I.

Republic of the Philippines

Cagayan State University
Carig Campus
COLLEGE OF ENGINEERING

CSU Vision DEPARTMENT OF CHEMICAL ENGINEERING

Transforming
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the BEST.
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i
 TABLE OF CONTENTS

Title Page i

Table of Contents 1

Introduction 2-4

Section 1.1 5-11

Section 1.2 12-13

Section 1.3 14-19

Section 1.4 20-22

Section 1.5 23-24

Section 1.6 25-26

Section 1.7 27-29

Conclusion 30

References 31

1
 INTRODUCTION

In our modern era, Chemistry became a huge part of mankind. It is a

cornerstone of technology and quantitative research in many disciplines. One of the

pillars of Chemistry is measurement. It is important for us to measure stuffs correctly

such as distance, time, and accuracy are all great things to measure. Measurements

is one of the key why we have this kind of life, the way we think was shaped by

knowing the concept of measurement.

Dimensions are physical quantities that can be measured while units are

arbitrary names that correlate to particular dimensions to make it relative.

Almost all engineering problems must include units. All units for the same

dimensions are related to each other through a conversion factor. There are two unit

system that was in measuring the English unit system and the metric unit system.

The most widely used is metric unit system for the reason that it is much easier to

use than English unit system. When you convert it with other physical quantity the

measurement remains the same.

There are seven primary dimensions to be exact those are: mass, length,

time, temperature, electric current, amount of light, and amount of matter. Primary

dimensions are independent or fundamental dimensions, from which other

dimensions can be obtained.

2
 Unit operation is a physical change to which material is subjected especially in

coordination with unit process like filtration, distillation, or extraction. Chemical

engineering usually used this as the basic step in a process. Properly executing the

operations: adding, subtracting, multiplying, and dividing are first to be consider in

measurement.

Converting from a certain unit to a new unit of measurement but still have the

same quantity is the sole definition of conversion of units. It is usually done through

multiplicative conversion factors. There are three steps to convert a unit from a

different unit. First, write the conversion as a fraction then multiply it out and lastly,

cancel any units that are both top and bottom.

In every equations it should always be dimensionally consistent so that it will

justify the equal sign in the equation. Not only the quantity should be the same but

also the unit that was used should also be the same. This is an engineering

computational tool. An important part of scientific and mathematical calculations is

the significant figures. It deals mainly with the precision and accuracy of numbers.

We cannot attain this type of society without the existence of measurement.

Our everyday life depends on measuring tools. With this regard, to highlight the

various purposes of Units and Dimensions this computer laboratory is done to be

able to understand the concept of units and dimensions, conversion of units,

3
dimensional consistency and to apply significant figures to find the final answers of

the following engineering problems.

Section 1.1 Units and Dimensions

Objectives:

4
  Understand the difference between Units and Dimensions.

 Describe the relationship between the SI system of units and the

English system of units.

 Solve problems involving units of length, mass, and volume.

1.1A Convert the following to the desired units:

(a) 60 mi/hr to m/s

Solutio

n:

60mi 1609. 344m 1hr m

⋅ ⋅ =26 . 82
1hr 1mi 3600s s

2 2
(b) 30 N/m to lbf/ft

30N 0 .223lb f 1m2 lb f

Solutio ⋅ ⋅ =0. 623
1m 2 1N 10 .76ft 2 ft 2
n:

(c) 16.3 J to Btu

9 . 4782×10−4 BTU
16J⋅ =0 . 015BTU
IJ

Solutio (d) 4.21 kW to J/s

n:

5
Solutio
1000W 1J/s
4 . 21kW⋅ ⋅ =4210J/s
n: 1kW 1W

1.1B Change the following to the desired units:

(a) 235 g to pounds.

Solutio

n: 0 . 0022lb
235g⋅ =0 . 517lb
1g

(b) 610 L to cubic feet.

Solutio 0 . 03531467ft3
610L⋅ =21. 54ft3
1L
n:

(c) 30 g/L to pounds/cubic feet.

30g 0 . 0022lb 1L lb
⋅ ⋅ =1 . 8689 3
L 1g 0 . 03531467ft 3
ft

6
Solutio

n: 2 2
(d) 14.7 lb/in to kg/cm

14 . 71lb 1in 2 0. 45359237kg kg

2
⋅ 2
⋅
2 1lb
=1. 034 2
in (2 . 54 ) cm cm

Find the kinetic energy of a ton of water moving at 60 mi/hr expressed as

(ft)(lbf).
Solutio

n: GIVEN
m=1ton
v =60mi/hr

1
KE= mv2
1.1C 2
W 2000lbs
m= =
Solutio g 32 .2ft/s 2
n: 60mi 5280ft 1hr ft
v= ⋅ ⋅ =88
hr 1mi 3600s s

Execute:
1 2000lbs ft 2
KE=
(
2 32 . 2ft/s2 )( )
88
s
=240496 .89ft . lbf

1.1D An elevator which weights 10,000 lb is pulled up 10 ft between the first

and second floors of a building 100 ft high. The greatest velocity the

elevator attains is 3 ft/s. How much kinetic energy does the elevator

have in (ft)(lbf) at this velocity?

7
 GIVEN
m=10000lb
v =3ft/s

1
KE= mv 2
2
W 10000lb
m= =
g 32 .2ft/s 2

Execute:
1 10000lb
KE=
(
2 32 . 2ft/s2 )
( 3ft/s )2

KE=1397 .51 ( ft )( lbf )

1.1E The Colburn equation for heat transfer is

2
 h  Cμ  3 0.023
   
 CG  k   DG/μ 0.2

where C = heat capacity, Btu/(lb of fluid)(°F); µ = viscosity, lb./(hr)(ft.); k =

2
thermal conductivity, Btu/(hr)(ft )(°F)/ft; D = pipe diameter, ft; and G =

2
mass velocity, lb/(hr)(ft ) of cross section. What are the units of the heat

transfer coefficient h?

2
 h  Cμ  3 0.023
   
The Colburn equation is 
CG  k   DG/μ 0.2

Solutio

n: Where:
8
 Btu
C  heat capacity,
(lb of fluid)( F)
lb
μ  viscosity,
(hr)(ft)
Btu
k  thermal conductivi ty,
(hr)(ft 2
)( F)
D  pipe diameter, ft
lb
G  mass velocity, 2
(hr)(ft )

Execute:

Btu lb m
 h    CG   
(lb m )( F)  h  ft2 
Btu
h 
( F) h   ft 2 

1.1F 235
Countercurrent gas centrifuges have been used to separate U from

238
U. The rate of diffusive transport is K = 2πDρr . If K = rate of

transport of light component to the center of the centrifuge, in g mol/(s)

3
(cm of height); D = diffusion coefficient; ρ = molar density, g moles/cm ;

and r = log mean radius, (r2 – r1)/ln (r2/r1), with r in cm, what are the units

of D?
Solutio

n:

Given:

9
 K  2Dr
g  mol
K
s  cm

g  mol
ρ
cm 3
r  cm

Execute:

g  mol g  mol
  cm  D
s  cm cm 3
g  mol
D  s  cm
g  mol
 cm
cm 3
g  mol cm 2
D 
s  cm g  mol
cm
D
s

1.1G The density of a certain liquid is given an equation of the following

Form.

ρ   A  BT e Cp

where ρ = density in g/cm3, t = temperature in °C, and P = pressure in atm.

a.) The equation is dimensionally consistent. What are the units of A, B, and C?

10
Solution:

3
 A is in g / cm

B is in g / cm (C)
3


1
 Since the exponent of e must be dimensionless, the unit of C is atm

b.) In the units above, A = 1.096, B = 0.00086, and C = 0.000953. Find A, B, and

C if ρ is expressed in lb/ft3, T in °R, and p in lbf/in 2.

Solution:

1 . 096g 1lb 1cm 3

A= 3
⋅ ⋅ 3
=68 . 43lb/ft3
cm 453 .59237g 0 . 00003531ft

0 . 00086g 1lb 1cm 3 ( 1° C )

B= 3 ⋅ ⋅ ⋅ =0 . 0298lb/ft 3 ( ° R )
cm ( ° C ) 453 .59237g 0 . 00003531ft ( 1. 8 ° R )
3

0 . 000953 1atm 0. 00006485

C= ⋅ =
atm 14 . 69595lb/in lb/in 2
2

11
 Section 1.2 Operations with Units

Objectives:

 Understand the purpose of each operation.

 Understand how each operations involving units work.

 Perform operations such as addition, subtraction, multiplication, and

division with corresponding units.

1.2A (a) How many g moles are represented by 100 g of CO 2?

Solutio

n: 1mol
gmole=100g⋅ =2. 273gmolCO2
44g

(b) Calculate the weight in pounds of 3.5 g moles of nitrogen.

lb
Solutio 3 .5gmol⋅14g⋅ =0 .11lb
454g
n:

1.2B Convert the following:

(a) 120 lb mol of NaCl to g.

Solutio

n:

12
 58 . 436lb 454gNaCl
120lbmolNaCl⋅ ⋅ =9 . 18×106 gNaCl
1lbmolNaCl 1lb

(b) 120 g mol of NaCl to lb.

Solutio

n:
58 . 436g 1lb
120gmol⋅ ⋅ =15 . 445lb
1gmol 454g

(c) 120 lb of NaCl to g mol.

Solutio
454g 1gmol
120lb⋅ ⋅ =932. 30gmol
n: 1lb 58. 436g

(d) 120 g of NaCl to lb mol.

1lb 1lbmol
120g⋅ ⋅ =0 .0045lbmol
454g 58. 436lb
Solutio

n:

Section 1.3 Conversion of Units and Conversion Factors

13
 Objectives:

 Convert one set of units in an equation into another equivalent sets

using conversion factors.

 Convert from one temperature scale to the other using conversion

formulas.

 Perform conversion among metric units.

A solution of sulfuric acid at 60°F is found to have a sp gr of 1.22. From the

tables in Perry's Chemical Engineer's Handbook, the solution is found to be

1.3A 30 percent by weight H2SO4. What is the concentration of H 2SO4 in the

3
following units: (a) lb mol/gal, (b) lb/ft , (c) g/L, (d) lb H2SO4/lb H2O, and (e)

lb mol H2O/lb mol total solution?

(a) lb mol/gal

Sp=1 .22
Solutio H 2 SO 4 density=(1. 22)(0 . 0624 )=0 . 076128lb/ft 3

n:

14
 density 0.0762128
cocentrati on of H 2 SO 4 
mw 98.079
 0.0007762 lb mol/ft 3
 1ft 3 
(0.0007762 lb mol/ft 3 ) 
 7.481gal 
 0.0001038 lb mol/gal

(b) lb/ft3

Answer
3
Concentration of H2SO4 = 0.076128lb /ft
:

(c) g/L

 g/L 
Solutio (0.076128l bmol/ft 3 )16.018 
 1lb/ft 3 
n:  1.220g/L

(d) lb H2SO4/lb H2O

lbH 2 SO 4 0.076128lb H2 SO 4 /ft 3


lbH 2 O 0.0624lbH 2 SO 4 /lbH 2 O/ft 3 1.22lbH 2 SO 4

lbH 2 O
Solutio

n:
A mixture of liquid hydrocarbons contains 10.0 percent n-heptane, 40.0

percent n-octane, and 50.0 percent i-pentane by weight. The specific

15
  60 F 
 
 60 F 
gravities of the pure components are

1.3B n-heptane = 0.685 n-octane = 0.705 i-pentane = 0.622

 60 F 
 
 60 F 
a.) What is the specific gravity of 93 lb of this mixture.

To find for the volume of each compound and the specific gravity of the

mixture use the eqn.

 wt fraction of compound  mass of mixture 

V
density of compound

ρ mixture
Specific gravity 
ρ reference

Solutio
Execute:
n:
Volume of n-heptane

V1 
 0.1  93 lb  
13.58 lb
 0.685  ρ ref   ρ ref 
Volume of n-octane

V2 
 0.4  93 lb  
52.77 lb
 0.705  ρ ref   ρ ref 
Volume of i-pentane

V3 
 0.5  93 lb  
74.76 lb
 0.622  ρ ref   ρ ref 
Therefore, specific gravity of the mixture is:

16
 ρ mixture
Specific gravity 
ρ reference
93 lb

V1  V2  V3
ρ ref
93 lb

 13.58 lb   52.77 lb   74.76 lb 
       
  ρ ref     ρ ref     ρ ref  
ρ ref
 0.659

1.3C Convert 17.2 ppm NH3 in water to the equivalent number of ppm of

NH3 gas in water vapor.

Solution:

We are given with the concentration of NH3 in ppm unit and we are asked to

convert the ppm of NH3 in water to the ppm of NH3 gas in water vapor.

Execute:

1mg
 0.494 ppm
cm 3

Converting :
17.2 ppm  0.494pm   8.4968 ppm

17.2 ppm NH3 is equivalent to 8.4968 ppm NH3 gas in water vapor.

1.3D Five thousand barrels of 28°API gas oil are blended with 20,000 bbl of 15°

API fuel oil. What is the °API (API gravity) of the mixture? What is the density

17
 3
in lb/gal and lb/ft ?

Given: Volume=5000 bbl

API=28 Degrees

Solutio ρ gas oil 141.5

Specific gravity  
ρ water API  131.5
n:

ρ water  0.999 g/cm 3  999kg/m 3

ρ gas oil 141.5


kg 28  131.5
999 3
m
ρ 1  ρ gas oil  886.26 kg/m 3
 kg  1 lb  1m 3 
ρ 1   886.26 3   
 m  0.453592kg  264.172gal 
1lb  0.453592kg ; 1m  264.172gal
3

lb
ρ 1  7.396
gal

 42 gal 
Volume(V )   5000 barrel  
1 
 1 barrel 
V1  5000  42gal 1 bbl  42 gal
mass(m )
density( ρ1 )  1

volume(v 1)

m1  ρ 1  v1
 lb 
  7.396  5000  42gal 
 gal 
m 1  7.396  5000  42 lb

18
Fuel oil
volume (v 2 )  2000bbl
API  15 
ρ fueloil 141.5

ρ water API  131.5

141.5
ρ 2  ρ fueloil  ( ρ water )
15  131.5
141.5  kg 
  999 3 
15  131.5  m 
kg
 964.904 3
m
 kg   1lb  1m 3 
  964.904 3    
 m   0.453592kg  264.172gal 
lb
ρ 2  8.05252
gal
 42gal 
volume(v 2 )  20000bbl    20000  42gal
 1bbl 
m2  ρ 2  v2
m 2  8.0525  20000  42lb

Total volume of mixture (v t )  v1  v 2

v t (5000  42  20000  42)gal
total mass (m t )  m1  m 2
m t (7.896  5000  8.0525  20000  42)lb
Total mass
mixture density (  ) 
Total volume
7.896  5000  8.0525  20000  42 lb
ρ
5000  42  20000  42 gal
lb
ρ  7.92
gal
lb
3
 1gal  0.133681ft 3
ft
lb 1gal
 7.92 
gal 0.133681ft 3
lb
 59.25 3
ft

19
 Section 1.4 Dimensional Consistency (Homogeneity)

Objectives:

 Determine whether the equations are dimensionally homogeneous.

 Use dimensional analysis to determine the appropriate units for an

unknown quantity in an equation.

 Check the units of a solution they have reached and verify that they are

precise.

1.4A A mixture of gases is analyzed and found to have the following composition:

CO2 12.0%

CO 6.0

CH4 27.3

H2 9.9

N2 44.8

How much will 3 lb moles of this gas mixture weight?

Solutio

n:
Component Mole % Mole Fraction Molar Mass
CO2 12.0% 0.12 44
CO 6.0 0.06 28
CH4 27.3 0.273 16
H2 9.9 0.099 2
N2 44.8 0.448 28

Avg. Mol.Wt.= (mole fraction) (molecular weight)

20
 (0.12)(44) (0.06)(28) (0.273)(16 ) (0.099)(2) (0.448)(28 )
 24.07g/mol
 24.07lb/lb mol

Moles of gas  3lb - mol

Mass of gas (Moles of gas)(Avg. molar weight of gas)
(3)(24.07) lb
 72.21 lb
1.4B You have 100 lb of gas of the following composition:

CH4 30%

H2 10%

N2 60%
What is the average molecular weight of this gas?

Solution:

w total
Average weight 
n total
where w total is the total weight of sample and n total is the total number of moles

Basis 100 kg mol

Component Mole % Molar mass

CH4 30 16
H2 10 2
N2 60 28

21
Weight (w) = (molar mass)(mole)

(16) (30) + (2) (10) + (28) (60) = 2180 kg

n total  30 kg mol  60 kg mol  10 kg mol  100 kg mol

w total  2180 kg
w total 2180 kg
Average weight  
n total 100 kg mol
21.80 kg /kg mol

Section 1.5 Significant Figures

Objectives:

 Recognize the number of significant figures in a given quantity.

22
  Apply the concept of significant figures to limit a measurement to

the appropriate number of digits.

 Limit results to the proper number of significant figures.

1.5A Two thermometers are to be checked against a standard thermometer.

The standard reads –22°F. What should the other two thermometers

read if they are calibrated in °C and K, respectively?

GIVEN:−22 ° F
Fahrenheit→Celsius
−22° F−32
=−30° C
1. 8

Celsius→Kelvin
−30 °C +273 .15=243 .15K

1.5B Mercury boils at 630K. What is its boiling temperature expressed in °C?

In °F? In °R?

GIVEN:630K
Kelvin→Celsius
630K−273 .15=356 .85°C

Kelvin→Fahrenheit
630K ( 1.8 )−459 .67=674 .33 ° F

Kelvin→Rankine
630K ( 1.8 )=1134 ° R

23
 Section 1.6 Validation of Problem Solutions

Objectives:

 Use various approaches to verify if your answer is correct.

 Recognize errors in your solutions.

 Express your answers correctly.

24
1.6B The pressure gauge on the steam condenser for a turbine indicates

26.2 in. Hg of vacuum. The barometer reading is 30.4 in. Hg. What is

the pressure in the condenser in psia?

Soluti

on:

Basis: 26.2 in Hg vacuum

 14.7psia 
p (30.4  26.2) in Hg 
 29.92 in Hg 
 2.06 psia

1.6C Examine the figure

Open

Tank
∆h = 20 in Hg

The barometer reads 740 mm Hg. Calculate tank pressure in psia.

Solution:

Basis : 20 in Hg gauge pressure

 14.7 psia 
20 in Hg   9.82620 psia
 29.92 in Hg 
 14.7 psi 
740 mmHg    14.31316 psia
 760 mmHg 
(14.31316 ) -( 9.82620)
 4.49 psia

25
 Section 1.7: Stoichiometry

Objectives:

 Understand the concept of stoichiometry and calculate the mass of a

product of chemical reaction given the starting masses of reactants.

 Determine the limiting and excess reactants in a chemical reaction.

1.7A How many grams of chromic sulfide will be formed from 0.718 grams

of chromic oxide according to the equation

2Cr2O3 + 3CS2 → 2Cr2S3 + 3CO2

26
Soluti

on.

Chromic Sulfide = Cr2S3

Chromic Oxide = Cr2O3

 1 mole Cr2 O 3 
0.718g Cr2 O 3    4.723995  10 3 mole Cr2 O 3
 151.99 g Cr2 O 3 
 2 mol Cr2 S 3 
4.723995  10 3 mole Cr2 O 3    4.723995  10 3 mole Cr2 S 3
 2 mol Cr2 O 3 
 200.19 g Cr2 S 3 
4.723995  10 3 mole Cr2 S 3    0.945 g Cr2 S 3
 1 mol Cr2 S 3 

1.7B A barytes composed of 100 percent BaSO4 is fused with carbon in the

form of coke containing 6 percent ash (which is infusible). The

composition of the fusion mass is

BaSO4 11.1%

BaS 72.8

C 13.9

Ash 2.2

100.%

Reaction:

BaSO4 + 4C → BaS+ 4CO

Find the excess reactant, the percentage of the excess reactant, and the

degree of completion of the reaction.

Solution:

27
Basis: 100 kg of fusion mass

Composition %=kg Molecular weight Kg mol

BaSO4 11.1 233.3 0.0476
BaS 72.8 169.3 0.430
C 13.9 12 1.16
Ash 2.2

BaSO 4  0.0476 mol  0.430 mol  0.4776 mol  BaSO 4 is the limiting reactant
C  1.16 mol  4(0.430 mol)  2.88 mol

4(0.4776)  1.9104 mol

2.88 - 1.91
% excess   100  50.79% excess C
1.91
0.4776 - 0.0476
Degree of Completion   0.90
0.4776

28
 Conclusion

The problems that were solved in this chapter defines and aligns the basic

knowledge learned about the concepts with the advanced points that moves these

concepts around chemical engineering. Units and dimensions including

stoichiometry are important and therefore must be handled critically, applying the

concept of dimensional consistency in your work is also essential, and reporting of

numerical values with an appropriate number of significant figures must be done

thoroughly.

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REFERENCES:

Balmer,R.T.(2011).Modern Engineering Thermodynamics.Elsevier Inc.,Edinburgh,

London.

Himmelblau,D.M. & Riggs,J.B.(2004).Basic Principles and Calculations in Chemical

Engineering.7th edition.Pearson Education Inc.,University of Texas,US

Houghton,E.L. et, al.(2016).Aerodynamics for Engineering Students.Butterworth-

Heinemann Publication.Oxford,United Kingdom.

Rubenstein,D.,Yin,W. & Frame,M.(2015).Biofluid Mechanics. Academic

Press.Cambridge,Massachusetts,United State.

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