British Gravitational (BG) System
U. S. Customary Units
In the United States, most engineers still use the U.S. Customary system of units. The unit of length is a foot (ft), which is equal to 0.3048 meter; the unit of mass is a pound mass (lbm), which is equal to 0.453592 kg; and the unit of time is a second (s). In U.S. Customary system, the unit of force is pound force (lbf) and 1 lbfis defined as the weight of an object having a mass of 1 lbmat sea level and at a latitude of 45, where acceleration due to gravity is 32.2 ft/s2. One pound force is equal to 4.448 newton (N). Because the pound force is not defined formally using Newton’s second law and instead is defined at a specific location, a correction factor must
T 1°R2 9 5T 1K2
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Chapter 6 Fundamental Dimensions and UnitsTABLE 6.3 Examples of Derived Units in Engineering
Name of Symbol for Expression in Terms Physical Quantity SI Unit SI Unit of Base Units
Acceleration m/s2
Angle Radian rad
Angular acceleration rad /s2
Angular velocity rad/s
Area m2
Density kg /m3
Energy, work, heat Joule J N
m or kgm2/s2Force Newton N kg
m/s2Frequency Hertz Hz s1
Impulse N
S or kgm/sMoment or torque N
m or kgm2/s2Momentum kg
m/sPower Watt W J/s or N
m/s or kgm2/s3Pressure, stress Pascal Pa N/m2or kg/m
s2Velocity m/s
Volume m3
Electric charge Coulomb C A
sElectric potential Volt V J/C or m2
kg/(s3A2)Electric resistance Ohm V/A or m2
kg/(s3A2)Electric conductance Siemens S 1/ or s3
A2/(m2kg)Electric capacitance Farad F C / V or s4
A2/(m2kg)Magnetic flux density Tesla T V
s /m2or kg/(s2A)Magnetic flux Weber Wb V
s or m2kg/(s2A)Inductance Henry H V
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be used in engineering formulas when using U.S. Customary units. The units of temperature in the U.S. Customary system are identical to the BG system, which we discussed earlier.
Finally, by comparing the units of mass in the BG system (that is one slug) to the U.S.
system pound mass, we note 1 slug⬇32.2 lbm. For those of us who might be slightly massive (overweight), it might be wiser to express our mass in slugs rather than in pound mass or kilogram. For example, a person who has a mass of 200 lbmor 90.7 kg would sound skinny if he were instead to express his mass as 6.2 slugs. Note that 200 lbm90.7 kg6.2 slugs, and therefore, he is telling the truth about his mass! So you don’t have to lie about your mass; knowl- edge of units can bring about instant results without any exercise or diet!
Examples of SI and U.S. Customary units used in our everyday lives are shown in Tables 6.4 and 6.5, respectively. The conversion factors for the fundamental and derived dimensions commonly encountered in engineering are shown in Table 6.6. Detailed conversion tables are also given on the back endpapers of this book.
6.2 Systems of Units
137
TABLE 6.4 Examples of SI Units in Everyday Use
Examples of SI Unit Usage SI Units Used
Camera film 35 mm
Medication dose such as pills 100 mg, 250 mg, or 500 mg Sports:
swimming 100 m breaststroke or butterfly stroke
running 100 m, 200 m, 400 m, 5000 m, and so on
Automobile engine capacity 2.2 L (liter), 3.8 L, and so on
Light bulbs 60 W, 100 W, or 150 W
Electric consumption kWh (killo-Watt-hour)
Radio broadcasting signal frequencies 88 –108 MHz (FM broadcast band) 0.54 –1.6 MHz (AM broadcast band)
Police, fire 153 –159 MHz
Global positioning system signals 1575.42 MHz and 1227.60 MHz
TABLE 6.5 Examples of U.S. Customary Units in Everyday Use
Examples of U.S. Customary Unit Usage U.S. Customary Units Used Fuel tank capacity 20 gallons or 2.67 ft3
(1 ft37.48 gallons) Sports (length of a football field) 100 yd or 300 ft Power capacity of an automobile 150 hp or 82500 lb
ft/s(1 hp550 lb
ft/s)Distance between two cities 100 miles (1 mile5280 ft)
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Chapter 6 Fundamental Dimensions and UnitsTABLE 6.6 Systems of Units and Conversion Factors
System of Units
Dimension SI BG U.S. Customary Conversion Factors
Length Meter (m) Foot (ft) Foot (ft) 1 ft 0.3048 m
1 m 3.2808 ft
Time Second (s) Second (s) Second (s)
Mass Kilogram (kg) Slugs* Pound mass (lbm) 1 lbm 0.4536 kg
1 kg 2.2046 lbm 1 slug 32.2 lbm
Force Newton (N) 1 lbf** One pound mass*** 1 N 224.809E-3lbf
weighs one pound force 1 lbf 4.448N at sea level
Temperature Degree Celsius (°C) Degree Fahrenheit (°F) or Degree Fahrenheit (°F) or or Kelvin (K)**** degree Rankine (°R) degree Rankine (°R)
K °C 273.15 °R °F 459.67 °R °F 459.67
Work, Energy Joule ( J) (1N)(1m) lbf ft (1 lbf)(1 ft) 1 J 0.7375 ft lbf
Commonly written lbf ft (1 lbf)(1 ft) 1 ftlbf 1.3558 J
as ftlbf 1 Btu 778.17 ftlbf
Power
kW 1000 W
1 hp 0.7457 kW
*Derived or secondary dimension
**Fundamental dimension
***Note unlike SI and BG systems, the relationship between pound force and pound mass is not defined using Newton’s second law
****Note a Temperature value expressed in K reads Kelvin not degree Kelvin
1 hp550 ft#lbf s
1 W0.7375 ft#lbf
s
lbf# ft
second 11lbf1 second2 11 ft2 lbf#
ft
second 11lbf1 second2 11 ft2
Watt1W2 1 second1 Joule
#
#
°R 95 K K 59 °R
°F 95 °C32
°C 59 3°F324
1N11kg2 a1m s2b
11 slug2 a1ft
s2b
66..33 U Un niitt C Co on nvve errssiio on n
Some of you may recall that not too long ago NASA lost a spacecraft calledMars Climate Orbiterbecause two groups of engineers working on the project neglected to communicate correctly their calculations with appropriate units. According to an internal review conducted by NASA’s Jet Propulsion Laboratory, “a failure to recognize and correct an
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6.3 Unit Conversion
139
I. Launch
• Delta II 7425
• Launched January 3, 1999
• Launch mass: 574 kilograms
II. Cruise
• Thruster attitude control
• Four trajectory-correction maneuvers (TCM); site- adjustment maneuver September 1, 1999;
contingency 5th TCM at entry –24 hours
• Eleven month cruise
• Near-simultaneous tracking with Mars Climate Orbiter or Mars Global Surveyor during approach
III. Entry, Descent, Landing
• Arrival: December 3, 1999
• Jettison cruise stage;
microprobes separate from cruise stage
• Hypersonic entry
• Parachute descent;
propulsive landing
• Descent imaging of landing site
IV. Landed Operations
• Lands in Martian spring at 76 degrees South latitude, 195 degrees West longitude (76S, 195W)
• 90-day landed mission
• Meteorology, imaging, soil analysis, trenching
• Data relay via Mars Climate Orbiter, Mars Global Surveyor, or direct- to-Earth high-gain antenna I.
II.
III.
IV.
■
Figure 6.1 Mars polar lander mission overview.Source:Courtesy of NASA.
error in a transfer of information between theMars Climate Orbiter spacecraft team in Colorado and the mission navigation team in California led to the loss of the spacecraft.”
The peer review findings indicated that one team used U.S. Customary units (e.g., foot and pound) while the other used SI units (e.g., meter and kilogram) for a key spacecraft oper- ation. According to NASA, the information exchanged between the teams was critical to the maneuvers required to place the spacecraft in the proper Mars orbit. An overview of the Mars polar lander mission is given in Figure 6.1.
As you can see, as engineering students, and later as practicing engineers, when perform- ing analysis, you will find a need to convert from one system of units to another. It is very important for you at this stage in your education to learn to convert information from one sys- tem of units to another correctly. It is also important for you to understand and to remember to show all your calculations with proper units. This point cannot be emphasized enough!
Always show the appropriate units that go with your calculations. Example 6.1 shows the steps that you need to take to convert from one system of units to another.
E xa m p l e 6 .1 A person who is 6 feet and 1 inch tall and weighs 185 pound force (lbf) is driving a car at a speed of 65 miles per hour over a distance of 25 miles. The outside air temperature is 80F and has a density of 0.0735 pound mass per cubic foot (lbm/ft3). Convert all of the values given in this example from U.S. Customary Units to SI units. Conversion tables are given on the back endpapers of this book.
Person’s height,H
or
Person’s weight,W
Speed of the car,S
or
Distance traveled,D
Temperature of air,T
T 1°C2 5
9180322 26.7°C T 1°C2 5
93T 1°F2 324 D 125 miles2 a5280 ft
1 mile b a0.3048 m
1 ft b a 1 km
1000 mb 40.233 km S a104,607 m
hb a 1 h
3600 sb 29.057 m/s S a65 miles
h b a5280 ft
1 mile b a0.3048 m
1 ft b 104,607 m/h104.607 km/h W 1185 lbf2 a4.448 N
1 lbf b 822.88 N H 11.854 m2 a100 cm
1 m b 185.4 cm H a6 ft 11 in.2 a 1 ft
12 in.b b a0.3048 m
1 ft b 1.854 m
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6.4 Dimensional Homogeneity
141
Density of air, r
E xa m p l e 6 . 2 For the following problems, use the conversion factors given on the front and back end covers
of this book.
a) Convert the given value of area,A, from cm2to m2.
b) Convert the given value of volume,V, from mm3to m3.
c) Convert the given value of atmospheric pressure,P, from N/m2to lbf/in2.
d) Convert the given value of the density of water, r, from kg/m3to lbm/ft3.
66..44 D Diim me en nssiio on na all H Ho om mo og ge en ne eiittyy
Another important concept that you need to understand completely is that all formulas used in engineering analysis must be dimensionally homogeneous. What do we mean by “dimen- sionally homogeneous?” Can you, say, add someone’s height who is 6 feet tall to his weight of 185 lbfand his body temperature of 98F? Of course not! What would be the result of such a calculation? Therefore, if we were to use the formulaLabc, in which the variableL on the left-hand side of the equation has a dimension of length, then the variablesa,b, andc on the right-hand side of equation should also have dimensions of length. Otherwise, if vari- ablesa,b, andchad dimensions such as length, weight, and temperature, respectively, the given
r a1000 kg
m3b a 1 lbm
0.4536 kgb a 1 m
3.28 ftb362.5 lbm/ft3 r1000 kg/m3
P a105N
m2b a 1 lbf
4.448 Nb a0.0254 m
1 in. b214.5 lbf/in2 P105 N/m2
V 11000 mm32 a 1 m
1000 mmb3106 m3 V1000 mm3
A 1100 cm22 a 1 m
100 cmb20.01 m2 A100 cm2
r a0.0735 lbm
ft3 b a0.453 kg
1 lbm b a 1 ft
0.3048 mb31.176 kg/m3
formula would be inhomogeneous, which would be like adding someone’s height to his weight and body temperature! Example 6.3 shows how to check for homogeneity of dimensions in an engineering formula.
E xa m p l e 6 . 3 (a) When a constant load is applied to a bar of constant cross section, as shown in Figure 6.2,
the amount by which the end of the bar will deflect can be determined from the following relationship:
(6.1)
where
dend deflection of the bar in meter (m) P applied load in newton (N)
Llength of the bar in meter (m) Across-sectional area of the bar in m2 E modulus of elasticity of the material What are the units for modulus of elasticity?
For Equation (6.1) to be dimensionally homogeneous, the units on the left-hand side of the equation must equal the units on the right-hand side. This equality requires the modulus of elasticity to have the units of N/m2, as follows:
Solving for the units ofEleads to N/m2, newton per squared meter, or force per unit area.
(b) The heat transfer rate through a solid material is governed by Fourier’s law:
(6.2)
where
qheat transfer rate
kthermal conductivity of the solid material in watts per meter degree Celsius, W/m
CAarea in m2
T1T2temperature difference, C Lthickness of the material, m
What is the appropriate unit for the heat transfer rateq?
Substituting for the units ofk,A,T1,T2, andL, we have
From this you can see that the appropriate SI unit for the heat transfer rate is the watt.
qkAT1T2
L a W
m
#
°Cb 1m22 a°C
mb W
qkAT1T2 L d PL
AE 1 m 1N2 1m2 m2E d PL
AE
142
Chapter 6 Fundamental Dimensions and UnitsL
d A
P (a)
■
Figure 6.2(a) The bar in Example 6.3 and (b) heat transfer through a solid material.
(b) T1
T2
L
A
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6.5 Numerical versus Symbolic Solutions
143
66..55 N Nu um me erriic ca all vve errssu uss S Syym mb bo olliic c S So ollu uttiio on nss
When you take your engineering classes, you need to be aware of two important things: (1) under- standing the basic concepts and principles associated with that class, and (2) how to apply them to solve real physical problems (situations). In order to gain an understanding of the basic concepts, you need to study carefully the statement of governing laws and the derivations of engineering for- mulas and their limitations. After you have studied the underlying concepts, you then need to apply them to physical situations by solving problems. After studying a certain concept initially, you may think that you completely understand the concept, but it is through the application of the concept (by doing the homework problems) that you really can test your understanding.
Moreover, homework problems in engineering typically require either a numerical or a symbolic solution. For problems that require numerical solution, data is given. In contrast, in the symbolic solution, the steps and the final answer are presented with variables that could be substituted with data, if necessary. The following example will demonstrate the difference between numerical and symbolic solutions.
E xa m p l e 6 . 4 Determine the load that can be lifted by the hydraulic system shown. All of the necessary infor-
mation is shown in Figure 6.3.
The general relationship among force, pressure, and area is explained in detail in Chapter 10.
At this time, don’t worry about understanding these relationships. The purpose of this example is to demonstrate the difference between a numerical and a symbolic solution. The concepts that are used to solve this problem are: F1m1g,F2m2g, and F2(A2/A1)F1, whereFdenotes force,mis mass,gis acceleration due to gravity (g9.81 m/s2), andArepresents area.