Transcript Class 10.1 - Gordon State College

Class 9.2
Units
&
Dimensions
Objectives
Know the difference between units and
dimensions
Understand the SI, USCS, and AES
systems of units
Know the SI prefixes from nano- to gigaUnderstand and apply the concept of
dimensional homogeneity
Objectives
What is the difference between an
absolute and a gravitational system of
units?
What is a coherent system of units?
Apply dimensional homogeneity to
constants and equations.
RAT 13
Dimensions & Units
Dimension - abstract quantity (e.g. length)
Unit - a specific definition of a dimension based upon a
physical reference (e.g. meter)
What does a “unit” mean?
How long is the rod?
Rod of unknown length
Reference:
Three rods of 1-m length
The unknown rod is 3 m long.
unit
number
The number is meaningless without the unit!
How do dimensions behave
in mathematical formulae?
Rule 1 - All terms that are added or subtracted
must have same dimensions
D  A B C
All have identical dimensions
How do dimensions behave
in mathematical formulae?
Rule 2 - Dimensions obey rules of
multiplication and division
 [M]  [T 2 ] 

 2 
AB  [T ]  [L] 
D

 [L]
C
 [M] 
 2 
 [L ] 
How do dimensions behave
in mathematical formulae?
Rule 3 - In scientific equations, the arguments of
“transcendental functions” must be dimensionless.
A  ln( x)
C  sin( x)
B  exp( x)
D3
x
x must be dimensionless
Exception - In engineering correlations, the argument may have
dimensions
Transcendental Function - Cannot be given by algebraic expressions
consisting only of the argument and constants. Requires an infinite
2
3
series
x
x
e  1  x   ···
2! 3!
x
Dimensionally
Homogeneous Equations
An equation is said to be dimensionally
homogeneous if the dimensions on both
sides of the equal sign are the same.
Dimensionally
Homogeneous Equations
Volume of the frustrum of a right pyramid
with a square base
b
h


B
h 2
V  B  Bb  b2
3
L 2
3
L     L  L2  L2
 1
 
       L .
3
Dimensional Analysis
Pendulum - What is the period?
pk m
a
g
b
[M]
[T]
0
1


a
0
L
 T 2 

0

2b
[L]
0

0

[T]  [M]a
p  km g
0
1 / 2 1 / 2
L
b
L
c
L
m
b
p
[L]c
g


0
0

a0
 b  1 / 2

c
 c  1 / 2
L
 pk
g
Absolute and Gravitational Unit Systems
F m a

L
F  M 2
T 
[F]
[M]
[L]
[T]
Absolute
—
×
×
×
Gravitational
×
—
×
×
× = defined unit
— = derived unit
Coherent and Noncoherent Unit Systems
Coherent Systems - equations can be written without
needing additional conversion factors
F m a
Noncoherent Systems - equations need additional
conversion factors
a
F m
gc
Conversion
Factor
(see Table 14.3)
Noncoherent Systems
F m a

L
F  M 2
T 
Noncoherent
[F]
[M]
[L]
[T]
×
×
×
×
× = defined unit
— = derived unit
The noncoherent system results when all four quantities are defined
in a way that is not internally consistent
Examples of Unit Systems
See Table 14.1
The International System of
Units (SI)
Fundamental Dimension
Base Unit
length [L]
meter (m)
mass [M]
kilogram (kg)
time [T]
second (s)
electric current [A]
ampere (A)
absolute temperature [q]
luminous intensity [l]
amount of substance [n]
kelvin (K)
candela (cd)
mole (mol)
The International System of
Units (SI)
Supplementary Dimension
Base Unit
plane angle
radian (rad)
solid angle
steradian (sr)
Fundamental Units (SI)
Mass:
“a cylinder of platinum-iridium
(kilogram) alloy maintained under vacuum
conditions by the International
Bureau of Weights and
Measures in Paris”
Fundamental Units (SI)
Time:
“the duration of 9,192,631,770 periods
(second) of the radiation corresponding to the
transition between the two hyperfine levels
of the ground state of the cesium-113
atom”
Fundamental Units (SI)
Length or
Distance:
(meter)
“the length of the path traveled
by light in vacuum during a time
interval of 1/299792458 seconds”
photon
Laser
1m
t=0s
t = 1/299792458 s
Fundamental Units (SI)
Electric
Current:
(ampere)
“that constant current which, if
maintained in two straight parallel
conductors of infinite length, of
negligible circular cross section, and
placed one meter apart in a vacuum,
would produce between these
conductors a force equal to 2 × 10-7
newtons per meter of length”
(see Figure 13.3 in Foundations of Engineering)
Fundamental Units (SI)
Temperature:
(kelvin)
The kelvin unit is 1/273.16 of the
temperature interval from absolute
zero to the triple point of water.
Water Phase Diagram
Pressure
273.16 K
Temperature
Fundamental Units (SI)
AMOUNT OF “the amount of a substance that
SUBSTANCE: contains as many elementary enti(mole)
ties as there are atoms in 0.012
kilograms of carbon 12”
Fundamental Units (SI)
LIGHT OR
LUMINOUS
INTENSITY:
(candela)
“the candela is the luminous
intensity of a source that emits
monochromatic radiation of
frequency 540 × 1012 Hz and that
has a radiant intensity of 1/683 watt
per steradian.“
See Figure 13.5 in Foundations of Engineering
Supplementary Units (SI)
PLANE
ANGLE:
(radian)
“the plane angle between two radii
of a circle which cut off on the
circumference an arc equal in
length to the radius:
Supplementary Units (SI)
SOLID
“the solid angle which, having its
ANGLE: vertex in the center of a sphere,
(steradian) cuts off an area of the surface of the
sphere equal to that of a
square with sides of length equal
to the radius of the sphere”
Derived Units
See Foundations, Table
13.4
Most important...
J, N, Hz, Pa, W, C, V
The International System of
Units (SI)
Prefix
Decimal Multiplier
Symbol
nano
10-9
n
micro
10-6
m
milli
10-3
m
centi
10-2
c
deci
10-1
d
deka
10+1
da
hecto
10+2
h
kilo
10+3
k
mega
10+6
M
giga
10+9
G
(SI)
Force = (mass) (acceleration)
m
1 N  1 kg· 2
s
U.S. Customary System of
Units (USCS)
Fundamenal Dimension
length [L]
foot (ft)
force [F]
pound (lb)
time [T]
second (s)
Derived Dimension
mass [FT2/L]
Base Unit
Unit
slug
Definition
lb f  s2 /ft
(USCS)
Force = (mass) (acceleration)
1 lb f  1 slug  ft/s
2
American Engineering System
of Units (AES)
Fundamenal Dimension
Base Unit
length [L]
foot (ft)
mass [m]
pound (lbm)
force [F]
pound (lbf)
time [T]
second (sec)
electric change [Q]
coulomb (C)
absolute temperature [q
degree Rankine (oR)
luminous intensity [l]
candela (cd)
amount of substance [n]
mole (mol)
(AES)
Force = (mass) (acceleration)
1 lb f  1 lbm  ft/s 2
lbm
lbf
ma
F 
gc
lb m ft
32.174
lb f s 2
ft/s2
Conversions Between Systems
of Units
1 ft  0.3048 m
1 ft
 1  conversion factor  F
0.3048 m
0.3048 m
 1  conversion factor  F
1 ft
0.3048 m
5 ft  F  5 ft 
 1.524 m
1 ft
2
 0.3048 m 
2
5 ft  F  5 ft  

0
.
4676
m

 1 ft

2
2
2
Temperature Scale vs
Temperature Interval
212oF
32oF
DT = 212oF - 32oF=180 oF
Scale
Interval
Temperature Conversion
Temperature Scale
o
o
 
R  1.8K 
F  1.8 o C  32
1
C 
1.8
F
o
 32
o

1
K 
1.8
Temperature Interval Conversion Factors
o
o
o
o
1.8 F 1.8 R 1 F 1 C
F o

 o 
C
K
R
K
 R
o
Pairs Exercise 1
 The force of wind acting on a body can be computed
by the formula:
F = 0.00256 Cd V2 A
where:
F = wind force (lbf)
Cd= drag coefficient (no units)
V = wind velocity (mi/h)
A = projected area(ft2)
 To keep the equation dimensionally homogeneous,
what are the units of 0.00256?
Pair Exercise 2
Pressure loss due to pipe friction
2 f L r v2
Dp 
d
Dp = pressure loss (Pa)
d = pipe diameter (m)
f = friction factor (dimensionless)
r = fluid density (kg/m3)
L = pipe length (m)
v = fluid velocity (m/s)
(1) Show equation is dimensionally homogeneous
(2) Find D p (Pa) for
d = 2 in, f = 0.02, r = 1 g/cm3, L = 20 ft, & v = 200 ft/min
Pair Exercise 2 (con’t)
(3) Using AES units, find D p (lbf/ft2)
for d = 2 in, f = 0.02, r = 1 g/cm3, L = 20
ft, & v = 200 ft/min
Formula Conversions
Some formulas have numeric constants that are not
dimensionless, i.e. units are hidden in the constant.
As an example, the velocity of sound is expressed by
the relation,
c  49.02 T
where
c = speed of sound (ft/s)
T = temperature (oR)
Formula Conversions
Convert this relationship so that c is in meters per
second and T is in kelvin.
Step 1 - Solve for the constant
c
49.02 
T
Step 2 - Units on left and right must be the same
ft
ft
c
s
s
49.02 o 1/ 2 
o 1/ 2
R
T R
Formula Conversions
Step 3 - Convert the units
ft
1/ 2
o
ft
0.3048 m  1.8 R 
m
s
  20.04
49.02 o 1/ 2  49.02 o 1/ 2 
 
1/2
R
s· R
ft
K
s·K


So
c  20.05 T
where
c = speed of sound (m/s)
T = temperature (K)
F
Pair Exercise 3
The flow of water over a weir can be computed
by:
Q = 5.35LH3/2
where: Q = volume of water (ft3/s)
L = length of weir(ft)
H = height of water over weir (ft)
Convert the formula so that Q is in gallons/min and L
and H are measured in inches.