Engineering-Fluid-Mechanics-9th-Edition-Crowe-Solution-Manual.pdf
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2.1: PROBLEM DEFINITION Find: How density diff ers ers from specific weight PLAN Consider their definitions (conceptual and mathematical) SOLUTION Densit Density y is a [ma [mass]/ ss]/[un [unit it volum volume], e], and specific weight is a [weight]/[unit volume]. Therefore, they are related by the equation γ = ρg, ρg , and density diff ers ers from specific weight weight by the the factor factor g , , the acceleration of gravity.
1
2.2: PROBLEM DEFINITION Find: Fluids for which we can (usually) assume density to be nearly constant Fluids 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, mercury cury and oil. oil. Ho Howe weve ver, r, even the densit density y of a liquid liquid varie variess sligh slightly tly as a functi function on of either either pressur pressuree or temperatu temperature. re. Sligh Slightt chang changes es in the volum volumee occupie occupied d by a given given mass of a liquid as a function of pressure can be calculated using the equation for elasticity. One must know the temperature and the pressure to determine the density of a gas .
2
2.3: PROBLEM DEFINITION Find: Where in this text you can find density data for such fluids as oil and mercury. SOLUTION Table A.4 in the Appendix contains density data for such fluids as oil and mercury .
3
2.4: PROBLEM DEFINITION Situation: An engineer needs to know the local density for an experiment with a glider. z = 2500 2500 ft. ft. Find: Calculate density using local conditions. Compare calculated density with the value from Table A.2, and make a recommendation. Properties: From Table A.2, R A.2, R air = 287 kgJ· K , ρ = ρ = 1.22kg/ 22kg/ m3 . Local temperature = 74.3 F = 296. 296.7 K. ◦
Local pressure = 27.3 in.-Hg = 92. 92.45 kPa kPa. PLAN Apply the ideal gas law for local conditions. SOLUTION Ideal gas law ρ =
p RT
92, 92, 450N/ 450N/ m2 = (287kg/ (287kg/ m3 )(296. )(296.7 K) = 1.086 kg/m 086 kg/m3 ρ = 1.09 kg/m 09 kg/m3 (local conditions) Table value. value. From Table Table A.2 ρ = 1.22 kg/m 22 kg/m3 (table value)
The density diff erence erence (local (local conditions conditions versus versus table value) value) is about 12%. 12%. Most Most of this diff erence erence is due to the eff ect ect of elevation on atmospheric pressure. Recommendation—use the local value of density because the eff ects ects of elevation are signi ficant . REVIEW Note: Note: Always Always use absolute absolute pressure pressure when working working with the ideal gas law.
4
2.5: PROBLEM DEFINITION Situation: Carbon dioxide. Find: Density and specific weight of CO2 . Properties: From Table A.2, R A.2, R CO = 189 J/kg 189 J/kg··K. p = p = 300 kPa kPa, T = 60 C. 2
◦
PLAN 1. First, apply the ideal gas law to find density. 2. Then, calculate specific weight using γ using γ = = ρg. ρg. SOLUTION 1. Idea Ideall gas gas law law ρCO
2
= =
P RT
300, 300, 000 kPa kPa (189J/ (189J/ kg K)(60 + 273 273)) K ρCO = 4.767 kg/m3 2
2. Spec Speciific weight γ = ρg = ρg Thus γ CO
2
= ρCO × g = 4.767kg/ 767kg/ m3 × 9. 9.81 m/ s2 2
γ CO = 46.764 N/m 3 2
REVIEW Always use absolute pressure when working with the ideal gas law.
5
2.6: PROBLEM DEFINITION Situation: Methane gas. Find: Density (kg/m3 ). Specific weight ( N/ m3 ). Properties: From Table A.2, R A.2, R Methane = 518 p = p = 300 kPa kPa, T = 60 C. ◦
J kg· K
.
PLAN 1. Apply Apply the the ideal ideal gas gas law law to find density. 2. Calcu Calculat latee specific weight using γ using γ = = ρg. ρg. SOLUTION 1. Idea Ideall gas gas law law ρMethane =
P RT
300, 300, 000 mN = 518 kgJ· K (60 + 273 K) 2
ρMethane = 1.74 kg/m3 2. Spec Speciific weight γ = ρg = ρg Thus γ Methane = ρMethane × g = 1.74kg/ 74kg/ m3 × 9. 9.81 m/ s2 γ Methane = 17.1 N/m3
REVIEW Always use absolute pressure when working with the ideal gas law.
6
2.7: PROBLEM DEFINITION Situation: Natural gas is stored in a spherical tank. Find: Ratio of final mass to initial mass in the tank. Properties: patm = 100 kPa kPa, p 1 = 100 kPa kPa-gage -gage.. p2 = 200 kPa kPa-gage, T -gage, T = 10 C. ◦
PLAN Use the ideal gas law to develop a formula for the ratio of final mass to initial mass. SOLUTION 1. Ma Mass ss in terms terms of of densit density y M = ρV
(1)
p RT
(2)
2. Idea Ideall gas gas law law ρ = 3. Comb Combine ine Eqs. Eqs. (1) and and (2) (2)
M = ρV = ( p/RT ) p/RT )V 4. Volume olume and gas temperature temperature are constant, constant, so M 2 p2 = M 1 p1 and M 2 300 kPa 300 kPa = M 1 200 kPa 200 kPa M 2 M 1
7
=1.5
2.8: PROBLEM DEFINITION Situation: Wind and water at 100 C and 5 and 5 atm atm. ◦
Find: Ratio of density of water to density of air. Properties: Air, Table A.2: Rair = 287 J/kg 287 J/kg··K. o Water (100 (100 C ), ), Table A.5: ρwater = 958 kg/m 958 kg/m3 . PLAN Apply the ideal gas law to air. SOLUTION Ideal gas law ρair =
p RT
506, 506, 600 kPa kPa (287J/ (287J/ kg K) (100 (100 + 273 273)) K = 4.73 kg/m 73 kg/m3 =
For water
ρwater = 958 kg/m 958 kg/m3
Ratio ρwater 958kg/ 958kg/ m3 = ρair 4.73kg/ 73kg/ m3 ρwater = 203 ρair REVIEW Always use absolute pressures when working with the ideal gas law.
8
2.9: PROBLEM DEFINITION Situation: Oxygen fills a tank. 3 V tank lbf . tank = 10 ft , W tank tank = 150 lbf Find: Weight (tank plus oxygen). Properties: From Table A.2, R A.2, R O2 = 1555 ft 1555 ft··lbf/( lbf/(slug · slug ·o R) . p = p = 500 psia, 500 psia, T T = 70 F. ◦
PLAN Apply the ideal gas law to find density of oxygen. Find the weight of the oxygen using speci fic weight ( weight (γ γ ) and add this to the weight of the tank. SOLUTION 1. Idea Ideall gas gas law law pabs. = 500 psia × psia × 144 144 psf/psi psf/psi = = 72, 72, 000 psf T = 460 460 + 70 = 530 530 R p ρ = RT 72, 72, 000 psf 000 psf = (155 (15555 ft lbf lbf / slugo R)(530o R) ρ = 0.087 slugs/ft 087 slugs/ft3 ◦
2. Spec Speciific weight γ = ρg slug ft × 32. 32 . 2 s2 ft3 γ = 2.80 lbf/ft 80 lbf/ft3 = 0.087
3. Weight eight of filled tank W oxygen oxygen = = W total total = =
2.80 lbf/ft 80 lbf/ft3 × 10 ft 10 ft 3 28 lbf W oxygen oxygen + W tank tank 28. 28.0 lbf lbf + 150 150 lbf lbf W total 178 lbf total = 178 lbf
REVIEW 9
1. For compre compresse ssed d gas in a tank, tank, pressure pressuress are often very very high high and the ideal gas gas assumptio assumption n is invalid. invalid. For this problem problem the pressure is about 34 atmospheres— atmospheres—it it is a good idea to check a thermodynamics reference to analyze whether or not real gas eff ects ects are significant. 2. Always Always use absolute absolute pressure pressure when working working with the ideal gas law. law.
10
2.10: PROBLEM PROBLEM DEFINITION Situation: Oxygen is released from a tank through a valve. V = = 10 10 m3 . Find: Mass of oxygen that has been released. Properties: RO = 260 kgJ· K . p1 = 800 kPa kPa, T 1 = 15 C. p2 = 600 kPa kPa, T 2 = 20 C. 2
◦
◦
PLAN 1. Use ideal gas law, expressed expressed in terms of density density and the gas-specific (not universal) gas constant. 2. Find Find the densi densitty for for the the case case before before the gas is releas released ed;; and and then then mass mass from from density, given the tank volume. 3. Find the density density for the case case after the gas is released, released, and the corresponding corresponding mass. mass. 4. Calcu Calculat latee the mas masss diff erence, erence, which is the mass released. SOLUTION 1. Idea Ideall gas gas law law ρ =
p RT
2. Densit Density y and mass mass for case case 1 ρ1 ρ1
800, 800, 000 mN = N· m (260 kg · K )(288K) kg = 10. 10.68 3 m 2
M 1 = ρ1V kg × 10 m3 3 m M 1 = 106.8 kg = 10. 10.68
3. Densit Density y and mass mass for case case 2 ρ2 ρ2
600, 600, 000 mN = N· m (260 kg )(288K) ·K kg = 8.01 3 m 2
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M 2 = ρ1 V kg × 10 m3 3 m = 80. 80.1 kg = 8.01
M 1 4. Ma Mass ss releas released ed from from tank tank
M 1 − M 2 = 106.8 − 80. 80.1 M 1 − M 2 = 26. 26.7 kg
12
2.11: PROBLEM PROBLEM DEFINITION Situation: Properties of air. Find: Specific weight (N/m3 ). Density (kg/m3 ). Properties: From Table A.2, R A.2, R = 287 p = p = 600 kPa kPa, T = 50 C. ◦
J kg· K
.
PLAN First, apply the ideal gas law to find densit density y. Then, Then, calcula calculate te specific weight using γ = ρg. SOLUTION 1. Idea Ideall gas gas law law ρair = =
P RT
600, 600, 000 kPa kPa (287J/ (287J/ kg K) (50 (50 + 273 273)) K ρair = 6.47 kg/m3
2. Spec Speciific weight γ air = ρair × g = 6.47kg/ 47kg/ m3 × 9. 9.81 m/ s2 γ air = 63.5 N/ m3 REVIEW Always use absolute pressure when working with the ideal gas law.
13
2.12: PROBLEM PROBLEM DEFINITION 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, ρ A.2, ρ air = 0.00237 slugs/ft 00237 slugs/ft3 . Assumptions: The density of air is the value at sea level for standard conditions. SOLUTION Units of slugs M = ρV M = 0. 0 .00237 slug × (5280)3 ft3 ft 3
M = 3. 3 .49 × 49 × 10 108 slugs Units of kg kg M = 3.49 × 49 × 10 10 slug × 14. 14.59 slug
¡
¢ µ
8
M = 5.09 × 09 × 10 109 kg
¶
REVIEW The mass will probably be somewhat less than this because density decreases with altitude.
14
2.13: PROBLEM PROBLEM DEFINITION Situation: For a cyclist, temperature changes aff ect ect air density, thereby aff ecting ecting both aerodynamic drag and tire pressure. Find: a.) Plot air density versus temperature for a range of -10 o C to 50o C. b.) Plot tire pressure versus temperature for the same temperature range. Properties: From Table A.2, R A.2, R air = 287 J/kg/K. 287 J/kg/K. Initial conditions for part b: p = 450kPa, 450kPa, T = 20 2 0 C. ◦
Assumptions: For part b, assume that the bike tire volume does not change. PLAN Apply the ideal gas law. SOLUTION a.) a.) Ideal Ideal ga gass law law ρ =
p 101000 kPa kPa = RT (287J/ (287J/ kg K)(273 K)(273 + T ) T )
1.40 1.35 ) m / g k ( y t i s n e D
3
1.30 1.25 1.20 1.15 1.10 1.05 -20
-10
0
10
20
30
40
50
60
o Temperature ( C )
b.) If the volume is constant, constant, since mass can’t change, change, then density must must be constant. Thus p po = T T o p = p = 450 kPa kPa
15
T 20 C
µ ¶ ◦
520 a P k , e r u s s e r p e r i T
500 480 460 440 420 400 380 -20
-10
0
10
20
30
40 o
Temperature, C
16
50
60
2.14: PROBLEM PROBLEM DEFINITION Situation: A design team needs to know how much CO 2 is needed to inflate a rubber raft. Inflation pressure is 3 psi above local atmospheric pressure, or a total of 17.7 psia. Since dimensio dimensions ns are in m, switch switch to kPa. kPa. Thus, Thus, inflation pressure is 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: Assume that the CO2 in the raft is at 62 F = 290 290 K. Assume that the volume of the raft can be approximated by a cylinder of diameter 0.45 m and a length of 16 m (8 meters for the length of the sides and 8 meters for the lengths of the ends plus center tubes). ◦
Properties: From Table A.2, R CO = 189 J/kgK. 189 J/kgK. 2
PLAN Mass is related to volume by M = ρ ∗ V . Densit Density y can be found found using using the ideal ideal gas gas law. SOLUTION Volume contained in the tubes. πD 2 ∆V = ×L 4 π × 0 × 0..452 = × 16 m 3 4
µ
∆V = 2. 2 .54
17
m3
¶
Ideal gas law p RT
ρ =
122, 122, 000N/ 000N/ m2 (189J/ (189J/ kg · kg · K)(290K) = 2.226 kg/m 226 kg/m3 =
Mass of CO2 M = ρ × V = 2.226 kg/m 226 kg/m3 × 2.54 m3
¡
M = 5.66kg
¢ ¡
¢
REVIEW The final mass mass (5.66 kg = 12.5 lbm) lbm) is large. This would would require require a large and potentially expensive CO2 tank. tank. Thus, Thus, this design idea idea may be impractica impracticall for a product that is driven by cost.
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2.15: PROBLEM PROBLEM DEFINITION Situation: A helium filled balloon is being designed. r = 1.3 m, z = 80, 80, 000ft. 000ft. Find: Weight of helium inside balloon. Properties: From Table A.2, R He = 2077 J/kg 2077 J/kg··K. p = p = 0.89 bar bar = 89 89 kPa kPa, T = 22 C = 295. 295.2 K. ◦
PLAN Weight is given by W = mg. Mass Mass is related related to volume by M by M = ρ ∗ V . Dens Densit ity y can can be found using the ideal gas law. SOLUTION Volume in a sphere 4 3 πr 3 4 = π (1. (1.3 m)3 3 = 9.203m3
V =
Ideal gas law ρ =
p RT
89, 89, 000N/ 000N/ m2 (207 (20777 J/ kg · kg · K) (295 (295..2 K) = 0.145 kg/m 145 kg/m3 =
Weight of helium W = ρ × V × × g = 0.145 kg/m 145 kg/m3 × 9.203m3 × 9.81 m/ s2 = 13. 13.10 N
¡
¢ ¡
Weight = 13 = 13..1 N
19
¢ ¡
¢
2.16: PROBLEM PROBLEM DEFINITION 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: C 6 H 12 12 O6
→
2(C 2(C H 3 C H 2OH ) + 2(CO 2(CO 2)
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: S alcohol 0 .80, 80, S s = 1.59, 59, S w = 1.08. 08. alcohol = 0. PLAN Imagine that the initial mixture is pure water plus saturated sugar solution and then use this visualization to find the mass of sugar that is initially present (per unit of volum volume). e). Next, Next, apply apply conserv conservat ation ion of mass mass to find the mass of alcohol that is produced produced (per unit of volume). volume). Then, solve solve for the problem unkno unknowns. wns. SOLUTION The initial density of the mixture is ρmix =
ρw V w + ρs V s V o
where ρw and ρs are the densities of water and sugar solution (saturated), V o is the initial volume of the mixture, and V s is the volu volume me of sugar sugar solu soluti tion on.. The The tota totall volume of the mixture is the volume of the pure water plus the volume of saturated solution V w + V s = V = V o The specific gravit gravity y is initially initially 1.08. 1.08. Thus Thus ρmix V s ρ V s = (1 − ) + s ρw V o ρw V o V s V s 1.08 = (1 − ) + 1. 1 .59 V o V o V s = 0.136 V o S i =
20
Thus, the mass of sugar per unit volume of mixture M s = 1.59 × 59 × 0 0..136 V o = 0.216 kg/m 216 kg/m3 The molecula molecularr weigh weightt of glucose glucose is 180 and ethyl ethyl alcohol alcohol 46. Thus Thus 1 kg of glucos glucosee converts to 0.51 kg of alcohol so the final density of alcohol is M a = 0.216 × 216 × 0 0..51 V o = 0.110 kg/m 110 kg/m3 The density of the final mixture based on the initial volume is M f f = (1 − 0.136) + 0. 0.110 V o = 0.974 kg/m 974 kg/m3 The final volume is altered because of conversion V f f M w M a = + V o ρw V o ρa V o V w 0. 0 .51M 51M s = + V o ρa V o V w 0. 0 .51ρ 51ρs V s = + V o ρa V o 0. 0 .51 × 51 × 1 1..59 = 0.864 + × 0. 0.136 0.8 = 1.002 The final density is M f f M f f V o = × V f f V o V f f 1 1.002 = 0.972 kg/m 972 kg/m3 = 0.974 × 974 ×
The final specific gravity is S f f = 0.972 The alcohol content by volume V a M a = V f f ρa V f f M a 1 V o = V o ρa V f f = 0.110 × 110 × = 0.137 21
1 1 × 0.8 1.002
Thus, Percent alcohol by volume = 13.7%
22
2.17: PROBLEM PROBLEM DEFINITION 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 Primary dimension dimensionss of viscosity viscosity are [ M LT ] . Five common units are: ·s i) Nm· s ; ii) dyn ; iii) iii) poise poise;; iv) centipo centipoise ise;; and and v) cm 2
2
lbf · s ft2
(b) To find the viscosity of SAE 10W-30 motor oil at 115 F, there are no tablular data in the text. text. Theref Therefore ore,, one should should use Figure Figure A.2. A.2. For tradit tradition ional al units units (because (because the temperature is given in Fahrenheit) one uses the left-hand axis to report that ·s μ = 1.2 × 10 × 10 3 lbf . ft Note: Note: one should should be careful careful to identify identify the correct factor factor of 10 for the log cycle cycle that contain containss the correct correct data point. For example, example, in this problem, problem, the answer is between between 3 2 1 × 10 and 1 and 1 × 10 . One should be able to determine that the answer is 1. 1 .2 × 10 3 and not 1 not 1 × × 10 10 2 . ◦
−
2
−
−
−
−
(c) The viscos viscosit ity y of water water decreas decreases es with with increas increasing ing temper temperatu ature re . This This is true for all liquids, and is because the loose molecular lattice within liquids, which provides a given resistance to shear at a relatively cool temperature, has smaller energy barriers resisting movement at higher temperatures. (d) The viscos viscosit ity y of air increa increases ses with with increa increasin sing g temperat temperature ure . This This is true for all all gases, gases, and is because b ecause gases gases do not have have a loose molecular molecular lattice. lattice. The only resistance resistance to shear provided in gases is due to random collision between di ff eren erentt laye layers rs.. As the temperature increases, there are more likely to be more collisions, and therefore a higher viscosity.
23
2.18: PROBLEM PROBLEM DEFINITION Situation: Change in viscosity and density due to temperature. T 1 = 10 C, T 2 = 70 C. ◦
◦
Find: Change in viscosity and density of water. Change in viscosity and density of air. Properties: p = p = 101 kN/ kN/ m2. PLAN For water, water, use data from Table Table A.5. For air, use data from Table Table A.3 SOLUTION Water μ70 = 4.04 × 04 × 10 10 4 N·s/m2 μ10 = 1.31 × 31 × 10 10 3 N·s/m2 ∆μ = −9. 06 × 06 × 10 10 4 N· s/ m2 −
−
−
ρ70 = 978 kg/m 978 kg/m3 ρ10 = 1000 kg/m 1000 kg/m3 ∆ρ = −22 kg/ kg/ m3 Air μ70 = 2.04 × 04 × 10 10 5 N · s/m · s/m2 μ10 = 1.76 × 76 × 10 10 5 N · s/m · s/m2 ∆μ = 2. 8 × 10 × 10 6 N · s/ m2 −
−
−
ρ70 = 1.03 kg/m 03 kg/m3 ρ10 = 1.25 kg/m 25 kg/m3 ∆ρ = −0.22 kg/ kg/ m3
24
2.19: PROBLEM PROBLEM DEFINITION Situation: Air at certain temperatures. T 1 = 10 C, T 2 = 70 C. ◦
◦
Find: Change in kinematic viscosity. Properties: From Table A.3, ν A.3, ν 70 = 1.99 × 99 × 10 10
5
−
m2 /s, ν /s, ν 10 = 1.41 × 41 × 10 10
5
−
m2 /s.
PLAN Use properties found in Table A.3. SOLUTION ∆vair,10 70 →
= (1. (1.99 − 1.41) × 41) × 10 10 ∆vair,10
70
→
5
−
= 5.8× 5.8×10 6 m2 /s −
REVIEW Sutherland’s equation could also be used to solve this problem.
25
2.20: PROBLEM PROBLEM DEFINITION 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) μ(N · s/m2 ) 6.7× 6.7 ×10 2 ρ(kg/m3) 880 ν (m2 /s) /s) 7.6× 7.6 ×10 5
kerosene 1.4× 1.4 ×10 3 (Fig (Fig.. AA-2) 2) 814 1.7× 1.7 ×10 6 (Fig (Fig.. AA-2) 2)
−
−
−
−
26
water 6.8 6.8 ×10 993 6.8 6.8 ×10
4
−
7
−
2.21: PROBLEM PROBLEM DEFINITION 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, μ A.3, μ air,20 C = 1.81 × 81 × 10 10 5 N·s/m2 ; ν = 1. 1 .51 × 51 × 10 10 5 m2 /s From Table A.5, μ A.5, μ water,20 C = 1.00 × 00 × 10 10 3 N·s/m2 ; ν = 1.00 × 00 × 10 10 6 m2 /s −
−
◦
−
−
◦
SOLUTION Dynamic viscosity μair 1.81 × 81 × 10 10 5 N · s/ m2 = μwater 1.00 × 00 × 10 10 3 N · s/ m2 μair = 1.81× 1.81×10 2 μwater −
−
−
Kinematic viscosity ν air 1.51 × 51 × 10 10 5 m2 / s air = ν water 1.00 × 00 × 10 10 6 m2 / s water ν air air = 5.1 ν water water −
−
27
2.22: PROBLEM PROBLEM DEFINITION Situation: Sutherland’s equation and the ideal gas law describe behaviors of common gases. Find: Develop an expression for the kinematic viscosity ratio ν /ν o , where ν where ν is is at temperature T ature T and pressure p. pressure p. Assumptions: Assume a gas is at temperature T o and pressure p pressure po , where the subscript ”o” de fines the reference state. PLAN Combine the ideal gas law and Sutherland’s equation. SOLUTION The ratio of kinematic viscosities is ν μ ρo = = ν o μo ρ
T T o
3/2
µ¶ µ¶
ν po = ν o p
T T o
28
T o + S po T T + S p T o
5/2
T o + S T + S
2.23: PROBLEM PROBLEM DEFINITION Situation: The dynamic viscosity of air. μo = 1.78 × 78 × 10 10 5 N·s/m2 . T o = 15 C, T = T = 100 C. −
◦
◦
Find: Dynamic viscosity μ viscosity μ.. Properties: From Table A.2, S A.2, S = = 111K 111K . SOLUTION Sutherland’s equation μ = μo =
3/2
T T o
T o + S T + S
µ¶ µ ¶ 373K 288K
3/2
288 288 K + 111 111 K 373 373 K + 111 111 K
μ = 1.21 μo Thus
μ = 1.21μ 21μo = 1.21 × 21 × 1.78 × 78 × 10 10
¡
μ = 2.15 × 15 × 10 10
5
−
29
5
−
N · s/ m2
N·s/m2
¢
2.24: PROBLEM PROBLEM DEFINITION Situation: Methane gas. vo = 1.59 × 59 × 10 10 5 m2 / s. T o = 15 C, T = T = 200 C. po = 1 atm atm, p = p = 2 atm atm. −
◦
◦
Find: Kinematic viscosity ( m2 / s). Properties: From Table A.2, S A.2, S = = 198K. 198K. PLAN Apply the ideal gas law and Sutherland’s equation. SOLUTION μ ρ ν μ ρo = ν o μo ρ ν =
Ideal-gas law
ν μ po T = ν o μo p T o
Sutherland’s equation ν po = ν o p
5/2
µ¶ T T o
T o + S T + S
so 5/2
µ ¶
ν 1 473K = ν o 2 288K = 1.252
288 288 K + 198 198 K 473 473 K + 198 198 K
and ν = 1.252 × 252 × 1 1..59 × 59 × 10 10
5
−
ν =
1.99 × 99 × 10 10
5
−
30
m2/ s
m2/s
2.25: PROBLEM PROBLEM DEFINITION Situation: Nitrogen gas. μo = 3.59 × 59 × 10 10 7 lbf · · s/ ft2 . T o = 59 F, T = T = 200 F. −
◦
◦
Find: μ using Sutherland’s equation. Properties: From Table A.2, S A.2, S =192o R. SOLUTION Sutherland’s equation μ = μo
T T o
3/2
T o + S T + S
µ¶ µ ¶ µ
660o R = 519o R = 1.197
3/2
519o R + 192o R 660o R + 192o R
lbf · · s μ = 1.197 × 197 × 3.59 × 59 × 10 10 ft2 = 4. 297 × 297 × 10 10 7 7
−
−
μ = 4.30 × 30 × 10 10
31
7
−
lbf-s/ft2
¶
2.26: PROBLEM PROBLEM DEFINITION Situation: Helium gas. vo = 1.22 × 22 × 10 10 3 ft2 / s. T o = 59 F, T = 30 F. po = 1 atm atm, p = p = 1.5atm. 5atm. −
◦
◦
Find: Kinematic viscosity using Sutherland’s equation. Properties: From Table A.2, S A.2, S =143o R. PLAN Combine the ideal gas law and Sutherland’s equation. SOLUTION ν po = ν o p
5/2
µ¶ µ ¶ µ T T o
T o + S T + S
1.5 490o R = 1 519o R = 1.359
5/2
519o R + 143o R 490o R + 143o R
2 3 ft ν = 1.359 × 359 × 1.22 × 22 × 10 10 s 2 ft = 1. 658 × 658 × 10 10 3 s −
−
ν = = 1.66 × 66 × 10 10
32
3
−
ft2 / s
¶
2.27: PROBLEM PROBLEM DEFINITION Situation: Absolute viscosity of propane. T o = 100 C, μ o = 1 × 10 × 10 5 N s/ m2. T = T = 400 C, μ = μ = 1.72 × 72 × 10 10 5 N s/ m2 . ◦
◦
−
−
Find: Sutherland’s constant. SOLUTION Sutherland’s equation μ μ
S = T o 1− o
T 1/2 T
¡¢ ¡¢ o
μ μ
o
Also
−
1
T 3/2 T o
μ = 1.72 μo T o 373K = T 673K Thus S = 0.964 T o S = = 360 K 360 K
33
2.28: PROBLEM PROBLEM DEFINITION Situation: Ammonia at room temperature. T o = 68 F, μ o = 2.07 × 07 × 10 10 7 lbf s/ s/ ft2 . T = T = 392 F, μ = μ = 3.46 × 46 × 10 10 7 lbf s/ s/ ft2 . ◦
−
◦
−
Find: Sutherland’s constant. SOLUTION Sutherland’s equation μ μ
S = T o 1− o
T 1/2 T
¡¢ ¡¢ o
μ μ
o
Calculations
−
1
T 3/2 T
(1)
o
μ 3.46 × 46 × 10 10 7 lbf s/ s/ ft2 = = 1. 1 .671 μo 2.07 × 07 × 10 10 7 lbf s/ s/ ft2 T o 528 R = = 0.6197 T 852 R −
−
(a)
◦
◦
Substitute (a) and (b) into Eq. (1) S = 1.71 T o S = = 903 o R
34
(b)
2.29: PROBLEM PROBLEM DEFINITION Situation: SAE 10W30 motor oil. T o = 38 C, μ o = 0.067Ns/ 067Ns/ m2 . T = 99 C, μ = μ = 0.011Ns/ 011Ns/ m2 . ◦
◦
Find: The viscosity of motor oil, oil , μ(60o C), using the equation μ = μ = C C eb/T . PLAN Use algebra and known values of viscosity (μ) to to solv solve for the the consta constan nt b. solve for the unknown value of viscosity.
Then Then,,
SOLUTION Viscosity variation of a liquid can be expressed as μ = C eb/T . Thus, evaluate μ at temperatures T temperatures T and T and T o and take the ratio:
∙
¸
μ 1 1 = exp b( − ) μo T T o Take the logarithm and solve for b. b =
ln (μ/μo ) ( T 1 − T 1 ) o
Data 0.011Ns/ 011Ns/ m2 = 0.164 0.067Ns/ 067Ns/ m2 T = 372 K T o = 311 K
μ/μo =
Solve for b for b b = 3430 (K) Viscosity ratio at 60 o C
∙
µ
μ 1 1 − = exp 3430 μo 333K 311K = 0.4833 μ = 0.4833 × 4833 × 0 0..067Ns/ 067Ns/ m2 μ =
0.032 N · N · s/ m2
35
¶¸
2.30: PROBLEM PROBLEM DEFINITION Situation: Viscosity of grade 100 aviation oil. T o = 100 F, μ o = 4.43 × 43 × 10 10 3 lbf s/ s/ ft2 . T = T = 210 F, μ = μ = 3.9 × 10 × 10 4 lbf s/ s/ ft2 . ◦
−
◦
−
Find: μ(150o F), using the equation μ equation μ = = C C eb/T . PLAN Use algebra and known values of viscosity (μ) to to solv solve for the the consta constan nt b. solve for the unknown value of viscosity.
Then Then,,
SOLUTION Viscosity variation of a liquid can be expressed as μ = C eb/T . Thus, evaluate μ at temperatures T temperatures T and T and T o and take the ratio:
∙
¸
μ 1 1 = exp b( − ) μo T T o Take the logarithm and solve for b b =
ln (μ/μo ) ( T 1 − T 1 ) o
Data μ = μo T = T o = Solve for b for b
0.39 × 39 × 10 10 4.43 × 43 × 10 10 670o R 560o R
lbf s/ s/ ft2 = 0.08804 3 lbf s/ s/ ft2 3
−
−
b = 8293 ( 8293 ( o R)
Viscosity ratio at 150 o F μ 1 1 = exp 8293 − μo 610oR 560oR = 0.299 lbf · · s μ = 0.299 × 299 × 4.43 × 43 × 10 10 3 ft2
∙
µ
µ
−
μ = 1.32 × 32 × 10 10
3 lbf · s ft2
−
36
¶¸ ¶
2.31: PROBLEM PROBLEM DEFINITION Situation: Oil (SAE 10W30) 10W30) fills the space between two plates. ∆y = 1/8 = 0.125in, 125in, u = u = 25ft/ 25ft/ s. Lower plate is at rest. Find: Shear stress in oil. Properties: Oil (SAE 10W30 @ 150 @ 150 F) from F) from Figure A.2: μ = 5.2 × 10 × 10 ◦
4
−
Assumptions: 1.) Assume oil is a Newtonian fluid. 2.) Assume Couette flow (linear velocity pro file). SOLUTION Rate of strain du = dy
∆u ∆y
25ft/ 25ft/ s (0. (0.125/ 125/12) ft
=
du = 2 4 00 s dy
1
−
Newton’s law of viscosity τ = μ
µ ¶ du dy
lbf · · s = 5.2 × 10 × 10 4 ft2 lbf = 1. 248 2 ft
µ
−
τ = 1. 1 .25
37
1 s
¶µ ¶
lbf ft2
× 2400
lbf ·s/ft2 .
2.32: PROBLEM PROBLEM DEFINITION Situation: Properties of air and water. T = 40 C, p = p = 170 kPa kPa. ◦
Find: Kinematic and dynamic viscosities of air and water. Properties: Air data from Table A.3, μ air = 1. 1 .91 × 91 × 10 10 5 N·s/m2 Water data from Table A.5, μ water = 6.53 × 53 × 10 10 4 N·s/m2, ρwater = 992 kg/m 992 kg/m3 . −
−
PLAN Apply the ideal gas law to find densit density y. dynamic and absolute viscosity.
Find Find kinemati kinematicc viscosit viscosity y as the ratio of
SOLUTION A.) Air Ideal gas law ρair =
p RT
170, 170, 000 kPa kPa (287J/ (287J/ kg K)(313 K)(313.2 K) = 1.89 kg/m 89 kg/m3 =
μair = 1. 1 .91 × 91 × 10 10
5 N· s m2
−
μ ρ 1.91 × 91 × 10 10 5 N s/ m2 = 1.89kg/ 89kg/ m3
ν =
−
ν air 10.1 × 10 × 10 air = 10.
6
m2 / s
5
N·s/m2
−
B.) water μwater = 6. 6 .53 × 53 × 10 10
−
μ ρ 6.53 × 53 × 10 10 4 N s/ m2 ν = 992kg/ 992kg/ m3 ν =
−
ν water 6 .58 × 58 × 10 10 water = 6.
7
−
38
m2/s
2.33: PROBLEM PROBLEM DEFINITION Situation: Sliding plate viscometer is used to measure fluid viscosity. viscosity. A = 50 × 50 × 10 1000 mm, mm, ∆y = 1 mm. u = 10 10 m/ s, F = 3 N. Find: Viscosity of the fluid. Assumptions: Linear velocity distribution. PLAN 1. The shear shear force force τ τ is is a force/area. 2. Use equation equation for viscosity viscosity to relate shear force force to the velocity velocity distribu distribution. tion. SOLUTION 1. Calcu Calculat latee shear shear force force τ =
Force Area
3N 50mm × 50mm × 10 1000 mm τ = 600 N τ =
2. Find Find visc viscosi osity ty μ = μ =
τ
³´ du dy
600N [10m/ [10m/ s] / [1 mm] mm]
μ = 6 × 10 × 10
39
2 N · s m2
−
2.34: PROBLEM PROBLEM DEFINITION Situation: Water flows ows near a wall. wall. The velocit velocity y distribution distribution is y u(y) = a b
³´
1/6
a = 10 10 m/ s, b = b = 2 mm and y 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): μ = 1.00 × 00 × 10 10 ◦
3
−
N · s/ m2 .
SOLUTION Rate of strain (algebraic equation)
∙³´
du d y = a dy dy b a 1 = 1/6 5/6 b 6y a = 6b
b y
1/6
¸
5/6
µ¶
Rate of strain (at y = 1 mm) du a = dy 6b
5/6
µ¶ b y
10 m/ s = 6 × 0 × 0..002m = 1 4 85 s 1 −
5/6
2 mm 1 mm
µ ¶
Shear Stress τ y=1mm = μ
du dy
N · N · s = 1.00 × 00 × 10 10 3 m2 = 1.485 Pa
µ
−
¶¡
τ (y (y = 1 mm) = 1. 1.49Pa
40
1485s
1
−
¢
2.35: PROBLEM PROBLEM DEFINITION Situation: Velocity distribution of crude oil between two walls. μ = 8 × 10 × 10 5 lbf s/ s/ ft2 , B = 0.1 ft. ft. u = 100y 100y(0. (0.1 − y ) ft/ ft/ s, T = T = 100 F. −
◦
Find: Shear stress at walls. SOLUTION Velocity distribution u = 100y 100y(0. (0.1 − y ) = 10y 10y − 100y 100y2 Rate of strain du/dy = 10 − 200y 200y (du/dy) du/dy)y=0 = 10 s 2 (du/dy) du/dy)y=0.1 = −
Shear stress τ 0 = μ
du = (8 × (8 × 10 10 5 ) × 10 × 10 dy −
τ 0 = 8×10
4
lbf/ft2
−
τ 0.1 = 8×10
4
−
Plot 0.10
0.08 e c 0.06 n a t s i D
0.04
0.02
0.00
Velocity
41
lbf/ft2
10 s
−
1
−
2.36: PROBLEM PROBLEM DEFINITION Situation: A liquid flows between parallel boundaries. y0 = 0.0 mm, mm, V 0 = 0.0 m/ s. y1 = 1.0 mm, mm, V 1 = 1.0 m/ s. y2 = 2.0 mm, mm, V 2 = 1.99 m/ s. y3 = 3.0 mm, mm, V 3 = 2.98 m/ s. Find: (a) Maximum shear stress. (b) Location where minimum shear stress occurs. SOLUTION (a) Maximum shear stress τ = μdV/dy τ max ≈ μ(∆V /∆y) next to wall τ max = (10 3 N · s/m2 )((1 m/s) )((1 m/s)//0.001 m 001 m)) −
τ max = 1.0 N/m2 (b)The (b)The minim minimum um shear shear stress stress will will occur midw midwa ay between between the the two two walls walls . nitude will be zero because the velocity gradient is zero at the midpoint.
42
Its mag-
2.37: PROBLEM PROBLEM DEFINITION Situation: Glycerin is flowing owing in betw b etween een two stationa stationary ry plates. The velocity velocity distribution distribution is u = −
1 dp By − y2 2μ dx
¡
dp/dx = dp/dx = −1.6kPa/ 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. at y = 0 mm). Properties: Glycerin (20 (20 C), C), Table A.4: μ = 1.41 N · s/ m2 . ◦
PLAN Find velocity by direct substitution into the speci fied velocity distribution. Find shear stress using the de finition of viscosity: τ = μ (du/dy) du/dy ), where the rate-ofstrain (i.e. the derivative du/dy derivative du/dy)) is found by diff erentiating erentiating the velocity distribution. SOLUTION a.) Velocity (at y = 12 mm) mm) 1 dp By − y2 2μ dx 1 = − 2 (1. (1.41 N · s/ m2 ) m = 0.2587 s
u =
−
¡
¢
¡
1600N/ 1600N/ m3
−
(0. (0.05m)(0. 05m)(0 .012 012 m) − (0. (0.012m)2
¢¡
u (y = 12 mm) mm) = 0. 0.259m/ 259m/ s Rate of strain (general expression)
¡ ¢¶ µ ¶µ ¶ ¡ ¢ µ ¶µ ¶
du d = dy dy = =
µ
1 dp By − y2 − 2μ dx 1 dp d By − y2 − 2μ dx dy 1 dp (B − 2y) − 2μ dx
Rate of strain (at y = 12mm) 12mm) du = dy
1 2μ
dp (B − 2 y ) dx 1 N = − −1600 2 (1. (1.41 N · s/ m2 ) m3 = 14. 14.75 s 1
µ ¶µ ¶ µ ¶µ −
−
43
¶
(0. (0.05 m − 2 × 0 × 0..012m)
¢
Definition of viscosity τ = μ
du dy
µ
N · N · s = 1.41 2 m = 20. 20. 798 Pa
¶¡
14. 14.75 s
1
−
¢
τ (y ( y = 12 mm) mm) = 20. 20.8 Pa b.) Velocity (at y = 0 mm) 1 dp By − y2 2μ dx 1 = − 2 (1. (1.41 N · s/ m2 ) m = 0.00 s
u =
¡
−
¢
¡
1600N/ 1600N/ m3
−
¢¡
(0. (0.05m)(0m) − (0m)2
¢
u (y = 0 mm) = 0m/ 0m / s Rate of strain (at y = 0 mm) du = dy
1 2μ
dp (B − 2y) dx 1 N = − −1600 2 (1. (1.41 N · s/ m2 ) m3 = 28. 28.37 s 1
µ ¶µ ¶ µ ¶µ −
−
¶
(0. (0.05 m − 2 × 0 × 0 m)
Shear stress (at y (at y = 0 mm) τ = μ
du dy
N · N · s = 1.41 2 m = 40. 40.00Pa
µ
¶¡
28. 28.37 s
1
−
¢
τ (y (y = 0 mm) mm) = 40. 40.0 Pa REVIEW 1. As expected expected,, the veloci velocitty at the wall wall (i.e. (i.e. at y = 0) is 0) is zero due to the no slip condition. 2. As expected, expected, the shear shear stress stress at the wall wall is larger larger than the shear shear stress stress away away from from the wall. wall. This This is because because shear shear stress stress is ma maxim ximum um at the wall wall and zero along the centerline (i.e. at y = B = B//2).
44
2.38: PROBLEM PROBLEM DEFINITION Situation: Laminar flow occurs occurs between two two horizon horizontal tal parallel parallel plates. plates. The velocity velocity distridistribution is 1 dp y u = − H y − y2 + ut 2μ ds H Pressure p Pressure p decreases with distance s distance s,, and the speed of the upper plate is u is u t . Note that u that u t has a negative value to represent that the upper plate is moving to the left. Moving plate: y = H. = 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: u
t
u y
H
s
PLAN By inspection, the rate of strain (du/dy) du/dy) or slope of the velocity pro file is larger at the movin moving g plate. plate. Thus, Thus, we expect expect shear shear stress stress τ to τ to be larger at y = H. To check this idea, find shear stress using the definition of viscosity: τ = μ (du/dy) du/dy). Evalu Evaluate ate and compare the shear stress at the locations y = H = H and y and y = 0. SOLUTION Part (a) 1. Shear Shear stre stress, ss, from from definition of viscosity du dy d = μ dy
τ = μ
1 dp y − H y − y2 + ut 2μ ds H H dp y dp u t = μ − + + 2μ ds μ ds H (H − 2y) dp μu t τ (y (y) = − + 2 ds H
∙
¡
∙
45
¢ ¸ ¸
Shear stress at y at y = H = H (H − 2H ) dp μu t + 2 ds H H dp μu t = + 2 ds H
τ (y (y = H = H )) =
−
µ¶
(1)
2. Shea Shearr stre stress ss at at y = 0 (H − 0) dp μu t + 2 ds H H dp μu t = − + 2 ds H
τ (y ( y = 0) =
−
µ¶
(2)
Since pressure decreases with distance, the pressure gradient dp/ds gradient dp/ds is negative. negative. Since the upper wall moves to the left, u t is negativ negative. e. Thus, Thus, maximum maximum shear shear stress stress occurs at y at y = = H H because because both terms in Eq. (1) have the same sign (they are both negative.) In other words, |τ (y ( y = H = H ))| > | > |τ τ (y (y = 0)| 0)| . Maximum shear stress occur at y = H = H . Part (b) Use definition of viscosity to find the location ( location (yy ) of zero shear stress
τ = μ
du dy
dp u t μ (H − 2y) + ds H dp u t μ = −(1/ (1/2) (H − 2y) + ds H =
μ(1/ (1/2μ)
−
Set τ Set τ = 0 and solve for y for y
0 =
dp u t μ (H − 2y) + ds H H μut − y = 2 Hdp/ds (1/ (1/2)
−
Part (c)
46
du = 0 at y = 0 dy du dp u t = −(1/ (1/2μ) (H − 2y ) + dy ds H dp ut Then, at y at y = 0 : du/dy = du/dy = 0 = −(1/ (1/2μ) H + + ds H dp Solve for u for u t : ut = (1/ (1/2μ) H 2 ds dp < 0, 0 , ut 1. choi choice ce is (c) (c) .
48
Correct ect
2.40: PROBLEM PROBLEM DEFINITION Situation: A cylinder falls inside a pipe filled with oil. d = 100mm, 100mm, D = 100. 100.5 mm. mm. = 200mm, 200mm, W = W = 15 15 N. Find: Speed at which the cylinder slides down the pipe. Properties: SAE 20W oil (10 oil (10 o C) from Figure A.2: μ A.2: μ = = 0.35 N· N·s/m2 . SOLUTION dV dy W μV fall fall = πd (D − d)/2 W ( W (D − d) V fall fall = 2πdμ 15N(0. 15N(0.5 × 10 × 10 3 m) V fall fall = (2π (2π × 0 × 0..1 m × 0 × 0..2 m × 3 × 3..5 × 10 × 10 τ = μ
−
V fall 17m/s fall = 0.17m/s
49
1
−
N s/ m2 )
2.41: PROBLEM PROBLEM DEFINITION Situation: A disk is rotated very close to a solid boundary with oil in between. ω a = 1 rad rad/ s, r 2 = 2 cm, r 3 = 3 cm. ω b = 2 rad rad/ s, r b = 3 cm. yc = 2 mm, μ c = 0.0 1 N s/ 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 = dV/dy = V V /y = /y = ωr/y. ωr/y. SOLUTION (a) Ratio of shear stresses dV μωr = dy y μ × 1 × 1 × 2 2/y /y = μ × 1 × 1 × 3 3/y /y
τ = μ τ 2 τ 3
τ 2 2 = τ 3 3 (b) Speed of oil V = ωr = ωr = 2 × 0 × 0..03 V = 0.06m/s 06m/s (c) Shear stress at surface
τ = μ
dV 0. 0 .06 m/ s = 0. 0 .0 1 N s/ s/ m2 × dy 0.002m
τ = 0.30N/m 30N/m2
50
2.42: PROBLEM PROBLEM DEFINITION 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 of D,S of D,S,, ω and μ. and μ. PLAN Apply the Newton’s law of viscosity. SOLUTION Shear stress dV dy μrω = s
τ = μ
Find diff erential erential torque–on an elemental strip of area of radius r the diff erential erential shear force will be τ dA or τ (2 τ (2πrdr πrdr)). The diff erential erential torque will be the product of the diff erential erential shear force and the radius r radius r.. dT one τ (2πrdr πrdr)] )] o ne side = r [τ (2 μrω = r (2πrdr (2πrdr)) s 2πμω 3 = r dr s rπμω 3 dT both r dr b oth sides = 4 s
h i ³ ´
Integrate D/2
T =
Z
4πμω 3 r dr s
0
1 πμωD4 T = 16 s
51
2.43: PROBLEM PROBLEM DEFINITION Situation: One type of viscometer involves the use of a rotating cylinder inside a fixed cylinder. T min min = 50 F, T max max = 200 F. ◦
◦
Find: (a) Design a viscometer that can be used to measure the viscosity of motor oil. Assumptions: Motor Motor oil is SAE 10W-30. 10W-30. Data from from Fig A-2: A-2: μ will vary from about 2 about 2 × 10 4 lbfs/ft2 to 8 to 8 × × 10 10 3 lbf-s/ft2 . Assume the only significant shear stress develops between the rotating cylinder and the fixed cylinder. cylinder. Assume we want the maximum rate of rotation (ω ( ω ) to be 3 rad/s. Maximum spacing is 0.05 in. −
−
SOLUTION One possible design solution is given below. Design decisions: 1. Let h Let h = = 4.0 in. = 0.333 ft 2. Let I.D. I.D. of of fixed cylinder = 9.00 in. = 0.7500 ft. 3. Let O.D. O.D. of rotating rotating cylinder = 8.900 8.900 in. = 0.7417 0.7417 ft. Let the applied torque, which drives the rotating cylinder, be produced by a force from a thread or small diameter mono filament line acting at a radial distance rs. Here rs is the radius radius of a spool on which which the thread thread of line line is wound wound.. The appli applied ed force is produced by a weight and pulley system shown in the sketch below. Pulley h
r c
W Δr
The relationship between μ, between μ, rs,ω,h, and ,ω,h, and W W is is now developed. T = r c F s where T where T = applied torque rc = outer radius of rotating cylinder 52
(1)
F s = shearing force developed at the outer radius of the rotating cylinder but F s = τ As where A where A s = area in shear = 2πr 2πrch τ = μdV/dy
≈
μ∆V /∆r where ∆V = r c ω and ∆r = spacing
Then T Then T = rc (μ∆V /∆r)(2πr )(2πrc h) rc ω )(2πr )(2πrc h) ∆r But the applied torque T torque T = W rs so Eq. (2) become = r c μ(
W rs = r = r c3 μω(2 μω(2π π)
(2)
h ∆r
Or W rs∆r (3) 2πωhrc3 The weight W weight W will will be arbitrarily chosen (say 2 or 3 oz.) and ω will ω will be determined by measuring measuring the time time it takes the weight weight to travel travel a given given distance. distance. So rs ω = V fall fall or ω = V = V fall Equation (3) then becomes fall /rs . Equation μ =
μ =
W V f f
rs2 rc3
∆r
µ ¶µ ¶µ ¶ 2πh
In our design let r let r s = 2 in. = 0.1667 ft. Then μ = μ = μ =
W (0. (0.1667)2 0.004167 V f f (.3708)3 (2π (2π × .3333) W 0.02779 V f f 0.05098 W (1. (1.085 × 085 × 10 10 3 ) lbf · s · s//ft2 V f f
µ ¶ µ ¶µ µ ¶
¶
−
Exampl Example: e: If W W = 2oz. oz. = 0.1 0.125 25lb lb.. and and V f f is measured to be 0.24 ft/s then 0.125 (1. (1.085 × 085 × 10 10 3 ) lbf lbf s/ s/ ft2 0.24 = 0.564 × 564 × 10 10 4 lbf · · s/ ft2
μ =
−
−
REVIEW Other things that could be noted or considered in the design: 1. Specify dimensions dimensions of all parts of the instrument. instrument. 2. Neglect Neglect friction in bearings bearings of pulley and on shaft of cylinder. cylinder. 3. Neglect Neglect weight weight of thread thread or monofilament line. 4. Consider Consider degree of accuracy accuracy.. 5. Estimate Estimate cost of the instrumen instrument. t. 53
2.44: PROBLEM PROBLEM DEFINITION Situation: Elasticity of ethyl alcohol and water. E ethyl 1 .06 × 06 × 10 109 Pa Pa.. ethyl = 1. 9 E water 15 × 10 10 Pa Pa.. water = 2.15 × 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
−∆
∆ p V = dρ/ρ ∆V
This means that elasticity is inversely related to change in density, and to the negative change 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 elasticity is inversely related to change in density.
54
2.45: PROBLEM PROBLEM DEFINITION Situation: Pressure is applied to a mass of water. water. 3 6 2 V = = 2000 2000 cm , p = p = 2 × 10 × 10 N/ m . Find: Volume after pressure applied (cm3 ). Properties: From Table A.5, E A.5, E = 2. 2 .2 × 10 × 109 Pa PLAN 1. Use modulus modulus of elasti elasticit city y equat equation ion to calcul calculat atee volum volumee chang changee result resulting ing from from pressure change. 2. Calcu Calculat latee final volume based on original volume and volume change. SOLUTION 1. Elasticit Elasticity y equation equation V ∆V ∆ p ∆V = − V E (2 × (2 × 10 106 ) Pa = − 2000 2000 cm3 9 (2. (2.2 × 10 × 10 ) Pa = −1.82cm3 E =
p
−∆
∙
¸
2. Final Final volum volumee V final final
= V + + ∆V = (2000 2000 − 1.82) cm3 V final 1998 cm3 final = 1998
55
2.46: PROBLEM PROBLEM DEFINITION Situation: Water is subjected to an increase in pressure. Find: Pressure increase needed to reduce volume by 2%. Properties: From Table A.5, E A.5, E = 2. 2 .2 × 10 × 109 Pa Pa.. PLAN Use modulus of elasticity equation to calculate pressure change required to achieve the desired volume change. SOLUTION Modulus of elasticity equation V ∆V ∆V ∆ p = E V E =
=
p
−∆
−
9
¡
¢µ
2.2 × 10 × 10 Pa
= 2.2 × 10 × 109 Pa (0. (0.02) 7 = 4.4 × 10 × 10 Pa
¡
¢
∆ p = p =
44 MPa MPa
56
0.01 × 01 × V V
−
¶
2.47: PROBLEM PROBLEM DEFINITION Situation: Open tank of water. water. T 20 20 = 20 C, T 8 80 0 = 80 C. V = = 400 400 l, d = d = 3 m. Hint: Hint: Volume olume change change is due to temperature. temperature. ◦
◦
Find: Percentage change in volume. Water level rise for given diameter. Properties: kg From Table A.5: ρ20 = 998 m ,and ρ and ρ 80 = 972 mkg . 3
3
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 in pres pressu sure re.. The The tank tank is open, open, so the press pressur uree at the surfa surface ce of the the tank tank is alw always atmospheric. SOLUTION a. Percen Percentag tagee chang changee in volume volume must must be calcula calculated ted for this mass mass of water water at two two temperatures. For the first temperature, the volume is given as V 2200 = 400 400 l = 0. 0.4 m3.Its density is kg ρ20 = 998 m . Therefore, the mass for both cases is given by. 3
m
kg × 0. 0.4 m3 m3 = 399.2 kg
= 9 98
For the second temperature, that mass takes up a larger volume: m 399. 399.2 kg = ρ 972 mkg = 0.411kg
V 8800 =
3
Therefore, the percentage change in volume is 0.411kg − 0.4 kg 0.4 kg
= 0.0275
volume % change =
= 2.8%
b. If the the tan tank k has has D = 3m, then V then V = = π πrr2 h = 7.68h. 68h.Therefore: Therefore:
57
h20 = .052m h80 = .054m And water level rise is 0 is 0..054 − 0.52m = 0. 0.002 002 m = 2 mm. water level rise is = is = 0.002 002 m = 2 mm REVIEW Density changes can result from temperature changes, as well as pressure changes.
58
2.48: PROBLEM PROBLEM DEFINITION Situation: Surface tension is an energy/area. Find: Force Show that Energy Area equals Length . Energy force · force · distance distance = Area area M T L · L = L2
"
=
#
2
∙ M ¸ T 2
Force = Length =
" # ∙ ¸ M T L L
2
M T 2
Force The primary dimensions for Energy and Length are both Area
59
M T 2
£¤
, so they are equal.
2.49: PROBLEM PROBLEM DEFINITION Situation: Very small spherical droplet of water. Find: Pressure inside. SOLUTION Refer to to Fig. Fig. 2-6(a). 2-6(a). The surface surface tension tension force, 2 force, 2πrσ πrσ,, will be resisted by the pressure force acting on the cut section of the spherical droplet or p( p(πr 2 ) = 2πrσ 2σ p = r p = p =
60
4σ d
2.50: PROBLEM PROBLEM DEFINITION Situation: A spherical soap bubble. Inside radius R radius R,, wall-thickness t wall-thickness t,, surface tension σ tension σ.. Special case: R = 4 mm. Find: Derive a formula for the pressure diff erence erence across the bubble Pressure diff erence erence for bubble with R with R = 4 mm. Assumptions: The eff ect ect of thickness is negligible, and the surface tension is that of pure water. PLAN Apply equilibrium, then the surface tension force equation. SOLUTION Force balance 2 x 2πRσ
p
Surface Surface tension tension force
∆ pπR
2
−
X
F = 0
2(2πRσ 2(2πRσ)) = 0
Formula for pressure di ff erence erence ∆ p = p =
4σ R
Pressure diff erence erence 4 × 7 × 7..3 × 10 × 10 2 N/m = 0.004 m 004 m ∆p4mm rad. = 73. 73.0 N/m2 −
∆ p4mm rad.
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2.51: PROBLEM PROBLEM DEFINITION Situation: A water bug is balanced on the surface of a water pond. n = 6 legs, legs, = = 5 mm/leg. Find: Maximum mass of bug to avoid sinking. Properties: Surface tension of water, from Table A.4, σ = 0.073 N/m. 073 N/m. PLAN Apply equilibrium, then the surface tension force equation. SOLUTION Force equilibrium Upward force due to surface tension = Weight of Bug F T T = mg To find the force of surface tension (F T T ), consider the cross section of one leg of the bug: Cross section of bug leg
Surface tension force on one side of leg
θ
F F Assume Assume θ is small small Then cos θ =1; F cos θ= F
Surface Surface tension tension force F T T = = = =
(2/ (2/leg)(6 leg)(6 legs legs))σ 12σ 12σ 12(0.073 N/m 073 N/m)(0 )(0..005 m) 0.0043 004388 N
Apply equilibrium F T T − mg = 0 F T 0.0043 004388 N T m = = g 9.81 m2 / s = 0.4465 × 4465 × 10 10 3 kg −
m = 0.447 × 447 × 10 10
62
3
−
kg
2.52: PROBLEM PROBLEM DEFINITION Situation: A water column in a glass tube is used to measure pressure. d1 = 0.25in, 25in, d 2 = 1/8 in, in, d 3 = 1/32in. 32in. Find: Height of water column due to surface tension e ff ects ects for all diameters. Properties: From Table Table A.4: surface surface tension of water water is 0.00 0.005 5 lbf/ft. SOLUTION Surface Surface tension tension force 4σ 4 × 0 × 0..005 005 lbf lbf / ft 3.21 × 21 × 10 10 4 = = ft. γd d 62. 62.4lbf / ft3 × d 1 1 3.21 × 21 × 10 10 4 ft in. = in. = ft.; ∆h = = 0. 0 .0154 ft. ft. = 0.18 0.185 5 in. in. 4 48 1/48 1 1 3.21 × 21 × 10 10 4 ft in. = in. = ft.; ∆h = = 0. 0 .0308 0308 ft. ft. = 0.36 0.369 9 in. in. 8 96 1/96 1 1 3.21 × 21 × 10 10 4 ft in. = in. = ft.; ∆h = = 0.123 ft.= 123 ft.= 1.48 in. 32 384 1/384 −
∆h
=
−
d =
−
d =
−
d =
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2.53: PROBLEM PROBLEM DEFINITION Situation: Two vertical glass plates y = 1 mm Find: Capillary rise ( rise (h h) between the plates. Properties: From Table A.4, surface tension of water is 7. 7 .3 × 10 × 10
2
−
N/m.
PLAN Apply equilibrium, then the surface tension force equation. SOLUTION θ y
σ
σ
y
Equilibrium
X
F y = 0
Force due to surface tension = Weight of fluid that has been pulled upward (2 (2) σ = (ht) ht) γ Solve for capillary rise (h ( h) 2σ − htγ = 0 2σ h = γt 2 × (7 × (7..3 × 10 × 10 2 N/ m) h = 9810N/ 9810N/ m3 × 0. 0.001m = 0.0149 m 0149 m h = 14.9 mm −
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2.54: PROBLEM PROBLEM DEFINITION 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. 7 .3 × 10 × 10
2
−
N/m
PLAN Apply equilibrium, then the surface tension force equation. SOLUTION Equilibrium (half the water droplet) Force due to pressure = Force due to surface surface tension pA = σL 2 ∆ pπR = 2πRσ Solve for pressure 2σ R 2 × 7 × 7..3 × 10 × 10 2 N/ m ∆ p = (0. (0.5 × 10 × 10 3 m) ∆ p
=
−
−
p = ∆ p =
292 N/m 292 N/m2
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2.55: PROBLEM PROBLEM DEFINITION Situation: A tube employing capillary rise is used to measure temperature of water T 0 = 0 C, T 100 100 = 100 C σ0 = 0.0756N/ 0756N/ m, σ 100 = 0.0589N/ 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 =
4σ γd
where γ where γ is is the specific weight and d is the tube diameter. diameter. For the change in surface tension due to temperature, the change in column elevation would be ∆h =
4∆σ 4 × 0 × 0..0167N/ 0167N/ m 6.8 × 10 × 10 = = γd 9810N/ 9810N/ m3 × d d
6
−
The cha change nge in colum column n elevati elevation on for for a 1-m 1-mm m diamete diameterr tube woul would d be 6.8 mm . Special equipment, such the optical system from a microscope, would have to be used to measure such a small change in de flection ection It is unlik unlikely ely that smalle smallerr tubes tubes ma made de of transparent material can be purchased to provide larger de flections.
66
2.56: PROBLEM PROBLEM DEFINITION Situation: A soap bubble and a droplet of water of equal diameter falling in air d = 2 mm, σ bubble = σ = σ droplet Find: Which has the greater pressure inside. SOLUTION The soap bubble will have the greatest pressure because there are two surfaces (two surface tension forces) creating the pressure within the bubble. The correct choice is a)
67
2.57: PROBLEM PROBLEM DEFINITION Situation: A hemispherical drop of water is suspended under a surface Find: Diameter of droplet just before separation Properties: Table A.5 (20 (20 C): γ = = 9790 9790 N/ m3 , σ = 0.073N/ 073N/ m. ◦
SOLUTION Equilibrium Weight of droplet = Force due to surface surface tension πD 3 γ = (πD) πD ) σ 12
µ ¶
Solve for D for D 12σ 12σ γ 12 × 12 × (0 (0..073 N/m 073 N/m)) = = 8. 8 . 948 × 948 × 10 10 9790 N/m 9790 N/m3 D = 9. 459 × 459 × 10 10 3 m
D2 =
−
D = 9.46mm
68
5
−
m2
2.58: PROBLEM PROBLEM DEFINITION Situation: Surface tension is being measured by suspending liquid from a ring Di = 10 cm, cm, D o = 9.5 cm W = W = 10 10 g, F = F = 16 16 g Find: Surface tension PLAN 1. Force orce equili equilibri brium um on on the fluid suspend suspended ed in the ring. ring. For force force due to surfac surfacee tension, use the form of the equation provided in the text for the special case of a ring being pulled out of a liquid. 2. Solve Solve for surface surface tension - all the other other forces are are known. known. SOLUTION 1. Force equilibrium equilibrium (Upward force) force) = (Weight of fluid) uid) + (Force ( Force due to surface tension) tension) F = W + σ(πDi + πDo ) 2. Solve Solve for for surface surface tensi tension on F − W π(Di + Do ) (0. (0.016 − 0.010)kg × 010)kg × 9 9..81 m/ s2 σ = π(0. (0.1 + 0. 0.095)m kg = 9.61 × 61 × 10 10 2 2 s σ =
−
σ = 0.0961 N/m 0961 N/m
69
2.59: PROBLEM PROBLEM DEFINITION Situation: A liquid reaches the vapor pressure Find: What happens to the liquid SOLUTION If a liquid reaches reaches its vapor vapor pressure pressure for a given temperatur temperature, e, it boils .
70
2.60: PROBLEM PROBLEM DEFINITION Find: How does vapor pressure change with increasing temperature? SOLUTION The vapor pressure pressure increases with increasing increasing temperatur temperaturee . To get an everyda everyday y feel for this, note from the Appendix that the vapor pressure of water at 212 F (100 C) is 101 kPa (14 kPa (14.7 .7 psia). To get water to boil at a lower lower temperature, temperature, you would would have have to exert a vacuum vacuum on on the water. water. To keep it from from boiling unti untill a higher temperatur temperature, e, you would have to pressurize it. ◦
71
◦
2.61: PROBLEM PROBLEM DEFINITION Situation: Watar at 60 F ◦
Find: The pressure that must be imposed for water to boil Properties: Water (60 (60 F), Table A.5: P v = 0.363 psia 363 psia ◦
SOLUTION The pressure to which the fluid uid must must be expose exposed d is P = 0.3 0.363 63 psia psia.. This This is low lower than atmospher atmospheric ic pressure. pressure. Therefore, Therefore, assuming assuming atmosphe atmospheric ric pressure pressure is 14.7 psia gage, gag e, or 14.7 psig, psig, the pressure needed could could also be reported as P = -14.34 psig .
72
2.62: PROBLEM PROBLEM DEFINITION Situation: T = 20 C,fluid is water. ◦
Find: The pressure that must be imposed to cause boiling Properties: Water (60 (60 F), Table A.5: P v = 2340 Pa 2340 Pa abs ◦
SOLUTION Bubbles will be noticed to be forming when P = P v . P = P = 2340 Pa 2340 Pa abs
73
2.63: PROBLEM PROBLEM DEFINITION Situation: Water in a closed tank T = 20 C, p = p = 10400 10400 Pa ◦
Find: Whether water will bubble into the vapor phase (boil). Properties: From Table A.5, at T = 20 C, P v = 2340 Pa 2340 Pa abs ◦
SOLUTION The tank pressure is 10,400 Pa abs, and P v = 2340 2340 Pa Pa abs. abs. So the the tank tank pressur pressuree is higher than the P the P v . Therefo Therefore re the water water will will not boil . REVIEW The water can be made to boil at this temperature only if the pressure is reduced to 2340 Pa Pa abs. abs. Or, Or, the the water ater can can be made ade to boil boil at this this press pressur uree only only if the the temperature is raised to approximately 50 C. ◦
74
2.64: PROBLEM PROBLEM DEFINITION Situation: The boiling temperature of water decreases with increasing elevation ∆ p kPa = 3.1C kPa . ∆T −
o
Find: Boiling temperature at an altitude of 3000 m Properties: T = T = 100o C, p C, p = = 101kN/ 101kN/ m2 . z3000 = 3000 3000 m, p 3000 = 69 69 kN/ kN/ m2 . 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 Let B Let BT T = "Boi "Boiling ling Temperature." emperature." Then, B Then, B T as T as a function of elevation is BT (3000 BT (3000 m) = m) = BT BT (0 m) +
∆BT
µ ¶ ∆ p
∆ p
Thus, BT (3000 BT (3000 m) = 100 C + ◦
µ
= 89. 89. 677 C ◦
1.0 C (101 − 69) kPa 3.1kPa
−
◦
¶
Boiling Temperature (3000 m) = m) = 89. 89.7 C ◦
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