrotational ﬂow around a circular cylinder which is creating no lift.

(a) Derive an expression for the velocity along the positive y axis in terms of U∞ , a,
and y.
(b) Sketch this function.
(c) Consider the streamline labelled A. Far upstream of the cylinder this streamline
is a distance L from the horizontal plane of symmetry. At what distance from the
origin does the streamline cross the y axis?

(Solution: page 104)

y
g
A
U∞

L                                                          x

a

45                                                             ME 215.3 Example Problems
90. Consider the ﬂow of air (ρ = 1.2 kg/m3 ) over a 20 cm diameter circular cylinder. The
freestream velocity is 20 m/s. The volume ﬂowrate per unit length between the two
streamlines labelled A and B is 1.19 m2 /s. The streamline labelled B passes through
the point (x, y) = (0, 0.12 m). What is the lift force per unit length on the cylinder?
(Solution: page 107)

0.5

0.4

0.3

0.2
(x, y) = (0, 0.12 m)

0.1
y
0
x

−0.1
B
−0.2         A

−0.3

−0.4

−0.5
−0.5       −0.4   −0.3   −0.2   −0.1    0           0.1    0.2     0.3      0.4   0.5

91. An inviscid ﬂow of air (ρ = 1.2 kg/m3 ) is produced by a freestream (U∞ = 50 m/s)
aligned with the x axis, a source at the origin, and a second source of equal strength
located at (x, y) = (0.1 m, 0). The stagnation streamline crosses the y axis at y = 9 cm.
Determine the gauge pressure at the point where the stagnation streamline crosses the
y axis. Determine the acceleration 2 cm upstream of the ﬁrst stagnation point.

ME 215.3 Example Problems                                                                        46
92. An inviscid ﬂow is produced by a freestream (U∞ = 10 m/s) aligned with the x axis,
three sources placed as shown, and a single sink with a strength appropriate to form
a closed streamline in the ﬂow. The streamline shown on the diagram is a distance a
from the x axis far upstream of the origin and passes through the point (x, y) = (0, 2a).
Determine the thickness of the object at x = 0 if a = 0.2 m.

y
sources        sink
U∞ = 10 m/s
a

a
x

a
2a

93. The maximum pressure diﬀerence between any two points on the surface of the Rankine
oval shown below is measured to be 500 Pa. The source and the sink are placed 40 cm
apart. The maximum thickness of the oval is 20 cm. Determine the freestream velocity
if the density of the air is 0.8 kg/m3 .

source
sink
U∞
20 cm

40 cm

47                                                                  ME 215.3 Example Problems
Solution: Problem # 1

Given:

• Air trapped over oil in a closed tank as shown

Open to atmosphere
Air
Patm = 101.3 kPa

h3 = 4 cm

h2 = 6 cm

h1 = 15 cm
Air
g          1                                                  Water        z
Oil                    B
s.g = 0.89                           A

Find:

• Calculate the air pressure in the tank.

Assumptions:

• ﬂuid is in a hydrostatic state

• no acceleration

• density of the air is negligible

• ρw = 998 kg/m3 (constant)

• ρoil is constant

• g is constant

Analysis:

∇P = ρ(g − a)
ˆ
g = −g k

ME 215.3 Example Problems                                                                    48
a=0
ˆ
∇P = −ρg k
dP
= −ρg
dz
dP = −ρgdz
Since ρ and g are constant,

∆P = −ρg∆z

Now,

P1 = (P1 − PB ) + (PB − PA ) + (PA − Patm ) + Patm

P1 = −ρoil g(z1 − zB ) − ρair g(zB − zA ) − ρw g(zA − zatm ) + Patm
z1 − zB = h3
zA − zatm = −h1
P1 = −ρoil gh3 + ρw gh1 + Patm
P1 = ρw g(h1 − sgoil h3 ) + Patm
kg      m
P1 = 998    3
9.81 2 (0.15[m] − 0.89(0.04)[m]) + 101, 300[Pa]
m       s
P1 = 102.4 kPa(abs)

The pressure in the air chamber is 102.4 kPa (abs).

49                                                                  ME 215.3 Example Problems
Solution: Problem # 2

Given:

• A rectangular gate holding a pool of water.

x             Patm = 100 kPa

z

3m
ρ = 1000 kg/m3
hinge

g                        gate (5 m wide)
3m

stop

Find:

• The force which the water exerts on the gate

Assumptions:

• ﬂuid is in a hydrostatic state

• no acceleration

• g is constant

• ρ is constant

ME 215.3 Example Problems                                            50
Analysis:

Establish a coordinate system (see diagram).

Fs = −               ˆ
P ndA
S

∇P = ρ(g − a)
ˆ
g = gk
a=0
ˆ
∇P = ρg k
dP
= ρg
dz
dP = ρgdz

dP = ρg              dz

P = ρgz + C
at z = 0, P = Patm . Therefore, C = Patm .

P = ρgz + Patm

ˆ ı
n =ˆ
z2        w
Fs = −                                   ı
(ρgz + Patm )ˆdydz
z1        0
z2                              w
ı
Fs = −ˆ             (ρgz + Patm )dz                dy
z1                           0
z2
ρgz 2
ı
Fs = −ˆ w       + Patm z
2                          z1

kg      m (62 − 32 ) 2
ı
Fs = −ˆ 5[m] 1000    9.81 2           [m ] + 100, 000[Pa](6 − 3)[m]
m3      s     2
ı
Fs = −2.16 ˆ MN
The water will exert a horizontal force of 2.16 MN on the gate.

51                                                            ME 215.3 Example Problems
Solution: Problem # 3

Given:

• A Tainter gate holding back a pool of water

z

m
22
length=2.44 m                                                          g

1.
gate

=
R
x                   ρ = 1000 kg/m3

Find:

• The net hydrostatic force on the gate

Assumptions:

• atmosphere is constant pressure

• ﬂuid is in a hydrostatic state

• density of water is constant

• g is constant

• no acceleration

Analysis:

Establish coordinate system (see diagram).

Fs = −         ˆ
P ndA
S

∇P = ρ(g − a)
ˆ
g = −g k
a=0

ME 215.3 Example Problems                                                      52
dP
= −ρg
dz
P = −ρgz + C
√
at z = R/ 2, P = 0 (gauge).
√
Therefore, C = ρgR/ 2.
√
P = ρg(R/ 2 − z)

ˆ        ı        ˆ
n = cos θˆ + sin θk

ˆ
n

θ

dA = Rdθdy
w       θ2        √
Fs = −                                        ı        ˆ
ρg(R/ 2 − z)(cos θˆ + sin θk)Rdθdy
0       θ1

z = R sin θ
θ2
cos θˆ sin θk
ı       ˆ
Fs = −ρgwR2                                                       ˆ
√ + √ − sin θ cos θˆ − sin2 θk dθ
ı
θ1         2       2
Recall that
1 − cos 2θ
sin2 θ =
2
θ2
ˆ
sin θˆ cos θk sin2 θ
ı                  1        sin 2θ
Fs = −ρgwR         2
√ − √ −             ı
ˆ−       θ−           ˆ
k
2       2    2       2          2
θ1
π/4
sin θ sin2 θ    ˆ   cos θ θ sin 2θ
Fs = −ρgwR2 ˆ
ı                      √ −           −k    √ + −
2    2              2  2   4          −π/4

kg      m                             1 1         1 1
Fs = − 1000                     9.81 2 2.44[m]1.222 [m2 ] ˆ
ı         −       − − −
m3      s                             2 4         2 4
ˆ                1 π 1        1 π 1
− k                  + −      −   − +
2 8 4        2 8 4

53                                                                         ME 215.3 Example Problems
π 1
ı ˆ
Fs = −35630[N] ˆ − k       −
4 2

ı       ˆ
Fs = −35.6ˆ + 10.2k kN

With respect to the coordinate system shown on the diagram, the net hydrostatic

ı       ˆ
force on the gate is (−35.6ˆ + 10.2k) kN.

ME 215.3 Example Problems                                                          54
Solution: Problem # 4

Given:

• A rectangular gate hinged at the top holding back a pool of water

Patm = 100 kPa

3m
ρ = 1000 kg/m3
hinge
x

g                      gate (5 m wide)
3m

z
stop

Find:

• The force which the stop exerts on the gate

Assumptions:

• ﬂuid is in a hydrostatic state

• no acceleration

• ρ is constant

• g is constant

• hinge is frictionless

Analysis:

Establish coordinate system (see diagram)

Draw a free body diagram of the gate.

55                                                            ME 215.3 Example Problems
x
Ft

z

Fs

Fb

Mo = 0

ˆ
(H k × Fb ) + Ms = 0

Ms = −                   ˆ
P (r × n)dA
S

ˆ
r = zk
ˆ ı
n =ˆ
ˆ     ˆ ı      
r × n = z k × ˆ = zˆ
∇P = ρ(g − a)
a=0
ˆ
g = gk
dP
= ρg
dz
P = ρgz + C
at z = 0, P = ρgh0 . Therefore, C = ρgh0 .

P = ρg(z + h0 )

dA = dydz
w    z2
Ms = −                    
(zˆ)ρg(z + h0 )dzdy
0       z1
z2
Ms = −ρgwˆ
                  (z 2 + h0 z)dz
z1
z2
z 3 h0 z 2

Ms = −ρgwˆ            +
3    2            z1

kg      m       33 3     3[m]32 2
Ms = −1000                       
9.81 2 5[m]ˆ   [m ] +       [m ]
m3      s       3          2

ME 215.3 Example Problems                                      56

Ms = −1104ˆkN · m
Recall that,
ˆ
(H k × Fb ) + Ms = 0

ˆ        1104ˆ
k × Fb =        [kN · m]
3[m]
ı
Fb = 368 kNˆ
The stop exerts a force of 368 kN on the gate.

57                                                ME 215.3 Example Problems
Solution: Problem # 22

Given:

• An object suspended in a liquid by a wire

10 cm cube, m = 35 kg
z

g

Find:

• The speciﬁc gravity of the ﬂuid

Assumptions:

• ﬂuid is in a hydrostatic state
• no acceleration
• block is in equilibrium
• mass and volume of wire are zero
• ρ is constant
• g is constant

Analysis:

Establish coordinate system (see diagram).

Fz = 0

Fb + T − W = 0

ME 215.3 Example Problems                                                58
ρgV + T − mg = 0
m                                   m
ρ9.81 2 (0.1[m])3 + 335.5[N] − 35[kg]9.81 2 = 0
s                                   s
3
ρ = 800 kg/m
sg = 800/998

The speciﬁc gravity of the liquid is 0.801.

59                                                     ME 215.3 Example Problems
Solution: Problem # 23

Given:

• A U–tube manometer used to measure linear acceleration
d

1                       a

h              z
g                                    2

x

L

Find:

• The magnitude of a in terms of geometry, g, and h

Assumptions:

• a constant

• ﬂuid is in a hydrostatic state

• ρ and g are constant

Analysis:

Establish coordinate system (see diagram).

Points 1 and 2 are at the same pressure. Therefore,
2                  2
dP =               dR · ∇P = 0
1                  1

ı      ˆ
dR = dxˆ + dz k
∇P = ρg − ρa

ME 215.3 Example Problems                                             60
ˆ
g = −g k
ı
a = aˆ
ˆ     ı
∇P = −ρg k − ρaˆ
dR · ∇P = −ρadx − ρgdz
2
0=           (ρadx + ρgdz) = ρa(x2 − x1 ) + ρg(z2 − z1 )
1

0 = ρaL + ρg(−h)
a = gh/L

The acceleration of the tube is gh/L.

61                                                              ME 215.3 Example Problems
Solution: Problem # 26

Given:

• A partially full container rotating around its axis.

z
ω = 50 rpm

2
g
30 cm
1
r

50 cm

Find:

• The depth of the ﬂuid on the axis

Assumptions:

• ω is constant

• ﬂuid is in a hydrostatic state

• ρ is constant

• g is constant

Analysis:

Establish coordinate system (see diagram).

The pressure at 1 and 2 are equal. Therefore,
2                  2
dP = 0 =           dR · ∇P
1                  1

r      ˆ
dR = drˆ + dz k

ME 215.3 Example Problems                                                  62
∇P = ρ(g − a)
ˆ
g = −g k
a = −ω 2 rˆ
r
ˆ
∇P = −ρg k + ρω 2 rˆ
r
2
0=            (−ρgdz + ρω 2 rdr)
1

ρω 2 2     2
ρg(z2 − z1 ) =    (r2 − r1 )
2
(5π/3[1/s])2 (0.25[m])2
0.3[m] − z1 =
2(9.81)[m/s2 ]
z1 = 21.3 cm
The ﬂuid is 21.3 cm deep on the centreline.

63                                             ME 215.3 Example Problems
Solution: Problem # 27

Given:

• A cylindrical container rotating about its axis.

z

ω

g
Ho

r

R

Find:

• Derive an equation for the shape of the liquid surface if the container spins at ω about
the z axis.

Assumptions:

• ω is constant

• ﬂuid is in a hydrostatic state

• g is constant

• ρ is constant

Analysis:

Establish coordinate system (see diagram).

The surface is a line of constant pressure. Therefore, dP = 0 along the surface.

0 = dR · ∇P

ME 215.3 Example Problems                                                                 64
r      ˆ
dR = drˆ + dz k
a = −ω 2 rˆ
r
ˆ
∇P = ρ(ω 2 rˆ − g k)
r
0 = ρω 2 rdr − ρgdz
ω2r
dr =          dz
g
ω 2r2
z=             +C
2g
Equate the initial and ﬁnal volumes.
R
z2πrdr = H0 πR2
0

R
ω 2r3
H0 πR2 = 2π                          + Cr dr
0        2g
R
ω 2 r 4 Cr 2
2
H0 R = 2        +
8g      2                    0
2    4
ω R
H0 R 2 =         + CR2
4g
ω 2 R2
C = H0 −
4g

The surface is deﬁned by

ω2              R2
z = H0 +                  r2 −
2g              2

65                                                  ME 215.3 Example Problems
Solution: Problem # 31

Given:

• A closed cylindrical container rotating about its axis.

z
60 rpm

vent
r

ρ = 1000 kg/m3
g
cylindrical can

Find:

• The force on the top of the can (net hydrostatic force)

Assumptions:

• ﬂuid is in a hydrostatic state
• ρ is constant
• g is constant

Analysis:

Establish a coordinate system (see diagram).

Fs = −         ˆ
P ndA
S

∇P = ρ(g − a)
ˆ
g = −g k

ME 215.3 Example Problems                                                 66
a = −ω 2 rˆ
r
ˆ
∇P = −ρg k + ρω 2 rˆ
r
Need P as a function of r. Therefore,
∂P
= ρω 2 r
∂r
ρω 2 r 2
P =            +C
2
at r = 0, P = 0. Therefore, C = 0.

ˆ    ˆ
n = −k

dA = rdrdθ
2π       R
ρω 2 r 2 ˆ
Fs = −                             (−k)rdrdθ
0        0         2
R
2πρω 2 ˆ                               ˆ   R4
Fs =          k                r 3 dr = πρω 2 k
2             0                          4
2
kg                 1        0.44 [m4 ] ˆ
Fs = π1000                       2π                       k
m3                 s            4
ˆ
Fs = 794k N
The net hydrostatic force on the lid of the can is 794 N upwards.

67                                                                 ME 215.3 Example Problems
Solution: Problem # 35

Given:

• A jet of water hitting a ﬂat barricade.

2

V = 15 m/s                               g

Aj = 0.01 m2
y

1                           x

2

Find:

• The force that the water exerts on the barricade.

Assumptions:

• jet is horizontal

• jet deﬂects to vertical plane

• pressure uniform on control surface

• neglect body forces

• neglect ﬂuid shear

• uniform ﬂow

ME 215.3 Example Problems                                           68
Analysis:

Since we are asked to ﬁnd a force we should probably consider the linear momentum equation.

d
ρgdV −
−             ˆ
P ndA + Fv + R =            (ρV )dV +
−                  ˆ
ρV (Vr · n)dA
dt
V
−                S                           V
−               S

Consider the control volume shown.

Each term of the momentum equation will now be discussed.

Body Force:

ˆ
ρgdV = −W k
−
V
−

W is the weight of EVERYTHING in the control volume, water, air, barricade etc.

Pressure Force:

ˆ
P ndA = 0
S

P is uniform over the entire control surface. Therefore, the net pressure force is ZERO.

Viscous Force:

Fv = 0

The ﬂow is perpendicular to the control surface everywhere.

Reaction:

R=0

We have chosen a control surface which “cuts” the support so we can ﬁnd R.

69                                                                  ME 215.3 Example Problems

d
(ρV )dV = 0
−
dt
V
−

Momentum transport term:

ˆ
ρV (Vr · n)dA =        ˙
(m V )d −        ˙
(mV )i
S                     d               i

The properties at all inlets and discharges are assumed to be uniform.

With these simpliﬁcations, the momentum equation becomes

ˆ
−W k + R = (ρw A2 V2 )V2 + (ρw A3 V3 )V3 − (ρw A1 V1 )V1

Consider only the x direction

Rx = −ρw A1 V12

kg              m             2
Rx = −998     3
0.01[m2 ] 15                   = −2.25 kN
m               s

This is the force exerted on the control volume by the “mounting strut”. The force exerted
on the barricade is in the opposite direction.

The water exerts a force of 2.25kN to the right on the barricade.

ME 215.3 Example Problems                                                              70
Solution: Problem # 36

Given:

• A jet of water deﬂecting oﬀ a vane mounted on a moving cart

45o

Aj = 0.01 m2                                          Vc = 5 m/s constant
Vj = 15 m/s

cart

Find:

• Force which the water exerts on the cart

Assumptions:

• neglect body forces

• neglect friction on vane

• uniform ﬂow

• cart travelling at constant velocity

Analysis:

Since we are asked to ﬁnd a force we should probably consider the linear momentum equa-
tion.
d
ρgdV −
−            ˆ
P ndA + Fv + R =              (ρV )dV +
−                  ˆ
ρV (Vr · n)dA
dt
V
−             S                            V
−                 S

We must choose a control volume.

71                                                                   ME 215.3 Example Problems
Should it be stationary or moving? (we will try both) What should it look like?

First consider what the control volume should look like. It is important to consider what we
are being asked to solve for. In this case we are looking for the force that the water exerts
on the cart. Therefore, one possible choice for a control volume is

.

C.S. between water and
plate (gap shown only
for clarity)

Consider the left-hand side of the momentum equation for this choice.

ρgdV −
−           ˆ
P ndA + Fv + R = ...
V
−            S

• The body force term is assumed to be zero. We are neglecting the weight of the water
on the vane.

• The pressure term is unknown.

– the pressure everywhere except on the surface of the plate is zero
– the integrated eﬀect of the pressure acting on the plate is what we were asked to
ﬁnd in this problem
– this term gives the force ON the CV. Therefore, it is equal and opposite to the
unknown in this problem

• The viscous force term is zero since we are assuming a frictionless surface.

• The reaction force is zero since this CS does not cut through any solid objects.

ME 215.3 Example Problems                                                                 72
Another choice for a CV may be as follows.

Consider the LHS of the momentum equation now.

ρgdV −
−           ˆ
P ndA + Fv + R = ...
V
−             S

• The body force term is zero as before.

• The pressure is now zero everywhere on the control surface.

• The viscous force term is zero as before.

• Now the control surface cuts through the cart. Therefore, the reaction force is NOT
zero. If we consider a free–body–diagram of the cart, we realise that R is opposite to
the force which the water exerts on the cart.

Moral of the Story!!
Either choice for a CV is ﬁne. It only changes which term on the LHS gives us the information
we want.

We will continue the problem with the second option.

1. Stationary Control Volume

73                                                             ME 215.3 Example Problems
Attach the coordinate system AND the control volume to the cart.

z
x

• This is considered a stationary control volume because it is not moving with respect
to the coordinate system.
• The coordinate system is inertial because it is moving at constant velocity.

We have already simpliﬁed the LHS so now,
d
R=            ρV dV +
−                  ˆ
ρV (Vr · n)dA
dt
Since nothing in the CV is changing with time, it is a steady problem. Also, assume uniform
ﬂow.
˙          ˙
R = (mV )out − (mV )in
˙      ˙
or, because mout = min
˙
R = m(Vout − Vin )
˙
m = ρ(Vj − Vc )Aj
ı
Vin = (Vj − Vc )ˆ
Assuming that the vane is frictionless, the magnitude of this velocity will not change at the
outlet.
√      √
2      2ˆ
Vout = (Vj − Vc )     ı
ˆ+    k
2       2

ME 215.3 Example Problems                                                                 74
√
2
R = ρ(Vj − Vc )Aj (Vj − Vc )                      ı ˆ                ı
(ˆ + k) − (Vj − Vc )ˆ
2
√         √
2          2ˆ
R = ρ(Vj − Vc )2 Aj                 ı
− 1 ˆ+    k
2          2
√                   √
kg             m         2
2       2          2ˆ
R = 998             10               0.01[m ]              ı
− 1 ˆ+    k
m3             s                            2         2

ı      ˆ
R = (−292ˆ + 705k) N
Remember, this is opposite to the force the water exerts on the cart.

ı      ˆ
The water exerts a force of (292ˆ − 705k) N on the cart. The coordinate system is
shown on the diagram.

2. Moving Control Volume

Now, attach the coordinate system to the pipe but attach the control volume to the cart.
The ﬂow inside the control volume is still steady. Therefore,

˙
R = m Vout − Vin

is still true and

˙
m = ρ(Vj − Vc )Aj

is still true. However, Vout and Vin are diﬀerent. Now,

ı
Vin = Vjˆ

Vout still has a magnitude of (Vj − Vc ) when viewed with respect to the vane. However, in
our coordinate system, the forward velocity of the cart must be added to this.

Vcˆ
i
kˆ)
i+
2( ˆ
2/
√

Vout
c)
V
−
j
(V

√
2
Vout = (Vj − Vc )       ı ˆ        ı
(ˆ + k) + Vcˆ
2

75                                                                             ME 215.3 Example Problems
Now,
√
2
R = ρ(Vj − Vc )Aj     (Vj − Vc )       ı ˆ        ı     ı
(ˆ + k) + Vcˆ − Vjˆ
2
√
2
R = ρ(Vj − Vc )Aj     (Vj − Vc )       ı ˆ                ı
(ˆ + k) − (Vj − Vc )ˆ
2

This intermediate result is identical to the stationary CV. Therefore, the answer will be
identical.

3. Stationary Control Volume

Try again with a diﬀerent stationary control volume.

Attach the coordinate system to the supply pipe.

Draw a stationary CV such that the cart is in it at a certain instant in time.

How is the analysis diﬀerent now?

Again, the LHS is the same.
d
R=                       ˙          ˙
ρV dV − + (mV )out − (mV )in
−
dt
V
−

The unsteady term is NOT ZERO anymore. The uniform ﬂow assumption is still valid.

Consider a sketch showing the CV at two times.

The total momentum (         ρV dV ) in (ii) is greater than in (i) because the column of liquid
−
ı
going at Vjˆ is longer. This column is getting longer at speed Vc . Therefore,
d
ı
ρV dV = ρVjˆAj Vc
−
dt
Now, at the inlet

˙                   ı
(mV )in = (ρVj Aj )Vjˆ

˙     ˙
However, note that min = mout for this analysis because mass is accumulating in the CV.

˙      ˙
mout = min − ρAj Vc

˙
mout = ρVj Aj − ρVc Aj = ρAj (Vj − Vc )
Back to the momentum equation.
√
2
ı
R = ρVj Vc Ajˆ + ρAj (Vj − Vc )                        ı ˆ
(Vj − Vc )(ˆ + k) + Vcˆ − ρVj2 Ajˆ
ı          ı
2

ME 215.3 Example Problems                                                                    76
z
(i)
x

z

(ii)             x

√
2
R = ρ(Vj − Vc )Aj (Vj − Vc )       ı ˆ                ı
(ˆ + k) − (Vj − Vc )ˆ
2
Again, this is the same intermediate result so the answer will be the same.

77                                                               ME 215.3 Example Problems
Solution: Problem # 37

Given:

• A nozzle which discharges water from a pipe

10 kPa

z
D2 = 25 mm
x
1             2

D1 = 50 mm
V1 = 1 m/s

Find:

• The force on the bolts at the ﬂange connection

Assumptions:

• neglect body forces
• no viscous force on CV
• uniform ﬂow

Analysis:

Since we are asked to ﬁnd a force, consider the linear momentum equation.

Choose a control volume that will help us ﬁnd what we are after. Therefore, cut the bolts
with the control surface.

A coordinate system is also shown on the diagram.

Consider the linear momentum equation.
d
ρgdV −
−               ˆ
P ndA + Fv + R =            (ρV )dV +
−                  ˆ
ρV (Vr · n)dA
dt
V
−            S                               V
−               S

ME 215.3 Example Problems                                                                78
Consider each term in this equation.

1. Neglect the body force. We have no information about weight of the nozzle.
2. Use gauge pressures so that the only non–zero pressure is at 1.
P1 = 10 kPa
ˆ    ı
n = −ˆ
ˆ
Since P1 is uniform and n doesn’t vary,

−         ˆ           ı           ı
P ndA = −P1 (−ˆ)A1 = P1 A1ˆ

3. Neglect the viscous force term. There is no ﬂow parallel to the CS.
4. Since our CS cuts the bolts, the reaction force will give us what we want to ﬁnd.
5. The unsteady term on the right-hand-side is zero because this is a steady problem.
6. Assume uniform ﬂow to simplify the last term in the equation.

So, the momentum equation becomes
ı       ˙                ˙
P1 A1ˆ + R = m(Vout − Vin ) = m(V2 − V1 )

ı
V1 = 1ˆ m/s
Find the magnitude of V2 from conservation of mass
d
(ρ)dV +
−                ˆ
ρ(V · n)dA = 0
dt
V
−             S

For steady, uniform ﬂow this becomes
ρV1 A1 = ρV2 A2
2
A1        50
V2 = V1      = V1
A2        25
ı
V2 = 4ˆ m/s
π                     kg   m π                        m
R = −10 × 103 [Pa] (0.05)2 [m2 ]ˆ + 998
ı        3
1     (0.05)2 [m2 ](4ˆ − 1ˆ)
ı    ı
4                     m    s 4                        s
ı
R = −13.8ˆ N
ı
This means the bolts are exerting a force ON the CV in the −ˆ direction. Therefore, the
bolts are in tension.

There is a tensile force on the bolts of 13.8 N.

79                                                              ME 215.3 Example Problems
Solution: Problem # 38

Given:

• The inlet section of a laminar pipe ﬂow

Uo                                u(r) = Umax (1 − (r/R)2)

r
1                          z          2

circular pipe

Find:

• Frictional drag on the ﬂuid between 1 and 2 in terms of P1 , P2 , ρ, Uo , and R

Assumptions:

• neglect body forces
• uniform ﬂow at 1
• ρ is constant

Analysis:

Since we are asked for a force (drag) on the ﬂuid we should consider the linear momentum
equation.

A coordinate system has already been given in the problem.
d
ρgdV −
−                ˆ
P ndA + Fv + R =            (ρV )dV +
−                  ˆ
ρV (Vr · n)dA
dt
V
−                  S                           V
−               S

The dashed line on the diagram is chosen as the CV. Consider each term of the momentum
equation.

ME 215.3 Example Problems                                                                      80
1. Since no information has been give about the orientation of the pipe and we are only
interested in frictional drag, ignore the body force term.

2. The only signiﬁcant pressure forces on our CS are at 1 and 2. Although pressure is
certainly acting on the rest of the CS, it is equal around the circumference and therefore
yields no net force.

3. The viscous force term is the unknown in this problem. We have chosen a CV including
just the water on the pipe so that the friction at the pipe walls enters the momentum
equation through this term.

4. R = 0 since no solid member penetrates the CS.

5. Since this is apparently a steady ﬂow, assume the unsteady term is zero.

6. The ﬁnal term in the equation must be handled carefully. Although we do have uniform
ﬂow at the inlet, we certainly don’t at the outlet. Therefore, we can apply the uniform
ﬂow assumption at 1, but not at 2.

ˆ
(P1 − P2 )πR2 k + Fv =                               ˆ       ˙
ρV (Vr · n)dA − (mV )1
2

At 2,

ˆ
(Vr · n) = u(r)
r    2
ˆ
V2 = Umax 1 −                       k
R
dA = rdθdr
R           2π                                  2
2ˆ                                              r      2
ˆ                   ˆ
Fv = (P2 − P1 )πR k +                                    ρ Umax     1−                      krdrdθ − Uo πR2 ρUo k
0           0                      R
R
ˆ       2 ˆ                                          2r 2  r4                     ˆ
Fv = (P2 − P1 )πR2 k + 2πρUmax k                                   1−        + 4              2
rdr − Uo πR2 ρk
0            R4    R
R
2ˆ            2 ˆ                   r2   2r 4   r6                        ˆ
Fv = (P2 − P1 )πR k +            2πρUmax k                    −      +                    2
− Uo πR2 ρk
2    4R4 6R4            0
2
ρUmax                2 ˆ
Fv = πR2 (P2 − P1 ) +                                − ρUo k
3
Eliminate Umax with conservation of mass (steady).
d
(ρ)dV +
−                    ˆ
ρ(V · n)dA = 0
dt
V
−                 S

r       2
0=           ρUmax 1 −                          rdrdθ − Uo ρπR2
2                      R

81                                                                                                  ME 215.3 Example Problems
R
r3
0 = ρUmax 2π            r−         dr − Uo ρπR2
0            R2
R
r2    r4
0 = ρUmax 2π       −               − Uo ρπR2
2    4R2       0
2
R
0 = ρUmax 2π     − Uo ρπR2
4
Umax = 2Uo
ρ(2Uo )2     2 ˆ
Fv = πR2 (P2 − P1 ) +                   − ρUo k
3

The frictional drag on the ﬂuid is

2
ρUo
πR2 (P2 − P1 ) +
3

ME 215.3 Example Problems                              82
Solution: Problem # 39

Given:

• A nozzle mounted at a pipe exit

?
z

x     D1 = 50 mm                    1                2
V1 = 1 m/s
D2 = 25 mm

Find:

• Force on the bolts assuming frictionless ﬂow

Assumptions:

• frictionless ﬂow

• uniform inlet and outlet

• no shaft work or heat transfer

• no elevation change

Analysis:

The control volume and coordinate system are shown above. Since we are asked for a force,
consider the momentum equation.
d
ρgdV −
−              ˆ
P ndA + Fv + R =             (ρV )dV +
−                  ˆ
ρV (Vr · n)dA
dt
V
−               S                             V
−               S

With the assumptions listed this becomes (see solution on page 79 for more details)

A1
R = ρV1 A1       V1      ı     ı        ı
ˆ − V1ˆ − P1 A1ˆ
A2

83                                                                    ME 215.3 Example Problems
Bernoulli’s equation is

ρV12              ρV 2
P1 +        + ρgz1 = P2 + 2 + ρgz2
2                 2
Since z1 = z2

ρ 2      2    998 kg   2   2  m2
P1 = (V2 − V1 ) =        (4 − 1 ) 2
2              2 m3           s

P1 = 7485 Pa

2
kg     m π                   m   50              m
R=         1000   3
1        (0.05)2 [m2 ] 1              ˆ− 1
ı        ı
ˆ
m      s 4                   s   25              s
π
−7485[Pa] (0.05)2[m2 ] = −8.82ˆ N
ı
4

The tensile force in the bolts is 8.82 N.

ME 215.3 Example Problems                                                   84
Solution: Problem # 40

Given:

• A nozzle mounted at the exit of a pipe
10 kPa

z

x    D1 = 50 mm                     1             2
V1 = 1 m/s
D2 = 25 mm

Find:

• The loss coeﬃcient for the nozzle

Assumptions:

• no shaft work, no heat transfer
• no elevation change

Analysis:

Bernoulli’s equation
ρV12                 ρV 2
P1 +     + ρgz1 = P2 + 2 + ρgz2 + ∆PL
2                    2
ρ
∆PL = P1 +     V 2 − V22
2 1
998 kg      m2     m2
∆PL = 100000[Pa] +            12 2 − 42 2
2 m3       s      s
∆PL = 2515 Pa
ρV 2
∆Pv = Km 1
2
Km = 5.04
The loss coeﬃcient of this nozzle based on the inlet velocity is 5.04.

85                                                              ME 215.3 Example Problems
Solution: Problem # 41

Given:

• A transition piece as shown

D2 = 80 mm
20 kPa

y                                                          2           45o
.

x       4 kg/s
1
Km = 0.50 (based on
D1 = 50 mm                                      discharge velocity)

Find:

• Force required to hold transition in place

Assumptions:

• neglect body forces

• uniform inlet and outlet

• no shaft work or heat transfer

Analysis:

Since we are asked to ﬁnd a force, consider the linear momentum equation.

Begin with the complete equation
d
ρgdV −
−            ˆ
P ndA + Fv + R =            (ρV )dV +
−                  ˆ
ρV (Vr · n)dA
dt
V
−            S                            V
−               S

• Since we have no information about the mass of the transition, or even the direction
of g, we will neglect body forces.

ME 215.3 Example Problems                                                                       86
• Choose a CV which isolates the transition piece. Therefore, R becomes the unknown
in the problem.

• A uniform pressure exists at 1 and 2.

• The ﬂow is steady so
d
ρV dV = 0
−
dt

• Use the uniform ﬂow assumption for the last term since we have no information about
how the velocity varies across the pipe.

Now, we have

−         ˆ         ˙
P ndA + R = m(V2 − V1 )

˙
m        4[kg/s]          4
V1 =     ı
ˆ=            3 ] π(0.05)2 [m2 ]
ı
= 2.04ˆ m/s
ρA1     998[kg/m
√                 √
˙
m      2                 2 4[kg/s]            4
V2 =              ı ˆ
(ˆ + ) =                 3 ] π(0.08)2 [m2 ]
ı ˆ             ı ˆ
(ˆ + ) = 0.5638(ˆ + ) m/s
ρA2    2                 2 998[kg/m
Now, let’s work on the pressure force term. At 1,

P1 = 20 kPa

ˆ    ı
n = −ˆ
At 2
√
2
ˆ
n=             ı ˆ
(ˆ + )
2

P2 =?
We must ﬁnd P2 using Bernoulli’s equation. Write Bernoulli’s equation (with pressure losses)
from 1 to 2.
ρV12       ρV 2
P1 +        = P2 + 2 + ∆PL
2          2
ρV22
∆PL = Km
2
ρ                 ρV 2
P2 = P1 + (V12 − V22 ) − Km 2
2                  2

87                                                                   ME 215.3 Example Problems
998 kg                                m2
P2 = 20, 000[Pa] +              (2.04)2 − 2(0.5638)2
2 m3                                 s2
998 kg               m2
− 0.5          2(0.5638)2 2
2 m3                 s

P2 = 21.6 kPa
√
2
ı
R = −P1 A1ˆ + P2 A2      ı ˆ      ˙
(ˆ + ) + m(V2 − V1 )
2

√
π      2   2              π      2  2    2
ı
R = − 20, 000[Pa] (0.05) [m ]ˆ + 21, 600[Pa] (0.08) [m ]      ı ˆ
(ˆ + )
4                         4            2
kg                               m
+ 4              ı                ı
(0.5638ˆ + 0.5638ˆ − 2.04ˆ)
s                               s

ı       
R = (31.6ˆ + 79.0ˆ) N

ı      
The force required to hold the transition in place is (31.6ˆ+ 79.0ˆ) N in the coordinate
system shown.

ME 215.3 Example Problems                                                                   88
Solution: Problem # 64

Given:

• A piping system as shown below

2

open, ﬂanged, globe valve

90o , regular, ﬂanged elbows
30 m
g

1

water, 20o C            50 m, 2” I.D. steel pipe
Q = 0.04 m3 /s

Find:

• P1 − P2

Assumptions:

• no shaft work or heat transfer

• uniform ﬂow at 1 and 2

• no diameter change

Analysis:

Write Bernoulli’s equation with losses from 1 to 2.

ρV12              ρV 2          L ρV 2             ρV 2
P1 +        + ρgz1 = P2 + 2 + ρgz2 + f        +      Km
2                 2            D 2                 2

89                                                              ME 215.3 Example Problems
Since there is no diameter change, V1 = V2 . Also, let z1 = 0.

L ρV 2              ρV 2
P1 − P2 = ρgz2 + f          +       Km
D 2                  2
ρV 2       L
P1 − P2 = ρgz2 +          f     +    Km
2         D
Q        0.04[m3 /s]
V =     =                      = 19.74 m/s
A   (2(0.0254))2π/4[m2 ]
We need to ﬁnd f . It is a function of roughness (ǫ) and Reynolds number (Re). For steel
pipe, ǫ = 0.046 mm (Table 6.1). Therefore,
ǫ    0.046
=         = 0.000906
D   2(25.4)

ρV D   VD   19.74[m/s](2(0.0254)[m])
Re =        =    =             −6 [m2 /s]
= 1 × 106
µ      ν        1 × 10
From the Colebrook equation with this (ǫ/D) and Re we get f = 0.0195. Now ﬁnd the
minor loss coeﬃcients from table 6–5.

Kvalve = 8.5

Kelbow = 0.39

kg      m         998 kg                           m   2
P1 − P2 = 998     3
9.81 2 30[m] +                          19.74
m       s          2 m3                            s
50
0.0195             + 8.5 + 2(0.39)
2(0.0254)

P1 − P2 = 5.83 MPa
The pressure drop between 1 and 2 is 5.83 MPa at this ﬂowrate.

ME 215.3 Example Problems                                                                 90
Solution: Problem # 65

Given:

• A pipe connecting two reservoirs as shown

1
ρ = 1000 kg/m3
µ = 0.001 kg/(m · s)
20 m
2

b
a

D = 2 cm                      D = 4 cm
L = 5m                        L = 5m
ǫ = 0.05 mm

Find:

• Volume ﬂowrate between the reservoirs

Assumptions:

• V1 = V2 = 0
• P1 = P2
• no heat transfer or shaft work

Analysis:

Write Bernoulli’s equation (with losses) from 1 to 2.
ρV12              ρV 2          L ρV 2                            ρV 2
P1 +        + ρgz1 = P2 + 2 + ρgz2 + f        +                     Km
2                 2            D 2                                2
L       ρVa2            L       ρVb2          ρV 2
ρgz1 = fa                  + fb                    +     Km
D   a    2              D   b    2             2

91                                                                          ME 215.3 Example Problems
Minor losses:
Km         source
entrance        0.5        Figure 6.21
expansion       0.56       Figure 6.22
exit            1.0        Figure 6.21

L        ρVa2          L       ρVb2 ρVa2                   ρV 2
ρgz1 = fa                        + fb                 +     (Kent. + Kexp. ) + b (Kexit )
D    a    2            D   b    2    2                      2

From conservation of mass,
2
Aa                   Da
Vb = Va    = Va
Ab                   Db
4                                  4
ρVa2            L             L        Da                                 Da
ρgz1 =                  f           + f                     + Kent. + Kexp. + Kexit
2              D   a         D   b    Db                                 Db
ǫ             0.05[mm]
=            = 0.0025
D     a        20[mm]
ǫ             0.05[mm]
=            = 0.00125
D     b        40[mm]
125
392.4 = Va2 250fa +                   fb + 1.1225
16
Initial guess, fully rough zone fa = 0.025, fb = 0.021.

Procedure:

• calculate Va from the previous equation

• use Va to get Rea and Reb

• use Rea and Reb to get new fa and fb

• repeat if necessary

Va   Rea     fa    Reb    fb
7.22 144,300 0.026 72,200 0.024
7.09 141,770 0.026 70,900 0.024
This result is converged because the f ’s have stopped changing.

The volume ﬂowrate between the reservoirs is 0.00223 m3/s.

ME 215.3 Example Problems                                                                           92
Solution: Problem # 69

Given:

• The head versus ﬂowrate characteristics of a pump

• Water is pumped through 100 m of 20 cm diameter cast iron pipe

• Water temperature is 20o C

100

80

60                                  Parabola

40

20

0
0   0.5    1             1.5              2      2.5
3
Flowrate, m /s

Find:

• Flowrate

Assumptions:

• no elevation change

• no minor losses

• pump inlet the same diameter as exit pipe

• pump inlet pressure same as pipe exit pressure

• no heat transfer

93                                                                    ME 215.3 Example Problems
Analysis:

Writing Bernoulli’s equation from the inlet of the pump (1) to the exit of the pipe (2) gives,

ρV12              ρV 2
P1 +        + ρgz1 = P2 + 2 + ρgz2 + ∆Pf + ∆Pm − ∆Pp
2                 2

With the assumptions stated above this reduces to

∆Pf = ∆Pp

or

hf = hp

The pump curve can be described by,

hp = 80 − 20Q2

where Q is in m3 /s and hp is in m. The frictional head loss is

L Q2
hf = f          .
D 2gA2
2
2             100[m]        Q2              4
(80 − 20Q )[m] = f
0.2[m]   2(9.81[m/s2 ])   π(0.2[m])2
80 = 25, 821f Q2 + 20Q2
80
Q=
25, 821f + 20
For cast iron pipe, ǫ = 0.26 mm so ǫ/D = 0.0013. The fully rough friction factor for this
ǫ/D is f = 0.0210. Therefore,

Q = 0.3772 m3 /s

Check the Reynolds number.
ρV D   VD   QD
Re =        =    =    = 2.4 × 106
µ      ν   Aν
This is fully rough so we can accept the answer.

The ﬂow rate is 0.377 m3 /s.

ME 215.3 Example Problems                                                                  94
Solution: Problem # 81

Given:

• Flow around a bridge pier

• U∞ = 5 m/s

• P∞ = 50 kPa

• stagnation point is 1 m upstream of source

y

x

2m
A

1m

Find:

• Pressure at point A

Assumptions:

• two-dimensional ﬂow

• inviscid, irrotational ﬂow

95                                                           ME 215.3 Example Problems
Analysis:

Must ﬁnd the source strength required to give the stagnation point 1 m upstream of the
source. Place the source at the origin.

Therefore, for the source
m
vr =
r
At the stagnation point, u = 0. Therefore,
m
0 = U∞ −
r
m    m
0=5       −
s   1[m]
Therefore, m = 5 m2 /s. Now we need the velocity at point A.

ψ = ψfreestream + ψsource

ψ = U∞ r sin θ + mθ
Therefore,
−∂ψ
vθ =       = −U∞ sin θ
∂r
1 ∂ψ              m
vr =    = U∞ cos θ +
r ∂θ               r
We need the (r, θ) coordinates of point A to ﬁnd these velocities. At the stagnation point
θ = π, r = 1 m/s. Therefore,

ψsp = 0 + mπ = mπ

Since A is on the same streamline

ψA = mπ

In (x, y) coordinates.
y
ψ = U∞ y + m tan−1
x
at A,

2
5[m2 /s]π = 5[m/s]2[m] + 5[m2 /s] tan−1
x
xA = 0.9153

ME 215.3 Example Problems                                                              96
rA =    (0.9153)2 + (2)2 = 2.200 m
2
θA = tan−1            = 1.142 rad
0.9152
Therefore,
2    2    2
VA = vθ + vr = (−5 sin(1.142))2 + (5 cos(1.142) + 5/2.200)2
2
VA = 39.61 m2 /s2
1   2    2                   1000 kg              m
PA = P∞ + ρ(V∞ − VA ) = 50, 000[Pa] +       3
(52 − 39.61)
2                              2  m               s
PA = 42.7 kPa
The pressure at A is 42.7 kPa.

97                                                           ME 215.3 Example Problems
Solution: Problem # 82

Given:

• A pair of doublets as shown

• U∞ = 10 m/s

2m            2m

doublets
2m                      y

x
30o

U∞ = 10 m/s

Find:

• Doublet strength

• Pressure at the origin

Assumptions:

• two-dimensional ﬂow

• inviscid, irrotational ﬂow

ME 215.3 Example Problems                                             98
Analysis:

ψ = ψfs + ψd1 + ψd2
λ sin θ1 λ sin θ2
ψ = U∞ (y cos α − x sin α) −             −
r1       r2
λ(y − 2)              λ(y − 2)
ψ = U∞ (y cos α − x sin α) −            2 + (x − 2)2
−
(y − 2)               (y − 2)2 + (x + 2)2
ψ0,0 = ψ4,1
λ(−2)          λ(−2)
ψ0,0 = 0 −        2 + (−2)2
−
(−2)            (−2)2 + (2)2
2λ 2λ    λ
ψ0,0 =      +   =
8   8   2
λ(−1)          λ(−1)
ψ4,1 = 10(cos 30o − 4 sin 30o ) −       2 + (2)2
−
(−1)           (−1)2 + (6)2
λ   λ   λ
ψ4,1 = −11.34 +      +   =
5 37    2
1 1  1
λ    − −            = −11.34
2 5 37
λ = −41.54 m3 /s

The doublet strength must be −41.54 m3 /s.

By symmetry, only the freestream contributes to the velocity at the origin. Therefore, the
pressure there is the same as the pressure far away from the doublets.

The pressure at the origin is P∞ .

99                                                                ME 215.3 Example Problems
Solution: Problem # 83

Given:

• A landing gear strut

• U∞ = 45 m/s

• ρ = 1.2 kg/m3

B

A

source           sink
U∞
12 cm             12 cm

30 cm

Find:

• PA − PB

Assumptions:

• two-dimensional ﬂow

• inviscid, irrotational ﬂow

ME 215.3 Example Problems                                        100
Analysis:

A Rankine oval can be modelled by superimposing a source and a sink with the freestream.

ψ = ψfs + ψsource + ψsink

ψ = U∞ r sin θ + mθsource − mθsink
Find m.

At the stagnation point, u = 0.
m             m
0 = U∞ −                +
(0.15 − 0.12) (0.15 + 0.12)

m = 1.519 m2 /s
Need VB .

At B, v = 0. Therefore, we only need u at B.

y                          y
ψ = U∞ y + m tan−1                  − m tan−1
x+a                        x−a

∂ψ           m                  1                 m          1
u=      = U∞ +                  2             −              2
∂y             y
1 + x+a              x+a           1+    y        x−a
x−a

PROBLEM!!!!! Don’t know y at B.

ψB = ψsp = 0

y                          y
0 = 45y + 1.519 tan−1            − 1.519 tan−1
0.12                       −0.12
Note that
y                           y
tan−1             = π − tan−1
−0.12                       0.12
y
0 = 45y + 2(1.519) tan−1        − π(1.519)
0.12
y
45y = 1.519π − 2(1.519) tan−1
0.12
y = 0.07027 m
Now,

1           1
uB = 45 + 1.519                                   2
1+   ( 0.07027 )2
0.12
0.12

101                                                                      ME 215.3 Example Problems
uB = 63.85 m/s
1 2           1 2
PA + ρUA = PB = ρUB
2             2
1 2      1.2
PA − PB = ρUB =        (63.85)2
2         2
PA − PB = 2.45 kPa
The pressure diﬀerence between points A and B is 2.45 kPa.

ME 215.3 Example Problems                                     102
Solution: Problem # 84

Given:

• Flow over a circular cylinder creating no lift

y
g
A
U∞

L                                            x
a

Find:

• Derive an expression for the velocity along the positive y axis in terms of U∞ , a, and
y.

• Sketch this function.

• Consider the streamline labelled A. Far upstream of the cylinder this streamline is a
distance L from the horizontal plane of symmetry. At what distance from the origin
does the streamline cross the y axis?

Assumptions:

• two-dimensional ﬂow

• inviscid, irrotational ﬂow

Analysis:

Non-lifting ﬂow over a cylinder can be modelled by a doublet and a freestream.

ψ = ψfs + ψdoublet

103                                                           ME 215.3 Example Problems
λ sin θ
ψ = U∞ r sin θ −
r
λ
ψ = sin θ(U∞ r − )
r

At the stagnation point, θ = π and r = a. Therefore,

ψsp = 0

Therefore, ψ = 0 on the circle. Therefore,
λ
U∞ a −      =0
a
λ = U∞ a2
a2
ψ = U∞ sin θ(r −     )
r
Along the y axis, vr = 0
∂ψ                a2
vθ = −    = −U∞ sin θ 1 + 2
∂r                r
π
θ=
2
a2
vθπ/2 = −U∞ 1 +
r2
Along the positive y axis, u = −vθ . Therefore
u    a2
=1+ 2
U∞    r

The velocity along the positive y axis is given by

u      a2
= 1 + 2.
U∞      r

ψA − ψsp = U∞ L

ψA = U∞ L
π             a2
U∞ L = U∞ sin               r−
2             r
a2
L=r−
r
2
r − Lr − a2 = 0

ME 215.3 Example Problems                              104
u/U∞

2

1

1                                                   r/a

L        L
r=     ±        + a2
2        2
Since r > a, we want the positive root.

L        L
r=     +        + a2
2        2

The streamline labelled A crosses the y axis at a distance of

L         L
+         + a2
2         2

from the origin.

105                                                              ME 215.3 Example Problems
Solution: Problem # 85

Given:

• Flow around a circular cylinder

• U∞ = 20 m/s

• ρ = 1.2 kg/m3

• Volume ﬂowrate per unit length between streamlines A and B is 1.19 m2 /s.

• Streamline B passes through point (x, y) = (0, 0.12m)

0.5

0.4

0.3

0.2
(x, y) = (0, 0.12 m)

0.1
y
0
x

−0.1
B
−0.2        A

−0.3

−0.4

−0.5
−0.5       −0.4   −0.3   −0.2   −0.1   0           0.1    0.2     0.3      0.4   0.5

Find:

• Lift force per unit length

ME 215.3 Example Problems                                                                       106
Assumptions:

• two-dimensional ﬂow

• inviscid, irrotational ﬂow

Analysis:

This ﬂow can be modelled by combining a freestream, a doublet, and a vortex.

ψ = ψfs + ψdoublet + ψvortex
λ sin θ        r
ψ = U∞ r sin θ −           − K ln
r           a
For ψ = 0 on r = a,
λ sin θ
0 = U∞ a sin θ −
a
λ = U∞ a2
ψB − ψA = Q = 1.19 m2 /s
Therefore,

ψB = 1.19 m2/s

Now, at (x, y) = (0, 0.12 m)

m2      m             20[m/s](0.1)2 [m2 ]          0.12[m]
1.19      = 20   (0.12[m]) −                     − K ln
s      s                 0.12[m]                   0.1[m]

K = −2.505 m2 /s
Γ = 2πK = −15.74 m2 /s
Kutta-Joukowski Theorem

FL = −ρU∞ Γ

kg    m                    m2
FL = −1.2     3
20             −15.74
m     s                     s
FL = 378 N/m

The lift force per unit length is 378 N/m.

107                                                        ME 215.3 Example Problems

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