liquid liquid extraction

DKK1413: Chapter 1 FKKSA, UMP

Chapter 1 Introduction to Engineering

Calculations 1


Chemical Engineering Profession

CHEMIST

VS

CHEMICAL ENGINEER

2


Role of Chemical Engineer

• Exploit advances in chemistry to create new products

• Design chemical processes & operate plants

• Develop energy resources

• Protect the environment

• Develop new, environmentally design, and safer processes to make the chemical products that people depend on.

• Work in research and development laboratories, creating polymeric materials with improved performance and durability.

• Work in manufacturing, making vaccines and antibiotics.

• Invent new ways to keep our food and water supplies safe. 3


CHEMICAL PROCESS

4

SEPARATION PROCESS

REACTION PROCESS

SEPARATION PROCESS

RAW MATERIALS

INTERMEDIATE PRODUCT

INTERMEDIATE PRODUCT

FINAL PRODUCT

Mass

Energy

Balances


Chemical Engineer Task

You need to:

• Minimize production of unwanted byproducts

• Separate the good (product) from the bad (byproducts)

• Recover the unused reactants

• Maximize profit, minimize energy consumption, minimize impact on the environment

5


OPPORTUNITIES FOR CHEMICAL ENGINEERS

• Petroleum and petrochemical

• Pharmaceuticals

• Polymers

• Energy

• Food

• Consumer products

• Biotechnology

• Electronic and optical materials. 6


In designing a new process or analyzing an

existing one, calculation of amounts and

properties of raw materials and products is

crucial.

This chapter presents the calculation

techniques of expressing the values of

process variables.

7


Introduction to Engineering Calculations

Units and Dimensions

Conversion of Units

Systems of Units

What are in this chapter?

8


Topic Outcomes

At the end of Chapter 1, you should:

• Convert one set of units…

• …in a function or equation into another equivalent

set

• … for mass, length, area, volume, time, energy

and force using conversion factor tables.

9


Units and Dimensions

• Dimensions are:

• Properties that can be measured such as length,

time, mass, temperature.

• Properties that can calculated by multiplying or

dividing other dimensions, such as velocity

(length/time), volume, density.

• Units are used for expressing the dimensions such as

feet (ft) or meter (m) for length, hours/seconds (hr/s) for

time.

10


Conversion of Units • A measured quantity can be expressed in terms of any units

having the appropriate dimension

• To convert a quantity expressed in terms of one unit to equivalent in terms of another unit, multiply the given quantity by the conversion factor

• Conversion factor – a ratio of equivalent values of a quantity expressed in different units

• Let say to convert 36 mg to gram

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36 mg 1 g = 0.036 g

1000 mg Conversion factor


CONVERSION OF UNITS

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Dimensional Equation

1. Write the given quantity and units on left

2. Write the units of conversion factors that cancel the old unit and replace them with the desired unit

3. Fill the value of the conversion factors

4. Carry out the arithmetic value

13



• Convert 1 cm/s2 to km/yr2

14

1 cm s2 h2 day2 m km

s2 h2 day2 yr2 cm m

1 cm 36002 s2 242 h2 3652 day2 1 m 1 km

s2 12 h2 12 day2 12 yr2 100 cm 1000 m

(3600 x 24 x 365) 2 km = 9.95 x 109 km/ yr 2

100 x 1000 yr2


• Convert 4 kg/m3 to lbm/ft3


4 kg 0.028317 m3 1 lbm

m3 1 ft3 0.453593 kg

4 kg m3 lbm

m3 ft3 kg

4 x 0.028317 x 1 lbm = 0.2497 lbm/ft3

1x 0.453593 ft3

15


Systems of Units • Components of a system of units:

• Base units - units for the dimensions of mass, length, time, temperature, electrical current, and light intensity.

• Multiple units- multiple or fractions of base unit

• E.g.: for time can be hours, millisecond, year, etc.

• Derived units - units that are obtained in one or two ways;

a) By multiplying and dividing base units; also referred to as compound units

• Example: ft/min (velocity), cm2(area), kg.m/s2 (force)

b) As defined equivalent of compound unit (Newton = 1 kg.m/s2) 16


Systems of Units

• 3 systems of unit:

a) SI system

b) American engineering system

c) CGS system

17


Base Units

18

Base Units

Quantity SI Symbol American Symbol CGS Symbol

Length meter m foot ft centimeter cm

Mass kilogram kg pound mass lbm gram g

Moles

gram-mole mole pound mole lbmole gram-mole mole

Time second s second s second s

Temperature Kelvin K Rankine R Kelvin K


Multiple SI Units

19

Multiple Unit Preferences

tera (T) = 10 12 centi (c) = 10 -2

giga (G) = 10 9 milli (m) = 10 -3

mega (M) = 10 6 micro (μ) = 10 -6

nano (n) = 10 -9


Derivatives SI Units

20

Derived SI Units

Quantity Unit Symbol Equivalent to the Base Unit

Volume Liter L 0.001m3 = 1000 cm3

Force Newton (SI)

Dyne (CGS)

N 1 kg.m/s2

1 g.cm/s2

Pressure Pascal Pa 1 N/m2

Energy/ Work

Joule

Calorie

J

cal 1 N.m = 1 kg.m2/s2

4.184 J =4.184 kg.m2/2

Power Watt W 1 J/s = 1 kg.m2/s3


Force and Weight

• Force is proportional to product of mass and acceleration

• Usually defined using derived units ;

1 Newton (N) = 1 kg.m/s2

1 dyne = 1 g.cm/s2

1 Ibf = 32.174 Ibm.ft/s2

• Weight of an object is force exerted on the object by gravitational attraction of the earth i.e. force of gravity, g.

• Value of gravitational acceleration:

g = 9.8066 m/s2

= 980.66 cm/s2

= 32.174 ft/s2

21


Force and Weight

• gc is used to denote the conversion factor from a natural force unit to a derived force unit.

22

gc = 1 kg.m/s2 = 32.174 lbm.ft/s2

1N 1 lbf


EXAMPLE : WEIGHT AND MASS

Water has a density of 62.4 Ibm/ft3. How much does 2.0 ft3 of

water weight at sea level, where the gravitational acceleration

is 32.174 ft/s2 in lbf?

23


Weight & Mass See this example:

Given the density of 2 ft3 water is 62.4 lbm/ft3. At the sea level,

the gravitational acceleration is 32.174 ft/s2.

The mass of water is

The weight of water is

mm lbft

ft

lbM 8.12424.62 3

3

f

m

fm lbsftlb

lbsftft

ft

lbW 8.124

/174.32

1/174.3224.62

2

23

3

g Conversion factor

24


25

What is the force (N and lbf) on a

20.0 kg mass under normal

gravity? What is the force (N and lbf)

on a 20.0 lbm mass under normal

gravity?

Exercise


26

The Reynolds number is the dimensionless

quantity that occurs frequently in the analysis of

the flow of fluids. For flow in pipes it is defined

as DV/ µ, where D is the pipe diameter, V is the

fluid velocity, is the fluid density, and µ is the

fluid viscosity. For a particular system having

D = 4.0 cm, V = 10.0 ft/s, = 0.700 g/cm3,

and µ = 0.18 centipoise (cP) (where 1 cP =

6.72 x 10-4 Ibm/ft.s). Calculate the Reynolds

number.

Exercise

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