fluid mechanics tuts and answers

ME 215.3

Fluid Mechanics I

Example Problems

c© James D. Bugg

January 2009

Department of Mechanical Engineering

University of Saskatchewan

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

6cm

4cm

15cm

Air

Air

Water

Oil

Open to atmosphere

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

3m

3m

ρ = 1000 kg/m3

hinge

gate (5 m wide)

stop

Patm = 100 kPa

1 ME 215.3 Example Problems

3. A Tainter gate is constructed from a quarter-cylinder and is used to hold back a pool ofwater. The radius of the gate is 1.22 m and it is 2.44 m long. Calculate the hydrostaticforce on the gate. (Solution: page 53)

R=

1.22

m

gate

ρ = 1000 kg/m3

length=2.44 m

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

Patm = 100 kPa

ρ = 1000 kg/m3

hinge

gate (5 m wide)

stop

3m

3m

ME 215.3 Example Problems 2

5. A circular hatch on a submarine is hinged as shown. The radius of the hatch is 35 cmabove the centre of the hatch. Determine how hard a sailor has to push on the centreof the hatch to open it. The seawater density is 1025 kg/m3 and the centre of the hatchis 2 m below the surface of the ocean. The air pressure inside the submarine is equalto atmospheric pressure.

hinge

opening

hatchhatch

70 cm

A

A

View A-A

~g

submarineoutside of

submarineinside of

force

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

H

gate

45o

2m

hinge

~g


7. A pool of fluid has a density that varies linearly from 1000 kg/m3 at the surface to1600 kg/m3 at a depth of 4 m. A 2 m by 2 m square gate is hinged along its bottomedge and held in place by a force F at its top edge. Find the force F .

F

2 m

2 m

~g

hinge

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

hinge

wood0.6 m

water

1.2 m

1.2 m


9. A cylindrical gate with a radius of curvature of√

2m holds back a pool of water witha layer of oil floating on the surface. The gate is 8 m long. What force does the pinnedconnection at A exert on the gate?

1 m~gA

√ 2m

Water (ρ = 1000 kg/m3)

Oil (s.g. = 0.8)

1 m

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

C

A

B

y = x2Parabolic gate

0.5 m

2 m

1.5 m

~g

ρ = 1000 kg/m3

watery

x

Cable

Hinge


11. A partially submerged pipe rests against a frictionless wall at B as shown. The specificgravity of the pipe material is 2.0 and the pipe has an outside diameter of 2 m. Thefluid is water with ρ = 1000 kg/m3. Find the inside diameter of the pipe and the forcewhich the wall exerts on the pipe. The ends of the pipe are closed.

~g

s.g. = 2.0

0.5 m B

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

b

Water

hinge

1 m

30o


13. A vertical, plane wall holds back a pool of water (ρ = 1000 kg/m3) which is 1.5 mdeep. The wall has a triangular gate in it that is hinged along the bottom edge andheld closed by a horizontal force F applied at the top corner. Calculate the force Frequired to hold it closed.

2m

1 m

~g

gategate

View A-A

A

A

1.5

m

hinge

F

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

gate c. of g.

~g1.8 m

45o

45o

2 m

hinge

2 m


15. Determine H when the L-shaped gate shown is just about to open. Neglect the weightof the gate and let the density of the fluid be ρ.

hingeL

H

gate

~g

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

hinge

3 m

F

9 m


17. A semi-circular gate is hinged at the bottom as shown. The density of the fluid varieslinearly from 1000 kg/m3 at the surface of the reservoir to 1500 kg/m3 at the bottomof the reservoir. Find the force F required to hold the gate in place?

1m

A

A

~g

hinge

Section A-A

gate

hinge

F

3m

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 wereincreased beyond 1.5 m the cylinder would roll over the stop. Assuming that the watercontacts the cylinder surface right up to point A, what is the specific gravity of thecylinder.

Diameter = 2 m

A

0.5 m

1.5 m


19. A gate composed of a quarter circle portion and a straight portion holds back a poolof 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

~g

A

Width of Gate is 8 m

Water (ρ = 1000 kg/m3)

Hinge

1.5 m

0.5 m

R = 1 m


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 joiningthe centre of the cylinder and the point where the cylinder contacts the edge of theslot subtends an angle of α = 30o with the vertical. If h = 150 mm what is the forceexerted on the cylinder by the tank?

h

R

α = 30o

~g


21. Water at 20oC is retained in a pool by a triangular gate which is hinged along its topedge and held in place by a stop at its lowest point. What force does the stop exerton the gate? What force does the hinge exert on the gate?

stop 3.5

m

7m

~g

A

1 m

Section A-A

A

hinge

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 accelerationdue to gravity, and the displacement of the fluid in the tube. (Solution: page 61)


24. An open-top cart half full of water (ρ = 1000 kg/m3) is shown at rest in the figurebelow. The cart begins to accelerate to the right at a constant 3 m/s2. After sometime, the fluid reaches a hydrostatic state. Determine the net hydrostatic force on therear, vertical end of the cart. The cart is 0.8 m wide.

1m

3 m/s2

2 m

~g

0.5

m

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

0.75 m

~g

hole

A

30o

1 m

~a


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

30 cm

50 cm

ω = 50 rpm

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

ω

R

Ho


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

4 cm

~g

12 cm

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

1000 kg/m3

1200 kg/m3

~g

20 cmω

10cm


30. The U-tube shown is rotated about the vertical axis indicated on the diagram at60 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 pointA?

Diameter, 8 mm

5cm

A

~g

3 cm 10 cm

Axis of rotation

Diameter, 5 mm

Rest level

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

60 rpm

cylindrical can

ρ = 1000 kg/m3

vent


32. An upright, 10-cm diameter, cylindrical paint can 20 cm deep spins around its axis ofsymmetry at 500 rpm. It is completely full of mineral oil (sg = 0.87) and there is avery small hole in the lid at the rim of the can. Determine the net hydrostatic forceon the lid.

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

~ghinge1.

5m

F

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 topof the cup. You come to a curve in the highway which is not banked. Determine theminimum radius of the curve for which the coffee will not spill. If you go around acurve 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 forceF1 applied at the top of the gate holds it closed. The pool of water (ρ = 998 kg/m3) is1.75 m deep. Calculate the force F1 required to hold the gate closed.

1.75

m

~ggate

hinge

1.25

m

1m

F1


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 aroundits axis at an angular speed of ω until the water reaches a hydrostatic state. Thespinning is stopped and the water is allowed to come to rest. After coming to rest, thewater is 10 cm deep. Determine ω in revolutions per minute (rpm).

40cm

30 cm

~g

20cm

initial water level

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

Aj = 0.01 m2

V = 15 m/s


38. A jet of water from issues from a 0.01 m2 nozzle at 15 m/s. It impinges on a vanemounted on a movable cart and is deflected through 45o. If the cart is moving away fromthe 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 of25 mm is attached at the end of the pipe. If the pressure just upstream of the nozzleis 10 kPa, what is the force on the bolts at the flange connection attaching the nozzleto the pipe? (Solution: page 79)

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

21 z

u(r) = Umax(1 − (r/R)2)

radius = Rcircular pipe

Uo

r


41. A 50 mm diameter pipe carries water at 1 m/s. A nozzle with an exit diameter of25 mm is attached at the end of the pipe. Assuming frictionless flow, find the force onthe flange connection. (Solution: page 84)

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

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

4 kg/s

Km = 0.50 (based ondischarge velocity)

transition piece

20 kPa

D1 = 50 mm

D2 = 80 mm

45o


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 theplate 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 surfaceof the plate. The dashed line labeled “A” is a streamline in this flow.

A

170 mm

10 m

10 m/s

yflat plate

45. An object is placed in a 1 m diameter wind tunnel and the air velocity downstreamof the object is found to linearly vary from zero at the centreline of the wind tunnelto a maximum at the wind tunnel wall. The air velocity upstream of the object isa uniform 20 m/s. A mercury manometer indicates a 10 mmHg pressure difference asshown. 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 dragforce on the object.

10 mm

~g

1 m diameter20 m/s

mercury (s.g.=13.56)


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

Pump

5 kg/s

~g

Cable

1 cm dia.

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

10 cm

D = 2 mm

20 cm


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

10 cm

guides

~g aluminum block

2.5 cm

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

D1 = 5 cm

∆P

m

Pipe

Flange

~gElbow

D2 = 3 cm


50. A pipe 4 cm in diameter supplies water to a 2 cm diameter nozzle. The gauge pressurejust upstream of the nozzle is 60 kPa and the head loss in the nozzle is given byhL = 0.2V 2

e /2g where Ve is the velocity at the nozzle exit. The jet of water (notshown) 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 tohold the plate in this position.

water

30o

ρ = 1000 kg/m3

60 kPa

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 expressionfor F as a function of Q and indicate the range of Q for which it is valid. Sketch thefunction.

2 m

D = 5 cm 45o

F

~g

Q

0.8 m1 m

2 m


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 thepaper). The upstream air velocity is uniform at 10 m/s. The downstream velocity isuniform at 12 m/s in the lower half of the wind tunnel. The downstream velocity inthe upper half is uniform. The vertical component of velocity is zero at the beginningand end of the test section. The test section is 1 m long. The engineer measures thepressure 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 theaerofoil. Neglect shear on the walls of the wind tunnel.

ofTest section

ofBeginning

1 m

~g

12m/s

x

10 m/s

0.25 m

0.25 m

End

Test section

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

~g


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

m = 4 kg/s

10 cm diameter

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 endof one arm a 12 mm diameter nozzle is oriented perpendicular to the arm and is in thesame horizontal plane as the arm. The other arm has a 1 mm wide slot that also emitswater in the horizontal plane. Find the rotational speed of the sprinkler in revolutionsper minute assuming that the pivot is frictionless.

Section A-A

15 cm15 cm

A

A

Plan View

12 mm dia.

axis of rotation

1 mm

2 cm

(enlarged)


56. Two circular coaxial jets of incompressible liquid with speed V collide as shown. Theinteraction region is open to atmosphere. Liquid leaves the interaction region as aconical sheet. Obtain an expression for the angle θ of the resulting flow in terms of d1

and d2.

d1V

V

V

d2

θV

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 of0.5 m. The empty tank has a mass of 2 kg. At the instant when the water is 0.5 m deepin the tank, what force will the scale read? Carefully explain each of your assumptions.

5 kg/s

~g

0.5 m

0.5 m diameter

0.5 m

2 cm

Scale


58. Calculate the force of the water (ρ = 1000 kg/m3) on the frictionless vane if

(a) the blade is stationary,

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

(c) the blade moves to the left at 20 m/s.

5 cm

60o

40 m/s

dia.

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

vane

1 kg/s

D = 1 cm

splitter

x

z

45o


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

45o

VD

61. Water at 20oC flows through an elbow/nozzle arrangement and exits to atmosphere asshown 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 theelbow and water.

Water

flanged

45o

30 cm

~g

200 kPa

connection


62. A nozzle for a spray system is designed to produce a flat radial sheet of water. Thesheet 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 diameterand the inlet pressure P1 is 50 kPa above atmospheric. Calculate the force exerted bythe spray nozzle on the supply pipe through the flanged connection.

R

35 mmsupply pipe

spray nozzle

P1

diameter

Water

thickness, t

connection

flanged

V2

63. A ride at an amusement park consists of a wheeled cart that zooms down an inclinedplane 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 bya compressed air cylinder aboard the cart. The initial gross mass of the cart and itsoccupants is 500 kg. Air escapes from the braking jet at a constant mass flow rate of20 kg/s and a constant velocity (relative to the nozzle) of 150 m/s. At the instant whenthe braking jet is activated, the speed of the cart is 40 m/s. Determine how much airmust escape in order to stop the cart. What is the stopping distance?


64. Fluid enters a 5 cm diameter pipe with a uniform velocity of Uo. The fluid exits thepipe at two locations. One is a 5 cm diameter exit with a turbulent velocity profiledescribed by

uz = Uc1

(

1 − r

R1

)1/7

.

where uz is the axial component of velocity, Uc1 is the centreline velocity, r is the radialcoordinate, and R1 is the radius of the pipe. The second exit is 1.5 cm in diameter andhas a laminar, parabolic velocity profile described by

uz = Uc2

(

1 −(

r

R2

)2)

.

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

1.5 cm diameter

laminar velocity profile

turbulent velocity profile

5 cm diameteruniform velocity

65. A 6 mm diameter angled nozzle is attached to the end of a 2 cm diameter pipe witha flanged connection as shown below. The angle between the nozzle and the supplypipe 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 inthe flanged connection.

4 cm

200kPa

2 cm diameter

6 mm diameter

flanged connection


66. A two-arm lawn sprinkler, as viewed from above, is shown in the sketch. Water isdelivered to the sprinkler at a volume flow rate of 5 L/min and it can be assumed thatthis flow is split evenly between the two nozzles. The nozzles both have a diameterof 2 mm and are both angled upwards from the horizontal plane at 30o. The densityof the water is 998 kg/m3. Assume the pivot is frictionless. Calculate the rotationalspeed of the sprinkler.

15 cm

45o

15 cm

67. Fluid (ρ = 850 kg/m3) enters a 5-cm diameter pipe with a uniform velocity of 3 m/sat location A. At location B, the fluid has a turbulent velocity profile described by

uz = Uc

(

1 − r

R

)1/5

where uz is the axial component of velocity, Uc is the centreline velocity, r is the radialcoordinate, and R is the radius of the pipe. The pressure at location A is 4 kPa largerthan at location B. Determine the viscous force on the wall of the pipe between locationA and B.

BA

5 cm diameteruniform velocity

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 is30 m. Calculate the pressure difference between the ends of the pipe. (Solution: page90)


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

20 m

ρ = 1000 kg/m3

L = 5 mD = 2 cm

ǫ = 0.05 mm

D = 4 cmL = 5 m

ba

µ = 0.001 kg/(m · s)

70. Calculate the frictional pressure drop in 100 m of 3-cm diameter smooth pipe withwater (ρ = 1000 kg/s, µ = 0.001 Pa · s) flowing at 1 kg/s. What would the frictionalpressure 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 pipereads as shown. The right leg of the manometer is open to atmosphere. The pipe issmooth and its diameter is 5 cm. Determine the volume flowrate through the pipe inL/min.

~g

20 cm

20 cm



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)

Flowrate(m3/s)

Pum

pH

ead

(m)

2.52.01.51.00.50.0

100

80

60

40

20

0

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

25 mm, ǫ = 0.5 mm ~g

20 mm, smooth


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

Stagnation tap

~gQ


5 cm

1 m

Static tap


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

5 m

D = 4 cm

~g

PB

PA


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

Flow Rate (m3/min)

Hea

d(m

)

5.04.54.03.53.02.52.01.51.00.50.0

60

55

50

45

40

35

30

25

20

15

10

5

0


78. An engineer needs to measure the minor loss coefficient of a reducing elbow whichhas already been installed in a piping system. Because of limited access to the pipingsystem, the only available pressure taps are 1 m upstream and 0.75 m downstream ofthe 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 andµ = 0.001 Pa · s.

2 cm diameter

1 m

7.5 cm

0.75 mReducing elbow

not to scale


~g

m = 1 kg/s

4 cm diameter


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 systemshown below. The sum of all minor loss coefficients is 12 and the total length of tubeis 8 m. The tube has an inside diameter of 5 mm and roughness ǫ = 0.025 mm. Whatis the flowrate achieved?

2.0 m Diameter D = 5 mmTotal length L = 8 m

Roughness ǫ = 0.05 mm

atmosphereDischarge to

80. A centrifugal pump draws water (ρ = 1000 kg/m3, µ = 0.001 Pa · s) from a largereservoir and pumps it through 1335 m of 30 cm inside diameter pipe with roughnessǫ = 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 Bare constants. When the pipe exit is at the same elevation as the reservoir surfacethe flowrate is 17.0 m3/min. However, when the exit of the same pipe is raised to anelevation of 50 m the flow reduces to 13.3 m3/min. How high can the exit be raisedbefore the flow will be zero? For all conditions the pipe discharges to atmosphere.Neglect minor losses in this problem.


81. A 6 mm internal diameter, thin-walled, smooth rubber hose 12 m long is used to siphonwater (ρ = 1000 kg/m3, ν = 10−6 m2/s) from a large tank. The outlet of the hose is6 m below the water surface in the tank. What will the volume flowrate be? Neglectminor losses due to bends in the hose.

6 m

6 mm I.D., 12 m long

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

Gate valve

All pipes have ǫ = 0.15 mm

All elbows shown are 90o, regular, flanged.

Pump

L = 340 m

D = 200 mmL = 170 m

D = 100 mm

Sudden

contraction

expansion

Sudden

Fully open


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

New valveD = 2 cm

5 mL = 4 m

84. A centrifugal pump is connected to a piping system as shown below. The pipe is 5 cmin diameter and has a roughness of ǫ = 0.5 mm. The two valves are identical and haveminor loss coefficients of 1.0 when fully open and 20 when 50% open. With both valvesfully 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 freesurface in the tank is at the same elevation as the pipe exit. Let ν = 10−6 m2/s andρ = 998 kg/m3. Determine PA if the first valve is fully open and the second valve isfully closed. Neglect losses which occur on the suction side of the pump.

25 m25 m 25 m 25 m

ValveValve

PumpPA


85. A portion of a piping system is shown below. At point B, the piping system dischargesto the atmosphere. The two elbows each have a minor loss coefficient of 2.4. The valvehas a minor loss coefficient of 6.9. The pipe is 25 mm in diameter and has a roughnessof 0.05 mm. There is a total of 15 m of pipe between point A and point B. Point B is3 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.

B

A

PA

Valve


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 mupstream of the source. (Solution: page 96)

A

y

x

2 m


87. A pair of doublets of equal strength are placed in a freestream as indicated below. Thestreamline 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)

x

U∞

= 10 m/s

doublets

30o

2 m

2 m2 m

y

88. The landing gear strut on a small aircraft has a cross–section as indicated below. Thisshape 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 differencein static pressure between points A and B? (Solution: page 101)

A

U∞

12 cm12 cm

B

sinksource

30 cm


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 streamlineis a distance L from the horizontal plane of symmetry. At what distance from theorigin does the streamline cross the y axis?

(Solution: page 104)

a

U∞

~g

L

A

x

y


90. Consider the flow of air (ρ = 1.2 kg/m3) over a 20 cm diameter circular cylinder. Thefreestream velocity is 20m/s. The volume flowrate per unit length between the twostreamlines labelled A and B is 1.19 m2/s. The streamline labelled B passes throughthe 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 −0.1 0 0.1 0.2 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

B

A

y

x

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

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 strengthlocated 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 they axis. Determine the acceleration 2 cm upstream of the first stagnation point.


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 forma closed streamline in the flow. The streamline shown on the diagram is a distance afrom 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.

a

sinksources

x

2a

y

aa

U∞

= 10 m/s

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

U∞

sourcesink

40 cm

20cm


Solution: Problem # 1

Given:

• Air trapped over oil in a closed tank as shown

~gz1

B

A

Patm = 101.3 kPa

h2

=6

cm

h3

=4

cm

h1

=15

cm

Air

Air

Water

s.g = 0.89Oil

Open to atmosphere

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)

~g = −gk


~a = 0

~∇P = −ρgkdP

dz= −ρg

dP = −ρgdzSince ρ and g are constant,

∆P = −ρg∆z

Now,

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

P1 = −ρoilg(z1 − zB) − ρairg(zB − zA) − ρwg(zA − zatm) + Patm

z1 − zB = h3

zA − zatm = −h1

P1 = −ρoilgh3 + ρwgh1 + Patm

P1 = ρwg(h1 − sgoilh3) + Patm

P1 = 998

[

kg

m3

]

9.81[m

s2

]

(0.15[m] − 0.89(0.04)[m]) + 101, 300[Pa]

P1 = 102.4 kPa(abs)

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



Given:

• A rectangular gate holding a pool of water.

~g

x

z

gate (5 m wide)

hingeρ = 1000 kg/m3

Patm = 100 kPa

3m

3m

stop

Find:

• The force which the water exerts on the gate

Assumptions:


• no acceleration

• g is constant

• ρ is constant


Analysis:

Establish a coordinate system (see diagram).

~Fs = −∫∫

S

P ndA

~∇P = ρ(~g − ~a)

~g = gk

~a = 0

~∇P = ρgk

dP

dz= ρg

dP = ρgdz∫

dP = ρg

∫

dz

P = ρgz + C

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

P = ρgz + Patm

n = ı

~Fs = −∫ z2

z1

∫ w

0

(ρgz + Patm)ıdydz

~Fs = −ı∫ z2

z1

(ρgz + Patm)dz

∫ w

0

dy

~Fs = −ı w[

ρgz2

2+ Patmz

]z2

z1

~Fs = −ı 5[m]

(

1000

[

kg

m3

]

9.81[m

s2

] (62 − 32)

2[m2] + 100, 000[Pa](6 − 3)[m]

)

~Fs = −2.16 ıMN

The water will exert a horizontal force of 2.16 MN on the gate.



Given:

• A Tainter gate holding back a pool of water

R=

1.22

m

~g

z

x ρ = 1000 kg/m3

gatelength=2.44 m

Find:

• The net hydrostatic force on the gate

Assumptions:

• atmosphere is constant pressure


• density of water is constant

• g is constant

• no acceleration

Analysis:

Establish coordinate system (see diagram).

~Fs = −∫∫

S

P ndA

~∇P = ρ(~g − ~a)

~g = −gk~a = 0


dP

dz= −ρg

P = −ρgz + C

at z = R/√

2, P = 0 (gauge).

Therefore, C = ρgR/√

2.

P = ρg(R/√

2 − z)

n = cos θı+ sin θk

θ

n

dA = Rdθdy

~Fs = −∫ w

0

∫ θ2

θ1

ρg(R/√

2 − z)(cos θı+ sin θk)Rdθdy

z = R sin θ

~Fs = −ρgwR2

∫ θ2

θ1

(

cos θı√2

+sin θk√

2− sin θ cos θı− sin2 θk

)

dθ

Recall that

sin2 θ =1 − cos 2θ

2

~Fs = −ρgwR2

[

sin θı√2

− cos θk√2

− sin2 θ

2ı− 1

2

(

θ − sin 2θ

2

)

k

]θ2

θ1

~Fs = −ρgwR2

[

ı

(

sin θ√2

− sin2 θ

2

)

− k

(

cos θ√2

+θ

2− sin 2θ

4

)]π/4

−π/4

~Fs = − 1000

[

kg

m3

]

9.81[m

s2

]

2.44[m]1.222[m2]

(

ı

{(

1

2− 1

4

)

−(

−1

2− 1

4

)}

− k

{(

1

2+π

8− 1

4

)

−(

1

2− π

8+

1

4

)})


~Fs = −35630[N]

(

ı− k

(

π

4− 1

2

))

~Fs =(

−35.6ı+ 10.2k)

kN

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

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



Given:

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

~g

x

z

gate (5 m wide)

hingeρ = 1000 kg/m3

3m

3m

Patm = 100 kPa

stop

Find:

• The force which the stop exerts on the gate

Assumptions:


• 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.


x

~Fs

~Fb

~Ft

z

∑

~Mo = 0

(Hk × ~Fb) + ~Ms = 0

~Ms = −∫∫

S

P (~r × n)dA

~r = zk

n = ı

~r × n = zk × ı = z

~∇P = ρ(~g − ~a)

~a = 0

~g = gk

dP

dz= ρg

P = ρgz + C

at z = 0, P = ρgh0. Therefore, C = ρgh0.

P = ρg(z + h0)

dA = dydz

~Ms = −∫ w

0

∫ z2

z1

(z)ρg(z + h0)dzdy

~Ms = −ρgw∫ z2

z1

(z2 + h0z)dz

~Ms = −ρgw[

z3

3+h0z

2

2

]z2

z1

~Ms = −1000

[

kg

m3

]

9.81[m

s2

]

5[m]

(

33

3[m3] +

3[m]32

2[m2]

)


~Ms = −1104 kN · mRecall that,

(Hk × ~Fb) + ~Ms = 0

k × ~Fb =1104

3[m][kN · m]

~Fb = 368 kNı

The stop exerts a force of 368 kN on the gate.



Given:

• An object suspended in a liquid by a wire

~g

z10 cm cube, m = 35 kg

Find:

• The specific gravity of the fluid

Assumptions:


• no acceleration

• block is in equilibrium

• mass and volume of wire are zero

• ρ is constant

• g is constant

Analysis:


∑

Fz = 0

Fb + T −W = 0


ρgV + T −mg = 0

ρ9.81[m

s2

]

(0.1[m])3 + 335.5[N] − 35[kg]9.81[m

s2

]

= 0

ρ = 800 kg/m3

sg = 800/998

The specific gravity of the liquid is 0.801.



Given:

• A U–tube manometer used to measure linear acceleration

~g

L

z

x

~a

2

1

d

h

Find:

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

Assumptions:

• ~a constant


• ρ and g are constant

Analysis:


Points 1 and 2 are at the same pressure. Therefore,

∫ 2

1

dP =

∫ 2

1

~dR · ~∇P = 0

~dR = dxı + dzk

~∇P = ρ~g − ρ~a


~g = −gk~a = aı

~∇P = −ρgk − ρaı

~dR · ~∇P = −ρadx − ρgdz

0 =

∫ 2

1

(ρadx+ ρgdz) = ρa(x2 − x1) + ρg(z2 − z1)

0 = ρaL+ ρg(−h)a = gh/L

The acceleration of the tube is gh/L.



Given:

• A partially full container rotating around its axis.

1

2

z

r

~g

ω = 50 rpm

50 cm

30 cm

Find:

• The depth of the fluid on the axis

Assumptions:

• ω is constant


• ρ is constant

• g is constant

Analysis:


The pressure at 1 and 2 are equal. Therefore,

∫ 2

1

dP = 0 =

∫ 2

1

~dR · ~∇P

~dR = drr + dzk


~∇P = ρ(~g − ~a)

~g = −gk~a = −ω2rr

~∇P = −ρgk + ρω2rr

0 =

∫ 2

1

(−ρgdz + ρω2rdr)

ρg(z2 − z1) =ρω2

2(r2

2 − r21)

0.3[m] − z1 =(5π/3[1/s])2(0.25[m])2

2(9.81)[m/s2]

z1 = 21.3 cm

The fluid is 21.3 cm deep on the centreline.



Given:

• A cylindrical container rotating about its axis.

~g

r

z

ω

R

Ho

Find:

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

Assumptions:

• ω is constant


• g is constant

• ρ is constant

Analysis:


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

0 = ~dR · ~∇P


~dR = drr + dzk

~a = −ω2rr

~∇P = ρ(ω2rr − gk)

0 = ρω2rdr − ρgdz∫

ω2r

gdr =

∫

dz

z =ω2r2

2g+ C

Equate the initial and final volumes.

∫ R

0

z2πrdr = H0πR2

H0πR2 = 2π

∫ R

0

(

ω2r3

2g+ Cr

)

dr

H0R2 = 2

[

ω2r4

8g+Cr2

2

]R

0

H0R2 =

ω2R4

4g+ CR2

C = H0 −ω2R2

4g

The surface is defined by

z = H0 +ω2

2g

(

r2 − R2

2

)



Given:

• A closed cylindrical container rotating about its axis.

r

60 rpm

z

~gcylindrical can

ρ = 1000 kg/m3

vent

Find:

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

Assumptions:


• ρ is constant

• g is constant

Analysis:

Establish a coordinate system (see diagram).

~Fs = −∫∫

S

P ndA

~∇P = ρ(~g − ~a)

~g = −gk


~a = −ω2rr

~∇P = −ρgk + ρω2rr

Need P as a function of r. Therefore,

∂P

∂r= ρω2r

P =ρω2r2

2+ C

at r = 0, P = 0. Therefore, C = 0.

n = −k

dA = rdrdθ

~Fs = −∫ 2π

0

∫ R

0

ρω2r2

2(−k)rdrdθ

~Fs =2πρω2

2k

∫ R

0

r3dr = πρω2kR4

4

~Fs = π1000

[

kg

m3

](

2π

[

1

s

])20.44[m4]

4k

~Fs = 794kN

The net hydrostatic force on the lid of the can is 794 N upwards.



Given:

• A jet of water hitting a flat barricade.

Aj = 0.01 m2

V = 15 m/s

2

2

1

y

x

~g

Find:

• The force that the water exerts on the barricade.

Assumptions:

• steady state

• jet is horizontal

• jet deflects to vertical plane

• pressure uniform on control surface

• neglect body forces

• neglect fluid shear

• uniform flow


Analysis:

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

∫∫∫

V−

ρ~gdV− −∫∫

S

P ndA+ ~Fv + ~R =d

dt

∫∫∫

V−

(ρ~V )dV− +

∫∫

S

ρ~V (~Vr · n)dA

Consider the control volume shown.

Each term of the momentum equation will now be discussed.

Body Force:

∫∫∫

V−

ρ~gdV− = −Wk

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

Pressure Force:

∫∫

S

P ndA = 0

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 6= 0

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


Unsteady term:

d

dt

∫∫∫

V−

(ρ~V )dV− = 0

Steady state is assumed.

Momentum transport term:

∫∫

S

ρ~V (~Vr · n)dA =∑

d

(m~V )d −∑

i

(m~V )i

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

With these simplifications, the momentum equation becomes

−Wk + ~R = (ρwA2V2)~V2 + (ρwA3V3)~V3 − (ρwA1V1)~V1

Consider only the x direction

Rx = −ρwA1V21

Rx = −998

[

kg

m3

]

0.01[m2](

15[m

s

])2

= −2.25 kN

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

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



Given:

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

45o

Vc = 5 m/s constantAj = 0.01 m2

Vj = 15 m/s

cart

Find:

• Force which the water exerts on the cart

Assumptions:


• 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.

∫∫∫

V−

ρ~gdV− −∫∫

S

P ndA+ ~Fv + ~R =d

dt

∫∫∫

V−

(ρ~V )dV− +

∫∫

S

ρ~V (~Vr · n)dA

We must choose a control volume.


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 weare being asked to solve for. In this case we are looking for the force that the water exertson the cart. Therefore, one possible choice for a control volume is

.

plate (gap shown onlyfor clarity)

C.S. between water and

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

∫∫∫

V−

ρ~gdV− −∫∫

S

P ndA+ ~Fv + ~R = ...

• The body force term is assumed to be zero. We are neglecting the weight of the wateron 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 tofind in this problem

– this term gives the force ON the CV. Therefore, it is equal and opposite to theunknown 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.


Another choice for a CV may be as follows.

Consider the LHS of the momentum equation now.

∫∫∫

V−

ρ~gdV− −∫∫

S

P ndA+ ~Fv + ~R = ...

• 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 NOTzero. If we consider a free–body–diagram of the cart, we realise that ~R is opposite tothe 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 informationwe want.

We will continue the problem with the second option.

1. Stationary Control Volume


Attach the coordinate system AND the control volume to the cart.

zx

• This is considered a stationary control volume because it is not moving with respectto the coordinate system.

• The coordinate system is inertial because it is moving at constant velocity.

We have already simplified the LHS so now,

~R =d

dt

∫∫∫

ρ~V dV− +

∫∫

ρ~V (~Vr · n)dA

Since nothing in the CV is changing with time, it is a steady problem. Also, assume uniformflow.

~R = (m~V )out − (m~V )in

or, because mout = min

~R = m(~Vout − ~Vin)

m = ρ(Vj − Vc)Aj

~Vin = (Vj − Vc)ı

Assuming that the vane is frictionless, the magnitude of this velocity will not change at theoutlet.

~Vout = (Vj − Vc)

(√2

2ı +

√2

2k

)


~R = ρ(Vj − Vc)Aj

(

(Vj − Vc)

(√2

2

)

(ı+ k) − (Vj − Vc)ı

)

~R = ρ(Vj − Vc)2Aj

((√2

2− 1

)

ı +

√2

2k

)

~R = 998

[

kg

m3

]

(

10[m

s

])2

0.01[m2]

((√2

2− 1

)

ı +

√2

2k

)

~R = (−292ı+ 705k) N

Remember, this is opposite to the force the water exerts on the cart.

The water exerts a force of (292ı − 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,

~Vin = Vjı

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

(Vj− V

c)√ 2/

2(i+k)

~Vout

Vci

~Vout = (Vj − Vc)

√2

2(ı + k) + Vcı


Now,


(

(Vj − Vc)

√2

2(ı+ k) + Vcı− Vjı

)


(

(Vj − Vc)

√2

2(ı+ k) − (Vj − Vc)ı

)

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

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.

~R =d

dt

∫∫∫

V−

ρ~V dV− − + (m~V )out − (m~V )in

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 (∫∫∫

ρ~V dV− ) in (ii) is greater than in (i) because the column of liquidgoing at Vjı is longer. This column is getting longer at speed Vc. Therefore,

d

dt

∫∫∫

ρ~V dV− = ρVjıAjVc

Now, at the inlet

(m~V )in = (ρVjAj)Vjı

However, note that min 6= mout for this analysis because mass is accumulating in the CV.

mout = min − ρAjVc

mout = ρVjAj − ρVcAj = ρAj(Vj − Vc)

Back to the momentum equation.

~R = ρVjVcAjı + ρAj(Vj − Vc)

((√2

2

)

(Vj − Vc)(ı+ k) + Vcı

)

− ρV 2j Ajı


(i)

(ii)

z

x

z

x


(

(Vj − Vc)

√2

2(ı+ k) − (Vj − Vc)ı

)

Again, this is the same intermediate result so the answer will be the same.



Given:

• A nozzle which discharges water from a pipe

1x

z

D2 = 25 mm

V1 = 1 m/sD1 = 50 mm

10 kPa

2

Find:

• The force on the bolts at the flange connection

Assumptions:


• no viscous force on CV

• uniform flow

• steady state

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 boltswith the control surface.

A coordinate system is also shown on the diagram.

Consider the linear momentum equation.∫∫∫

V−

ρ~gdV− −∫∫

S

P ndA+ ~Fv + ~R =d

dt

∫∫∫

V−

(ρ~V )dV− +

∫∫

S

ρ~V (~Vr · n)dA


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

n = −ıSince P1 is uniform and n doesn’t vary,

−∫∫

P ndA = −P1(−ı)A1 = P1A1 ı

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

P1A1 ı+ ~R = m(~Vout − ~Vin) = m(~V2 − ~V1)

~V1 = 1ım/s

Find the magnitude of ~V2 from conservation of mass

d

dt

∫ ∫

V−

∫

(ρ)dV− +

∫

S

∫

ρ(~V · n)dA = 0

For steady, uniform flow this becomes

ρV1A1 = ρV2A2

V2 = V1

A1

A2

= V1

(

50

25

)2

~V2 = 4ım/s

~R = −10 × 103[Pa]π

4(0.05)2[m2 ]ı+ 998

[

kg

m3

]

1[m

s

] π

4(0.05)2[m2](4ı− 1ı)

[m

s

]

~R = −13.8ıN

This means the bolts are exerting a force ON the CV in the −ı direction. Therefore, thebolts are in tension.

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



Given:

• The inlet section of a laminar pipe flow

1

u(r) = Umax(1 − (r/R)2)

radius= Rcircular pipe

Uo

rz 2

Find:

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

Assumptions:


• uniform flow at 1

• steady state

• ρ is constant

Analysis:

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

A coordinate system has already been given in the problem.∫∫∫

V−

ρ~gdV− −∫∫

S

P ndA+ ~Fv + ~R =d

dt

∫∫∫

V−

(ρ~V )dV− +

∫∫

S

ρ~V (~Vr · n)dA

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


1. Since no information has been give about the orientation of the pipe and we are onlyinterested in frictional drag, ignore the body force term.

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

3. The viscous force term is the unknown in this problem. We have chosen a CV includingjust the water on the pipe so that the friction at the pipe walls enters the momentumequation 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 uniformflow at the inlet, we certainly don’t at the outlet. Therefore, we can apply the uniformflow assumption at 1, but not at 2.

(P1 − P2)πR2k + ~Fv =

∫∫

2

ρ~V (~Vr · n)dA− (m~V )1

At 2,

(~Vr · n) = u(r)

~V2 = Umax

(

1 −( r

R

)2)

k

dA = rdθdr

~Fv = (P2 − P1)πR2k +

∫ R

0

∫ 2π

0

ρ

(

Umax

(

1 −( r

R

)2))2

krdrdθ − UoπR2ρUok

~Fv = (P2 − P1)πR2k + 2πρU2

maxk

∫ R

0

(

1 − 2r2

R4+r4

R4

)

rdr − U2oπR

2ρk

~Fv = (P2 − P1)πR2k + 2πρU2

maxk

[

r2

2− 2r4

4R4+

r6

6R4

]R

0

− U2oπR

2ρk

~Fv = πR2

(

(P2 − P1) +ρU2

max

3− ρU2

o

)

k

Eliminate Umax with conservation of mass (steady).

d

dt

∫ ∫

V−

∫

(ρ)dV− +

∫

S

∫

ρ(~V · n)dA = 0

0 =

∫∫

2

ρUmax

(

1 −( r

R

)2)

rdrdθ − UoρπR2


0 = ρUmax2π

∫ R

0

(

r − r3

R2

)

dr − UoρπR2

0 = ρUmax2π

[

r2

2− r4

4R2

]R

0

− UoρπR2

0 = ρUmax2πR2

4− UoρπR

2

Umax = 2Uo

~Fv = πR2

(

(P2 − P1) +ρ(2Uo)

2

3− ρU2

o

)

k

The frictional drag on the fluid is

πR2

(

(P2 − P1) +ρU2

o

3

)



Given:

• A nozzle mounted at a pipe exit

D2 = 25 mm

x

z?

V1 = 1 m/sD1 = 50 mm 21

Find:

• Force on the bolts assuming frictionless flow

Assumptions:

• frictionless flow

• uniform inlet and outlet

• steady, incompressible flow

• 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.

∫∫∫

V−

ρ~gdV− −∫∫

S

P ndA+ ~Fv + ~R =d

dt

∫∫∫

V−

(ρ~V )dV− +

∫∫

S

ρ~V (~Vr · n)dA

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

~R = ρV1A1

((

V1

A1

A2

)

ı− V1 ı

)

− P1A1 ı


Bernoulli’s equation is

P1 +ρV 2

1

2+ ρgz1 = P2 +

ρV 22

2+ ρgz2

Since z1 = z2

P1 =ρ

2(V 2

2 − V 21 ) =

998

2

[

kg

m3

]

(42 − 12)

[

m2

s2

]

P1 = 7485 Pa

~R = 1000

[

kg

m3

]

1[m

s

] π

4(0.05)2[m2]

(

1[m

s

]

(

50

25

)2

ı− 1[m

s

]

ı

)

−7485[Pa]π

4(0.05)2[m2] = −8.82ıN

The tensile force in the bolts is 8.82 N.



Given:

• A nozzle mounted at the exit of a pipe

D2 = 25 mmV1 = 1 m/s

D1 = 50 mm

10 kPa

z

x 21

Find:

• The loss coefficient for the nozzle

Assumptions:


• no shaft work, no heat transfer


Analysis:

Bernoulli’s equation

P1 +ρV 2

1

2+ ρgz1 = P2 +

ρV 22

2+ ρgz2 + ∆PL

∆PL = P1 +ρ

2

(

V 21 − V 2

2

)

∆PL = 100000[Pa] +998

2

[

kg

m3

](

12

[

m2

s2

]

− 42

[

m2

s2

])

∆PL = 2515 Pa

∆Pv = Km

ρV 21

2Km = 5.04

The loss coefficient of this nozzle based on the inlet velocity is 5.04.



Given:

• A transition piece as shown

Km = 0.50 (based on

.

discharge velocity)

y

x

20 kPa

D1 = 50 mm

D2 = 80 mm

45o

4 kg/s

2

1

Find:

• Force required to hold transition in place

Assumptions:


• uniform inlet and outlet



Analysis:

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

Begin with the complete equation

∫∫∫

V−

ρ~gdV− −∫∫

S

P ndA+ ~Fv + ~R =d

dt

∫∫∫

V−

(ρ~V )dV− +

∫∫

S

ρ~V (~Vr · n)dA

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


• Choose a CV which isolates the transition piece. Therefore, ~R becomes the unknownin the problem.

• A uniform pressure exists at 1 and 2.

• The flow is steady so

d

dt

∫∫∫

ρ~V dV− = 0

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

Now, we have

−∫∫

P ndA + ~R = m(~V2 − ~V1)

~V1 =m

ρA1

ı =4[kg/s]

998[kg/m3]

4

π(0.05)2[m2]= 2.04ım/s

~V2 =m

ρA2

(√2

2

)

(ı+ ) =

√2

2

4[kg/s]

998[kg/m3]

4

π(0.08)2[m2](ı+ ) = 0.5638(ı+ ) m/s

Now, let’s work on the pressure force term. At 1,

P1 = 20 kPa

n = −ıAt 2

n =

(√2

2

)

(ı + )

P2 =?

We must find P2 using Bernoulli’s equation. Write Bernoulli’s equation (with pressure losses)from 1 to 2.

P1 +ρV 2

1

2= P2 +

ρV 22

2+ ∆PL

∆PL = Km

ρV 22

2

P2 = P1 +ρ

2(V 2

1 − V 22 ) −Km

ρV 22

2


P2 = 20, 000[Pa] +998

2

[

kg

m3

]

(

(2.04)2 − 2(0.5638)2)

[

m2

s2

]

− 0.5998

2

[

kg

m3

]

2(0.5638)2

[

m2

s2

]

P2 = 21.6 kPa

Now, return to the momentum equation

~R = −P1A1 ı+ P2A2

√2

2(ı + ) + m(~V2 − ~V1)

~R = − 20, 000[Pa]π

4(0.05)2[m2 ]ı+ 21, 600[Pa]

π

4(0.08)2[m2]

√2

2(ı + )

+ 4

[

kg

s

]

(0.5638ı+ 0.5638− 2.04ı)[m

s

]

~R = (31.6ı+ 79.0) N

The force required to hold the transition in place is (31.6ı+79.0) N in the coordinate

system shown.



Given:

• A piping system as shown below

90o, regular, flanged elbows

~g

Q = 0.04 m3/swater, 20oC

30 m

open, flanged, globe valve

1

2

50 m, 2” I.D. steel pipe

Find:

• P1 − P2

Assumptions:


• uniform flow at 1 and 2

• no diameter change


Analysis:

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

P1 +ρV 2

1

2+ ρgz1 = P2 +

ρV 22

2+ ρgz2 + f

L

D

ρV 2

2+∑

Km

ρV 2

2


Since there is no diameter change, V1 = V2. Also, let z1 = 0.

P1 − P2 = ρgz2 + fL

D

ρV 2

2+∑

Km

ρV 2

2

P1 − P2 = ρgz2 +ρV 2

2

(

fL

D+∑

Km

)

V =Q

A=

0.04[m3/s]

(2(0.0254))2π/4[m2]= 19.74 m/s

We need to find f . It is a function of roughness (ǫ) and Reynolds number (Re). For steelpipe, ǫ = 0.046 mm (Table 6.1). Therefore,

ǫ

D=

0.046

2(25.4)= 0.000906

Re =ρV D

µ=V D

ν=

19.74[m/s](2(0.0254)[m])

1 × 10−6[m2/s]= 1 × 106

From the Colebrook equation with this (ǫ/D) and Re we get f = 0.0195. Now find theminor loss coefficients from table 6–5.

Kvalve = 8.5

Kelbow = 0.39

P1 − P2 = 998

[

kg

m3

]

9.81[m

s2

]

30[m] +998

2

[

kg

m3

]

(

19.74[m

s

])2

(

0.019550

2(0.0254)+ 8.5 + 2(0.39)

)

P1 − P2 = 5.83 MPa

The pressure drop between 1 and 2 is 5.83 MPa at this flowrate.



Given:

• A pipe connecting two reservoirs as shown

20 m

ρ = 1000 kg/m3

L = 5 mD = 2 cm

2

1

ǫ = 0.05 mm

D = 4 cmL = 5 m

ba

µ = 0.001 kg/(m · s)

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.

P1 +ρV 2

1

2+ ρgz1 = P2 +

ρV 22

2+ ρgz2 + f

L

D

ρV 2

2+∑

Km

ρV 2

2

ρgz1 = fa

(

L

D

)

a

ρV 2a

2+ fb

(

L

D

)

b

ρV 2b

2+∑

Km

ρV 2

2


Minor losses:

Km sourceentrance 0.5 Figure 6.21expansion 0.56 Figure 6.22exit 1.0 Figure 6.21

ρgz1 = fa

(

L

D

)

a

ρV 2a

2+ fb

(

L

D

)

b

ρV 2b

2+ρV 2

a

2(Kent. +Kexp.) +

ρV 2b

2(Kexit)

From conservation of mass,

Vb = Va

Aa

Ab

= Va

(

Da

Db

)2

ρgz1 =ρV 2

a

2

(

(

fL

D

)

a

+

(

fL

D

)

b

(

Da

Db

)4

+Kent. +Kexp. +Kexit

(

Da

Db

)4)

( ǫ

D

)

a=

0.05[mm]

20[mm]= 0.0025

( ǫ

D

)

b=

0.05[mm]

40[mm]= 0.00125

392.4 = V 2a

(

250fa +125

16fb + 1.1225

)

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 0.0247.09 141,770 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.



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 20oC

Flowrate, m3/s

Parabola

Pum

phea

d,m

2.50

100

80

60

40

20

01 1.5 20.5

Find:

• Flowrate

Assumptions:


• no minor losses

• pump inlet the same diameter as exit pipe

• pump inlet pressure same as pipe exit pressure

• no heat transfer



Analysis:

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

P1 +ρV 2

1

2+ ρgz1 = P2 +

ρV 22

2+ ρgz2 + ∆Pf + ∆Pm − ∆Pp

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

hf = fL

D

Q2

2gA2.

(80 − 20Q2)[m] = f

(

100[m]

0.2[m]

)

Q2

2(9.81[m/s2])

(

4

π(0.2[m])2

)2

80 = 25, 821fQ2 + 20Q2

Q =

√

80

25, 821f + 20

For cast iron pipe, ǫ = 0.26 mm so ǫ/D = 0.0013. The fully rough friction factor for thisǫ/D is f = 0.0210. Therefore,

Q = 0.3772 m3/s

Check the Reynolds number.

Re =ρV D

µ=V D

ν=QD

Aν= 2.4 × 106

This is fully rough so we can accept the answer.

The flow rate is 0.377 m3/s.



Given:

• Flow around a bridge pier

• U∞

= 5 m/s

• P∞

= 50 kPa

• stagnation point is 1 m upstream of source

2 m

x

1 m

A

y

Find:

• Pressure at point A

Assumptions:

• two-dimensional flow

• incompressible, steady flow

• inviscid, irrotational flow


Analysis:

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

Therefore, for the source

vr =m

r

At the stagnation point, u = 0. Therefore,

0 = U∞− m

r

0 = 5[m

s

]

− m

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,

vθ =−∂ψ∂r

= −U∞

sin θ

vr =1

r

∂ψ

∂θ= U

∞cos θ +

m

r

We need the (r, θ) coordinates of point A to find these velocities. At the stagnation pointθ = π, r = 1 m/s. Therefore,

ψsp = 0 +mπ = mπ

Since A is on the same streamline

ψA = mπ

In (x, y) coordinates.

ψ = U∞y +m tan−1

(y

x

)

at A,

5[m2/s]π = 5[m/s]2[m] + 5[m2/s] tan−1

(

2

x

)

xA = 0.9153


rA =√

(0.9153)2 + (2)2 = 2.200 m

θA = tan−1

(

2

0.9152

)

= 1.142 rad

Therefore,

V 2A = v2

θ + v2r = (−5 sin(1.142))2 + (5 cos(1.142) + 5/2.200)2

V 2A = 39.61 m2/s2

PA = P∞

+1

2ρ(V 2

∞− V 2

A) = 50, 000[Pa] +1000

2

[

kg

m3

]

(52 − 39.61)[m

s

]

PA = 42.7 kPa

The pressure at A is 42.7 kPa.



Given:

• A pair of doublets as shown

• U∞

= 10 m/s

x

U∞

= 10 m/s

doublets

30o

2 m

2 m2 m

y

Find:

• Doublet strength

• Pressure at the origin

Assumptions:





Analysis:

ψ = ψfs + ψd1 + ψd2

ψ = U∞

(y cosα− x sinα) − λ sin θ1r1

− λ sin θ2r2

ψ = U∞

(y cosα− x sinα) − λ(y − 2)

(y − 2)2 + (x− 2)2− λ(y − 2)

(y − 2)2 + (x+ 2)2

ψ0,0 = ψ4,1

ψ0,0 = 0 − λ(−2)

(−2)2 + (−2)2− λ(−2)

(−2)2 + (2)2

ψ0,0 =2λ

8+

2λ

8=λ

2

ψ4,1 = 10(cos 30o − 4 sin 30o) − λ(−1)

(−1)2 + (2)2− λ(−1)

(−1)2 + (6)2

ψ4,1 = −11.34 +λ

5+

λ

37=λ

2

λ

(

1

2− 1

5− 1

37

)

= −11.34

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, thepressure there is the same as the pressure far away from the doublets.

The pressure at the origin is P∞

.



Given:

• A landing gear strut

• U∞

= 45 m/s

• ρ = 1.2 kg/m3

A

U∞

12 cm12 cm

B

sinksource

30 cm

Find:

• PA − PB

Assumptions:





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.

0 = U∞− m

(0.15 − 0.12)+

m

(0.15 + 0.12)

m = 1.519 m2/s

Need VB.

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

ψ = U∞y +m tan−1

(

y

x+ a

)

−m tan−1

(

y

x− a

)

u =∂ψ

∂y= U

∞+

m

1 +(

yx+a

)2

(

1

x+ a

)

− m

1 +(

yx−a

)2

(

1

x− a

)

PROBLEM!!!!! Don’t know y at B.

ψB = ψsp = 0

0 = 45y + 1.519 tan−1( y

0.12

)

− 1.519 tan−1

(

y

−0.12

)

Note that

tan−1

(

y

−0.12

)

= π − tan−1( y

0.12

)

0 = 45y + 2(1.519) tan−1( y

0.12

)

− π(1.519)

45y = 1.519π − 2(1.519) tan−1( y

0.12

)

y = 0.07027 m

Now,

uB = 45 + 1.519

(

1

1 + (0.070270.12

)2

(

1

0.12

))

2


uB = 63.85 m/s

PA +1

2ρU2

A = PB =1

2ρU2

B

PA − PB =1

2ρU2

B =1.2

2(63.85)2

PA − PB = 2.45 kPa

The pressure difference between points A and B is 2.45 kPa.



Given:

• Flow over a circular cylinder creating no lift

a

U∞

~g

L

A

x

y

Find:

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

, a, andy.

• Sketch this function.

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

Assumptions:




Analysis:

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

ψ = ψfs + ψdoublet


ψ = U∞r sin θ − λ sin θ

r

ψ = sin θ(U∞r − λ

r)

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

ψsp = 0

Therefore, ψ = 0 on the circle. Therefore,

U∞a− λ

a= 0

λ = U∞a2

ψ = U∞

sin θ(r − a2

r)

Along the y axis, vr = 0

vθ = −∂ψ∂r

= −U∞

sin θ

(

1 +a2

r2

)

θ =π

2

vθπ/2 = −U∞

(

1 +a2

r2

)

Along the positive y axis, u = −vθ. Therefore

u

U∞

= 1 +a2

r2

The velocity along the positive y axis is given by

u

U∞

= 1 +a2

r2.

ψA − ψsp = U∞L

ψA = U∞L

U∞L = U

∞sin(π

2

)

(

r − a2

r

)

L = r − a2

r

r2 − Lr − a2 = 0


1

u/U∞

r/a1

2

r =L

2±

√

(

L

2+ a2

)

Since r > a, we want the positive root.

r =L

2+

√

(

L

2+ a2

)

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

L

2+

√

(

L

2+ a2

)

from the origin.



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 −0.1 0 0.1 0.2 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

B

A

y

x

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

Find:

• Lift force per unit length


Assumptions:




Analysis:

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

ψ = ψfs + ψdoublet + ψvortex

ψ = U∞r sin θ − λ sin θ

r−K ln

(r

a

)

For ψ = 0 on r = a,

0 = U∞a sin θ − λ 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)

1.19

[

m2

s

]

= 20[m

s

]

(0.12[m]) − 20[m/s](0.1)2[m2]

0.12[m]−K ln

(

0.12[m]

0.1[m]

)

K = −2.505 m2/s

Γ = 2πK = −15.74 m2/s

Kutta-Joukowski Theorem

FL = −ρU∞

Γ

FL = −1.2

[

kg

m3

]

20[m

s

]

(

−15.74

[

m2

s

])

FL = 378 N/m

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


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