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Chapter 2 Properties of Fluids

PROPRIETARY MATERIAL

2-43

Solution A block is moved at constant velocity on an inclined surface. The force that needs to be applied in the

horizontal direction when the block is dry, and the percent reduction in the required force when an oil film is applied on thesurface are to be determined.

Assumptions 1 The inclined surface is plane (perfectly flat, although tilted). 2 The friction coefficient and the oil film

thickness are uniform. 3 The weight of the oil layer is negligible.

Properties The absolute viscosity of oil is given to be = 0.012 Pa s = 0.012 N s/m2.

Analysis (a) The velocity of the block is constant, and thus itsacceleration and the net force acting on it are zero. A free body diagram of the

block is given. Then the force balance gives

200

F 1

V = 0.8 m/s

W = 150 N

F

F N 1

y

x

200

200

:0 x F 020sin20cos 11 N f F F F (1)

:0 y F 020sin20cos1 W F F f N (2)

Friction force: (3)1 N f fF F

Substituting Eq. (3) into Eq. (2) and solving for F N 1 gives

N0.17720sin27.020cos

N15020sin20cos

1 f

W F N

Then from Eq. (1):

N 105.520sin) N177(20cos) N17727.0(20sin20cos 11 N f F F F

(b) In this case, the friction force is replaced by the shear force

applied on the bottom surface of the block due to the oil. Because

of the no-slip condition, the oil film sticks to the inclined surfaceat the bottom and the lower surface of the block at the top. Then

the shear force is expressed as

50 cm 0.4 mm

F shear = w A s

200

F 2

V = 0.8 m/s

W = 150 N

F 2

2 2

-4

0 8 m/s 0 012 N s/m 0 5 0 2 m

4 10 m

2 4 N

shear w s

s

F A

V A

h

.. . .

.Replacing the friction force by the shear force in part (a),

:0 x F 020sin20cos 22 N shear F F F (4)

:0 y F 020sin20cos2 W F F shear N (5)

Eq. (5) gives N60.5120cos/)] N150(20sin) N4.2[(20cos/)20sin(2 W F F shear N

Substituting into Eq. (4), the required horizontal force is determined to be

N 57.220sin) N5.160(20cos) N4.2(20sin20cos 22 N shear F F F

Then, our final result is expressed as

Percentage reduction in required force = 1 2

1

105 5 57 2100% 100%

105 5

F F . .

F .45.8%

Discussion Note that the force required to push the block on the inclined surface reduces significantly by oiling thesurface.

. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-17



Chapter 2 Properties of Fluids

PROPRIETARY MATERIAL

2-51

l = 0.12 cmfluid

R

Solution The torque and the rpm of a double cylinder viscometer are given. The viscosity of the fluid is to be

determined.

Assumptions 1 The inner cylinder is completely submerged in oil. 2 Theviscous effects on the two ends of the inner cylinder are negligible. 3 The fluid

is Newtonian.

Analysis Substituting the given values, the viscosity of the fluid is

determined to be

2s/mN0.0231

T

m)75.0)(s60/200(m)075.0(4

m)m)(0.0012 N8.0(

4 1-3232 Ln R

Discussion This is the viscosity value at the temperature that existed duringthe experiment. Viscosity is a strong function of temperature, and the values

can be significantly different at different temperatures.

2-52E

Solution The torque and the rpm of a double cylinder viscometer are

given. The viscosity of the fluid is to be determined.

. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to

teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-25

Assumptions 1 The inner cylinder is completely submerged in the fluid. 2

The viscous effects on the two ends of the inner cylinder are negligible. 3 The

fluid is Newtonian.

Analysis Substituting the given values, the viscosity of the fluid is determined

to be

25 s/ftlbf 109.97ft)3)(s60/250(ft)12/6.5(4

ft)2ft)(0.05/1lbf 2.1(

4 1-3232 Ln R

T

Discussion This is the viscosity value at temperature that existed during

the experiment. Viscosity is a strong function of temperature, and the values can be significantly different at differenttemperatures.

l = 0.05 in

fluid

R



Chapter 3 Pressure and Fluid Statics

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to


3-26

3-47

Solution Two chambers with the same fluid at their base are separated by a piston. The gage pressure in each air

chamber is to be determined.

Assumptions 1 Water is an incompressible substance. 2 The

variation of pressure with elevation in each air chamber is

negligible because of the low density of air.

Properties We take the density of water to be =1000

kg/m3.

Analysis The piston is in equilibrium, and thus the net

force acting on the piston must be zero. A vertical force

balance on the piston involves the pressure force exerted by

water on the piston face, the atmospheric pressure force, and

the piston weight, and yields

piston pistonatm piston W A P A P C piston

piston

atm A

W P P C

The pressure at the bottom of each air chamber is determined

from the hydrostatic pressure relation to be

CE g A

W P CE g P P P C E

piston

piston

atmAair CE g A

W P

piston

piston

gageA,air

CD g A

W P CD g P P P C D

piston

piston

atmBair CD g A

W P

piston

piston

gageB,air

Substituting,

3 2 2

air A, gage 2 2

25 N 1 N

(1000 kg/m )(9.81 m/s )(0.25 m) 2806 N/m0 3 m) 4 1 kg m/s P ( . / 2.81 kPa

3 2 2

air B, gage 2 2

25 N 1 N(1000 kg/m )(9.81 m/s )(0.25 m) 2099 N/m

0 3 m) 4 1 kg m/s P

( . / 2.10 kPa

Discussion Note that there is a vacuum of about 2 kPa in tank B which pulls the water up.

air air

water

Piston

50 cm

25 cm30 cm

A

B

E

D

30 cm

90 cm

C






3-59

3-97E

Solution A vertical cylindrical tank open to the atmosphere is rotated about the centerline. The angular velocity at

which the bottom of the tank will first be exposed, and the maximum water height at this moment are to be determined.

Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid

body. 2 Water is an incompressible fluid.

Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the

equation for the free surface of the liquid is given as

)2(4

)( 222

0 r R g

hr z s

where h0 = 1 ft is the original height of the liquid before rotation. Just before dry spot appear at the center of bottom

surface, the height of the liquid at the center equals zero, and thus z s(0) = 0. Solving the equation above for and

substituting,

2

0

2 2

4 32 2 ft/s 1 ft411 35 rad/s

1 ft

. gh.

R11.4 rad/s

Noting that one complete revolution corresponds to 2 radians, the rotational speed of the container can also be expressed

in terms of revolutions per minute (rpm) as

rpm108min1

s60

rad/rev2

rad/s35.11

2n

Therefore, the rotational speed of this container should be limited to 108 rpm to avoid any dry spots at the bottom surface

of the tank.

The maximum vertical height of the liquid occurs a the edges of the tank (r = R = 1 ft), and it is

ft2.00)ft/s2.32(4

ft)1(rad/s)35.11()ft1(

4)(

2

2222

0 g

Rh R z s

Discussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid

property.

2 ft

z

r 0






3-65

3-103

Solution Milk is transported in a completely filled horizontal cylindrical tank decelerating at a specified rate. The

maximum pressure difference in the tanker is to be determined.

Assumptions 1 The acceleration remains constant. 2 Milk is an incompressible substance.

Properties The density of the milk is given to be1020 kg/m3.

Analysis We take the x- and z- axes as shown.

The horizontal deceleration is in the x direction, and

thus a x is positive. Also, there is no acceleration in

the vertical direction, and thus a z = 0. The pressure

difference between two points 1 and 2 in an

incompressible fluid in linear rigid body motion is

given by

))(()( 121212 z z a g x xa P P z x )()( 121212 z z g x xa P P x

The first term is due to deceleration in the horizontal direction and the resulting compression effect towards the front of the

tanker, while the second term is simply the hydrostatic pressure that increases with depth. Therefore, we reason that the

lowest pressure in the tank will occur at point 1 (upper front corner), and the higher pressure at point 2 (the lower rear

corner). Therefore, the maximum pressure difference in the tank is

kPa47.92

2

223

1212121212max

kN/m)0.309.17(

m/skg1000

kN1m)3)(m/s81.9(m)7)(m/s5.2()kg/m1020(

)]()([)()( z z g x xa z z g x xa P P P x x

since x1 = 7 m, x2 = 0, z 1 = 3 m, and z 2 = 0.

Discussion Note that the variation of pressure along a horizontal line is due to acceleration in the horizontal direction

while the variation of pressure in the vertical direction is due to the effects of gravity and acceleration in the vertical

direction (which is zero in this case).

3-104

Solution A vertical U-tube partially filled with

alcohol is rotated at a specified rate about one of its arms.

The elevation difference between the fluid levels in the two

arms is to be determined.

Assumptions 1 Alcohol is an incompressible fluid.

Analysis Taking the base of the left arm of the U-tube

as the origin (r = 0, z = 0), the equation for the free surface

of the liquid is given as

)2(4

)( 222

0 r R g

hr z s

where h0 = 0.20 m is the original height of the liquid before rotation,

and = 4.2 rad/s. The fluid rise at the right arm relative to the fluid

level in the left arm (the center of rotation) is

m0.056)m/s81.9(2

m)25.0(rad/s)2.4(

244)0()(

2

222222

0

22

0 g

R

g

Rh

g

Rh z R z h s s

Discussion The analysis is valid for any liquid since the result is independent of density or any other fluid property.

z

x2

1 g

a x = 3 m/s2

R = 25 cm

h0 =

20 cm

r0



Chapter 8 Flow in Pipes



8-80

8-110

Solution The flow rate of water is to be measured with flow nozzle equipped with an inverted air-water manometer.For a given differential height, the flow rate and head loss caused by the nozzle meter are to be determined.

Assumptions 1 The flow is steady and incompressible. 2 The discharge coefficient of the nozzle meter is C d = 0.96.

Properties The density and dynamic viscosity of water are given to be = 998 kg/m3 and = 1.002 10-3 kg/m s,

respectively.

Analysis The diameter ratio and the throat area of the meter are

24220 m10142.34/m)02.0(4/

50.04/2/

d A

Dd

Noting that P = 4 kPa = 4000 N/m2, the flow rate becomes

/sm100.781 33

4

224

44421

50.01

m)32.0)(m/s81.9(2)96.0)(m10142.3(

1

2

)1(

2

)1(

)(2 ghC A

ghC A

P P C A d o

w

wd od oV

which is equivalent to 0.781 L/s. The average flow velocity in the pipe is

s/m621.04/m)04.0(

s/m10781.0

4/ 2

33

2 D AV

c

V V

The percent pressure (or head) loss for nozzle meters is given in Fig. 8-59 for = 0.5 to be 62%. Therefore,

OHm0.20 2m)32.0(62.0)lossheadTotal)(fractionlossPermanent( Lh

The head loss between the two measurement sections can be determined from the energy equation, which simplifies to ( z 1= z 2)

4 2 242 211 2 2 1

221 (4/2) 1 0 621 m/s0 32 m 0 025 m H O

2 2 2 9 81 m/s L w

f

D / d V . P P V V h h . . g g g .

Discussion The Reynolds number of flow through the pipe is

4

3-

3

1047.2skg/m101.002

m)m/s)(0.04)(0.621kg/m998(Re

VD

Substituting the and Re values into the orifice discharge coefficient relation gives

968.0)1047.2(

)50.0(53.69975.0

Re

53.69975.0

5.04

5.0

5.0

5.0

d C

which is almost identical to the assumed value of 0.96.

32 cm

4 cm2 cmWater



Chapter 8 Flow in Pipes



8-94

8-123 [ Also solved using EES on enclosed DVD]

Solution A pipeline that transports oil at a specified rate branches out into two parallel pipes made of commercialsteel that reconnects downstream. The flow rates through each of the parallel pipes are to be determined.

Assumptions 1 The flow is steady and incompressible. 2 Entrance effects are negligible, and thus the flow is fully

developed. 3 Minor losses are disregarded. 4 Flows through both pipes are turbulent (to be verified).

Properties The density and dynamic viscosity of oil at 40 C are = 876 kg/m3 and = 0.2177 kg/m s. The roughness

of commercial steel pipes is = 0.000045 m.

Analysis This problem cannot be solved directly since the velocities (or flow rates) in the pipes are not known.Below we will set up the equations to be solved by an equation solver. The head loss in two parallel branches must be the

same, and the total flow rate must the sum of the flow rates in the parallel branches. Therefore,

(1)2,1, L L hh

(2)32121 V V V V V

We designate the 30-cm diameter pipe by 1 and the 45-cm diameter pipe by 2. The average velocity, the relative roughness,

the Reynolds number, friction factor, and the head loss in each pipe are expressed as

(4)4/m)45.0(4/

(3)4/m)30.0(4/

2

222

2

2

2,

22

21121

1

1,

11

V V V

V V V

V D A

V

V D AV

c

c

101m45.0

m105.4rf

105.1m30.0

m105.4rf

45

2

22

45

1

11

D

D

(6) kg/m2177.0

m)(0.45)kg/m876(ReRe

(5) kg/m2177.0

m)(0.30)kg/m876(ReRe

23

222

2

13

111

1

s

DV

s

DV

V

V

Re

51.2

7.3

105.1log0.2

1

Re

51.2

7.3

/log0.2

1

11

4

111

1

1 f f f

D

f (7)

Re

51.2

7.3

101log0.2

1

Re

51.2

7.3

/log0.2

1

22

4

222

2

2 f f f

D

f (8)

(9))m/s81.9(2m30.0

m500

2 2

21

11,

21

1

111,

V f h

g

V

D

L f h L L

(10))m/s81.9(2m45.0

m8002 2

2

222,

2

2

2

222, V f h g

V D L f h L L

This is a system of 10 equations in 10 unknowns, and solving them simultaneously by an equation solver gives

/sm2.09/sm0.91 3321 , V V ,

V 1 = 12.9 m/s, V 2 = 13.1 m/s, m3922,1, L L hh

Re1 = 15,540, Re2 = 23,800, f 1 = 0.02785, f 2 = 0.02505

Note that Re > 4000 for both pipes, and thus the assumption of turbulent flow is verified.

Discussion This problem can also be solved by using an iterative approach, but it would be very time consuming.

Equation solvers such as EES are invaluable for theses kinds of problems.

30 cm 500 m

A

B

800 m45 cm

3 m3/s

1

2

solucion ayudantia examen

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