Download CM2110 Chapter 2 - Chemical Engineering

CM2110 Chapter 2
2.1 to 2.3) Units and Dimensions, Conversions of Units, Systems of Units
American
System
CGS
Engineering System
Internationale
(English)
(SI)
Length
ft
m
cm
Mass
lbm
kg
g
Time
s
s
s
These are the base units.
Some conversion factors are on the front cover of the text and on p. 11 Table 2.3-1.
Another reference is the back cover of Perry’s Chemical Engineers Handbook.
We’ll spend a lot of time in CM2110 converting between different units. Unit conversion
is VERY IMPORTANT! You’ll find that in industry all three systems of units are used.
2.4) Force and Weight
a) Force
Newton’s 2nd law of motion
ma
F
gc
m = mass
a = acceleration
gc= conversion factor
F= force
SI: Force= kg .m/s2 = Newton (or N): mass in kg and acceleration in m/s2
CGS: Force= g .cm/s2 = dyne
: mass in g and acceleration in cm/s2
2
.
American Engr System: lbm ft/s
1 Newton = 1N =
1 dyne =
1kg  m
= 1kg .m/s2
2
s
1g  cm
= 1g .cm/s2
s2
American Engr System:
Force= lbf = lbforce= defined as 1 unit mass (1 lbmass) times acceleration of gravity (ft/s2)
a = g = acceleration of gravity = 32.174 ft/s2 = 9.8066 m/s2= 980.66 cm/s2
1 lbf = 32.174 lbm .ft/s2
1
by definition :
SI
CGS
Amer. Engr.
2
2
32.174lbm  ft / s 2
1kg  m / s
1g  cm / s


gc =
N
dyne
lb f
b) Weight
Weight of an object is force exerted on that object by gravity.
ma
F
gc
m = mass
a = acceleration g = acceleration of gravity
gc= conversion factor
F= force
W  Weight  F 
mg
gc
2.5) Dimensional Homogeneity
Quantities can be added/subtracted only if their units are the same.
Example 1) V (m/s) = Vo (m/s) + g (m/s2) t (s)
Velocity
(m/s) = (m/s) + (m/s) Units are OK!
Example 2) V (m/s) = Vo (m/s) + g (m/s2) Units are NOT OK!
NOT a valid equation. Need all units to be the same!
2.6) Significant Figures
I’ll typically use 3 significant figures in CM2110.
Significant Figure Examples
2300. = 4 significant figures
0.03500 = 4 significant figures = 3.500 x 10-2
Chapter 2 Problems
Use the front cover of your text book to get conversion factors!
1. Verify that 1gal = 3.785 L = 0.003785 m3 = 231 in3
Add these to the front cover of your book.
2
 28.317 L 
 = 3.785 L
1 gal 
 7.4805 gal 
3
 1 ft 3
 0.3048m 

 = 0.003785 m3
1 gal 
7
.
4805
gal
1
ft



3
 1 ft 3
 12in 

 = 231 in3
1 gal 
 7.4805 gal  1 ft 
2.
Verify that 1 g/cm3 = 62.4 lbm/ft3 = 8.34 lbm/gal = 1000 kg/ m3
Add these to the front cover of your book.
1g
cm 3
3
 1lbm  30.48cm 


 = 62.4 lbm/ft3
 453.6 g  1 ft 




3
1g  62.4lbm / ft  1 ft



3 
3
 = 8.34 lbm/gal
7
.
4805
gal
cm  1g / cm






3
1g
cm 3
 1kg  100cm 


 = 1,000. kg/m3
1000
g
1
m




3
3. Convert 57.5 lbm/ft3 to kg/m3
57.5lbm
ft 3
 0.4536kg 

1 ft 3


 = 921 kg/m3
3 
 1lbm  0.028317m 
4. Re = Reynolds Number = dimensionless quantity used a lot by CM in fluid flow to
determine if we have laminar or turbulent flow in a pipe.
Re< 2100 means laminar flow
Re> 2100 means turbulent flow
Re =
Dv

D = pipe diameter = 2.067 in
v = fluid velocity = 4.8 ft/s
 = fluid density = 0.805 g/cm3
 = fluid viscosity = 0.43 cP
1cP = 1 centipoise = 10-3 kg/(m.s)
3
What is Re? Is flow laminar or turbulent?
62.4lbm / ft 3
1g / cm 3
 2.067in (4.8 ft / s)(0.805g / cm 3 
 1 ft  1 ft  453.6 g  1kg 
1cP


 3

 =


0.43cP

 10 kg / m  s  12in  0.3048m  1lbm  1000 g 
143,700 >2100 so turbulent flow
5. A storage tank contains 10,000 gal of a liquid with a density of 1 g/cm3. How many
m3 and ft3 and kg and lbm of liquid are in the tank? How much does the liquid weigh (lbf
and N)?
 0.003785m 3 
 = 37.9 m3
10,000 gal 
1
gal


 1 ft 3

 = 1,337 ft3
10,000 gal 
 7.4805 gal 
Density = Mass/Volume
Mass = Density . Volume
Mass:
3
 1g  1000kg / m 
37.9 m3  3 
3
 = 37,900 kg
 cm  1g / cm

3
 1g  62.4lbm / ft 
1337 ft3  3 
3
 = 83,429 lbm
 cm  1g / cm

W  Weight 
mg
gc
Amer. Engr Units:
W=
(83,429lb m )(32.174 ft / s 2 )
= 83,429 lbf
lb m ft / s 2
32.174
lb f
SI Units
4
1 Newton = 1N =
W=
1kg  m
s2
(37,900kg)(9.8066m / s 2 )
= 371,670 N
kg  m
1
N  s2
6. You can do additional Ch 2 problems on your own. Answers to selected problems are
in the back of your book. In addition, there are more problems on the ‘Instructional
Tutorials’ that came on the CD with your textbook.
7. Some selected answers to Ch 2 problems. You are to do these on your own. Do not
turn them in. They are to help you study for tests and quizzes.
2.1 a) 1.1844 x 109 ms b) 26.0 mi/h c) 3.85 x 10 4 cm4/min. g
2.2 a) 340 m/s b) 57.5 lbm/ft3 c) 120 hp
2.8 a) 25.0 lbf b) 2.6 kg c) 9x 10 9 dynes
2.23 Re = 2x104
5