**Unformatted text preview: **Notes to instructors
Introduction
The following ideas and information are provided to assist the instructor in the design and implementation
of the course. Traditionally this course is taught at Washington State University and the University of Idaho as a
three-credit semester course which means 3 hours of lecture per week for 15 weeks. Basically the first 11 chapters
and Chapter 13 (Flow Measurements) are covered in Mechanical Engineering. Chapters 12 (Compressible Flow)
and Chapter 14 (Turbomachinery) may be covered depending on the time available and exposure to compressible
flow in other courses (Thermodynamics). Open channel flow (Chapter 15) is generally not covered in Mechanical
Engineering. When the text is used in Civil Engineering, Chapters 1-11 and 13 are nominally covered and Chapters
14 and 15 may be included if time permits and exposure to open channel flow may not be available in other courses.
The book can be used for 10-week quarter courses by selecting the chapters, or parts of the chapters, most appropriate
for the course. Author Contact
Every eﬀort has been made to insure that the solution manual is error free. If errors are found (and they
will be!) please contact Professors Crowe or Elger.
Donald Elger
Mechanical Engineering Dept
University of Idaho
Moscow, ID 83844-0902
Phone (208) 885-7889
Fax (208) 885-9031
e-mail: [email protected] Clayton Crowe
School of Mechanical Eng. & Matl. Science
Washington State University
Pullman, WA 99164-2920
Phone (509) 335-3214
Fax (509) 335-4662
e-mail: [email protected] Design and Computer Problems
Design problems (marked in the text in blue) are those problems that require engineering practices such
as estimation, making asummptions and considering realistic materials and components. These problems provide a
platform for student discussion and group activity. One approach is to divide the class into small groups of three or
four and have these groups work on the design problems together. Each group can then report on their design to
the rest of the class. The role of the professor is to help the student learn the practices of the design review—that is,
teach the student to ask in-depth questions and teach them how to develop meaningful and in-depth answers. This
dialogue stimulates interest and class discussion. Solutions to most design problems are included in the solution
manual.
Computer-oriented problems (marked in the text is blue) are those problems may best be solved using
software such as spreadsheets, TK Solver or MathCad. The choice is left to the student. The answer book also
includes the results for the computer-oriented problems. 1 PROBLEM 2.1
Situation: An engineer needs density for an experiment with a glider.
Local temperature = 74.3 ◦ F = 296.7 K.
Local pressure = 27.3 in.-Hg = 92.45 kPa.
Find: (a) Calculate density using local conditions.
(b) Compare calculated density with the value from Table A.2, and make a recommendation.
Properties: From Table A.2, Rair = 287 J
,
kg· K ρ = 1.22 kg/ m3 . APPROACH
Apply the ideal gas law for local conditions.
ANALYSIS
a.) Ideal gas law
ρ = p
RT 92, 450 N/ m2
(287 kg/ m3 ) (296.7 K)
= 1.086 kg/m3
= ρ = 1.09 kg/m3 (local conditions)
b.) Table value. From Table A.2
ρ = 1.22 kg/m3 (table value) COMMENTS
1. The density diﬀerence (local conditions versus table value) is about 12%. Most
of this diﬀerence is due to the eﬀect of elevation on atmospheric pressure.
2. Answer ⇒ Recommendation—use the local value of density because the eﬀects
of elevation are significant. 1 PROBLEM 2.2
Situation: Carbon dioxide is at 300 kPa and 60o C.
Find: Density and specific weight of CO2 .
Properties: From Table A.2, RCO2 = 189 J/kg·K.
APPROACH
First, apply the ideal gas law to find density. Then, calculate specific weight using
γ = ρg.
ANALYSIS
Ideal gas law
P
RT
300, 000
=
189(60 + 273) ρCO2 = = 4.767 kg/m3 Specific weight
γ = ρg
Thus
γ CO2 = ρCO2 × g
= 4.767 × 9.81
= 46.764 N/m3 2 PROBLEM 2.3
Situation: Methane is at 500 kPa and 60o C.
Find: Density and specific weight.
Properties: From Table A.2, RMethane = 518 J
.
kg· K APPROACH
First, apply the ideal gas law to find density. Then, calculate specific weight using
γ = ρg.
ANALYSIS
Ideal gas law
P
RT
500, 000
=
518(60 + 273) ρHe = = 2.89 kg/m3 Specific weight
γ = ρg
Thus
γ He = ρHe × g
= 2.89 × 9.81
= 28.4 N/m3 3 PROBLEM 2.4
Situation: Natural gas (10 ◦ C) is stored in a spherical tank. Atmospheric pressure is
100 kPa.
Initial tank pressure is 100 kPa-gage. Final tank pressure is 200 kPa-gage.
Temperature is constant at 10 ◦ C.
Find: Ratio of final mass to initial mass in the tank.
APPROACH
Use the ideal gas law to develop a formula for the ratio of final mass to initial mass.
ANALYSIS
Mass
M = ρV (1) p
RT (2) Ideal gas law
ρ=
Combine Eqs. (1) and (2) M = ρV−
= (p/RT )V
−
Volume and gas temperature are constant so
p2
M2
=
M1
p1
and
300 kPa
M2
=
M1
200 kPa
= 1.5 4 PROBLEM 2.5
Situation: Water and air are at T = 100o C and p = 5 atm.
Find: Ratio of density of water to density of air.
Properties: From Table A.2, Rair = 287 J/kg·K. From Table A.5, ρwater = 958 kg/m3 .
APPROACH
Apply the ideal gas to air. Look up the density of water in Table A.5.
ANALYSIS
Ideal gas law
ρair = p
RT 506, 600
287(100 + 273)
= 4.73 kg/m3
= For water
ρwater = 958 kg/m3
Ratio
ρwater
958
=
ρair
4.73
= 5 202 PROBLEM 2.6
Situation: Oxygen (p = 400 psia, T = 70 ◦ F)fills a tank. Tank volume = 10 ft3 . Tank
weight =100 lbf.
Find: Weight (tank plus oxygen).
Properties: From Table A.2, RO2 = 1555 ft·lbf/(slug ·o R) .
APPROACH
Apply the ideal gas law to find density of oxygen. Then find the weight of the oxygen
using specific weight (γ) and add this to the weight of the tank.
ANALYSIS
Ideal gas law
pabs. = 400 psia × 144 psf/psi = 57, 600 psf
T = 460 + 70 = 530◦ R
p
ρ =
RT
57, 600
=
1555 × 530
= 0.0699 slugs/ft3
Specific weight (oxygen)
γ = ρg
= 0.0699 × 32.2
= 2.25 lbf/ft3
Weight of filled tank
Woxygen =
=
Wtotal =
=
Wtotal = 2.25 lbf/ft3 × 10 ft3
22.5 lbf
Woxygen + Wtank
22.5 lbf + 100 lbf
122.5 lbf COMMENTS
For compressed gas in a tank, pressures are often very high and the ideal gas assumption is invalid. For this problem the pressure is about 27 atmospheres—it is a good
idea to check a Thermodynamics reference to analyze whether or not real gas eﬀects
are significant. 6 PROBLEM 2.7
Situation: Air is at an absolute pressure of p = 600 kPa and a temperature of
T = 50o C.
Find: (a) Specific weight, and (b) density
Properties: From Table A.2, R = 287 J
.
kg· K APPROACH
First, apply the ideal gas law to find density. Then, calculate specific weight using
γ = ρg.
ANALYSIS
Ideal gas law
P
RT
600, 000
=
287(50 + 273) ρair = = 6.47 kg/m3 Specific weight
γ air = ρair × g
= 6.47 × 9.81
= 63.5 N/ m3 7 PROBLEM 2.8
Situation: Consider a mass of air with a volume of 1 cubic mile.
Find: Mass of air in a volume of 1 mi3 . Express the answer using units of slugs and
kg.
Properties: From Table A.2, ρair = 0.00237 slugs/ft3 .
Assumptions: The density of air is the value at sea level for standard conditions.
ANALYSIS
Units of slugs
M = ρV
= 0.00237 slug
× (5280)3 ft3
ft3
M = 3.49 × 108 slugs
Units of kg ¶
µ
¡
¢
kg
8
M = 3.49 × 10 slug × 14.59
slug
M = 5.09 × 109 kg COMMENTS
The mass will probably be somewhat less than this because density decreases with
altitude. 8 PROBLEM 2.9
Situation: This problem involves the eﬀects of temperature on the properties of air.
The application is a bicyclist.
Find: a.) Plot air density versus temperature for a range of -10o C to 50o C.
b.) Plot tire pressure versus temperature for the same temperature range.
Properties: From Table A.2, Rair = 287 J/kg/K.
Assumptions: For part b, assume that the bike tire was initially inflated to ptire = 450
kPa, abs at T = 20o C.
APPROACH
Apply the ideal gas law.
ANALYSIS
Ideal gas law
ρ= 101000
p
=
RT
287 × (273 + T ) 1.40 3 Density (kg/m ) 1.35
1.30
1.25
1.20
1.15
1.10
1.05
-20 -10 0 10 20 30
o T emperature ( C ) with density constant
p = po 9 T
To 40 50 60 520
Tire pressure, kPa 500
480
460
440
420
400
380
-20 -10 0 10 20 30 Temperature, oC 10 40 50 60 PROBLEM 2.10
Situation: A design team needs to know how much CO2 is needed to inflate a rubber
raft.
Raft is shown in the sketch below.
Inflation pressure is 3 psi above local atmospheric pressure. Thus, inflation pressure
is 17.7 psi = 122 kPa. Find: (a)Estimate the volume of the raft.
(b) Calculate the mass of CO2 in grams to inflate the raft.
Properties: From Table A.2, RCO2 = 189 J/kgK.
Assumptions: 1.) Assume that the CO2 in the raft is at 62 ◦ F = 290 K.
2.) 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).
APPROACH
Mass is related to volume by m = ρ∗Volume. Density can be found using the ideal
gas law.
ANALYSIS
Volume contained in the tubes.
πD2
∆V— =
×L
¶
µ4
π × 0.452
× 16 m3
=
4
= 2.54 m3
∆V— = 2.54 m3
Ideal gas law
ρ = p
RT 122, 000 N/ m2
(189 J/ kg · K) (290 K)
= 2.226 kg/m3
= 11 Mass of CO2
m = ρ × Volume
¢ ¡
¢
¡
= 2.226 kg/m3 × 2.54 m3
= 5.66 kg
m = 5.66 kg
COMMENTS
The final mass (5.66 kg = 12.5 lbm) is large. This would require a large and potentially expensive CO2 tank. Thus, this design idea may be impractical for a product
that is driven by cost. 12 PROBLEM 2.11
Situation: The application is a helium filled balloon of radius r = 1.3 m.
p = 0.89 bar = 89 kPa.
T = 22 ◦ C = 295.2 K.
Find: Weight of helium inside balloon.
Properties: From Table A.2, RHe = 2077 J/kg·K.
APPROACH
Weight is given by W = mg. Mass is related to volume by m = ρ∗Volume. Density
can be found using the ideal gas law.
ANALYSIS
Volume in a sphere
4 3
πr
3
4
π1.33 m3
=
3
= 9.203 m3 Volume = Ideal gas law
ρ = p
RT 89, 000 N/ m2
(2077 J/ kg · K) (295.2 K)
= 0.145 kg/m3 = Weight of helium
W = ρ × Volume × g
¢ ¡
¢ ¡
¢
¡
= 0.145 kg/m3 × 9.203 m3 × 9.81 m/ s2
= 13.10 N
Weight = 13.1 N 13 PROBLEM 2.12
Situation: In the wine and beer industries, fermentation involves glucose (C6 H12 O6 )
being converted to ethyl alcohol (CH3 CH2 OH) plus carbon dioxide gas that escapes
from the vat.
C6 H12 O6 → 2(CH3 CH2 OH) + 2(CO2 )
The initial specific gravity is 1.08.
Specific gravity of alcohol is 0.80.
Saturated solution (water + sugar) has a specific gravity of 1.59.
Find: (a.) Final specific gravity of the wine.
(b.) Percent alcohol content by volume after fermentation.
Assumptions: All of the sugar is converted to alcohol.
APPROACH
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 volume). Next, apply conservation of mass to find the mass of alcohol that is
produced (per unit of volume). Then, solve for the problem unknowns.
ANALYSIS
The initial density of the mixture is
ρmix = ρw Vw + ρs Vs
Vo where ρw and ρs are the densities of water and sugar solution (saturated), Vo is the
initial volume of the mixture, and Vs is the volume of sugar solution. The total
volume of the mixture is the volume of the pure water plus the volume of saturated
solution
Vw + Vs = Vo
The specific gravity is initially 1.08. Thus
ρmix
Vs
ρ Vs
= (1 − ) + s
ρw
Vo
ρw Vo
Vs
Vs
1.08 = (1 − ) + 1.59
Vo
Vo
Vs
= 0.136
Vo
Si = Thus, the mass of sugar per unit volume of mixture
ms
= 1.59 × 0.136
Vo
= 0.216 kg/m3
14 The molecular weight of glucose is 180 and ethyl alcohol 46. Thus 1 kg of glucose
converts to 0.51 kg of alcohol so the final density of alcohol is
ma
= 0.216 × 0.51
Vo
= 0.110 kg/m3
The density of the final mixture based on the initial volume is
mf
Vo = (1 − 0.136) + 0.110
= 0.974 kg/m3 The final volume is altered because of conversion
Vf
mw
ma
=
+
Vo
ρw Vo ρa Vo
Vw 0.51ms
=
+
Vo
ρa Vo
Vw 0.51ρs Vs
=
+
Vo
ρa Vo
0.51 × 1.59
× 0.136
= 0.864 +
0.8
= 1.002
The final density is
mf
Vf = mf
Vo
×
Vo
Vf 1
1.002
= 0.972 kg/m3 = 0.974 ×
The final specific gravity is Sf = 0.972
The alcohol content by volume
Va
ma
=
Vf
ρa Vf
ma 1 Vo
=
Vo ρa Vf
= 0.110 ×
= 0.137 1
1
×
0.8 1.002 Thus,
Percent alcohol by volume = 13.7% 15 PROBLEM 2.13
Situation: This problem involves the viscosity and density of air and water.
Find: (a)Change in viscosity and density of water for a temperature change of 10o C
to 70o C.
(b)Change in viscosity and density of air for a temperature change of 10o C to 70o C.
APPROACH
For water, use data from Table A.5. For air, use data from Table A.3
ANALYSIS
Water
µ70 = 4.04 × 10−4 N·s/m2
µ10 = 1.31 × 10−3 N·s/m2
∆µ=-9. 06×10−4 N · s/m2 ρ70 = 978 kg/m3
ρ10 = 1000 kg/m3
∆ρ=-22 kg/ m3
Air µ70 = 2.04 × 10−5 N · s/m2
µ10 = 1.76 × 10−5 N · s/m2
∆µ = 2. 8 × 10−6 N·s/m2
ρ70 = 1.03 kg/m3
ρ10 = 1.25 kg/m3
∆ρ = −0.22 kg/ m3 16 PROBLEM 2.14
Situation: Air at 10o C and 60o C.
Find: Change in kinematic viscosity from 10o C to 60o C.
Properties: From table A.3, ν 60 = 1.89 × 10−5 m2 /s, ν 10 = 1.41 × 10−5 m2 /s.
APPROACH
Use properties found in table A.3.
ANALYSIS
∆vair,10→60 = (1.89 − 1.41) × 10−5 = 4.8×10−6 m2 /s 17 PROBLEM 2.15
Situation: This problem involves viscosity of SAE 10W-30 oil, kerosene and water.
Find: Dynamic and kinematic viscosity of each fluid at 38o C.
APPROACH
Use property data found in Table A.4, Fig. A.2 and Table A.5.
ANALYSIS Oil (SAE 10W-30)
µ(N · s/m )
6.7×10−2
ρ(kg/m3 )
880
ν(m2 /s)
7.6×10−5 kerosene
1.4×10−3 (Fig. A-2) 2 1.7×10−6 (Fig. A-2) 18 water
6.8×10−4
993
6.8×10−7 PROBLEM 2.16
Situation: Air and water at 20o C.
Find: (a)Ratio of dynamic viscosity of air to that of water.
(b)Ratio of kinematic viscosity of air to that of water.
Properties: From Table A.3, µair,20◦ C = 1.81 × 10−5 N·s/m2 ; ν = 1.51 × 10−5 m2 /s
From Table A.5, µwater,20◦ C = 1.00 × 10−3 N·s/m2 ; ν = 1.00 × 10−6 m2 /s
ANALYSIS
1.81 × 10−5 N · s/ m2
= 1.81×10−2
1.00 × 10−3 N · s/ m2
1.51 × 10−5 m2 / s
=
= 15.1
1.00 × 10−6 m2 / s µair /µwater =
ν air /ν water 19 PROBLEM 2.17 Computer Problem - no solution is provided. 20 PROBLEM 2.18
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 is at
temperature T and pressure p.
Assumptions: Assume a gas is at temperature To and pressure po , where the subscript
”o” defines the reference state.
APPROACH
Combine the ideal gas law and Sutherland’s equation.
ANALYSIS
The ratio of kinematic viscosities is
µ ¶3/2
ν
To + S po T
µ ρo
T
=
=
νo
µo ρ
To
T + S p To
³ ´5/2
po
To +S
ν
T
=
νo
p
To
T +S 21 PROBLEM 2.19
Situation: The viscosity of air is µair (15o C) = 1.78 × 10−5 N·s/m2 . Find: Dynamic viscosity µ of air at 200 ◦ C using Sutherland’s equation. Properties: From Table A.2, S = 111K.
ANALYSIS
Sutherland’s equation
µ
=
µo µ µ T
To ¶3/2 473
=
288
= 1.438 To + S
T +S ¶3/2 288 + 111
473 + 111 Thus
µ = 1.438µo
¡
¢
= 1.438 × 1.78 × 10−5 N · s/ m2
µ = 2.56 × 10−5 N·s/m2 22 PROBLEM 2.20
Situation: Kinematic viscosity of methane at 15o C and 1 atm is 1.59 × 10−5 m2 / s. Find: Kinematic viscosity of methane at 200o C and 2 atm.
Properties: From Table A.2, S = 198 K.
APPROACH
Apply the ideal gas law and Sutherland’s equation.
ANALYSIS µ
ρ
µ ρo
=
µo ρ ν =
ν
νo
Ideal-gas law ν
µ po T
=
νo
µo p To
Sutherland’s equation
ν
po
=
νo
p µ T
To ¶5/2 To + S
T +S so
µ
¶5/2
288 + 198
ν
1 473
=
νo
2 288
473 + 198
= 1.252
and
ν = 1.252 × 1.59 × 10−5 m2 /s
= 1.99 × 10−5 m2 / s 23 PROBLEM 2.21
Situation: Nitrogen at 59o F has a dynamic viscosity of 3.59 × 10−7 lbf · s/ ft2 .
Find: µ at 200o F using Sutherland’s equation. Properties: From Table A.2, S =192o R.
ANALYSIS
Sutherland’s equation
µ
=
µo µ µ T
To ¶3/2 660
=
519
= 1.197 To + S
T +S ¶3/2 519 + 192
660 + 192 ¶
µ
−7 lbf · s
µ = 1.197 × 3.59 × 10
ft2
= 4. 297 × 10−7
µ = 4.30 × 10−7 lbf-s/ft2 24 PROBLEM 2.22
Situation: Helium at 59o F has a kinematic viscosity of 1.22 × 10−3 ft2 / s. Find: Kinematic viscosity at 30o F and 1.5 atm using Sutherland’s equation. Properties: From Table A.2, S =143o R.
APPROACH
Combine the ideal gas law and Sutherland’s equation.
ANALYSIS
¶5/2
To + S
T
To
T +S
µ
¶5/2
519 + 143
1.5 490
1 519
490 + 143
1.359 µ
2¶
−3 ft
1.359 × 1.22 × 10
s
2
ft
1. 658 × 10−3
s ν
po
=
νo
p
=
=
ν =
= µ ν = 1.66 × 10−3 ft2 / s 25 PROBLEM 2.23
Situation: Information about propane is provided in the problem statement.
Find: Sutherland’s constant.
ANALYSIS
Sutherland’s equation ¡ ¢
µ To 1/2
−1
S
µ
T
= o
¡
¢
3/2
To
1 − µµ TTo
o Also µ
= 1.72
µo
To
373
=
T
673
Thus
S
= 0.964
To
S = 360 K 26 PROBLEM 2.24
Situation: Information about ammonia is provided in the problem statement.
Find: Sutherland’s constant.
ANALYSIS
Sutherland’s equation ¡ ¢
µ To 1/2
−1
S
µ
T
= o
¡
¢
3/2
To
1 − µµ TTo (1) o Calculations 3.46 × 10−7
µ
=
= 1.671
µo
2.07 × 10−7
To
528
=
= 0.6197
T
852
Substitute (a) and (b) into Eq. (1)
S
= 1.71
To
S = 903 o R 27 (a)
(b) PROBLEM 2.25
Situation: Information about SAE 10W30 motor oil is provided in the problem statement.
Find: The viscosity of motor oil at 60 ◦ C, µ(60o C), using the equation µ = Ceb/T .
APPROACH
Use algebra and known values of viscosity (µ) to solve for the constant b.
solve for the unknown value of viscosity. Then, ANALYSIS
Viscosity variation of a liquid can be expressed as µ = Ceb/T . Thus, evaluate µ at
temperatures T and To and take the ratio:
·
¸
µ
1
1
= exp b( − )
µo
T
To
Take the logarithm and solve for b.
b= ln (µ/µo )
( T1 − T1o ) Data
µ/µo = 0.011/0.067 = 0.164
T = 372
To = 311
Solve for b
b = 3430 (K)
Viscosity ratio at 60o C
µ
1
1
−
)
= exp[3430(
µo
333 311
= 0.4833
µ = 0.4833 × 0.067
= 0.032 N · s/ m2 28 PROBLEM 2.26
Situation: Information about grade 100 aviation oil is provided in the problem statement
Find: µ(150o F), using the equation µ = Ceb/T .
APPROACH
Use algebra and known values of viscosity (µ) to solve for the constant b.
solve for the unknown value of viscosity. Then, ANALYSIS
Viscosity variation of a liquid can be expressed as µ = Ceb/T . Thus, evaluate µ at
temperatures T and To and take the ratio:
·
¸
µ
1
1
= exp b( − )
µo
T
To
Take the logarithm and solve for b
b= ln (µ/µo )
( T1 − T1o ) Data
µ
0.39 × 10−3
=
= 0.08804
µo
4.43 × 10−3
T = 670
To = 560
Solve for b
b = 8293 (o R)
Viscosity ratio at 150o F
1
1
µ
= exp[8293(
−
)
µo
610 560
= 0.299 µ
¶
−3 lbf · s
µ = 0.299 × 4.43 × 10
ft2
= 1.32 × 10−3 lbf-s/ft2 29 PROBLEM 2.27
Situation: This problem involves the creation of a computer program to find Sutherland’s constant and application to CO2 .
Find: Develop a computer program and carry out the activities described in the
textbook.
ANALYSIS
Sutherland’s constant
¡ ¢
µ 273 1/2
−1
S
µ
T
= o
¡
¢
3/2
273
1 − µµ 273
T (1) o Program Eq. (1), process data and take the average
S = 127 K
Define error ¯
¯µ
¯ − µ |calc ¯
µ
¯
¯µ
error = 100 × ¯ o µo
¯
¯
¯
µ
o The results are T(K)
260 270 280
290
300
350
500 1000 1500
|calc .9606 .991 1.021 1.050 1.079 1.217 1.582 2.489 3.168
error(%) .013 .039 .084 .118 .108 .366 .486 1.17 3.56
µ
µo COMMENTS
The error is less than 0.5% for temperatures up to 500 K. The error is greater than
3.5% for temperatures above 1500K. 30 PROBLEM 2.28
Situation: Oil (SAE 10W30) fills the space between two plates. Plate spacing is
∆y = 1/8 = 0.125 in.
Lower plate is at rest. Upper plate is moving with a speed u = 25 ft/ s.
Find: Shear stress.
Properties: Oil (SAE 10W30 @ 150 ◦ F) from Figure A.2: µ = 5.2 × 10−4 lbf·s/ft2 .
Assumptions: 1.) Assume oil is a Newtonian fluid. 2.) Assume Couette flow (linear
velocity profile).
ANALYSIS
Rate of strain
∆u
du
=
dy
∆y
25 ft/ s
=
(0.125/12) ft
= 2400 s−1
Newton’s law of viscosity
µ ¶
du
τ = µ
dy
¶ µ
¶
µ
1
−4 lbf · s
× 2400
=
5.2 × 10
s
ft2
lbf
= 1. 248 2
ft
τ = 1.25 lbf/ ft2 31 PROBLEM 2.29
Situation: Air and water at 40 ◦ C and absolute pressure of 170 kPa
Find: Kinematic and dynamic viscosities of air and water.
Properties: Air data from Table A.3, µair = 1.91 × 10−5 N·s/m2
Water data from Table A.5, µwater = 6.53 × 10−4 N·s/m2 , ρwater = 992 kg/m3 .
APPROACH
Apply the ideal gas law to find density.
dynamic and absolute viscosity. Find kinematic viscosity as the ratio of ANALYSIS
A.) Air
Ideal gas law
p
RT
170,...

View
Full Document

- Spring '16
- Not Good
- Shear Stress, Surface tension