Chapter_7_Basic_calculation

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BKF1513
INTRODUCTION TO CHEMICAL ENGINEERING
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Chap 7
Basic Calculation in Engineering
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What are in this chapter?
Unit and Dimensions 
Conversion of units
Mass, moles and composition 
Temperature and pressure 
Volume, density and concentration 
Flowrates 
Introduction to process flow diagrams
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2
Think ?
Answer the following questions YES or NO. Can you
Divide ft by s? 
Divide m by cm?
Multiply ft by s? 
Divide ft by cm? 
Add ft and s? 
Subtract m and (deg) K? 
Add cm and m2? 
Add 1 and 2 cm? 
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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.
Contain significant amount of information
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	36 mg	1 g	=	0.036 g
	 	1000 mg		
0.001 is a conversion factor
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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Conversion Factor
SMS2011
6
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Dimensional Equation
To convert a quantity with a unit to its equivalent in term of other units, set up a dimensional equation:
Write the given quantity and units on left
Write the units of conversion factors that cancel the old unit and replace them with the desired unit
Fill the value of the conversion factors
Perform the arithmetic value
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Convert 1 cm/s2 to km/yr2
	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
	36002 x 242 x 3652 	km	=	9.95 x 109 km/ yr 2
	100 x 1000 	yr2		
Dimensional Equation
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Dimensional Equation
Convert 8 mm2 to in2?
	8 mm2	(39.37)2 in2
		(1000)2 mm2
= 0.01240 in2
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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		
Dimensional Equation
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Exercises: Dimensional Equation
Nanotechnology is the generic term that refer to the synthesis and application of such small particles. An example of a semiconductor is ZnS with a particle diameter of 1.8nm. Meanwhile, our 5 cent coin has a diameter of 16mm. How many times 5 cent coin is bigger than this nano particle?
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Systems 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;
By multiplying and dividing base units; also referred to as compound units. Eg.: ft/min (velocity), cm2(area), kg.m/s2 (force) 
As defined equivalent of compound unit. (Newton = 1 kg.m/s2)
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Systems of Units
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Systems of Units
Standard International System
American Engineering System
Centimeter Gram Second (CGS)
System
Base Units
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System/
Dimension
Mass
Moles
SI
meter, m
kilogram, kg
American
foot, ft
pound mole, 
lb-mole
gram-mole, 
mole
CGS
gram-mole,
mole
gram, g
centimeter, cm
pound-mass, lbm
Length
Time
Temperature
second, s
Kelvin, k
Rankine, R
second, s
Kelvin, K
second, s
Multiple SI Units
	Prefix	Factor
	tera (T) 	10 12
	giga (G) 	10 9
	mega (M)	10 6
	kilo (k)	10 3
	centi (c)	10 -2
	milli (m)	10 -3
	micro (μ)	10 -6
	nano (n)	10 -9
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Derived SI Units
	Quantity	Unit	Symbol	Equivalent to the Base Unit
	Volume	Liter	L	0.001 m3 = 1000 cm3
	Force	Newton (SI)
Dyne (CGS)	N	1 kg m s-2
1 g cm s-2
	Pressure	Pascal	Pa	1 N m-2
	Energy/ Work	Joule
Calorie	J
cal	1 N m = 1 kg m2 s-2
4.184 J = 4.184 kg m2 s-2
	Power	Watt	W	1 J s-1 = 1 kg.m2 s-3
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Chemical Composition
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Chemical Composition
Moles & MW
Concentration
ppm & ppb
Mass & Moles Fraction
Moles
Moles – Latin word mean “heap” or “pile”
Definition by 1969 International Committee on Weights and Measures
the amount of substance that contains as many elementary entities (e.g. atoms, molecules, ions, electrons) (6.022 x 1023) as there are atoms in 0.012 kg of carbon-12 (12C)
Formula	: moles = mass/molecular weight
Unit 	: g-mole, kmol, lb-mole etc.
				 (g-mole is same as mol )
One g-mole of any species contains 6.02 x 1023 (Avogadro’s number) molecules of that species.
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Molecular Weight
Molecular weight of compound- sum of the atomic weights of atoms that constitute a molecule of the compound.
Atomic weight of element- mass of an atom based on carbon isotope 12C.
If the molecular weight of a substance is M, then there are M kg/kmol, M g/mol, and M lbm/lb-mole of this substance. 
At 1 mol of a substance, its mass in gram is equal to its molecular weight. Eg. CO has molecular weight of 28. Hence, at 1 mol, mass of CO is 28 g. At 1 kmol, mass of CO is 28 kg.
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Average Molecular Weight
The average molecular weight is mean molecular weight of a mixture (kg/kmol, lbm/lb-mole, etc.).
If yi is the mole fraction of the component i of the mixture and Mi is the molecular weight:
If xi is the mass fraction of the component i of the mixture and Mi is the molecular weight:
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Mass and Mole Fractions
Process streams consist of mixtures of liquids or gases, or solutions of one or more solutes in a liquid solvents.
The following terms may be used to define the composition of a mixture of substances, including a species A.
Mass fraction: xA= mass of A / total mass
	Unit: kg A/kg total; g A/g total; lbm A/lbm total
Mole fraction: yA= moles of A/ total moles
	Unit: kmol A/kmol total; lb-moles A/lb-mole total
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Exercise
2.4 kg of chlorine pass into a process every 3.1 minutes. Calculate the molar flow rate of the chlorine in kmol/hr?
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Exercises
What is molar flow rate for 100kg/h CO2 (M=44) fed to the reactor?
What is corresponding mass flow rate of 850lb-moles/min CO2? 
How many gram of O2 consist in 100g of CO2?
Find number of molecules of CO2 in 100g of CO2?
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2005/2006 II
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Solution
	a)	100kg CO2	1 kmol CO2	=	2.27 kmol CO2
		h	44 kg CO2		h
	b)	850 lb-moles CO2	44 lbm CO2	=	37 400 lbm CO2
		min	1 lb-moles CO2		min
	C)	100 g CO2	1mol CO2	1 mol O2	32 g O2	=	72.73 g O2
			44 g CO2	1 mol CO2	1 mol O2		
	d)	100 g CO2	1mol CO2	6.02 x 1023 Molecules	=	1.37 x 1024 Molecules
			44 g CO2	1 mol CO2		
a) What is molar flow rate for 100kg/h CO2 (M=44) fed to the reactor?
b) What is corresponding mass flow rate of 850lb-moles/min CO2? 
c) How many gram of O2 consist in 100g of CO2?
d) Find number of molecules of CO2 in 100g of CO2?
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Exercise
A mixture of gases has the following mass composition:
		O2		16%
		CO		4%
		CO2		17%
		N2		63%
What is the molar composition?
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Solution
4 steps to convert from mass fractions to moles fractions:
Assuming as a basis of calculation a mass of the mixture (e.g. 100 kg or 100 lbm).
Using the known mass fractions to calculate the mass of each component in the basis quantity. 
Convert these masses to moles using their molecular weights.
Taking the ratio of the moles of each component to the total number of moles.
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Solution
	Basis: 100g of mixture						
	Component	Mass Fraction	Mass	MW	Moles		Mole Fraction
	i	xi	mi	Mi	ni		yi
	O2	0.16	16	32		0.500	0.152
	CO	0.04	4	28		0.143	0.044
	CO2	0.17	17	44		0.386	0.118
	N2	0.63	63	28		2.250	0.686
	Total	1.00	100	 		3.279	1.000
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Group Discussion
If 100 lbm/min of A (MA=2) and 300 lbm/minof B (MB=28) flow through pipes, find
Mass flow rate of A
Total mass flow rate
Mass fractions of A and B
Molar flow rate of B
Total molar flow rate
Mole fractions of A and B
	
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Group Discussion
If 100 lbm/min of A (MA=2) and 300 lbm/min of B (MB=3) flow through pipes, find
Mass fractions of A and B
	0.25 lbm A/lbm; 0.75 lbm B/lbm
Mole fractions of A and B
	0.333 mole A/mole; 0.667 mole B/mole
Mass flow rate of A 
		(100 lbm A/min)
Molar flow rate of B 
		(100 lb-mole B/min)
Total mass flow rate 
		(400 lbm/min)
Total molar flow rate 
		(150 lb-moles/min)	
	
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Concentration
Mass concentration
Molar concentration
Molarity :
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Exercise
A 0.50 molar aqueous solution of sulfuric acid flows into a process unit at a rate of 1.25 m3/min. The SG of the solution is 1.03. calculate 
the mass concentration of H2SO4 in kg/m3, 
the mass flow rate of H2SO4 in kg/s and 
mass flow rate of solution 
the mass fraction of H2SO4.
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Quiz 1
100 kmol/hr of the reactants enters the reactor with the composition as given in the stoichiometric equation. Calculate
The mass fraction of each component in the inlet
The inlet mass flow rate 
The chemical reactions are:
G1. C2H4 + H2O  C2H5OH
G2. C2H6 + Cl2  C2H5Cl + HCl
G3. CH4 + 2O2  CO2 + 2H2O
G4. C3H8 + 5O2  3CO2 + 4H2O
G5. CO2 + 3H2  CH3OH + H2O
G6. 4NH3+5O2  4NO +6H2O
G7. 2H2S+SO2  3S+2H2O
G8. CH3OH + CH3COOH  CH3COOCH3+H2O
SMS2011
Membrane Technology @ FKKSA, UMP
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Parts per million (ppm)& 
Parts per billion (ppb)
To express the concentrations of trace species in mixtures of gases or liquids.
May refer to mass ratios (usual for liquids) or mole ratios (usual for gases). How many parts (in gram or moles) of the species are present per million or billion parts of the mixture.
15 ppm SO2 in air meaning that:
	every million moles of air contains 15 moles of SO2
	a.k.a. mole fractions of SO2 in air is 15 x 10-6
				
ppmi= yi x 106
ppbi = yi x 109
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More Exercise
commercial sulfuric acid is 98wt% H2SO4 and 2wt% H2O. what is the mole ratio of H2SO4 to H2O?
a compound contains 50% sulfur and 50% oxygen by mass. What is the constant x and y if it has an empirical formula of SxOy 
how many kg of activated carbon must be mixed with 38 kg of sand so that the final mixture is 28% activated carbon?
a gas mixture contains 40 lb of O2, 25 lb of SO2 ans 30 lb of SO3. What is the composition of the mixture in mole fractions?
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More Exercise
In the production of a drug having a MW of 192, the exit stream from the reactor flows at a rate of 10.5 L/min. The drug concentration is 41.2% (in water) and the SG of the solution is 1.024. Calculate the concentration of the drug in kg/L in the exit stream and the flow rate of the drug in kmol/min.
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More Exercises
100 kmol/hr of the reactants enters the reactor and it is fully converted to the products. The product is continuously withdrawn from the reactor. Calculate
The inlet and outlet mass flow rate
The mass fraction of each component in the inlet and outlet
The outlet molar flow rate
The chemical reactions are:
G1. C2H4 + H2O  C2H5OH
G2. C2H6 + Cl2  C2H5Cl + HCl
G3. CH4 + 2O2  CO2 + 2H2O
G4. C3H8 + 5O2  3CO2 + 4H2O
G5. CO2 + 3H2  CH3OH + H2O
G6. 4NH3+5O2  4NO +6H2O
G7. 2H2S+SO2  3S+2H2O
G8. CH3OH + CH3COOH  CH3COOCH3+H2O
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Force and Weight
Force is proportional to product of mass and acceleration (i.e.; F = m.a) 
Usually defined using derived units ;
		1 Newton (N)	= 1 kg m s-2
		1 dyne		= 1 g cm s-2
		1 Ibf		= 32.174 Ibm ft s-2
Weight of an object is force exerted on the object by gravitational attraction of the earth i.e. force of gravity, g.
gc is used to denote the conversion factor from a natural force unit to a derived force unit.
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 If you're working a problem and have mass units, but want force, divide by gc (or vice versa). 
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Value of g
9.8066 m s-2 
980.66 cm s-2 
32.174 ft s-2
gc
1 kg m s-2 / 
1 N
32.174 lbm ft s-2 / 
1 lbf
Force and Weight - 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. What is the weight of the water in lbf?
	2 ft3	62.4 lbm	32.174 ft	1 lbf
		ft3	s2	32.174 lbm ft s-2
= 124.8 lbf
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Volume and Density
Volume has dimension of [L3] i.e; cm3, m3, ft3 
The following conversion to convert between different units of volume;
	1cm	1mL = 0.001L = 0.033814 fl.oz (US) = 0.061023 in3 = 2.6418E-4 gallons(US) = 3.53E-5 ft3 
	1 liter(L)	1000 cm3 = 1dm3 = 0.001m3 = 61.02in3 = 33.81fl.oz(US) = 0.2641 gallons (US) = 0.2199 gallons(UK) 
	1ft3	28316.8cm3 = 28.31L = 0.02831m3 = 1728in3 = 7.480 gallons(US)
	1 barrel (oil)	42 gallon(US) = 1.333 barrels (US liquid) = 5.614ft3 =0.1589m3
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Density
Specific density is mass per volume [M/L3] 
i.e; lb/ft3, kg/m3
Molar density in molar per volume on of [N/L3] 
i.e; g mol/m3, lbmol/ft3 
Specific gravity (SG) is the ratio of the specific density of a substrate at the specific density of water at 40C.
Example; 
 What is the density of liquid benzene at 200C ? if SG of benzene is 0.8765 at 20 0C. 
Answer :
The density water at 40C is 1.0g/cm3. therefore the density of liquid benzene at 200C is 0.8765 g/cm3 = 54.72 lb/ft3
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Density
The density of gases is strongly dependent on pressure and temperature . The molar density of gases can be calculated with reasonable accuracy from the ideal gas law.
Specific volume and densities of gases often reported at Standard Temperature and Pressure (STP) which is 00C and 1 atm 
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Temperature
Temperature of a substance in a particular state of aggregation (solid, liquid, or gas) is a measure of the average kinetic energy possessed by the substance molecules.
Some temperature measuring devices: resistance thermometer, thermocouple, pyrometer and thermometer.
temperature conversion equation
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Exercise
Calculate the interval temperature from 20°F to 80°F in °C.
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Pressure
A pressure is the ratio of a force to the area on which the force acts (P= F/A). 
Pressure units: N/m2, dynes/cm2, lbf/in2, psi, Pa.
Hydrostatic pressure of the fluid- the pressure P of the fluid at the base of the column.
				
Head - the height of a hypothetical column of the fluid that would exert the given pressure at the top were zero.
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Atmospheric, Absolute & Gauge Pressure 
The atmospheric pressure can be thought of as the pressure at the base of a column of fluid (air) located at the point measurement (e.g. at sea level).
A typical value of the atmospheric pressure at sea level, 760.0 mm Hg, has been designated as a standard pressure of 1 atmosphere.
Pressure-measuring devices give gauge pressure of fluid.
 A gauge pressure of zero indicated that the absolute pressure (fluid pressure) is equal to atmospheric pressure.
 Relationship between absolute pressure and gauge pressure is:
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Fluid Pressure Measurement
Common pressure measurement devices –
Bourdon gauge: hollow tube closed at one end and bent into a C configuration.
Manometer: U-shaped tube partially filled with fluid of known density.
Manometer configuration
open-end
differential
sealed end
When the ends of the tubes are exposed to different pressures, the field level drops in the high pressure arm and rise in the low pressure arm.
The pressures difference can be calculated from the measured difference between the liquid level in each arm
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Flow rate
Continuous process involve movement of materials from one point to another with certain rate.
Flow rate- the rate at which a material is transported through a process line.
Flow rate can be expressed as :
mass flow rate (mass/time) 
volumetric flow rate (volume/time)
molar flow rate (mol/time)
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Flow rate
The density of a fluid can be used to convert a known volumetric flow rate of a process stream to the mass flow rate of that stream or vice versa
Flow meter is a device mounted in a process line that provides a continuousreading of the flow rate in the line.
Two commonly used flow meter are rotameter and orifice meter.
rotameter
orifice meter
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Exercise
40 gal/min of hydrocarbon fuel having a SG of 0.91 flow into a tank truck with a load limit of 40 000 lb fuel. How long will it take to fill the tank in the truck?
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Example
LPG Gas typically comprised of the following compositions:
	
a) Calculate the mass fraction for the above LPG gas.
b) Calculate the volumetric air required for the above combustion if the 100 m3/hr of LPG is been used. 
c) Find the stoichiometric air-LPG ratio for the above combustion?
	No	Compound	Volume fraction
	1.
2.	Propane
Butane	0.3
0.7
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Answer
a) Calculate the mass fraction for the above LPG gas.
b) Calculate the volumetric air required for the above combustion if the 100 m3/hr of LPG is been used. 
C3H6 + 4.5O2  3CO2 + 3H2O
 C4H8 + 6O2  4CO2 + 4H2O 
Oxygen required = 0.3(4.5)+ 0.7(6) x 100 m3/hr 
 	 =555 m3/h
Air required = 555 m3/hr ÷ 0.21 (air contain 21% O2, 79% N2)
 = 2642 m3/hr 
c) Find the stoichiometric air-LPG ratio for the above combustion?
 
	No	Compound	fraction 	MW	Fraction x MW	Mole fraction
	1.
2.	Propane 
Butane	0.3
0.7	42
56	12.6
39.2
-------------
51.8	0.24
0.76
---------
1.00
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Convert 1 mm Hg of vacuum to atm (absolute)
Example
	1 mm Hg	 1 atm
		760 mm Hg
	1 mm Hg	 atm
		 mm Hg
	1 x 1	atm	=	0.0013 atm
	760			
		P abs = P gauge + Patm
		P abs = 0.0013 atm + 1 atm = 1.0013 atm
Questions
In domestic application the intake pressure is specified at 11” wc(inch water column). What is the intake pressure in psi..?
The value of Y after a series of runs are given in tabel1
	
	Determine ;
		i) mean
		ii)range
		iii)variance
		iv)standard deviation 
	Run 	1	2	3	4	5
	Y(%)	70.2	73.0	69.5	67.4	71.0
Answer
In domestic application the intake pressure is specified at 11” wc(inch water column). What is the intake pressure in psi..?
	= 
	
i) mean=(70.2+73+69.5+67.4+71)/5=70.22
ii)range= 73.0-67.4 = 5.6
iii)Variance : S2=[(70.2-70.22)^2+(73.0-70.22)^2+(69.5-		 70.22)^2+(67.4-70.22)^2+(71.0-		 70.22)^2]÷ (5-1) 
		 = 4.2
iv)standard deviation = 4.2^0.5
			 = 2.04
	Run 	1	2	3	4	5
	Y(%)	70.2	73.0	69.5	67.4	71.0
Introduction to Process flow Diagram
The chemical process plant is a physical facility in which the raw materials undergo chemical and physical changes in order to make the desired product.
 Chemical process can be classify in to three methods ;
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Process Classification
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Batch
Feed is charge to the process and product is removed when the process is completed
Continuous
Input and output is continuously red and remove from the process
Semi-batch
Neither batch nor continuous
No mass is fed or removed from the process during the operation
Used for small scale production
Operate in unsteady state
Operate in steady state
Used for large scale production
During the process a part of reactant can be fed or a part of product can be removed.
Continues Process
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Advantages
better energy efficient – heat integration
Disadvantages
more complicated
easy for automation – less operators
lower operation cost
less flexibility –standard equipment in batch process can be produce a range of products
product quality is more consistent
Process Flow Sheet
Generally the flow of materials through a chemical process plant is shown visually on process flow sheets.
Generally a chemical plant contains the following processes;
 - Feed preparation facilities 
Reactor 
Separators
Environmental control facilities 
Product formulation facilities
Material transfer equipment 
Energy transfer equipment 
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Types of Chemical Process Flow Sheet
There are three types of process flow sheet namely ;
	Diagram 	Information 
	Input-output flow diagram	-Raw materials 
-Reaction stoichiometry 
-Products
	Block flow diagram	-Everything above , plus
-Material balances 
-Major process units
-Process unit performance specification 
	Process flow diagrams (PFD)	-Everything above , plus 
-Energy balance
-Process condition(T and P)
-Major process equipment specification 
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Example of Input – output Flow Diagrams
Mixing
200 g
150 g
m g
0.40 g MeOH/g
0.60 g H2O/g
0.70 g MeOH/g
0.30 g H2O/g
x g MeOH/g
(1-x) g H2O/g
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Example of Block Diagram
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The Process Flow Diagram
PFD Drawing Symbols
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Symbols for Stream Identification
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Engineering Drawing
Input-output structure
Block flow diagram
Process flow diagram
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Example of typical process equipment 
icons in PFD 
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Introduction to Chemical Engineering (BKF1513)
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Introduction to Chemical Engineering (BKF1513)
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