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Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

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Chapter 4 1 CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5
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Page 1: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 1CHEM ENG 1007

Reading Materials: Chapter 4

LECTURES 4-5

Page 2: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 2CHEM ENG 1007

Class Activity

1. How tall are you?

2. How long does it take you to get to Adelaide Uni every morning?

3. How many days are there in one year?

4. How heavy are you?

5. What is the distance from the earth to the moon?

6. What is the average weight of the new-born baby?

Page 3: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 3CHEM ENG 1007

Objective

At the end of these lectures you should be able to:

convert between sets of units by using dimensional equation

add, subtract, multiply and divide units use SI, American Engineering and cgs

units understand dimensional homogeneity

Page 4: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 4CHEM ENG 1007

Quick Quiz 1: True or FalseA quantity is meaningless without units!

§4.1 Units & Dimensions

"the distance from my house to school is six"

unless we follow that statement with "miles" or "kilometres", or whichever unit, the

above statement is meaningless.

Page 5: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 5CHEM ENG 1007

Quick Quiz 2: True or FalseAll additive/subtractive terms must have the same dimension!

§4.1 Units & Dimensions

Page 6: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 6CHEM ENG 1007

Every Monday you buy three litres of milk and five packages of Tim Tam. Since you have to carry your purchases home, you’d like to know how many kilograms of food you’ve bought.Can we just add together litres of milk plus packages of tim tam to get kilograms of food?

(3 L milk) + (5 packages of tim tam) =

? kg of food

Illustration 1

Page 7: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 7CHEM ENG 1007

a) 6.201 cm + 7.4 cm + 0.68 cm + 12.0 cm = ?b) 1.6 km + 1.62 m + 1200 cm = ?c) 8.264 g - 7.8 g = ?d) 10.4168 m - 6.0 m = ?e) 12.00 m + 15.001 kg = ?f) 131 cm x 2.3 cm = ?g) 5.7621 m x 6.201 m = ?h) 20.2 cm divided by 7.41 s = ?i) 40.002 g divided by 13.000005 ml = ?

Illustration 2

Assignment 1, Question 7

Page 8: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 8CHEM ENG 1007

Case Study 1: Feet or Meters?

SEOUL, South Korea – A mix up in the cockpit over whether altitude guidance was measured in feet or meters led to the crash of a Korean Air Lines McDonnell Douglas MD-11 freighter soon after takeoff in Shanghai in April 1999.

The crash killed all 3 crew-members. Five people on the ground were killed and 40 more were injured when the plane went down in light rain onto a construction site near Shanghai’s Hongqiao Airport.

Page 9: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 9CHEM ENG 1007

Wall Street Journal, June 6, 2001

A Chinese air-traffic controller directed the pilots to an altitude of 1,500 meters (4,950 feet). The plane was climbing rapidly to that level when the co-pilot told the pilot he thought the instructed height was 1,500 feet (455 meters). The international aviation industry commonly measures altitude in feet, and the confusion led the pilot to conclude the jet was almost 1,000 meters too high, so he quickly moved the controls to lower the plane. As the plane descended, the pilot realized the error but couldn’t correct the mistake in time.

South Korea’s Ministry of Construction and Transportation said Korean Air Lines would lose the right to serve the Seoul-Shanghai cargo route for at least two years because of errors by the pilots. Korean Air Lines said it would appeal the decision…..

Case Study 1: Feet or Meters?

Page 10: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 10CHEM ENG 1007

Mars Probe Lost Due to Simple Math Error

NASA lost its $125-million Mars Climate Orbiter because spacecraft engineers failed to convert from English to metric measurements when exchanging vital data before the craft was launched.

pound-seconds versus newton-seconds

Want to know more…..

http://articles.latimes.com/1999/oct/01/news/mn-17288

Case Study 2

Page 11: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 11CHEM ENG 1007

Dimensions versus Units

Dimensions are properties that can be measured.

Four basic dimensions: length [L], mass [M], time

[T], and temperature []

Units are specific magnitudes of dimensions.

Different units can measure the same dimension

Example: Dimension: length

Units: foot, metre, yard, mile, etc.

Such units are related by a Conversion factor

Page 12: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 12CHEM ENG 1007

Dimension versus units

Dimension: mass

Units: kilogram (kg) and pound (lbm)

0

1 kg

1 lbm

0.454 kg 2 kg

2 lbm

Conversion factor: 1 lbm = 0.453593 kg

1 kg = 2.2046 lbm

Illustration 3

Page 13: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 13CHEM ENG 1007

Systems of Units

Three Base Unit groupings are:

1. Metric system (also known as SI system)

2. American / British engineering system.

3. CGS (Centimetres-Grams-Seconds) system.

Should be

Kelvin

Page 14: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 14CHEM ENG 1007

Concept Inventory

1. Which of the following set of units is in CGS system

a) kg, m, s, N

b) lbm, ft, s, lbf

c) g, m, s, N

d) g, cm, s, dyne

e) kg, cm, s, dyne

Page 15: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 15CHEM ENG 1007

§4.1.1 Conversion Factors

A conversion factor is an equation that equates two quantities expressed in different units where, in fact, the two quantities are exactly equivalent.

Examples:

12 in = 1 ft 1000 g = 1 kg 60s = 1 min

Page 16: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 16CHEM ENG 1007

Conversion Factors Table

Or turn to page xi of the text-book

Page 17: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 17CHEM ENG 1007

§4.1.1 Conversion Factors

Conversion factors dimensionless quantities.

Enable conversion between different units systems.

Conversion factors do not change magnitude, just value of units.

Page 18: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 18CHEM ENG 1007

Is Conversion factor of metres to feet really dimensionless?

1m 3.281ft

Rearrange?:  Dimensionless?:

 

ft

1 3.281m

[L]ft

3.281m [L]

Dimensionless

Illustration 4

Page 19: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 19CHEM ENG 1007

PrefixesPrefixes are used in SI to indicate powers of ten. Multiples or fractions of basic units defined for convenience.

Factor Prefix Symbol Factor Prefix Symbol

109 giga G 10-1 deci d

106 mega M 10-2 centi c

103 kilo k 10-3 milli m

102 hecto h 10-6 micro

101 deka da 10-9 nano n

Commonly used for Joule, Pascal, Metre, Byte

Page 20: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 20CHEM ENG 1007

Quick Quiz

How many Megabytes in 1Gigabytes?

From the prefixes table, we get the following conversion factors

109 Bytes = 1 GB

106 Bytes = 1 MG

By dimensional equation

1 GB910 Bytes

1 GB 6

1 MB

10 Bytes 310 MB

Page 21: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 21CHEM ENG 1007

Quick Quiz

How many micrograms in 5 kilograms?

Again from the prefixes table, we get the following conversion factors

103 g = 1 kg

10-6 g = 1 g

5 kg 310 g

1 kg

6

1 g

10 g 95x10 g

Page 22: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 22CHEM ENG 1007

Quick Quiz

How many millimetres in 1 nanometres?

Again from the prefixes table, we get the following conversion factors

10-3 m = 1 mm

10-9 m = 1 nm

1 nm 9 10 m

1 nm

3

1mm

10 m 610 mm

Page 23: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 23CHEM ENG 1007

Write conversion factors that relate to each of the following pairs of units:

1. Litres and mL

2. Hours and seconds

3. mg and kg

4. Nanometers and kilometers

Quick Quiz

1 L = 1000 mL

1 h = 3600 s

106 mg = 1 kg

1012 nm = 1 km

Page 24: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 24CHEM ENG 1007

Unit Conversion

Conversion of units from one system to another is necessary in process calculations and analysis, when

Data is from different sources, or

Variables are measured from instruments of different standards

(starting quantity) x (conversion factor) = equivalent quantity

Page 25: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 25CHEM ENG 1007

Illustration 5

Convert 28 inches to its equivalent number of feet. 12 in = 1 ft

28 in 1

12

ft

in

2.333

12 28

1

ft = 2.3 ft (s.f)

inin

ft

When solving a problem, the idea is to set up an equation so that all unwanted units cancel, leaving only the desired units.

Convert mass of 60 lbm into kg?

 1 lbm = 0.454 kg

60 mlb 0.454

m

kg

lb

27 30 kg (1 s.f )kg

Page 26: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 26CHEM ENG 1007

Dimensional Equation A convenient method for unit conversion.How to set up a dimensional equation?

Write the given quantity and its units on the left, write the units of conversion factors that

cancel the old units and replace them with the described ones, fill in the values of the conversion factors, and carry out the

indicated arithmetic to find the desired value.

28in ft

in

Page 27: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 27CHEM ENG 1007

Convert a mass of 150 lbm into equivalent in kg.

Solution:

If 1 lbm is equivalent to 0.454 kg then

or by dimensional equation

m m

m

0.454kg150 lb 150 lb 68.1kg 68 kg (2 s.f )

1 lb

mm

150 lb150 lb

m

0.454 kg

1 lb68 kg

Illustration 6

Page 28: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 28CHEM ENG 1007

Using dimensional equation, convert a weight 90 N into lbf ?

1 N = 0.2281 lbf

90 N90 N f0.2281lb

N f f20.3 lb 20 lb

8 m 3.2808 ft 60 sm ft ft

8 1575 2000 (1 s.f )s m 1mins min min

Illustration 7

Using dimensional equation, convert 8 m/s into ft/min?

1 m = 3.2808 ft; 1 min = 60 s

Page 29: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 29CHEM ENG 1007

Convert a 40 mile/gal into km/L

Convert a force of 40 lbf into equivalent in N.

mile 40 mile

40 gal gal

1.6 km

1 mile

1 gal km14.2

4.5 L L

ff

40 lb40 lb

f

1 N

0.225 lb178 N

Quick Quiz

Page 30: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 30CHEM ENG 1007

Derived as combination of the base units: by multiplying or dividing of the base units

velocity (m/s), area (m2), flow rate (ft3s-1)

§4.1.5 Combined Units

Page 31: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 31CHEM ENG 1007

§4.1.5 Combined Units

To produce conversion factors for combined units,

conversion factors of base units can be raised to

any power.

Examples:

2 4

3

100 10area :

1 1

12 1728volume :

1 1

22

2

33

3

cm cm L

m m

in inL

ft ft

Page 32: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 32CHEM ENG 1007

Convert 25 lbm.ft/min2 to its equivalent in kg.cm/s2?  

2 m2

m 22m

m

25 lb .ft 0.4536 kg 30.48 cm 1 min25 lb .ft / min

min 1 lb 1 ft 60 s

25 lb

. ft 20.4536 kg 30.48 cm 1 min

2min m1 lb 1 ft 2 2

2

60 s

= 0.096 kg cm/s

Illustration 8

Page 33: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 33CHEM ENG 1007

Convert 45 m3/d2 to its equivalent in mm3/s2?  

3 2 23

3 23 2 22

45 m 1000 mm 1d 1h45 m / d

d 1 m 24h 3600 s

Quick Quiz

3 33 2 2

2 2 2 22 3

45 m 1000 mm 1 d 1h

d 1 m 24 h 3600 s

345 m 3 3 21000 mm 1 d 21h

2d 31 m 2 224 h

2 2

3 2

3600 s

6.0 mm /s

Page 34: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 34CHEM ENG 1007

Derived equivalents of combined units:

1 N = 1 kgm s-2

1 dyne = 1 gcm s-2

1 lbf = 32.174 lbmft s-2

1 Pa = 1 Nm s-2

1 gal = 0.16 ft3 (imperial)

1 gal = 0.133 ft3 (US)

§4.1.5 Combined Units

Coherent derived units

Non-coherent derived units

Page 35: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 35CHEM ENG 1007

Coherent versus Non-coherent derived Units

Coherent Derived Units:

Value of defined unit = Combined value of Basic Units

e.g. 1 J (joule) = 1 (kg.m2)/s2

Non-coherentValue of defined unit Value of base units

e.g. 1 Btu 1 (kg.m2)/s3 Conversion factor required

e.g. 9.486x10-4 Btu = 1 (kg.m2)/s3 Many American/British defined units are non-coherent (e.g. Btu, lbf)

Page 36: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 36CHEM ENG 1007

Summary for Units Conversion

To convert a quantity expressed in terms of one unit to its equivalent in terms of another unit, multiply the given quantity by the conversion factor (new unit/old unit).

1 g45 mg x 0.045 g

1000 mg

45 mg 1g

1000 mg0.045 g

Alternative way:

Page 37: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 37CHEM ENG 1007

Rules of dimensional consistency

§4.2.5 Dimensional Consistency

1. Terms that are added together (or subtracted) must have the same units. For example, in the equation Q = ab + c2, the units of ab must be the same as those of c2.

2. Exponents must be unitless (dimensionless). Thus, if an exponent consists of several terms, the units of all those terms must cancel. For example, in the equation y = xab/c, the units in the term ab/c must all cancel out to leave no units.

Page 38: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 38CHEM ENG 1007

Arrhenius equation

Illustration 9

Parameters Symbols Units

Gas constant R J / mol.K

Temperature T K

Activation energy Ea ?

aEk A exp

RT

a aE E

RT J

mol.K

K

a

1

JE

mol

Page 39: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 39CHEM ENG 1007

Ideal Gas Equation

Illustration 10

Parameters Symbols Dimensions Units

Pressure P [M] [L]-1 [T] -2 Pa

Volume V [L]3 m3

# of moles n [N] mol

Gas constant R [M] [L] 2 [T] -2 [N]-1 []-1 Pa.m3/mol.K

Temperature T [] K

PV nRT

In terms of these parameters, show that ideal gas equation obeys the laws of dimensional consistency

Page 40: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 40CHEM ENG 1007

Substitute units into Equation?

3LHS P V Pa . m

RHS n R T mol

3Pa.m

mol

.K K

3Pa . m

i.e., LHS = RHS

Illustration 10

Page 41: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 41CHEM ENG 1007

Substitute base dimensions into Equation?

23

2 2

[M] [M] [L]LHS . [L]

[L] [T] [T]

RHS [N]

2[M] [L]

. [N] [ ] 2

. [ ][T]

2

2

[M] [L]

[T]

LHS RHS

Illustration 10

Page 42: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 42CHEM ENG 1007

Consider this equation dv

dy

2shear stress N/m

v velocity m / s

y width m

 What are SI units of ?

Illustration 11

Page 43: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 43CHEM ENG 1007

Compare dimensional homogeneity!  

Rearrange yields units for .

2

N mLHS .

m

/ s

m

RHS

2

N . sPa.s

m

Illustration 11

dv

dy

Page 44: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 44CHEM ENG 1007

§4.1.6 Force

The force “F” to accelerate a mass “m” at an acceleration rate “a” is defined by Newton’s second law as

F = ma [M.L.T-2]

One form of acceleration is that associated with earth’s gravity. The rate of acceleration is described by the gravitational acceleration, g, and the force that an object exerts on the earth’s surface (its “weight”), is

W = mg

Page 45: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 45CHEM ENG 1007

§4.1.6 Force

Factors for Unit conversions of Force

1 N = 1 kg.m/s2 = 105 dynes = 105 g.cm/s2 = 0.22481 lbf

Page 46: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 46CHEM ENG 1007

Example 4.2

mm 2 2

m ff2

m2

lb ft32.174ftW mg 1 lb 32.174

s s

lb ft 1 lbW 32.174 1 lb

s lb ft32.174

s

An object has a mass equal to 1 lbm. What is its weight in pounds-force (lbf)

Solution:

From Table 4.3, we note that g = 32.174 ft/s2

Page 47: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 47CHEM ENG 1007

Example 4.3

2 2

2

2

a)

9.8066 m kg mW mg 8.41 kg 82.5

s s

kg m 1 NW 82.5 82.5 N

kg ms 1 s

An object has a mass equal to 8.41 kg. What is its weight

a) in Newtons (N)

b) pounds-force (lbf)

Solution:

ff

b)

0.22481 lbW 82.5 N 18.5 lb

1 N

Page 48: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 48CHEM ENG 1007

Quick Quiz

From Reading Question 4:

The weight of an astronaut is measured on a distant planet and found to be one-fifth of his weight on the earth’s surface.

Is his mass different on that distant planet than on earth?

What does the weight difference imply about the acceleration of gravity on the distant planet?

Page 49: Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURES 4-5.

Chapter 4 49CHEM ENG 1007

Summary2 case-studies

Identified the difference in SI, American Engineering, and CGS units system

Understand conversion factors dimensional equations dimensionless dimensional consistency coherent and non-coherent derived units Force/Weight

Able to convert between sets of units


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