fluid 9ed solution manual

7/25/2019 Fluid 9ed Solution Manual

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2.1: PROBLEM DEFINITION

Find:How density differs from specific weight

PLANConsider their definitions (conceptual and mathematical)

SOLUTION

Density is a [mass]/[unit volume], and specific weight is a [weight]/[unit volume].Therefore, they are related by the equation =g, and density differs from specificweight by the factor g, the acceleration of gravity.

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Find:Fluids for which we can (usually) assume density to be nearly constantFluids for which density should be calculated as a function of temperature and

pressure?

SOLUTION

Density can usually be assumed to be nearly constant for liquids , such as water, mer-cury and oil. However, even the density of a liquid varies slightly as a function ofeither pressure or temperature. Slight changes in the volume occupied by a givenmass of a liquid as a function of pressure can be calculated using the equation forelasticity.

One must know the temperature and the pressure to determine the density of a gas .

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Find:Where in this text you can find density data for such fluids as oil and mercury.

SOLUTIONTable A.4 in the Appendix contains density data for such fluids as oil and mercury .

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Situation:An engineer needs to know the local density for an experiment with a glider.z = 2500 ft.

Find:Calculate density using local conditions.Compare calculated density with the value from Table A.2, and make a recommen-

dation.

Properties:

From Table A.2,Rair = 287 Jkg K

, = 1.22kg/ m3.Local temperature = 74.3 F = 296.7 K.

Local pressure = 27.3 in.-Hg = 92.45 kPa.

PLANApply the ideal gas law for local conditions.

SOLUTION

Ideal gas law

= p

RT

= 92, 450N/ m2

(287kg/ m3)(296.7 K)

= 1.086kg/m3

= 1.09 kg/m3 (local conditions)

Table value. From Table A.2

= 1.22 kg/m3 (table value)

The density difference (local conditions versus table value) is about 12%. Mostof this difference is due to the effect of elevation on atmospheric pressure.

Recommendationuse the local value of density because the effects of elevation are signi

REVIEW

Note: Always use absolute pressure when working with the ideal gas law.

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Situation:Carbon dioxide.

Find:

Density and specific weight of CO2.Properties:

From Table A.2,RCO2 = 189 J/kgK.p= 300 kPa,T = 60 C.

PLAN

1. First, apply the ideal gas law to find density.2. Then, calculate specific weight using = g.

SOLUTION

1. Ideal gas law

CO2 = P

RT

= 300, 000 kPa

(189J/ kg K)(60 + 273) K

CO2 = 4.767 kg/m3

2. Specifi

c weight = g

Thus

CO2 = CO2 g

= 4.767kg/ m3 9.81 m/ s2

CO2 =46.764 N/m3

REVIEW

Always use absolute pressure when working with the ideal gas law.

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Situation:Methane gas.

Find:

Density (kg/m3

).Specific weight ( N/ m3).

Properties:

From Table A.2,RMethane= 518 Jkg K

.p= 300 kPa,T = 60 C.

PLAN

1. Apply the ideal gas law to find density.2. Calculate specific weight using = g.

SOLUTION1. Ideal gas law

Methane = P

RT

= 300, 000 N

m2

518 Jkg K

(60 + 273 K)

Methane= 1.74 kg/m3

2. Specific weight

= g

Thus

Methane = Methane g

= 1.74kg/ m3 9.81 m/ s2

Methane = 17.1 N/m3

REVIEW


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Situation:Natural gas is stored in a spherical tank.

Find:

Ratio offinal mass to initial mass in the tank.Properties:

patm= 100 kPa,p1 = 100 kPa-gage.p2= 200 kPa-gage, T = 10 C.

PLAN

Use the ideal gas law to develop a formula for the ratio offinal mass to initial mass.

SOLUTION

1. Mass in terms of density

M=V (1)

2. Ideal gas law

= p

RT (2)

3. Combine Eqs. (1) and (2)

M = V

= (p/RT)V

4. Volume and gas temperature are constant, so

M2M1

=p2p1

and

M2M1

= 300 kPa

200 kPaM2M1

=1.5

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Situation:Wind and water at 100 Cand 5 atm.

Find:

Ratio of density of water to density of air.Properties:

Air, Table A.2: Rair = 287 J/kgK.Water (100oC), Table A.5: water= 958 kg/m

3.

PLAN

Apply the ideal gas law to air.

SOLUTION

Ideal gas law

air = p

RT

= 506, 600 kPa

(287J/ kg K) (100 + 273) K

= 4.73 kg/m3

For waterwater= 958 kg/m

3

Ratio

waterair

= 958kg/ m3

4.73kg/ m3

waterair

= 203

REVIEW

Always use absolute pressures when working with the ideal gas law.

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Situation:Oxygen fills a tank.Vtank = 10 ft

3,Wtank = 150 lbf.

Find:Weight (tank plus oxygen).

Properties:From Table A.2,RO2 = 1555 ftlbf/(slug

o R) .p= 500psia,T= 70 F.

PLAN

Apply the ideal gas law to find density of oxygen.Find the weight of the oxygen using specific weight()and add this to the weight ofthe tank.

SOLUTION

1. Ideal gas law

pabs. = 500 psia144 psf/psi= 72, 000 psf

T = 460 + 70 = 530R

= p

RT

= 72, 000 psf

(1555 ft lbf/ slugoR)(530oR)

= 0.087 slugs/ft3

2. Specific weight

= g

= 0.087slug

ft3 32.2

ft

s2

= 2.80 lbf/ft3

3. Weight offilled tank

Woxygen = 2.80 lbf/ft3 10 ft3

= 28lbf

Wtotal = Woxygen+ Wtank

= 28.0lbf + 150lbf

Wtotal = 178 lbf

REVIEW

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1. For compressed gas in a tank, pressures are often very high and the ideal gasassumption is invalid. For this problem the pressure is about 34 atmospheresit isa good idea to check a thermodynamics reference to analyze whether or not real gaseffects are significant.2. Always use absolute pressure when working with the ideal gas law.

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Situation:Oxygen is released from a tank through a valve.V= 10 m3.

Find:Mass of oxygen that has been released.

Properties:

RO2 = 260 J

kg K.

p1= 800 kPa,T1= 15 C.p2= 600 kPa,T2= 20

C.

PLAN

1. Use ideal gas law, expressed in terms of density and the gas-specific (not universal)gas constant.

2. Find the density for the case before the gas is released; and then mass fromdensity, given the tank volume.3. Find the density for the case after the gas is released, and the corresponding mass.4. Calculate the mass difference, which is the mass released.

SOLUTION

1. Ideal gas law

= p

RT

2. Density and mass for case 1

1 = 800, 000 N

m2

(260 N mkg K

)(288K)

1 = 10.68kg

m3

M1 = 1V

= 10.68kg

m310 m3

M1 = 106.8 kg

3. Density and mass for case 2

2 = 600, 000 N

m2

(260 N mkg K

)(288K)

2 = 8.01kg

m3

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M2 = 1V

= 8.01kg

m310 m3

M1 = 80.1 kg

4. Mass released from tank

M1 M2 = 106.8 80.1

M1 M2 = 26.7 kg

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Situation:Properties of air.

Find:

Specific weight (N/m3

).Density (kg/m3).

Properties:

From Table A.2,R = 287 Jkg K

.p= 600 kPa,T = 50 C.

PLAN

First, apply the ideal gas law to find density. Then, calculate specific weight using=g.

SOLUTION1. Ideal gas law

air = P

RT

= 600, 000 kPa

(287J/ kg K) (50 + 273) K

air =6.47 kg/m3

2. Specific weight

air = air g

= 6.47kg/ m3 9.81 m/ s2

air = 63.5 N/ m3

REVIEW


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Situation:Consider a mass of air in the atmosphere.V= 1 mi3.

Find:Mass of air using units of slugs and kg.

Properties:From Table A.2,air = 0.00237 slugs/ft

3.

Assumptions:The density of air is the value at sea level for standard conditions.

SOLUTION

Units of slugs

M=VM= 0.00237 slug

ft3 (5280)3 ft3

M= 3.49108 slugs

Units of kg

M=

3.49108 slug

14.59

kg

slug

M= 5.09109 kg

REVIEW

The mass will probably be somewhat less than this because density decreases withaltitude.

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Situation:For a cyclist, temperature changes affect air density, thereby affecting both aero-

dynamic drag and tire pressure.

Find:a.) Plot air density versus temperature for a range of -10oC to 50oC.b.) Plot tire pressure versus temperature for the same temperature range.

Properties:From Table A.2,Rair = 287 J/kg/K.Initial conditions for part b: p= 450 kPa,T = 20 C.

Assumptions:For part b, assume that the bike tire volume does not change.

PLAN

Apply the ideal gas law.

SOLUTION

a.) Ideal gas law

= p

RT =

101000 kPa

(287J/ kg K)(273 + T)

Temperature (oC )

-20 -10 0 10 20 30 40 50 60

1.05

1.10

1.15

1.20

1.25

1.30

1.35

1.40

Density(kg/m

)3

b.) If the volume is constant, since mass cant change, then density must be constant.Thus

p

T =

poTo

p= 450 kPa

T

20 C

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Temperature,oC

-20 -10 0 10 20 30 40 50 60

380

400

420

440

460

480

500

520

Tirepressure,

kPa

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Situation:Design of a CO2 cartridge to inflate a rubber raft.Inflation pressure = 3 psi above patm = 17.7 psia = 122 kPa abs.

Find:Estimate the volume of the raft.Calculate the mass of CO2 (in grams) to inflate the raft.

Sketch:

Assumptions:CO2 in the raft is at 62 F = 290 K.Volume of the raft Volume of a cylinder withD = 0.45 m& L = 16 m(8 meters

for the length of the sides and 8 meters for the lengths of the ends plus center tubes).

Properties:

CO2, Table A.2, R = 189 J/kgK.

PLAN

Since mass is related to volume by m = V, the steps are:1. Find volume using the formula for a cylinder.2. Find density using the ideal gas law (IGL).3. Calculate mass.

SOLUTION

1. Volume

V = D2

4 L

=

0.452

4 16

m3

V= 2.54 m3

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2. Ideal gas law

= p

RT

= 122, 000N/ m2

(189J/ kg K)(290K)

= 2.226kg/m3

3. Mass of CO2

m = V

=

2.226kg/m3

2.54 m3

m= 5660 g

REVIEW

The final mass (5.66 kg = 12.5 lbm) is large. This would require a large and potentiallyexpensive CO2 tank. Thus, this design idea may be impractical for a product that isdriven by cost.

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Situation:A helium filled balloon is being designed.r= 1.3 m,z = 80, 000ft.

Find:Weight of helium inside balloon.

Properties:From Table A.2, RHe = 2077 J/kgK.

p= 0.89 bar = 89 kPa,T = 22 C = 295.2 K.

PLAN

Weight is given byW =mg. Mass is related to volume byM= V. Density canbe found using the ideal gas law.

SOLUTIONVolume in a sphere

V = 4

3r3

= 4

3 (1.3 m)3

= 9.203m3

Ideal gas law

= pRT

= 89, 000N/ m2

(2077 J/ kg K) (295.2 K)

= 0.145kg/m3

Weight of helium

W = V g

=

0.145kg/m3

9.203m3

9.81 m/ s2

= 13.10 NWeight =13.1N

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Situation:Hydrometers are used to measure alcohol content of wine and beer by measuring

specific weight at various stages of fermentation.Fermentation is described by the following equation:

C6H12O6 2(CH3CH2OH) + 2(CO2)

Find:Final specific gravity of the wine.Percent alcohol content by volume after fermentation.

Assumptions:All of the sugar is converted to alcohol.Initial liquid is only sugar and water.

Properties:Salcohol = 0.80,Ss= 1.59,Sw= 1.08.

PLAN

Imagine that the initial mixture is pure water plus saturated sugar solution and thenuse this visualization to find the mass of sugar that is initially present (per unitof volume). Next, apply conservation of mass to find the mass of alcohol that isproduced (per unit of volume). Then, solve for the problem unknowns.

SOLUTION

The initial density of the mixture is

mix=wVw+ sVs

Vo

where w ands are the densities of water and sugar solution (saturated), Vo is theinitial volume of the mixture, and Vs is the volume of sugar solution. The totalvolume of the mixture is the volume of the pure water plus the volume of saturatedsolution

Vw+ Vs= Vo

The specific gravity is initially 1.08. Thus

Si = mix

w= (1

VsVo

) + sw

VsVo

1.08 = (1 VsVo

) + 1.59VsVo

VsVo

= 0.136

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Thus, the mass of sugar per unit volume of mixture

MsVo

= 1.590.136

= 0.216 kg/m3

The molecular weight of glucose is 180 and ethyl alcohol 46. Thus 1 kg of glucoseconverts to 0.51 kg of alcohol so the final density of alcohol is

MaVo

= 0.2160.51

= 0.110 kg/m3

The density of the final mixture based on the initial volume is

MfVo

= (1 0.136) + 0.110

= 0.974 kg/m3

The final volume is altered because of conversion

VfVo

= Mw

wVo+

MaaVo

= Vw

Vo+

0.51MsaVo

= Vw

Vo+

0.51sa

VsVo

= 0.864 +0.511.59

0.8 0.136

= 1.002

The final density is

MfVf

= Mf

Vo

VoVf

= 0.974 1

1.002= 0.972 kg/m3

The final specific gravity isSf= 0.972

The alcohol content by volumeVaVf

= Ma

aVf

= Ma

Vo

1

a

VoVf

= 0.110 1

0.8

1

1.002= 0.137

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Thus,

Percent alcohol by volume = 13.7%

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24/917


Situation:Several preview questions about viscosity are answered.

Find:

(a) The primary dimensions of viscosity and five common units of viscosity.(b) The viscosity of motor oil (in traditional units).(c) How and why viscosity of water varies with temperature?(d) How and why viscosity of air varies with temperature?

SOLUTION

a) Primary dimensions of viscosity are [MLT] .

Five common units are:

i) N sm2

; ii) dyn scm2

; iii) poise; iv) centipoise; and v) lbf sft2

(b) Tofind the viscosity of SAE 10W-30 motor oil at 115 F, there are no tablular datain the text. Therefore, one should use Figure A.2. For traditional units (becausethe temperature is given in Fahrenheit) one uses the left-hand axis to report that

= 1.2103 lbf sft2

.

Note: one should be careful to identify the correct factor of 10 for the log cycle thatcontains the correct data point. For example, in this problem, the answer is between1 103 and1 102.One should be able to determine that the answer is 1.2 103

and not 1 102.

(c) The viscosity of water decreases with increasing temperature . This is true forall liquids, and is because the loose molecular lattice within liquids, which provides agiven resistance to shear at a relatively cool temperature, has smaller energy barriersresisting movement at higher temperatures.

(d) The viscosity of air increases with increasing temperature . This is true for allgases, and is because gases do not have a loose molecular lattice. The only resistanceto shear provided in gases is due to random collision between different layers. Asthe temperature increases, there are more likely to be more collisions, and thereforea higher viscosity.

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Situation:Change in viscosity and density due to temperature.T1= 10 C,T2= 70 C.

Find:Change in viscosity and density of water.Change in viscosity and density of air.

Properties:p= 101 kN/ m2.

PLAN

For water, use data from Table A.5. For air, use data from Table A.3

SOLUTION

Water

70= 4.04104 Ns/m2

10= 1.31103 Ns/m2

= 9. 06104 N s/ m2

70 = 978 kg/m3

10 = 1000 kg/m3

= 22 kg/ m3

Air

70 = 2.04105 N s/m2

10 = 1.76105 N s/m2

= 2. 8106 N s/ m2

70 = 1.03kg/m3

10 = 1.25kg/m3

= 0.22 kg/ m3

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Situation:Air at certain temperatures.T1= 10 C,T2= 70 C.

Find:Change in kinematic viscosity.

Properties:From Table A.3,70 = 1.9910

5 m2/s,10= 1.41105 m2/s.

PLAN

Use properties found in Table A.3.

SOLUTION

vair,1070 = (1.99 1.41)105

vair,1070 = 5.8106m2/s

REVIEW

Sutherlands equation could also be used to solve this problem.

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Situation:Viscosity of SAE 10W-30 oil, kerosene and water.T = 38 C = 100 F.

Find:Dynamic and kinematic viscosity of each fluid.

PLAN

Use property data found in Table A.4, Fig. A.2 and Table A.5.

SOLUTION

Oil (SAE 10W-30) kerosene water

(N s/m

2

) 6.7102

1.4103

(Fig. A-2) 6.8104

(kg/m3) 880 814 993

(m2/s) 7.6105 1.7106 (Fig. A-2) 6.8107

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Situation:Dynamic and kinematic viscosity of air and water.T = 20 C.

Find:Ratio of dynamic viscosity of air to that of water.Ratio of kinematic viscosity of air to that of water.

Properties:From Table A.3,air,20C= 1.8110

5 Ns/m2; = 1.51105 m2/sFrom Table A.5,water,20C= 1.0010

3 Ns/m2; = 1.00106 m2/s

SOLUTION

Dynamic viscosity

airwater

= 1.81105

N s/ m2

1.00103 N s/ m2

airwater

= 1.81102

Kinematic viscosity

airwater

= 1.51105 m2/ s

1.00106 m2/ s

air

water

= 5.1

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Situation:Sutherlands equation and the ideal gas law describe behaviors of common gases.

Find:

Develop an expression for the kinematic viscosity ratio /o, whereis at temper-atureT and pressurep.

Assumptions:Assume a gas is at temperature Toand pressurepo, where the subscript o defines

the reference state.

PLAN

Combine the ideal gas law and Sutherlands equation.

SOLUTION

The ratio of kinematic viscosities is

o=

o

o

=

T

To

3/2 To+ ST+ S

pop

T

To

o=

pop

T

To

5/2To+ S

T+ S

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30/917


Situation:The dynamic viscosity of air.o= 1.7810

5 Ns/m2.T

o= 15 C,T= 100 C.

Find:Dynamic viscosity.

Properties:From Table A.2,S= 111K.

SOLUTION

Sutherlands equation

o= T

To

3/2 To+ S

T+ S

=

373K

288K

3/2 288 K + 111 K373 K + 111 K

o= 1.21

Thus

= 1.21o= 1.21 1.7810

5 N s/ m2= 2.15105 Ns/m2

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31/917


Situation:Methane gas.vo= 1.5910

5 m2/ s.T

o= 15 C,T= 200 C.

po= 1 atm,p= 2 atm.

Find:Kinematic viscosity ( m2/ s).

Properties:From Table A.2,S= 198K.

PLAN

Apply the ideal gas law and Sutherlands equation.

SOLUTION

=

o=

o

o

Ideal-gas law

o=

o

pop

T

To


o=

pop

T

To

5/2To+ S

T+ S

so

o=

1

2

473K

288K

5/2288 K + 198 K

473 K + 198 K= 1.252

and

= 1.2521.59105 m2/s

= 1.99105 m2/ s

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32/917


Situation:Nitrogen gas.o= 3.5910

7 lbf s/ ft2.T

o= 59 F,T= 200 F.

Find: using Sutherlands equation.

Properties:From Table A.2,S=192oR.

SOLUTION


o= T

To

3/2 To+ S

T+ S

=

660oR

519oR

3/2 519oR + 192oR660oR + 192oR

= 1.197

= 1.197

3.59107

lbf s

ft2

= 4. 297107

= 4.30107 lbf-s/ft2

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Situation:Helium gas.vo= 1.2210

3 ft2/ s.T

o= 59 F,T = 30 F.

po= 1 atm,p= 1.5atm.

Find:Kinematic viscosity using Sutherlands equation.

Properties:From Table A.2,S=143oR.

PLAN

Combine the ideal gas law and Sutherlands equation.

SOLUTION

o=

pop

T

To

5/2To+ S

T+ S

= 1.5

1

490oR

519oR

5/2 519oR + 143oR490oR + 143oR

= 1.359

= 1.359

1.22103

ft2

s

= 1. 658103ft

2

s

= 1.66103 ft2/ s

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34/917


Situation:Absolute viscosity of propane.To= 100 C,o= 110

5 N s/ m2.T= 400 C, = 1.72105 N s/ m2.

Find:Sutherlands constant.

SOLUTION


S

To=

o

ToT

1/2 1

1 o

ToT

3/2Also

o

= 1.72

ToT

= 373K

673K

Thus

S

To= 0.964

S= 360 K

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Situation:Ammonia at room temperature.To= 68 F,o= 2.0710

7 lbf s/ ft2.T= 392 F, = 3.46107 lbf s/ ft2.

Find:Sutherlands constant.

SOLUTION


S

To=

o

ToT

1/2 1

1 o

ToT

3/2 (1)Calculations

o

= 3.46107 lbf s/ ft2

2.07107 lbf s/ ft2 = 1.671 (a)

ToT

= 528 R

852 R= 0.6197 (b)

Substitute (a) and (b) into Eq. (1)

S

To= 1.71

S= 903 oR

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36/917


Situation:SAE 10W30 motor oil.To= 38 C,o= 0.067Ns/ m

2.T = 99 C, = 0.011Ns/ m2.

Find:The viscosity of motor oil, (60oC), using the equation =Ceb/T.

PLAN

Use algebra and known values of viscosity () to solve for the constant b. Then,solve for the unknown value of viscosity.

SOLUTION

Viscosity variation of a liquid can be expressed as = Ceb/T. Thus, evaluate at

temperaturesT andTo and take the ratio:

o= exp

b(

1

T

1

To)

Take the logarithm and solve for b.

b= ln (/o)

( 1T 1To

)

Data

/o = 0.011Ns/ m2

0.067Ns/ m2 = 0.164

T = 372 K

To = 311 K

Solve forbb= 3430 (K)

Viscosity ratio at 60oC

o= exp3430 1

333K

1

311K

= 0.4833

= 0.48330.067Ns/ m2

= 0.032 N s/ m2

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Situation:Viscosity of grade 100 aviation oil.To= 100 F,o= 4.4310

3 lbf s/ ft2.T= 210 F, = 3.9104 lbf s/ ft2.

Find:(150oF), using the equation = Ceb/T.

PLAN

Use algebra and known values of viscosity () to solve for the constant b. Then,solve for the unknown value of viscosity.

SOLUTION

Viscosity variation of a liquid can be expressed as = Ceb/T. Thus, evaluate at

temperaturesT andTo and take the ratio:

o= exp

b(

1

T

1

To)

Take the logarithm and solve for b

b= ln (/o)

( 1T 1To

)

Data

o

= 0.39103 lbf s/ ft2

4.43103 lbf s/ ft2 = 0.08804

T = 670oR

To = 560oR

Solve forbb= 8293 (oR)

Viscosity ratio at 150oF

o = exp

8293 1

610oR

1

560oR

= 0.299

= 0.299

4.43103

lbf s

ft2

= 1.32103 lbf sft2

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38/917


Situation:Oil (SAE 10W30) fills the space between two plates.y= 1/8 = 0.125in,u = 25ft/ s.

Lower plate is at rest.

Find:Shear stress in oil.

Properties:Oil (SAE 10W30 @150 F) from Figure A.2: = 5.2104 lbfs/ft2.

Assumptions:1.) Assume oil is a Newtonian fluid.2.) Assume Couette flow (linear velocity profile).

SOLUTION

Rate of strain

du

dy =

u

y

= 25ft/ s

(0.125/12) ft

du

dy = 2400 s1

Newtons law of viscosity

=

du

dy

=

5.2104

lbf s

ft2

2400

1

s

= 1. 248lbf

ft2

= 1.25 lbfft2

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Situation:Properties of air and water.T = 40 C,p = 170 kPa.

Find:Kinematic and dynamic viscosities of air and water.

Properties:Air data from Table A.3, air = 1.9110

5 Ns/m2

Water data from Table A.5, water= 6.53104 Ns/m2, water= 992 kg/m

3.

PLAN

Apply the ideal gas law to find density. Find kinematic viscosity as the ratio ofdynamic and absolute viscosity.

SOLUTIONA.) AirIdeal gas law

air = p

RT

= 170, 000 kPa

(287J/ kg K)(313.2 K)

= 1.89 kg/m3

air = 1.91105 N s

m2

=

= 1.91105 N s/ m2

1.89kg/ m3

air = 10.1106 m2/ s

B.) water

water= 6.53105 Ns/m2

=

= 6.53104 N s/ m2

992kg/ m3

water= 6.58107 m2/s

38



40/917


Situation:Sliding plate viscometer is used to measure fluid viscosity.A= 50100 mm, y = 1 mm.u= 10 m/ s,F = 3 N.

Find:Viscosity of the fluid.

Assumptions:Linear velocity distribution.

PLAN

1. The shear forceis a force/area.2. Use equation for viscosity to relate shear force to the velocity distribution.

SOLUTION1. Calculate shear force

= Force

Area

= 3 N

50mm100 mm = 600 N

2. Find viscosity

=

dudy

=

600N

[10m/ s] / [1 mm]

= 6102 N sm2

39

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41/917


Situation:Water flows near a wall. The velocity distribution is

u(y) =ay

b1/6

a= 10 m/ s,b = 2 mm and y is the distance (mm) from the wall.

Find:Shear stress in the water at y = 1 mm.

Properties:Table A.5 (water at 20 C): = 1.00103 N s/ m2.

SOLUTION

Rate of strain (algebraic equation)

du

dy =

d

dy

ay

b

1/6

= a

b1/61

6y5/6

= a

6b

b

y

5/6

Rate of strain (at y = 1 mm)

du

dy =

a

6b

b

y

5/6

= 10 m/ s

60.002m

2 mm

1 mm

5/6= 1485 s1

Shear Stress

y=1mm = du

dy

=

1.00103

N s

m2

1485s1

= 1.485 Pa

(y= 1 mm) = 1.49Pa

40



42/917


Situation:Velocity distribution of crude oil between two walls.= 8105 lbf s/ ft2,B = 0.1 ft.u= 100y(0.1 y) ft/ s, T= 100 F.

Find:Shear stress at walls.

SOLUTION

Velocity distributionu= 100y(0.1 y) = 10y 100y2

Rate of strain

du/dy = 10 200y

(du/dy)y=0 = 10s2

(du/dy)y=0.1 = 10s1

Shear stress

0 = du

dy = (8 105)10

0 = 8104 lbf/ft2

0.1 = 8104 lbf/ft2

Plot

0.00

0.02

0.04

0.06

0.08

0.10

Distance

Velocity

41

-


43/917


Situation:A liquid flows between parallel boundaries.y0 = 0.0 mm,V0= 0.0 m/ s.y1 = 1.0 mm,V1= 1.0 m/ s.y2 = 2.0 mm,V2= 1.99 m/ s.y3 = 3.0 mm,V3= 2.98 m/ s.

Find:(a) Maximum shear stress.(b) Location where minimum shear stress occurs.

SOLUTION

(a) Maximum shear stress

= dV/dy

max (V /y)next to wall

max = (103N s/m2)((1m/s)/0.001m)

max = 1.0 N/m2

(b)The minimum shear stress will occur midway between the two walls . Its mag-nitude will be zero because the velocity gradient is zero at the midpoint.

42



44/917


Situation:Glycerin is flowing in between two stationary plates. The velocity distribution is

u= 1

2

dp

dxBy y2

dp/dx= 1.6kPa/ m,B = 5 cm.

Find:Velocity and shear stress at a distance of 12 mm from wall (i.e. at y = 12 mm).Velocity and shear stress at the wall (i.e. aty = 0 mm).

Properties:Glycerin (20 C), Table A.4: = 1.41 N s/ m2.

PLAN

Find velocity by direct substitution into the specified velocity distribution.Find shear stress using the definition of viscosity: = (du/dy), where the rate-of-strain (i.e. the derivativedu/dy) is found by differentiating the velocity distribution.

SOLUTION

a.) Velocity (at y = 12 mm)

u = 1

2

dp

dx

By y2

=

1

2 (1.41 N s/ m2) 1600N/ m3

(0.05m)(0.012m) (0.012m)2

= 0.2587

m

s

u (y= 12 mm) = 0.259m/ s

Rate of strain (general expression)

du

dy =

d

dy

1

2

dp

dx

By y2

=

1

2

dp

dx

d

dy

By y2

= 1

2dp

dx (B 2y)

Rate of strain (at y = 12mm)

du

dy =

1

2

dp

dx

(B 2y)

=

1

2 (1.41 N s/ m2)

1600

N

m3

(0.05 m 20.012m)

= 14.75 s1

43

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45/917

Definition of viscosity

= du

dy

= 1.41N s

m2 14.75 s1= 20. 798 Pa

(y= 12 mm) = 20.8 Pa

b.) Velocity (at y = 0 mm)

u = 1

2

dp

dx

By y2

=

1

2 (1.41 N s/ m2)

1600N/ m3

(0.05m)(0m) (0m)2

= 0.00

m

s

u (y= 0 mm) = 0 m/ s

Rate of strain (at y = 0 mm)

du

dy =

1

2

dp

dx

(B 2y)

=

1

2 (1.41 N s/ m2)

1600

N

m3

(0.05 m 20 m)

= 28.37 s1

Shear stress (at y = 0 mm)

= du

dy

=

1.41

N s

m2

28.37 s1

= 40.00Pa

(y= 0 mm) = 40.0 Pa

REVIEW

1. As expected, the velocity at the wall (i.e. aty = 0) is zero due to the no slipcondition.

2. As expected, the shear stress at the wall is larger than the shear stress awayfrom the wall. This is because shear stress is maximum at the wall and zeroalong the centerline (i.e. at y = B/2).

44



46/917


Situation:Laminar flow occurs between two horizontal parallel plates. The velocity distrib-

ution is

u=

1

2

dp

ds

Hy

y2

+ uty

H

Pressurep decreases with distance s, and the speed of the upper plate is ut. Notethat ut has a negative value to represent that the upper plate is moving to the left.

Moving plate: y = H.Stationary plate: y= 0.

Find:(a) Whether shear stress is greatest at the moving or stationary plate.(b) Location of zero shear stress.(c) Derive an expression for plate speed to make the shear stress zero at y = 0.

Sketch:

H

u

ut

y

s

PLAN

By inspection, the rate of strain (du/dy) or slope of the velocity profile is larger atthe moving plate. Thus, we expect shear stress to be larger at y = H. To checkthis idea, find shear stress using the definition of viscosity: = (du/dy). Evaluateand compare the shear stress at the locations y = Handy = 0.

SOLUTION

Part (a)1. Shear stress, from definition of viscosity

= dudy

= d

dy

1

2

dp

ds

Hy y2

+ ut

y

H

=

H

2

dp

ds+

y

dp

ds+

utH

(y) = (H 2y)

2

dp

ds+

utH

45

-


47/917

Shear stress at y = H

(y= H) = (H 2H)

2

dp

ds+

utH

= H

2

dp

ds +

ut

H

(1)

2. Shear stress aty = 0

(y= 0) = (H 0)

2

dp

ds+

utH

= H

2

dp

ds

+

utH

(2)

Since pressure decreases with distance, the pressure gradientdp/dsis negative. Sincethe upper wall moves to the left, ut is negative. Thus, maximum shear stress occursaty = Hbecause both terms in Eq. (1) have the same sign (they are both negative.)

In other words,|(y= H)|> |(y= 0)|

.

Maximum shear stress occur at y = H .

Part (b)Use definition of viscosity to find the location (y)of zero shear stress

= du

dy

= (1/2)dp

ds(H 2y) +

ut

H

= (1/2)dp

ds(H 2y) +

ut

H

Set= 0 and solve fory

0 =

(1/2)

dp

ds (H

2y) +

ut

H

y=H

2

utHdp/ds

Part (c)

46



48/917

= du

dy = 0at y = 0

du

dy = (1/2)

dp

ds(H 2y) +

utH

Then, at y = 0 :du/dy= 0 = (1/2)dpds

H+ utH

Solve forut : ut= (1/2)dp

dsH2

Note : because dp

ds


49/917


Situation:Oxygen at 50 F and 100 F.

Find:

Ratio of viscosities: 100

50 .

SOLUTION

Because the viscosity of gases increases with temperature 100/50 > 1. Correct

choice is (c) .

48



50/917


Situation:A cylinder falls inside a pipe filled with oil.d= 100mm,D = 100.5 mm.= 200mm,W= 15 N.

Find:Speed at which the cylinder slides down the pipe.

Properties:SAE 20W oil(10oC) from Figure A.2: = 0.35 Ns/m2.

SOLUTION

= dV

dyW

d =

Vfall(D d)/2

Vfall = W(D d)

2d

Vfall = 15N(0.5103 m)

(20.1 m0.2 m3.5101 N s/ m2)

Vfall = 0.17m/s

49

-


51/917


Situation:A disk is rotated very close to a solid boundary with oil in between.a= 1 rad/ s,r2= 2 cm,r3 = 3 cm.

b= 2 rad/ s,r

b= 3 cm.

H= 2 mm,c= 0.0 1 N s/ m2.

Find:(a) Ratio of shear stress at 2 cm to shear stress at 3 cm.(b) Speed of oil at contact with disk surface.(c) Shear stress at disk surface.

Assumptions:Linear velocity distribution: dV/dy= V /y= r/y.

SOLUTION

(a) Ratio of shear stresses

= dV

dy =

r

y

23

= 12/y

13/y

23

=2

3

(b) Speed of oil

V = r = 20.03

V = 0.06m/s

(c) Shear stress at surface

= dV

dy = 0.0 1 N s/ m2

0.06 m/ s

0.002m

= 0.30N/m2

50



52/917


Situation:A disk is rotated in a container of oil to damp the motion of an instrument.

Find:

Derive an equation for damping torque as a function ofD,S, and.

PLAN

Apply the Newtons law of viscosity.

SOLUTION

Shear stress

= dV

dy

= r

s

Find differential torqueon an elemental strip of area of radius r the differentialshear force will be dA or(2rdr). The differential torque will be the product ofthe differential shear force and the radius r.

dTone side = r[(2rdr)]

= rhr

s (2rdr)

i=

2

s r3dr

dTboth sides = 4r

s

r3dr

Integrate

T =

D/2Z0

4

s r3dr

T = 1

16

D4

s

51

-


53/917


Situation:One type of viscometer involves the use of a rotating cylinder inside a fixed cylinder.Tmin= 50 F,Tmax= 200 F.

Find:(a) Design a viscometer that can be used to measure the viscosity of motor oil.

Assumptions:Motor oil is SAE 10W-30. Data from Fig A-2: will vary from about2 104lbf-

s/ft2 to8 103lbf-s/ft2.Assume the only significant shear stress develops between the rotating cylinder and

the fixed cylinder.Assume we want the maximum rate of rotation ()to be 3 rad/s.Maximum spacing is 0.05 in.

SOLUTION

One possible design solution is given below.Design decisions:

1. Let h = 4.0in. = 0.333 ft

2. Let I.D. offixed cylinder = 9.00 in. = 0.7500 ft.

3. Let O.D. of rotating cylinder = 8.900 in. = 0.7417 ft.

Let the applied torque, which drives the rotating cylinder, be produced by a forcefrom a thread or small diameter monofilament line acting at a radial distance rs.

Here rs is the radius of a spool on which the thread of line is wound. The appliedforce is produced by a weight and pulley system shown in the sketch below.

hrc

rW

Pulley

The relationship between , rs,,h,and Wis now developed.

T =rcFs (1)

whereT =applied torque

52



54/917

rc= outer radius of rotating cylinderFs = shearing force developed at the outer radius of the rotating cylinder but Fs = As whereAs= area in shear = 2rch

=dV/dy V /r where V =rc and r= spacing

ThenT =rc(V /r)(2r

ch)

=rc(rc

r)(2rch) (2)

But the applied torqueT =W rs so Eq. (2) become

W rs = r3c (2)

h

rOr

=W rsr

2hr3c(3)

The weightWwill be arbitrarily chosen (say 2 or 3 oz.) and will be determined bymeasuring the time it takes the weight to travel a given distance. Sors =Vfall or= Vfall/rs. Equation (3) then becomes

=

W

Vf

r2sr3c

r

2h

In our design let rs= 2 in. = 0.1667 ft. Then

=

W

Vf

(0.1667)2

(.3708)30.004167

(2 .3333)

= W

Vf0.02779

0.05098 =

W

Vf

(1.085103)lbfs/ft2

Example: IfW = 2oz. = 0.125lb. andVfis measured to be 0.24 ft/s then

= 0.125

0.24(1.085103) lbf s/ ft2

= 0.564104 lbf s/ ft2

REVIEW

Other things that could be noted or considered in the design:

1. Specify dimensions of all parts of the instrument.

2. Neglect friction in bearings of pulley and on shaft of cylinder.

3. Neglect weight of thread or monofilament line.

4. Consider degree of accuracy.

5. Estimate cost of the instrument.

53

-


55/917


Situation:Elasticity of ethyl alcohol and water.Eethyl= 1.0610

9 Pa.E

water= 2.15109 Pa.

Find:Which substance is easier to compress, and why.

PLAN

Consider bulk density equation.

SOLUTION

The bulk modulus of elasticity is given by:

E= p V

V = p

d/

This means that elasticity is inversely related to change in density, and to the negativechange in volume.Therefore, the liquid with the smaller elasticity is easier to compress.Ethyl alcohol is easier to compress because it has the smaller elasticity , because elas-

ticity is inversely related to change in density.

54



56/917


Situation:Pressure is applied to a mass of water.V= 2000 cm3,p = 2106 N/ m2.

Find:Volume after pressure applied (cm3).

Properties:From Table A.5,E= 2.2109 Pa

PLAN

1. Use modulus of elasticity equation to calculate volume change resulting frompressure change.2. Calculate final volume based on original volume and volume change.

SOLUTION1. Elasticity equation

E = p V

V

V = p

E V

=

(2106) Pa

(2.2109) Pa

2000 cm3

= 1.82cm3

2. Final volume

Vfinal = V+ V

= (2000 1.82) cm3

Vfinal = 1998 cm3

55

-


57/917


Situation:Water is subjected to an increase in pressure.

Find:

Pressure increase needed to reduce volume by 2%.Properties:

From Table A.5,E= 2.2109 Pa.

PLAN

Use modulus of elasticity equation to calculate pressure change required to achievethe desired volume change.

SOLUTION Modulus of elasticity equation

E = p VV

p = EV

V

=

2.2109 Pa0.01 V

V

=

2.2109 Pa

(0.02)

= 4.4107 Pa

p= 44 MPa

56



58/917


Situation:Open tank of water.T20 = 20 C,T80 = 80 C.V= 400 l,d = 3 m.Hint: Volume change is due to temperature.

Find:Percentage change in volume.Water level rise for given diameter.

Properties:

From Table A.5: 20 = 998 kg

m3,and80 = 972

kg

m3.

PLAN

This problem is NOT solved using the elasticity equation, because the volume change

results from a change in temperature causing a density change, NOT a change inpressure. The tank is open, so the pressure at the surface of the tank is alwaysatmospheric.

SOLUTION

a. Percentage change in volume must be calculated for this mass of water at twotemperatures.For the first temperature, the volume is given as V20 = 400 L = 0.4 m

3.Its density is20 = 998

kg

m3. Therefore, the mass for both cases is given by.

m = 998kgm3

0.4 m3

= 399.2 kg

For the second temperature, that mass takes up a larger volume:

V80 = m

=

399.2 kg

972 kgm3

= 0.411m3

Therefore, the percentage change in volume is

0.411m3 0.4 m3

0.4 m3 = 0.0275

volume % change = = 2.8%

b. If the tank hasD = 3 m, thenV= r2h= 7.68h.Therefore:

57

-


59/917

h20 = .052m

h80 = .054m

And water level rise is0.054 0.52m = 0.002 m = 2 mm.

water level rise is = 0.002 m = 2 mm

REVIEW

Density changes can result from temperature changes, as well as pressure changes.

58



60/917


Situation:Surface tension is an energy/area.

Find:

Show that Energy

Area equals ForceLength .

Energy

Area =

forcedistance

area

=

"M LT2 L

L2

#

=

M

T2

Force

Length =

"M LT2

L

#

=

M

T2

The primary dimensions for EnergyArea

and ForceLength

are both

MT2

, so they are equal.

59

-


61/917


Situation:Very small spherical droplet of water.

Find:

Pressure inside.

SOLUTION

Refer to Fig. 2-6(a). The surface tension force,2r , will be resisted by the pressureforce acting on the cut section of the spherical droplet or

p(r2) = 2r

p = 2

r

p=4

d

60



62/917


Situation:A spherical soap bubble.Inside radiusR, wall-thicknesst, surface tension .Special case: R= 4 mm.

Find:Derive a formula for the pressure difference across the bubblePressure difference for bubble with R = 4 mm.

Assumptions:The effect of thickness is negligible, and the surface tension is that of pure water.

PLAN

Apply equilibrium, then the surface tension force equation.

SOLUTIONForce balance

p

2 x 2 R

Surface tension force

XF = 0

pR2 2(2R) = 0

Formula for pressure difference

p=

4

RPressure difference

p4mm rad. = 47.3102 N/m

0.004 m

p4mm rad. = 73.0 N/m2

61

-


63/917


Situation:A water bug is balanced on the surface of a water pond.n= 6legs, = 5 mm/leg.

Find:Maximum mass of bug to avoid sinking.

Properties:Surface tension of water, from Table A.4, = 0.073 N/m.

PLAN


SOLUTION

Force equilibrium

Upward force due to surface tension = Weight of BugFT = mg

To find the force of surface tension (FT), consider the cross section of one leg of thebug:

F F

Surface tensionforce on oneside of leg

Cross sectionof bug leg

Assume is smallThen cos =1; F cos = F


FT = (2/leg)(6legs)

= 12

= 12(0.073N/m)(0.005 m)

= 0.00438 N

Apply equilibriumFT mg = 0

m = FT

g =

0.00438 N

9.81 m2/ s

= 0.4465103 kg

m= 0.447103 kg

62



64/917


Situation:A water column in a glass tube is used to measure pressure.d1= 0.25in,d2= 1/8 in,d3= 1/32in.

Find:Height of water column due to surface tension effects for all diameters.

Properties:From Table A.4: surface tension of water is 0.005 lbf/ft.

SOLUTION


h = 4

d

=40.005 lbf/ ft

62.4lbf/ ft3

d

=3.21104

d

ft.

d = 1

4 in.=

1

48 ft.; h=

3.21104 ft

1/48 = 0.0154 ft. = 0.185 in.

d = 1

8 in.=

1

96 ft.; h=

3.21104 ft

1/96 = 0.0308 ft. = 0.369 in.

d = 1

32 in.=

1

384 ft.; h=

3.21104 ft

1/384 = 0.123 ft.= 1.48 in.

63

-


65/917


Situation:Two vertical glass platesy= 1 mm

Find:Capillary rise(h)between the plates.

Properties:From Table A.4, surface tension of water is 7.3102 N/m.

PLAN


SOLUTION

y

y

Equilibrium

XFy = 0

Force due to surface tension = Weight offluid that has been pulled upward

(2) = (ht)

Solve for capillary rise (h)

2 ht = 0

h = 2

t

h = 2(7.3102 N/ m)

9810N/ m3 0.001m= 0.0149 m

h = 14.9 mm

64



66/917


Situation:A spherical water drop.d= 1 mm

Find:Pressure inside the droplet (N/m2)

Properties:From Table A.4, surface tension of water is 7.3102 N/m

PLAN


SOLUTION

Equilibrium (half the water droplet)

Force due to pressure = Force due to surface tension

pA = L

pR2 = 2R

Solve for pressure

p = 2

R

p = 27.3102 N/ m

(0.5103 m)

p= 292 N/m2

65

-


67/917


Situation:A tube employing capillary rise is used to measure temperature of waterT0= 0 C,T100 = 100 C0 = 0.0756N/ m, 100 = 0.0589N/ m

Find:Size the tube (this means specify diameter and length).

PLAN

Apply equilibrium and the surface tension force equation.

SOLUTION

The elevation in a column due to surface tension is

h=4

d

whereis the specific weight and d is the tube diameter. For the change in surfacetension due to temperature, the change in column elevation would be

h=4

d =

40.0167N/ m

9810N/ m3 d =

6.8106

d

The change in column elevation for a 1-mm diameter tube would be 6.8 mm . Spe-cial equipment, such the optical system from a microscope, would have to be used to

measure such a small change in deflection It is unlikely that smaller tubes made oftransparent material can be purchased to provide larger deflections.

66



68/917


Situation:A soap bubble and a droplet of water of equal diameter falling in aird= 2 mm, bubble= droplet

Find:Which has the greater pressure inside.

SOLUTION

The soap bubble will have the greatest pressure because there are two surfaces (twosurface tension forces) creating the pressure within the bubble. The correct choice is

a)

67

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69/917


Situation:A hemispherical drop of water is suspended under a surface

Find:

Diameter of droplet just before separationProperties:

Table A.5 (20 C): = 9790 N/ m3, = 0.073N/ m.

SOLUTION

Equilibrium

Weight of droplet = Force due to surface tension

D312

= (D)

Solve forD

D2 = 12

= 12(0.073 N/m)

9790 N/m3 = 8. 948105 m2

D = 9. 459103 m

D= 9.46mm

68



70/917


Situation:Surface tension is being measured by suspending liquid from a ringDi= 10 cm,Do= 9.5 cmW= 10 g,F= 16 g

Find:Surface tension

PLAN

1. Force equilibrium on the fluid suspended in the ring. For force due to surfacetension, use the form of the equation provided in the text for the special case of aring being pulled out of a liquid.2. Solve for surface tension - all the other forces are known.

SOLUTION

1. Force equilibrium

(Upward force) = (Weight offluid) + (Force due to surface tension)

F = W+ (Di+ Do)

2. Solve for surface tension

= F W

(Di+ Do)

= (0.016 0.010)kg9.81 m/ s2

(0.1 + 0.095)m= 9.61102

kg

s2

= 0.0961 N/m

69

-


71/917


Situation:A liquid reaches the vapor pressure

Find:

What happens to the liquid

SOLUTION

If a liquid reaches its vapor pressure for a given temperature, it boils .

70



72/917


Find:How does vapor pressure change with increasing temperature?

SOLUTIONThe vapor pressure increases with increasing temperature . To get an everyday feel

for this, note from the Appendix that the vapor pressure of water at 212 F(100 C)is 101 kPa(14.7 psia). To get water to boil at a lower temperature, you would haveto exert a vacuum on the water. To keep it from boiling until a higher temperature,you would have to pressurize it.

71

-


73/917


Situation:Watar at 60 F

Find:

The pressure that must be imposed for water to boilProperties:

Water (60 F), Table A.5: Pv = 0.363 psia

SOLUTION

The pressure to which the fluid must be exposed is P = 0.363 psia. This is lowerthan atmospheric pressure. Therefore, assuming atmospheric pressure is 14.7 psiagage, or 14.7 psig, the pressure needed could also be reported as P = -14.34 psig .

72



74/917


Situation:T = 20 C,fluid is water.

Find:

The pressure that must be imposed to cause boilingProperties:

Water (60 F), Table A.5: Pv = 2340 Pa abs

SOLUTION

Bubbles will be noticed to be forming when P =Pv.

P= 2340 Pa abs

73

-


75/917


Situation:Water in a closed tankT = 20 C,p = 10400 Pa

Find:Whether water will bubble into the vapor phase (boil).

Properties:From Table A.5, at T= 20 C,Pv= 2340 Pa abs

SOLUTION

The tank pressure is 10,400 Pa abs, and Pv = 2340 Pa abs. So the tank pressure ishigher than thePv.Therefore the water will not boil .

REVIEW

The water can be made to boil at this temperature only if the pressure is reducedto 2340 Pa abs. Or, the water can be made to boil at this pressure only if thetemperature is raised to approximately50 C.

74



76/917


Situation:The boiling temperature of water decreases with increasing elevationpT

= 3.1 kPaoC

.

Find:Boiling temperature at an altitude of 3000 m

Properties:T= 100oC,p = 101kN/ m2.z3000 = 3000 m,p3000 = 69 kN/ m

2.

Assumptions:Assume that vapor pressure versus boiling temperature is a linear relationship.

PLAN

Develop a linear equation for boiling temperature as a function of elevation.

SOLUTION

LetBT = "Boiling Temperature." Then,BTas a function of elevation is

BT (3000 m)=BT (0 m) +

BT

p

p

Thus,

BT (3000 m) = 100 C + 1.0 C

3.1kPa (101 69) kPa= 89. 677 C

Boiling Temperature (3000 m)= 89.7 C

75

-


77/917


Apply the grid method to cases a, b, c and d.a.)Situation:

Pressure values need to be converted.

Find:Calculate the gage pressure (kPa) corresponding to 8 in. H2O (vacuum).

Solution:

b.)Situation:


Find:Calculate the gage pressure (psig) corresponding to 120 kPa-abs.

Properties:patm = 14.70psi.

Solution:

pabs =

120 kPa

1

14.70 psi

101.3 kPa

= 17.4 psia

pgage =pabs patm = (17.4psia) (14.70psia) = 2.71 psipgage = 2.71 psig

c.)Situation:

Pressure values need to be converted.Find:

Calculate the absolute pressure (psia) corresponding to a pressure of 0.5 bar (gage).

Properties:patm= 14.70psi.

Solution:

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pgage =

0.5bar

1

14.70psi

1.013 bar

= 7.25 psig

pabs =patm+pgage = (7.25 psig) + (14.70 psia) = 21.9psia

pabs = 21.9 psia

d.)Situation:


Find:Calculate the pressure (kPa-abs) corresponding to a blood pressure of 120 mm-Hg.

Properties:

Solution:

pgage =120 mm-Hg

1 101.3 kPa

760 mm-Hg

= 17.00 kPa-gage

pabs =patm +pgage = (101.3kPa) + (17.00 kPa-gage) = 118 kPagage

pabs = 118 kPa gage

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Apply the grid method to:a.)Situation:

A sphere contains an ideal gas.

Find:Calculate the density of helium at a gage pressure of 20 in. H2O.

Properties:From Table A.2: Rhelium = 2077 J/ kg K.

Solution:

pabs = patm +pgage = 101.3 kPa +

20 in. H2O

1

248.8 Pa

1.0 in. H2O

= 106.3kPa

Ideal gas law:

= p

RT =

106.3kPa

1

kg K

2077J

1

293.2 K

1000 Pa

1kPa

J

N m

N

Pa m2

= 0.175 kg/m3

b.)Situation:

A sphere contains an ideal gas.

Find:Calculate the density of argon at a vacuum pressure of 3 psi.

Properties:From Table A.2: Rmethane= 518J/ kg K.

Solution:

pabs = patm pvacuum = 101.3kPa

3psi

1

101.3kPa

14.696 psi

= 80.62 kPa

Ideal gas law:

= p

RT =

80.62 kPa

1

kg K

518J

1

293.2 K

1000 Pa

1kPa

J

N m

N

Pa m2

= 0.531 kg/m3

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Using Section 3.1 and other resources, answer the questions below. Strive for depth,clarity, and accuracy while also combining sketches, words and equations in ways thatenhance the effectiveness of your communication.

a. What are five important facts that engineers need to know about pressure?

Pressure is often expressed using "gage pressure," where gage pressure is thedifference between local atmospheric pressure and actual pressure.

Primary dimensions of pressure are M/LT2.

Vacuum pressure = negative gage pressure. Negative vacuum pressure = gagepressure.

Pressure is often expressed as length of a fluid column; e.g. the pressure of air

in a duct is 10 inches of water column.

pressure is defined using a derivative

b. What are five common instances in which people use gage pressure?

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car tire pressure is expressed as gage pressure.

blood pressure measured by a doctor is a gage pressure.

the pressure inside a pressure cooker is expressed as a gage pressure.

a Bourdon-tube pressure gage gives a pressure reading as a gage pressure.

the pressure that a scuba diver feels is usually expressed as a gage pressure; e.g.a diver at a depth of 10 m will experience a pressure of 1 atm.

c. What are the most common units for pressure?

Pa, psi, psf

length of a column of water(in. H20; ft H2O)

length of a column of mercury (mm Hg; in. Hg)

bar

d. Why is pressure defined using a derivative?Pressure is defined as a derivative because pressure can vary at every point along asurface.

e. How is pressure similar to shear stress? How does pressure differ from shear stress?

Similarities

Both pressure and shear stress give a ratio of force to area. Both pressure and shear stress apply at a point (they are defined using a

derivative.

Pressure and shear stress have the same units.

Both pressure and shear stress are types of "stress."

Differences: (the easy way to show differences is to make a table as shownbelow)

Attribute Pressure Shear Stressdirection of associ-

ated force

associated with force normal

to area

associated with force tan-

gent to an areapresence in a hydro-static fluid

pressure is non-zero shear stress is zero

typical magnitude much larger than shearstress

much smaller than pressure

main physical cause associated with weight offluid & motion offluid (non-viscous effects)

associated with motion offluid (viscous effects)

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Situation:A Crosby gage tester is applied to calibrate a pressure gage.Indicated pressure on the gage is p = 200 kPa.W= 140 N,D = 0.03 m.

Find:Percent error in gage reading.

PLAN

1. Calculate the pressure that the gage should be indicating (true pressure).2. Compare this true pressure with the actual pressure.

SOLUTION

1. True pressure

ptrue = FA

= 140N

(/40.032) m2

= 198, 049 kPa

2. Percent error

% Error = (precorded ptrue)100

ptrue

= (200 kPa 198 kPa) 100

198 kPa= 1.0101%

% Error= 1.01%

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Situation:A hydraulic machine is used to provide a mechanical advantage.m1= 0.025kg,m2= 7500 kg.

Find:

(a) Derive an algebraic equation for the mechanical advantage.(b) CalculateD1 andD2 so the mouse can support the elephant.

Assumptions:

Neglect the mass of the pistons.

Neglect the friction between the piston and the cylinder wall.

The pistons are at the same elevation; thus, the pressure acting on the bottomof each piston is the same.

A mouse can fit onto a piston of diameter D1= 70 mm.

PLAN

1. Define "mechanical advantage."2. Derive an equation for the pressure acting on piston 1.3. Derive an equation for the pressure acting on piston 2.4. Derive an equation for mechanical advantage by combining steps 2 and 3.5. Calculate D2 by using the result of step 4.

SOLUTION

1. Mechanical advantage. Mechanical

advantage

=

Weight "lifted" by the mouse

Weight of the mouse =

W2W1

(1)

whereW2 is the weight of the elephant, and W1 is the weight of the mouse.

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2. Equilibrium (piston 1):

W1 = p

D21

4

p = W1 4D21

(2)3. Equilibrium (piston 2):

W2 = p

D22

4

p = W2

4

D22

(3)

4. Combine Eqs. (2) and (3):

p= W1

4D21

=W2

4D22

(5)

Solve Eq. (5) for mechanical advantage:

W2W1

=

D2D1

25. Calculate D2.

W2

W1= D2

D12

(7500kg) (9.80 m/ s2)

(0.025kg)(9.80 m/ s2) = 300000 =

D2

0.07 m

2D2 = 38.3 m

The ratio of(D2/D1)needs to be

300, 000.IfD1= 70mm,then D2= 38.3 m.

REVIEW

1. Notice. The mechanical advantage varies as the diameter ratio squared.2. The mouse needs a mechanical advantage of 300,000:1. This results in a pistonthat is impractical (diameter = 38.3 m = 126 ft !).

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Situation:To work the problem, data was recorded from a parked vehicle. Relevant infor-

mation:

Left front tire of a parked VW Passat 2003 GLX Wagon (with 4-motion).

Bridgestone snow tires on the vehicle.

Inflation pressure = 36 psig. This value was found by using a conventional"stick-type" tire pressure gage.

Contact Patch: 5.88in 7.5 in. The 7.5 inch dimension is across the tread.These data were found by measuring with a ruler.

Weight on the front axle = 2514 lbf. This data was recorded from a sticker

on the driver side door jamb. The owners manual states that this is maximumweight (car + occupants + cargo).

Assumptions:

The weight on the car axle without a load is 2000 lbf. Thus, the load actingon the left front tire is 1000 lbf.

The thickness of the tire tread is 1 inch. The thickness of the tire sidewall is1/2 inch.

The contact path isfl

at and rectangular. Neglect any tensile force carried by the material of the tire.

Find:Measure the size of the contact patch.Calculate the size of the contact patch.Compare the measurement with the calculation and discuss.

PLAN

To estimate the area of contact, apply equilibrium to the contact patch.

SOLUTION

Equilibrium in the vertical direction applied to a section of the car tire

piAi= Fpavement

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wherepi is the inflation pressure,Ai is the area of the contact patch on the inside ofthe tire and Fpavement is the normal force due to the pavement. Thus,

Ai = Fpavement

pi

= 1000lbf36lbf/ in2

= 27.8 in2

Comparison. The actual contact patch has an areaAo= 5.88in7.5 in = 44.1 in2.

Using the assumed thickness of rubber, this would correspond to an inside contactarea ofAo= 4.88in5.5 in = 26.8 in

2.Thus, the predicted contact area

27.8 in2

andthe measured contact area

26.8 in2

agree to within about 1 part in 25 or about 4%.

REVIEW

The comparison between predicted and measured contact area is highly dependenton the assumptions made.

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Problem 3.7

Apply the grid method to calculations involving the hydrostatic equation:

p= z = gz

Note: Unit cancellations are not shown in this solution.a.)Situation:

Pressure varies with elevation.z= 10 ft.

Find:Pressure change (kPa).

Properties:

= 90 lb/ ft3.

Solution:Convert density to units of kg/m3:

=

90 lbm

ft3

35.315ft3

m3

1.0 kg

2.2046 lbm

= 1442

kg

m3

Calculate the pressure change:

p= gz=

1442 kg

m3

9.81 m

s2

10ft

1.0

m3.208ft

Pa m s2kg

p= 43.1kPa

b.)Situation:

Pressure varies with elevation.z= 22 m,S= 0.8.

Find:Pressure change (psf).

Properties:

= 62.4lbf/ ft3.

Solution:

p= z = S H2O z =

(0.862.4) lbf

ft3

22 m

1.0

3.2808 ft

m

p= 3600 psf

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88/917

c.)Situation:


Find:

Pressure change (in H2O).

Properties:air, = 1.2 kg/ m3.

Solution:

p= gz=

1.2 kg

m3

9.81 m

s2

1000 ft

1.0

m3.281ft

Pa m s2kg

in.-H2O

248.4 Pa

p= 14.8in H2O

d.)Situation:

Pressure varies with elevation.p= 1/6atm,S= 13.

Find:Elevation change (mm).

Properties:= 9810 N/ m3, patm= 101.3kPa.

Solution:d. Calculate z (mm) corresponding toS= 13 and p= 1/6atm.

z =p

=

p

SH2O=

1/6atm

1.0

m3

(139810) N

101.3103 Pa

atm

1000 mm

1.0 m

z= 132 mm

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Problem 3.8

Using Section 3.2 and other resources, answer the questions below. Strive for depth,clarity, and accuracy while also combining sketches, words and equations in ways thatenhance the effectiveness of your communication.

a. What does hydrostatic mean? How do engineers identify if a fluid is hydrostatic?

Each fluid particle within the body is in force equilibrium(z-direction) with thenet force due to pressure balancing the weight of the particle. Here, the z-direction is aligned with the gravity vector.

Engineers establish hydrostatic conditions by analyzing the forces acting in thez-direction.

b. What are common forms of the hydrostatic equation? Are the forms equivalentor are they different?

There are three common forms; these are given in Table F.2 (front of book).

These equations are equivalent because you can start with any of the equationsand derive the other two.

c. What is a datum? How do engineers establish a datum?

A datum is a fixed reference point from which elevations are measured.

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Engineers select a datum that makes calculations easy. For example, selecta datum on the free surface of a river below a dam so that all elevations arepositive.

d. What are the main ideas of Eq. (3.5)? That is, what is the meaning of thisequation?

pz =p + z = constant

This equation means that the sum of(p + z)has the same numerical value at everylocation within a body offluid.

e. What assumptions need to be satisfied to apply the hydrostatic equation?

pz =p + z = constantThis equation is valid when

the density of the fluid is constant at all locations.

equilibrium is satisfied in the z-direction (net force of pressure balances weightof the fluid particle.

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Problem 3.9

Apply the grid method to each situation below. Unit cancellations are not shown inthese solutions.a.)Situation:


Find:Pressure change (Pa).

Properties:air, = 1.2 kg/ m3.

Solution:

p= gz

p = gz

=

1.2 kg

m3

9.81 m

s2

8 ft

1.0

m3.281ft

Pa m s2kg

p= 28.7 Pa

b.)Situation:

Pressure increases with depth in the ocean.Pressure reading is 2 atm gage.

Find:Water depth (m).

Properties:Seawater, Table A.4, S= 1.03, = 10070 N/ m3.

Solution:

z =p

=

2.0atm

1.0

m3

10070 N

101.3103 Pa

atm

N

Pa m2

z = 20.1 m

c.)Situation:

Pressure decreases with elevation in the atmosphere.z= 1200 ft.

Find:Pressure (mbar).

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Assumptions:Density of air is constant.

Properties:Air, = 1.1 kg/ m3.

Solution:

p=gz=

1.1 kg

m3

9.81 m

s2

1200 ft1.0

m3.281ft

Pa m s2kg

= 3947 Pa

Pressure at summit:

psummit = pbase+ p= 940 mbar

3947 Pa

1.0

102 mbar

Pa

psummit = 901 mbar(absolute)

d.)Situation:

Pressure increases with depth in a lake.z= 350 m.

Find:Pressure (MPa).

Properties:Water,= 9810 N/ m3.

Solution:

p = z

=

9810N

m3

350m

1.0

Pa m2

N

MPa

106 Pa

pmax = 3.4MPa(gage) [about 34 atmospheres]

e.)Situation:

Pressure increase with water depth in a standpipe.z= 60 m.

Find:Pressure (kPa).

Properties:Water,= 9810 N/ m3.

Solution:

16

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93/917

p = z

=

9810N

m3

60 m

1.0

Pa m2

N

kPa

103 Pa

pmax = 589 kPa (gage) [nearly 6 atmospheres]

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Situation:Air above a long tube is pressurized.Initial state: pair1= 50kPa-vacuumFinal state: p

air2= 25 kPa-vacuum.

Find:Willh increase or decrease?The change in water column height (h)in meters.

Assumptions:Atmospheric pressure is 100 kPa.

Properties:Water (20 C), Table A.5,= 9790 N/ m3.

PLANSince pressure increases, the water column height will decrease. Use absolute pressurein the hydrostatic equation.1. Findh (initial state) by applying the hydrostatic equation.2. Findh (final state)by applying the hydrostatic equation.3. Find the change in height by h= h(final state) h (initial state) .

SOLUTION

1. Initial State. Locate point 1 on the reservoir surface; point 2 on the water surfaceinside the tube:

p1

+ z1 = p2

+ z2

100 kPa

9790N/ m3+ 0 =

50 kPa

9790N/ m3+ h

h (initial state) = 5.107m

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2. Final State:

p1

+ z1 = p2

+ z2

100 kPa

9790N/ m3+ 0 =

75 kPa

9790N/ m3+ h

h (final state) = 2.554m

3. Change in height:

h = h(final state) h (initial state)= 2.554m 5.107m = 2.55 m

The height has decreased by 2.55 m.

REVIEW

Tip! In the hydrostatic equation, use gage pressure or absolute pressure. Usingvacuum pressure will give a wrong answer.

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Situation:A closed tank contains air, oil, and water.

Find:

Specific gravity of oil.Pressure at C (kPa-gage).

Sketch:

0.5 m

1.0 m

Air

0.5 m

1.0 m

0.5 m

CROWE: Fluid Mechanics 8e

Prob. 3-7 w-55

A

B

C

pA= 50.0 kPa

pB= 58.53 kPa

pC= ?

Oil

Water

T= 10C


PLAN

1. Find the oil specific gravity by applying the hydrostatic equation from A to B.2. Apply the hydrostatic equation to the water.3. Apply the hydrostatic equation to the oil.

4. Find the pressure at C by combining results for steps 2 and 3.

SOLUTION

1. Hydrostatic equation (from oil surface to elevation B):

pA+ zA = pB+ zB

50, 000 N/m2 + oil(1 m) = 58, 530 N/m2 + oil(0m)

oil = 8530 N/m3

Specific gravity:

S= oilwater

=8530 N/m3

9810 N/m3

Soil = 0.87

2. Hydrostatic equation (in water):

pc = (pbtm of oil) + water(1 m)

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3. Hydrostatic equation (in oil):

pbtm of oil = (58, 530 Pa + oil 0.5 m)

4. Combine equations:

pc = (58, 530 Pa + oil0.5 m) + water(1 m)

=

58, 530 Pa + 8530 N/ m3 0.5 m

+ 9810 N/ m3 (1m)

= 72, 605 N/m2

pc= 72.6 kPa-gage

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Situation:A manometer is described in the problem statement.dleft= 1 mm, dright= 3 mm.

Find:Water surface level in the left tube as compared to the right tube.

SOLUTION

(a) The water surface level in the left tube will be higher because of greater surfacetension effects for that tube.

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Situation:A force is applied to a piston.F1= 200 N,d1= 4 cm,d2= 10 cm.

Find:Force resisted by piston.

Assumptions:Neglect piston weight.

PLAN

Apply the hydrostatic equation and equilibrium.

SOLUTION

1. Equilibrium (piston 1)

F1 = p1A1

p1 = F1

A1

= 4200 N

(0.04m)2 m2

= 1.592105 Pa

2. Hydrostatic equation

p2+ z2 = p1+ z1

p2 = p1+ (Swater) (z1 z2)= 1.592105 Pa +

0.859810 N/ m3

(2 m)

= 1.425105 Pa

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100/917

3. Equilibrium (piston 2)

F2 = p2A2

= 1.425105 N/ m2 (0.1 m)24

!= 1119 N

F2= 1120 N

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Situation:A diver goes underwater.z= 50 m.

Find:Gage pressure (kPa).Ratio of pressure to normal atmospheric pressure.


PLAN

1. Apply the hydrostatic equation.2. Calculate the pressure ratio (use absolute pressure values).

SOLUTION1. Hydrostatic equation

p = z = 9790N/ m3 50 m

= 489, 500 N/m2

p= 490 kPa gage

2. Calculate pressure ratio

p50

patm=

489.5 kPa + 101.3kPa

101.3kPa

p50patm

= 5.83

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Situation:Water and kerosene are in a tank.zwater = 1 m,zkerosene= 0.75 m.

Find:Gage pressure at bottom of tank (kPa-gage).

Properties:Water (20 C), Table A.5,w= 9790 N/m

3.Kerosene(20 C) , Table A.4,

k= 8010 N/m3.

SOLUTION

Manometer equation (add up pressure from the top of the tank to the bottom of thetank).

patm + k (0.75 m) + w (1.0 m) =pbtm

Solve for pressure

pbtm = 0 + k (0.75 m) + w (1.0 m)

=

8010N/ m3

(0.75 m) +

9790N/ m3

(1.0 m)

= 15.8kPa

pbtm = 15.8kPa gage

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Situation:A hydraulic lift is being designed.Wmax= 10 ton = 20000 lbf,Wparts= 1000 lbf.L= 6 ft, t= 20 s.Diameter range: 2 8 in.Pressure range: 200 3000 psig.Available pumping capacity: 5,10,15 gpm.

Find:Select a hydraulic pump capacity (gpm).Select a cylinder diameter (D).

PLAN

Apply equilibrium to find the smallest bore diameter (D) that works. Thenfind the

largest bore diameter that works by considering the lift speed requirement. Selectbore and pump combinations that meet the desired specifications.

SOLUTION

Equilibrium (piston)F =pA

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104/917

whereF = 21, 000 lbfis the load that needs to be lifted and p is the pressure on thebottom of the piston. Maximum pressure is 3000 psig so minimum bore area is

Amin = F

pmax

=

21, 000 lbf

3000 in2= 7.0 in2

Corresponding minimum bore diameter is

D =

r4

A

Dmin = 2.98 in

The pump needs to provide enough flow to raise the lift in 20 seconds.

AL= Vt

where A is the bore area, L is stroke (lift height),

V is the volume/time of fluidprovided by the pump, and t is the time. Thus, the maximum bore area is

Amax =Vt

L

Conversion from gallons to cubic feet

ft3

: 7.48 gal=1 ft3. Thus, the maximumbore diameter for three pumps (to meet the lift speed specification) is given in thetable below.

pump (gpm) pump (cfm) A (ft2) Dmax (in)5 0.668 0.037 2.61

10 1.337 0.074 3.6815 2.01 0.116 4.61

Since the minimum bore diameter is 2.98 in., the 5 gpm pump will not work. The 10gpm pump can be used with a 3 in. bore. The 15 gpm pump can be used with a 3or 4 in. bore.

1.) The 10 gpm pump will work with a bore diameter between 3.0 and 3.6 inches.

2.) The 15 gpm pump will work with a bore diameter between 3.0 and 4.6 inches.

REVIEW

1. These are preliminary design values. Other issues such as pressure drop in thehydraulic lines and valves would have to be considered.

2. We recommend selecting the 15 gpm pump and a 4.5 inch bore to providelatitude to handle pressure losses, and to reduce the maximum system pressure.

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Situation:Initial State: Water levels as shown. Valve in open.Final State: Water is added to the tank with the valve closed.

Find:

Increase of water level

in manometer (in meters).Properties:

Water (20 C), Table A.5,w=9790 N/m3.

patm= 100 kPa.

Assumptions: Ideal gas.

PLAN

Apply the hydrostatic equation and the ideal gas law.

SOLUTION

Ideal gas law (mole form; apply to air in the manometer tube)

pV= n


106/917

Substitute (a) and (b) into Eq. (1)

p1V1 = p2V2100, 000N/ m2

(1 m Atube) =

100, 000 N/m2 + w(1m )

(1 m )Atube

100, 000N/ m2 = 100, 000N/ m2 + 9790 N/ m3 (1) (1

)

Solving for = 0.0824 m

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Situation:A tank is fitted with a manometer.S= 3,z1= 0.15 m.

Find:Deflection of the manometer (cm).

Properties:

water=9810 N/m3.

PLAN

Apply the hydrostatic principle to the water and then to the manometer fluid.

SOLUTION

1. Hydrostatic equation (location 1 is on the free surface of the water; location 2 isthe interface)

p1

water

+ z1 = p2

water

+ z2

0 Pa

9810N/ m3+ 0.15 m =

p29810N/ m3

+ 0 m

p2 = (0.15m)

9810N/ m3

= 1471.5 Pa

2. Hydrostatic equation (manometer fluid; let location 3 be on the free surface)

p2man. fluid

+ z2 = p3man. fluid

+ z3

1471.5 Pa

3(9810N/ m3)

+ 0 m = 0 Pa

man. fluid+ h

3. Solve for h

h = 1471.5 Pa

3(9810N/ m3)= 0.0500m

h= 5.00 cm

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Situation:A mass sits on top of a piston situated above a reservoir of oil.

Oil

Weight

Piston

D1

h1

h2w

Find:Derive an equation forh2 in terms of the specified parameters.

Assumptions:Neglect the mass of the piston.Neglect friction between the piston and the cylinder wall.The pressure at the top of the oil column is 0 kPa-gage.

PLAN

1. Relatew to pressure acting on the bottom of the piston using equilibrium.2. Related pressure on the bottom of the piston to the oil column height using thehydrostatic equation.3. Findh2 by combining steps 1 and 2.

SOLUTION1. Equilibrium (piston):

w= p1

D21

4

(1)

2. Hydrostatic equation. (point 1 at btm of piston; point 2 at top of oil column):

p1

+ z1 = p2

+ z2

p1Swater

+ 0 = 0 + h2

p1 = S water h2 (2)


mg= S water h2

D21

4

Answer: h2= 4w

(S) (water) (D21)

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REVIEW

1. Notice. Column height h2 increases linearly with increasing weight w.Similarly,h2decreases linearly with Sand decreases quadratically with D1.2. Notice. The apparatus involved in the problem could be used to create an instru-ment for weighing an object.

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Situation:A mass sits on top of a piston situated above a reservoir of oil.m= 10kg,S= 0.8,h1= 42 mm.D

1= 42mm,D

2= 5 mm.

Oil

Weight

Piston

D1

h1

h2w

Find:

Calculateh2 (m).Assumptions:

Neglect the mass of the piston.Neglect friction between the piston and the cylinder wall.The pressure at the top of the oil column is 0 kPa-gage.

PLAN

1. Relate massm to pressure acting on the bottom of the piston using equilibrium.2. Related pressure on the bottom of the piston to the oil column height using thehydrostatic equation.

3. Findh2 by combining steps 1 and 2.

SOLUTION

1. Equilibrium (piston):

mg= p1

D21

4

(1)

2. Hydrostatic equation. (point 1 at btm of piston; point 2 at top of oil column):

p1

+ z1 = p2

+ z2

p1Swater+ 0 = 0 + h2

p1 = S water h2 (2)


mg= S water h2

D21

4

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111/917

h2= 4mg

(S) (water) (D21)

= 4(10kg)(9.81 m/ s2)

(0.8)(9810N/ m3) () (0.142 m2)

h2= 0.812m

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Situation:An odd tank contains water, air and a liquid.

.

Find:Maximum gage pressure (kPa).Where will maximum pressure occur.Hydrostatic force (in kN) on top of the last chamber, surface CD.

Properties:water= 9810 N/m

3.

PLAN

1. To find the maximum pressure, apply the manometer equation.

2. To find the hydrostatic force, multiply pressure times area.

SOLUTION

1. Manometer eqn. (start at surface AB; neglect pressure changes in the air; end atthe bottom of the liquid reservoir)

0 + 4 H2O+ 33H2O = pmax

pmax = 13 m9, 810N/ m3

= 127, 530 N/m2

pmax= 127.5 kPa

Answer Maximum pressure will be at the bottom of the liquid that has a specificgravity ofS= 3.

2. Hydrostatic force

FCD = pA

= (127, 530N/ m2 1 m39810 N/ m3)1 m2FCD = 98.1 kN

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Situation:A steel pipe is connected to a steel chamber.= 2.5 ft,W= 600lbf.D

1= 0.25,z

1= 5.

D2= ,S= 1.2.

Find:Force exerted on chamber by bolts (lbf).

Properties:

water= 62.4lbf/ ft3.

PLAN

Apply equilibrium and the hydrostatic equation.

SOLUTION

1. Equilibrium. (system is the steel structure plus the liquid within)

(Force exerted by bolts) + (Weight of the liquid) +

(Weight of the steel) = (Pressure force acting on the bottom of the free body )

FB+ Wliquid+ Ws= p2A2 (1)

2. Hydrostatic equation (location 1 is on surface; location 2 at the bottom)

p1

+ z1 = p2liquid

+ z2

0 + 5 = p21.2water

+ 0

p2 = 1.2water5

= 1.262.452.5

= 936 psfg

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3. Area

A2=D2

4 =

2

4 =

(2.5ft)2

4 = 4.909ft2

4. Weight of liquid

Wliquid =

A2 +d2

4 4

liquid=

A2 +3

16

(1.2) water

=

4.909ft2

(2.5 ft) +

(2.5ft)3

16

!(1.2)

62.4

lbf

ft3

fluid 9ed solution manual

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