Transcript Dimensions, Units, and Their Conversion

209201 Chemical Stoichiometry
References :
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Basic Principles and Calculations in Chemical Engineering, 7 edition,
David. M. Himmerblau and James B. Riggs.
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ปริ มาณสัมพันธ์สาหรับงานอุตสาหกรรม เล่ม 1, รศ. ดร. อนันต์เสวก เห่วซึงเจริ ญ,
วิทยาศาสตร์ , มหาวิทยาลัยเชียงใหม่.
ภาควิชาเคมีอตุ สาหกรรม, คณะ
Lecture 1. Introduction
1.1 Dimensions, Units and Their
Conversion
• Dimensions : basic concepts of measurement
e.g. length, time, mass, temperature, etc.
Fundamental (Basic) Dimensions
• Dimensions that can be measured
independently and are sufficient to describe
essential physical quantities.
• Length, Time, Mass, Temperature,
Molar amount
Derived Dimensions
• Dimensions that can be developed in terms of
the fundamental dimensions.
• Velocity, Acceleration, Force, Pressure,
Density, Heat Capacity, Energy, Power, etc.
• Important Note!
– “Fundamental” or “Derived” – depending also on
dimension/unit systems.
– (ดูตาราง 1.1, AH)
Units
• Units : means of expressing dimensions
– Length; feet, meters
– Time; hours, seconds
• Systems of Units
– SI : Le Systeme Internationale d’Unites (International
System)
– AE : American Engineering
– USCS : U.S. Conventional System
– Metric systems : CGS (centimeters, grams, seconds)
– English system
– (ดูตาราง 1.2, ตาราง 1.3 และภาคผนวก 1, AH)
SI Prefixes (Remember!)
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•
•
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•
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•
•
G, giga, 109
M, mega, 106
k, kilo, 103
h, hecto, 102
da, deka, 101
d, deci, 10-1
c, centi, 10-2
m, milli, 10-3
μ, micro, 10-6
n, nano, 10-9
Caution!
• camel’s hair brush :
camel’s – hair brush or camel’s hair – brush?
• Written with confusion
– ms – milliseconds or meter seconds?
– cm2 – square centimeters or centi square meters?
• Written without confusion
– m.s or (m)(s)
– (cm)2
1.2 Operations with Units
1.
Addition, Substraction, Equality
–
–
–
–
2.
Multiplication and Division
–
–
–
–
3.
only they are the same
2 kg + 1 kg = 3 kg
5 kg + 3 J
10 kg + 5 g
be canceled or merged if they are identical
m2/m -> m
(kg)(m)/(s)
m2/cm
Nonlinear Operations
–
–
–
–
sin, log, exp, …
for simplicity, transform or scale variables to be dimensionless before apply
nonlinear operations
log (r m) = log r + log m
log (r m / R m) = log r + log m – log R – log m = log r – log R = log(r/R)
1.3 Conversion of Units and
Conversion Factors
• Many sources – web sites, mobile
applications, CD, handbooks, etc. – find a very
good one and keep for your academic life and
career.
• For conversion or calculation, be systematic in
writing.
• Prefer format :
1100 ft 1 mile 60 s 60 min
s 5280 ft 1 min 1 hr
Caution!
• There are several “miles”, “pounds”, “gallon”,
“oz”, and “barrel”.
– U.S. frequent-flier mile (nautical mile) -> 1.85 km
– U.S. mile -> 1.61 km
– Ancient Italian mile -> 37 modern U.S. miles
Special Attention : Mass and Force
• Newton’s Law
F = ma or F = Cma
F = force
C = constant or conversion factor, whose numerical
value and units depend on those selected for F,
m and a
m = mass
a = acceleration
Special Attention : Mass and Force
• AE system
– Make the numerical value of the force and the mass be
essentially the same at the earth’s surface.
– 1 lbf : the unit of force when 1 lbm is accelerated at g ft/s2
– gc is a conversion factor (to cancel the numerical value of g).
1lb m g ft
F=
s2
(lb m )(ft)
g c = 32.174 2
(s )(lb f )
1
lb f
1lb m g ft
F=
= 1 lb f
2
gc (lb m )(ft)
s
s2
g
» 1
gc
Special Attention : Mass and Force
– 1 kgf : the unit of force when 1 kgm is accelerated at g m/s2
– gc is a conversion factor (to cancel the numerical value of
g).
1kg m g m
F=
s2
(kgm )(m)
gc = 9.80665 2
(s )(kg f )
1
kgf
1kg m g m
F=
= 1 kg f
2
gc (kg m )(m)
s
s2
g
» 1
gc
Special Attention : Mass and Force
• SI system
– 1 Newton (N) : the unit of force when 1 kg is
accelerated at 1 m/s2
1kg 1 m
F=
2
s
F=
1
N
1kg 1 m
=1N
2
(kg)(m)
s
s2
Examples Lecture 1.3
(P3, DMH)
Determine the kinetic energy of one pound of
fluid moving in a pipe at the speed of 3 feet per
second.
Ans. 0.14 (ft)(lbf)
Examples Lecture 1.3
(P4, DMH)
Convert the following from AE to SI units:
a. 4 lbm/ft to kg/m
b. 1.00 lbm/(ft3)(s) to kg/(m3)(s)
Ans. (a) 5.96 kg/m; (b) 16.0 kg/(m3)(s)
Examples Lecture 1.3
(P5, DMH)
Convert 1.57x10-2 g/(cm)(s) to lbm/(ft)(s)
Ans. 1.06x10-3 lbm/(ft)(s)
1.4 Dimensional Consistency
(Homogeneity)
d =16.2 -16.2e-0.021t
- Both terms must have the same unit as d.
- 0.021 must have the unit of time.
d
2ax
2
2
1+ (x / a ) =
dx
1+ (x 2 / a 2 )
- This cannot be correct.
- The lefthand side has units of 1/x, but the righthand
side has units of x2 (the product of ax).
Examples Lecture 1.4
(P1, DMH)
An orifice meter is used to measure the rate of flow of a fluid in pipes. The
flow rate is related to the pressure drop by the following equation
u=c
where
u
ΔP
ρ
c
DP
r
= fluid velocity
= pressure drop (force per unit area)
= density of the flowing fluid
= constant
Ans. c is dimensionless.
Examples Lecture 1.4
(P2, DMH)
The thermal conductivity k of a liquid metal is
predicted via the empirical equation
k = A exp (B/T)
where k is in J/(s)(m)(K) and A and B are constants.
What are the units of A and B?
Ans. A has the same units as k; B has the units of T
Significant Figures
• 1.43
– 1.43 ± 0.005, between 1.425 and 1.435
– 1.43 ± 0.01, between 1.44 and 1.42
• Engineering Calculation
– Cost of inaccuracy is high, knowledge of the
uncertainty is vital, and vice versa.
1.5 Significant Figures
• Decision criteria
1.
2.
3.
4.
Common Sense
Absolute Error
Relative Error
Statistical Analysis
Absolute Error
• Numbers with a decimal point
– the last significant figure in a number represents
the associated uncertainty.
– 100.3 -> 4 significant figures
– 100.300 -> normally 6 significant figures since
there usually was a reason for displaying the
trailing zeros e.g. rounding 100.2997 to 100.300
– 100. ? -> 3 significant figures
Absolute Error
• Numbers without a decimal point
– assume that the trailing zeros do not imply any
additional accuracy
– 458,300 -> 4 significant figures
– 0.23 or 0.230 or 0.2300 -> 2 significant figures
Absolute Error
• Multiplying or Dividing
– the lowest number of significant figures retained.
– (1.47)(3.0926) -> 4.54612 -> 4.55
• Adding or Subtracting
– final significant figures determined by the error
interval of the largest number
– 110.3 + 0.038 -> 110.338 -> 110.3
Relative Error
• 1.01/1.09 -> 0.9266 -> 0.927 (absolute error)
– Uncertainty by absolute error
(0.001/0.927)100 -> 0.1%
– Uncertainty of original numbers
(0.01/1.09)100 -> 1%
• Should the answer be truncated to 0.93,
rather than 0.927? -> The decision is yours.
Statistical Analysis
• more rigorous and complicated
• concept of confidence limits
What criterion would be chosen?
• Most answers are based on “absolute error”.
• In examples and problems, common sense should
often be used.
• Precision of measurement in practice and relation of
accuracy as compared to other values in the examples
or problems should be considered.
– 10 kg without a decimal point
• Accuracy is not trivial -> less significant figures
• Accuracy is trivial -> more significant figures since mass can be
accurately measure to a level of mg.
– 2/3 can be treated as 0.6667 in relation to the accuracy of
other values in the samples or problems.
Examples Lecture 1.5
A) If 20,100 kg is subtracted from 22,400 kg, is the
answer of 2,300 kg good to 4 significant figures?
- If note that 22,400, 20,100, and 2,300 have no decimal
point,
- 22,400 -> 3 significant figures
- 20,100 -> 3 significant figures
Scientific Notation will help.
2.24x10 4 - 2.01x10 4 = 0.23x10 4
-> 2 significant figures (by absolute error)
Examples Section 1.5
(E1.8, DMH)
If 20,100 kg is subtracted from 22,400 kg, is the answer of 2,300 kg good to 4 significant figures?
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If a decimal point point were placed in each number thus 22,400. and 20,100.,
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22,400. -> 5 significant figures
20,100. -> 5 significant figures
22,400. – 20,100. = 2,300.
-> 4 significant figures (by absolute error)
Relative Error
–
–
–
–
(1/224)(100), about 0.5%
(1/201)(100), about 0.5%
(1/23)(100), about 5%
(1/230)(100), about 0.5%
-> 3 significant figures (by relative error)
-> 0.230 x 104 ? No. -> 230. x 10
Examples Lecture 1.5
(E1.9, DMH)
If the DNA is stretched out, and a cut made with
a width of 3 μm, how many base pairs (bp)
should be reported in the fragment?
Note : 1 kb = 1000 base pairs, 3 kb = 1 μm
3m m 3kb 1000bp
= 9000bp
1m m 1kb
Precision in the 9000 value is determined by that of
3 μm , e.g. 3.0 or 3.00 μm.
1.6 Validation/Verification of
Problem Solutions
1. Repeat the calculations, possibly in a different order.
2. Start with the answer and perform the calculations in
reverse order.
3. Review your assumptions and procedures. Make sure
two errors do not cancel each other.
4. Compare numerical values with experimental data or
data in a database (handbooks, Internet, textbooks)
5. Examine the behavior of the calculation procedure.
For example, use another starting value and check
that the result changed appropriately.
6. Assess whether the answer is reasonable given what
you know about the problem and its background.
Problems : Chapter 1
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P1.4*
P1.11**
P1.12**
P1.20**
P1.21***
P1.30*
P1.33*