Material & Energy Balances

```Material & Energy
Balances
CHM 322
UNITS AND DIMENSIONS
• Chemical processes involve
variables that need to be
expressed in some way.
• A measured variable has a
numerical value and a unit.
Examples: 2 meters, 4.29
kilograms
• Measurable units are specific values
of dimensions that have been defined
by convention: grams for mass,
seconds for time, centimeters or feet for
length
• Numerical values of same units can be
3 cm -1 cm = 2 cm
• A dimension is a property that
can be measured : length, time,
mass, temperature
• OR it can be calculated by
multiplying or dividing other
dimensions : velocity
(length/time ), volume (length3),
density (mass /length3)
• Numerical values of different
units may always be combined by
multiplication or division
3 N X 4 m = 12 N . m
5.0 km / 2.0 h = 2.5 km /h
7.0 km /h X 4 h = 28 km
3 m X 4 m = 12 m2
CONVERSION OF UNITS
• To convert a variable expressed in
terms of one unit to its equivalent in
terms of another unit, multiply the
given variable by the conversion
factor (new unit/old unit):
36 mg x (1g/1000 mg) = 0.036 g
36 mg 1 g
1000 mg
Example 2.2-1 Conversion of
Units
Convert an acceleration of 1 cm/s2
to its equivalent in km /y2
1 cm
36002 s2
242 h2
3652
day2
1m
1 km
s2
12 h2
12 day2
12 yr2
102 cm
103 m
= (3600 X 24 X 365)2 km/ (102 X 103) yr2
=9.95 X 109 km /yr2
SYSTEMS OF UNITS
A system of units has the following
components :
1. Base units : mass, length, time,
temperature
2. Multiple units: multiples or
fractions of base units , ex: minutes,
hours, milliseconds
3. Derived units obtained by:
A. By multiplying and dividing base or
multiple units. These are called
compound units.
(cm2, ft /min, kg.m/s2,...)
B. As defined equivalents of compound
units
(1 erg ≡ (1g. cm / s2), 1 lbf ≡ 32.174
lbm.. ft/s2)
•
•
•
•
There are three systems of units:
S I, base units: m, kg, s
CGS, base units: cm, g, s
American engineering system, base
units: ft, lbm, s
Example 2.3-1 Conversion
Between Systems of Units
Convert 23 lbm.ft/min2 to its equivalent
in kg.cm/s2
Solution
• Set the dimensional equation as
before:
• Fill in the units of conversion factors
to cancel old units
• Fill in the numerical values of the
conversion factors
• Do the arithmetic
23 lbm. ft 0.453593 kg
100 cm 12 min2
min2
3.281 ft
1 lbm
602 s2
= (23 ) (0.453593)(100)/(3.281)(3600) = 0.088
kg.cm/s2
FORCE AND WEIGHT
• Newton’s second law of motion:
force is proportional to the
product of mass and acceleration
•Therefore , natural force units are
compound
• To avoid using compound units in
calculations involving forces, derived force
units are defined in each system of units
• In SI system, 1 N ≡ 1 kg. m/s2
• In CGS system, 1 dyne ≡ 1g. cm/s2
• In AES system, 1 lbf ≡ 32.174 lbm. ft/s2
• The conversion factor from natural to
derived force unit is denoted gc (gc is called
Newton’s conversion factor)
• The value of gc depends on the system of
units
• In SI system, gc = 1 kg . (m /s2) /1 N
• In CGS system, gc = 1 g . (cm /s2) / 1 dyne
• In AE system, gc = 32.174 lbm. (ft /s2) / lbf
• The weight of an object is the
force exerted on the object by the
gravitational attraction
• W = mg, m is the mass of the
object, g is the acceleration of
gravity usually measured at sea
level
Example 2.4-1 Weight and
Mass
Water has a density of 62.4 lbm/ft3.
How much does 2.000 ft3 of water
weigh at sea level and 45o latitude
and in Denver, Colarado, where the
altitude is 5374 ft and the
gravitational acceleration is 32.139
ft/s2?
Solution
• Mass of water (M) = (62.4 lbm /ft3 ) (2 ft3) =
124.8 lbm
• Weight of water ( W) =
124.8 (lbm) g (ft /s2) X
(1 lbf /32.174 lbm . ft /s2)
• At sea level g = 32.174 ft / s2, so W= 124.8
lbf
DIMENSIONAL
HOMOGENITY AND
DIMENSIONLESS
EQUATIONS
• Every valid equation must be
terms on both sides of the equation must
have the same dimensions
• u (m/s) = uo (m /s) + g (m /s2) t(s)
This equation is dimensionally
homogeneous, since the dimensions of u, uo
, and gt is (Length /Time)
• The above equation is dimensionally
homogenous and consistent in units
• Terms of inconsistent equations
applying the necessary
conversion factor(s)
• It is not necessarily true that each
homogenous equation must be
valid:
M = 2M
Example 2.6-1
Dimensional Homogeneity
•
Consider the valid equation
D(ft) = 3t(s) + 4
1. What are the dimensions of the
constants 3 and 4 ?
2. If the equation is consistent in units,
what are the units of 3 and 4?
3. Derive an equation for distance in
meters in terms of time in minutes
Solution
1. Because of dimensional
homogeneity, the constant 3
must have the dimension of
length/time. Also the constant 4
must have the dimension of
length
2. For consistency, the units of the
constants must be 3 ft/s and 4 ft
• Define new variables D’(m) and
t’(min) in terms of the old
variables
• Find the relations between the old
and the new variables
• Substitute the relations in the
orginal equation and simplify
D(ft) = D’ (m) 3.2808 ft
= 3.28 D’
1m
t(s) =
t’ (min)
60 s
1 min
= 60 t’
• Substituting in the original
equation
3.28 D’ = (3)(60 t’) + 4
• Simplify by dividing by 3.28
D’(m) = 55t’(min) + 1.22
Dimensionless Quantities
• Pure numbers are dimensionless
• Multiplicative combination of
variables with no net dimensions
D(cm) u(cm/s) ρ(g /cm3)/ μ[g/(cm . s)]
• Exponents, functions as log, exp, sin
and their arguments are
dimensionless
Example 2.6-2
Dimensional Homogeneity and
Dimensionless Groups
A quantity k depends on the temperature T in
the following manner:
k(mol /cm3.s)
= 1.2 X 105 exp (-20,000/(1.987 T))
The units of the quantity 20,000 are cal/ mol,
and T is in K( kelvin), what are the units of
1.2 X 105 and 1.987 ?
Solution
• Equation must be consistent in
units
• Exp part is dimensionless
• 1.2 X 105 should have the same
units as k (mol / cm3.s)
20,000 cal
mol
1
mol. K
T(K)
1.987 cal
• Therefore 1.987 has the units
cal /(mol.K)
Validating Results
•
A question arises as how to check
numerical solutions of problems
• There are several ways.
in to equations, to see if they are valid
2. Order-of-magnitude estimation: obtaining
a rough or an approximate answer to the
problem close enough to the exact answer
3.Testing by reason (solution makes sense):
for example, the temperature of a reactor is
higher than the interior temperature of the
sun
Order-of –magnitude Estimation Procedure:
• Substitute simple integers for all numerical
quantities
• Do the calculations by hand, rounding off
• The final estimated answer should be of the
same magnitude as the exact answer
Example 2.5-1
•Calculation of a stream volumetric flow
rate is given by:


254
13

V


(0.879)
(62.4)
(
0
.
866
)(
62
.
4
)


1

( 31.3145)(60)
Estimate Volumetric flow rate without
using a calculator. (exact solution is
0.00230)
 250
10 
1

V



1
(1)
(50)
(
1
)(
60
)

 (4  10 )(60)
5
2

 0.2  10  0.002
2
25  10
Processes and Process
Variables
MASS AND VOLUME
• Density of a substance is the mass per unit
volume of this substance (kg/m3, g/cm3,
lbm / ft3). It depends slightly on temperature
for liquids and solids.
• Specific volume is inverse of density
(volume occupied by a unit mass of the
substance)
• Density is the conversion factor to relate
mass and volume
Examples
For a volume of 20.0 cm3 of CCl4, if the
density of CCl4 is 1.595 g/ cm3, the mass is:
20 cm3
1.595 g
cm3
= 31.9 g
The volume of 6.20 lbm of CCl4 is:
6.20 lbm
454 g
1 cm3 = 1760 cm3
1 lbm
1.595 g
• The specific gravity of a substance is the
ratio of the density (ρ) of the substance to the
density (ρref) at a specific condition:
SG = ρ / ρref
• The reference most commonly used for
liquids and solids is water at 4oC
ρH2O (4 oC) = 1 g/ cm3
= 1000 kg / m3
= 62.43 lbm /ft3
Example 3.1-1 Mass , Volume,
Density
Calculate the density of mercury in
lbm/ ft3 from a tabulated specific
gravity and calculate the volume in
ft3 occupied by 215 kg of mercury
Solution
• Specific gravity of mercury at
20 oC is 13.546 (Table B.1)
• ρHg = 13.546 X 62.43 lbm/ ft3
= 845.7 lbm/ ft3
215
kg
1 lbm
1 ft3
0.454 kg
845.7 lbm
Example 3.1-2
Effect of Temperature on liquid
Density
In Example 3.1-1, 215 kg of mercury
was found to occupy 0.56 ft3 at 20 oC.
(1) What volume would the mercury
occupy at 100 oC?
(2) Suppose the mercury is contained in
a cylinder having a diameter of 0.25 in.
What change in height would be
observed as the mercury is heated from
20 oC to 100 oC?
Solution
1. V(100 oC) = Vo [1+ 0.18182 X 10-3
(100)
+ 0.0078 X 10-6 (100)2]
•
V(20 oC) = 0.560 ft3 = Vo [1+ 0.18182
X 10-3 (20) +0.0078 X 10-6 (20)2]
•
Solving for Vo (volume of mercury at
0 oC) from second equation and
substituting in first equation:
 V(100oC) = 0.568 ft3
2. The volume of the mercury equals πD2 H/
4 , where D is the cylinder diameter and
H is its height, Since D is constant = 0.25
in
D = 0.25 /12
H(100 oC) - H (20oC) =
(V(100oC) – V(20 oC) ) /( πD2/4) ,
= 23.5 ft
FLOW RATE
• Flow rate is the rate at which a
material is transported through a
process line
• Flow rate is expressed as mass flow
rate
or volumetric flow rate
• Mass flow rate and volumetric flow
rate are related through the density in
the same way as the mass and volume

m m
ρ 

V V
• 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
CHEMICAL COMPOSITION
There are different ways to express
mixture compositions
1.Mass fraction
2.Mole fraction
3.Concentration
Moles and Molecular Weight
• A gram- mole or mol of a species is the
amount of that species whose mass in grams
is numerically equal to its molecular weight
• Mol applies to both compounds and atoms
• Other units for moles are k mol,
lb-mol, ton - mol
• If molecular weight of a substance is then
there is M kg/ kmol, M lb/ lb-mol, M g/mol
• Molecular weight is used as a conversion
factor that relates mass and number of
moles of a substance
• 34 kg of NH3 is equivalent to:
34 kg NH3
1 kmol NH3 = 2 kmol
17 kg NH3
• 4 lb-moles of NH3 is equivalent to
40 lb-moles NH3
17 lbm NH3 = 68 lbm
1 lb- mole NH3
• Same factors used to convert mass units to
one another are used to convert molar units
to one another
• Factor to convert from lbm to g is 454 g
/lbm Also to convert from lb-moles to mols ,
the factor is the same, i.e. 454 mol / lb-mol
• One mol of any substance contains
6 X 1023 molecules
Example 3.3-1
Conversion between Mass and
Moles
How many of each of the following are
contained in 100 g of CO2
(M = 44)?
(2) lb-moles CO2
(3) mol C
(4) mol O
(5) mol O2
(6) g O
(7) g O2
(8) molecules of CO2
Solution
(1)
100 g CO2
1 mol CO2
44.01 g CO2
= 2.273 mol
CO2
(2)
2.273 mol CO2
1 lb-mol CO2
=
5.011 X 10-3 lb-mole
453.6 mol CO2
Each 1 mol CO2 contains 1 mol C, 1 mol
O2 , and 2 mol O
(3)
2.273 mol CO2
1 mol C
1 mol CO2
= 2.273 mol C
(4)
2.273 mol CO2
2 mol O
1 mol CO2
= 4.546 mol O
(5)
2.273 mol CO2
1 mol O2
1 mol CO2
= 2.273 mol O2
(6)
4.546 mol O
16.0 g O
1 mol O
= 72.7 g O
(7)
2.273 mol O2
32.0 g O2
1 mol O2
= 72.7 g O2
(8)
2.273 mol CO2
6.02 X 1023
Molecules = 1.37 X 1024
molecules
1 mol CO2
• The molecular weight of a
species can be used to relate
the mass flow rate of a stream
to its molar flow rate
For example, if carbon dioxide flows
through a pipeline at a rate of 100 kg/h,
the molar flow rate of CO2 is:
100 kg CO2
h
1 kmol CO2
44.0 kg CO2
=
2.27 kmol /h
If the output from a chemical reactor
contains CO2 flowing at a rate of 850 lbmoles/ min, then mass flow rate is:
85 lb-moles CO2
min
44.0 lbm CO2
= 37,400 lbm/ min
lb-mole CO2
Mass and Mole fractions
and Average Molecular
Weights
• Process streams are usually
mixtures of liquids or gases or
solutions of one or more solutes in
a liquid solvent:
• Composition of a mixture in
terms of a species A is given as:
mass of A
Mass fraction  x A 
total mass
Mole fraction  y A 
moles of A
total moles
Mass percent  100x A
Mole percent  100y A
Example 3.3-2
Conversions Using Mass and
Mole Fractions
A solution contains 15% by mass A
(xA = 0.15) and 20 mole% B (yB = 0.20)
1. Calculate the mass of A in 175 kg of the
solution
2. Calculate the mass flow rate of A in a
stream of solution flowing at a rate
of 53 lbm /h
3. Calculate the molar flow rate of B in a
stream flowing at a rate of 1000 mol/min
4. Calculate the total solution flow
rate that corresponds to a molar
flow rate of 28 kmol B/s
5. Calculate the mass of the
solution that contains 300 lbm of
A
Solution
(1)
175 kg solution
0.15 kg A = 26 kg A
kg solution
(2)
53 lbm
solution
h
0.15 lbm A
1 lbm solution
= 8 lbm A /h
(3)
1000 mol
solution
min
0.2 mol B
= 200 mol B/ min
1mol solution
(4)
28 kmol B
s
1 kmol solution =
140 kmol Solution /s
0.2 kmol B
(5)
300 lbm A
1 lbm solution =
2000 lbm solution
0.15 lbm A
Conversion from mass
fractions to mole fractions or
vice versa
1. Assume a mass of the mixture
as a basis of calculation
2. Calculate the mass of each component
using the known mass fraction
3. Convert the masses into moles
4. Calculate mole fractions
• A similar procedure is followed to
convert mole fractions to mass
fractions. In this case taking total
number of moles as a basis.
Example 3.3-3
Conversion from a
Composition by Mass to a
Molar Composition
A mixture of gases has the following
composition by mass :
O2 16 %
CO 4 %
CO2 17 %
N2
63 %
What is the molar composition?
Solution
Basis: 100 g of the mixture
Set up calculations in table format
comp.
i
xi
mi
Mi
ni
yi
O2
CO
0.16
0.04
16
4
32
28
0.5
0.143
0.150
0.044
CO2
0.17
17
44
0.386
0.120
N2
0.63
63
28
2.250
0.690
Total
1.00
100
3.279
1.00
Calculation of Average Molecular
Weight
•
The average molecular weight of a mixture
is the ratio of the mass of a sample of the
mixture to the total number of moles of all
species in the sample
•
If the mole fractions of components are
given, then :
M  y1M1  y 2 M 2  .... 
y M
i
all components
i
• If the composition of the mixture is
given in terms of mass fractions, then
the average molecular weight is given
by the equation:
1
x1
x2
xi


 ....  
M M1 M 2
all components M i
Example 3.3-4
Calculation of an Average
Molecular Weight
Calculate the average molecular weight
of air:
1. If the approximate molar composition
is 79% N2 , 21% O2
2. If the approximate mass composition
is 76.7% N2 , 23.3%O2
Solution
(1)
• Using yN2 = 0.79, yO2 = 0.21
and the equation:
M  y1M1  y 2 M 2  .... 
y M
i
all components
i
0.79 kmol N2
28 kg N2
kmol
kmol N2
M
+
0.21 kmol O2
32 kg O2
kmol
kmol O2
= 29 kg / kmol = 29 lbm /lb-mol
= 29 g / mol
(2)
From the Equation:
1
x1
x2
xi


 ....  
M M1 M 2
all components M i
= 0.767 g N2 + 0.2333 g O2/ g
28 g N2 / mol 32 g O2 / mol
M
= 29 g / mol
Concentration
• Mass concentration of a component in a
mixture or solution is the mass of this
component per unit volume of the mixture
(g/cm3, lbm / ft3, kg/ m3)
• Molar concentration of a component in a
mixture or solution is the number
of moles of the component per unit volume of
the mixture (kmol / m3, lb-moles / ft3,….)
• Molarity of a solution is the value of the
molar concentration of the solute expressed in
gram-moles /liter solution
Concentration of a component is used as a
conversion factor to relate:
1. Mass or moles of a component to a mixture
volume
2. Mass or molar flow rate of a
component to volumetric flow rate
of a stream
Examples
What is the number of moles in 5L of a
0.02 molar solution of NaOH?
5L
0.02 mol NaOH = 0.1 mol
L
2L
min
0.02 mol NaOH = 0.04 mol
L
min
Example 3.3-5
Conversion between mass,
molar and volumetric flow rates
of solution
• A 0.5- molar aqueous solution of
sulfuric acid ( SG =1.03 )flows into a
process unit at a rate of 1.25 m3/ min
Calculate:
(1) Mass concentration of H2SO4 in
kg/m3
(2) Mass flow rate of H2SO4 in kg/s
(3)Mass fraction of H2SO4
Solution
1.
cH2SO4 (kg H2SO4/ m3) =
0.5 mol H2SO4 98 g
1kg
103 L
L
mol 103g 1m3
= 49 kg H2 SO4
m3
2.
m H2 SO4 ( kg H2SO4 ) =
s
1.25 m3 49 kg H2SO4 1 min = 1 kg
min
m3
60 s
s
3. The mass fraction of H2SO4 equals the
ratio of the mass flow rate of H2SO4
to the total mass flow rate, which can
be calculated from the total
volumetric flow rate and the solution
density
 solution = 1.03 (1000 kg) = 1030 kg
m3
m3
• m solution (kg)
s
= 1.25 m3 solution 1030 kg
1 min
min
m3 solution 60s
= 21.46 kg
s
• xH2SO4 = mH2SO4 = 1 kg H2SO4 /s
msolution 21.46 kg solution /s
= 0.0 48 kg H2SO4
kg solution
Parts per Million and Parts per
Billion
• These units are used to express trace
concentrations of species in gas or liquid
mixtures
• The definitions usually mean grams of some
component per million grams or moles of
some component per million moles
• Conversions to mole or mass fractions can
be done through the following equations:
ppmi = yi X 106
ppbi = yi X 109
Temperature
• Temperature of a substance either
liquid, gas, or solid is defined as a
measure of the internal energy
possessed by the substance
• A temperature scale is obtained by
assigning two known temperatures
(freezing point of water = 0 oC ,
boiling point of water = 100 oC)
Temperature Scales
• Celsius scale: 0, 100
• Fahrenheit scale: 32, 212
• Kelvin scale: 0-273.15 oC
• Rankine scale: 0-459.67 oF
Conversion of Temperature
Units
Use the following relationships:
T(K) = T(oC) + 273.15
T(oR) = T(oF) + 459.67
T(oR) = 1.8 T(K)
T(oF) = 1.8 T(oC) + 32
• A degree is both a temperature and a
temperature interval
• There are conversion factors relating
temperature intervals on different
scales:
1.8 oF , 1.8 oR , 1oF , 1o C
1 oC 1 K
1oR
1K
Example 3.5-2
Temperature Conversion
Consider the interval from 20 oF to
80 o F
1. Calculate the equivalent
temperature in oC and the interval
between them
2. Calculate directly the interval in o C
between the temperatures.
Solution
1.T ( o C) = T ( o F) -32
1.8
T1 (20 o F) = (20 - 32)oC = -6.7 oC
1.8
T2 (80 o F) = (80 - 32) o C = 26.6oC
1.8
and
T2 - T1 = 26.6 – (-6.7 ) oC = 33.3oC
2. ∆T( oC) = (80 - 20) o F
1o C
1.8 oF
= 33.3 o C
Example 3.5-3
Temperature Conversion and Dimensional
Homogeneity
The heat capacity of ammonia is given
by the equation :
Cp ( Btu / lbm o F) =
0.487 + 2.29 X 10-4 T (o F)
Determine the expression for Cp in (J/g.oC) in terms
of T ( oC)
Solution
Two steps,
1. Substitute for T (o F) and simplify the resulting
equation
Cp (Btu )
lbm oF
= 0.487 + 2.29 X 10-4 [ 1.8 T (o C + 32]
= 0.494 +4.12 X 10-4 T (o C)
• Convert cP to the desired units and simplify
the resulting equation
Cp J 1.0 o C 9.486 X10-4 Btu 454 g
g. oC 1.8 o F 1 J
1 lbm
Cp (J/ g. oC)
= 2.06 + 1.72 X 10-3 T(o C)
Question (2)
A liquid mixture of water and phenol. The
mixture contains 125 ppb phenol by mass.
1. What is the mass fraction of phenol ?
2. How many milligrams of phenol are
contained in one kilogram of the liquid?
ppbi = yi * 10 9
Question (3)
• Which is a higher temperature 1oC or 1oF?
Question (4)
A solution with volume V(L) contains n (mol)
of a solute A with a molecular weight of MA
(g A/mol)
What is the molar concentration of A in terms
of V, n, and MA
Fundamentals of Material
Balances
• Law of conservation of mass is the basis of
mass or material balances
• Constraints imposed by material balances
must be taken into account when designing
a new process or analyzing an existing
process
• Inputs and outputs of an individual process
unit or the entire process should satisfy
material balance equations
General Procedure for
Performing Material Balances
1. First, given a process description, classify
it to either batch, continuous, or semibatch
process. Also classify the process as either
2. Draw and fully label a process flow block
diagram
3.
Choose a basis of calculation
4.
For a multiple- unit process, identify
units for which balances may be written
• Do a degree of freedom analysis for the
overall system and for each individual units
• Specify the necessary number of process
variables
• Solve the equations and obtain unknowns
• Systems involve recycle, purge, bypass
streams and chemical reactions
Process Classification
• Batch process: no continuous flow of inputs and
inputs. For example: adding reactants rapidly to a
tank, product is removed after some time.
• Continuous process: continuous flow of both inputs
and outputs. For example: Pumping a mixture of
liquids into a distillation column and withdraw
• Semibatch process: either the input or output flows
continuously but not both. For example: blending
two liquids in a tank with no withdrawal of liquid.
Processes
• Steady process: where all the process variables do
not change with time
• Transient process: if any process
variable changes with time
• Continuous processes are run at
• Batch processes are necessarily
transient
General Balance Equation
• Mass balance may be written in the
following general way:
Input (streams entering through system
boundaries) + generation ( mass produced
within system) - output (streams leaving
through system boundaries) – consumption
(consumed within system) = accumulation
(buildup within system)
Rules to Simplify General
Material Balance Equation
• If total mass is balanced, then
generation is 0 and consumption is 0
• If nonreactive species is balanced,
then generation is 0 and consumption is 0
• At steady state, accumulation is 0,
regardless of what is being balanced
Balances on Continuous
• Type of balance is differential
• This type of balance is done at an instant in time.
• Each term in the balance equation is a rate ( rate of
input, rate of output, rate of
generation, rate of consumption)
• The material balance equation
simplifies to:
input + generation = output +
consumption
Example 4.2-2 Material
Balances on a Continuous
Distillation Process
Given:
• 1000 kg benzene + toluene
• Two products
• Xbenzene mixture = 0.5
• mbenzene top = 450 kg /h
• mtoluene bottom = 475 kg /h
Required
• Write balances on benzene and
toluene
• Calculate unknown component
flow rates in the output streams
Solution
• Process at steady state there is no
accumulation
• No chemical reaction involved 
there is no generation or consumption
• Form of mass balance equation is
input = output
• The flow chart of the process is first drawn
450 kg B/ h
m1 kg T /h
500 kg B/ h
500 kg T /h
m2 kg B/ h
475 kg T /h
Balance Equations
Benzene Balance:
mbenzene feed = mbenzene top + mbenzene bottom
1000 X 0.5 = 450 kg /h + m2
m2 = 50 kg B /h
Toluene Balance:
mtoluene feed = mtoluene top + mtoluene bottom
1000 X 0.5 = m1+ 475 kg T/h
m1 = 25 kg T/h
Need to check solution:
• Substitute back the answers in the total mass balance
equation:
1000 kg/h = 450 + m1 + m2 + 475
Use m1 = 25 kg /h , m2 = 50 kg /h
 1000 kg/h = 1000 kg/h
Integral Balances on Batch
Processes
• An example of a batch process is the
production of NH3 from N2 and H2 in
a batch reactor.
• At to = 0  no mol NH3 in reactor
• At tf  nf mol NH3 in reactor
• Besides, no ammonia enters or leaves
the reactor
• Material balance equation simplifies
to:
generation = accumulation
• Also the accumulation of ammonia
itself is the difference nf – no
(final amount of ammonia - initial
amount)
• Therefore, In general for a batch
process where the balanced quantity
is total mass or some species mass, the
material balance equation is:
Accumulation = final output -initial
input = generation - consumption
initial input + generation = final output
+ consumption
Example 4.2-3
Balances on a Batch Mixing
Process
Given:
•
•
•
•
•
2 mixtures methanol –water
Mix1  40 wt % methanol
Mix2  70 wt % methanol
200 g Mix1 + 150 g Mix2  product
Solution
• Process is batch with no reaction involved, then no
generation and consumption terms in material
balance equation
200 g
0.4 CH3 OH
150 g
0.7 CH3 OH
mg
x CH3 OH
• Total Mass Balance
200 +150 = m = 350 g
Methanol Balance:
200 g 0.4 g CH3OH + 150 g 0.7 g CH3OH
g
g
=
m (g )
x (g CH3 OH)
g
• Substituting m =350 g in the last equation
 x = 0.529 g CH3OH / g
• Use water balance to check the solution
Input = Output
(200)(0.600) +(150)(0.300) =165 g H2O
= (350)(1-0.529) = 165 g H2 O
Integral Balances on
Semibatch Processes
Example 4.2-4
n (kmol /min)
0.1 kmol C6H14 / kmol
0.9 kmol air / kmol
0.1 kmol air / min
Given:
1. Semibatch process, air is bubbled
continuously through hexane ( air
rate = 0.1 kmol/min)
2. Output mixture of hexane vapor &
air (0.1 kmol hexane/kmol mixture)
Required
•
Time required to vaporize 10 m3 of
hexane
Solution:
1. Air Balance:
accumulation = 0 (air doesn’t
dissolve in hexane), generation =
0,consumption = 0 (air doesn’t
react)
ِ
Air Balance
Input = Output
0.1 kmol air = 0.9 kmol air n kmol
min
kmol
min
n = 0.111 kmol /min
• Integral Balance on hexane:
generation = 0, consumption = 0, Input = 0
accumulation = - output
OR
depletion = - accumulation = output
Depletion of hexane
= -10 m3 0.659 kg 103 L 1 k mol
L
m3 86.2 kg
= -76.45 kmol
• The balance on hexane is therefore:
-76.45 kmol C6 H14 = -0.1 n tf
 n = 0.111 k mol /min
tf = 6880 min
Example 4.3-1
Flowchart of an Air Humidification
and Oxygenation Process
Given:
•
•
•
•
Evaporation chamber
Three feed streams
Out put gas stream
(1.5 mol% H2O)
Required
1. Draw and label a flowchart
2. Calculate all unknown stream
variables
n3(mol gas /min)
0.015 mol H2 O /mol
0.2 n1(mol O2 /min)
y mol O2 / mol
(0.985 –y) mol N2 / mol
n1(mol air /min)
0.21 mol O2 /mol
0.79 mol N2 / mol
20 cm3H2 O/min
n2 (mol H2O/min)
• n2 = 20 cm3 H2O 1.00 g H2O 1 mol
min
cm3
18 g
= 1.11 mol H2O / min
H2O Balance
n2 (mol H2O) =n3 mol 0.015 mol H2O
min
min
mol
• Substitute the value of n2 = 1.11 mol/ min
n3 = 74.1 mol/ min
• Total mol balance
0.2 n1 + n1 + n2 = n3
• Substitute n2 and n3 and obtain
n1 = 60.8 mol
min
• N2 Balance
n1 mol 0.79 mol N2
min
mol
=
n3 mol (0.985-y) mol N2
min
mol
0.79 n1 = n3 (0.985 – y)
• Substitute the values of n1and n3
 y = 0.337 mol O2 / mol
Question (5)
A stream has a flow rate of 250 kg/h. The
stream contains x kg C6H6 / kg stream and the
rest is C7 H8 . Calculate the mass flow rate of
the C7H8 in terms of x by using a single
dimensional equation.
Flow Chart Scaling and Basis of
Calculation
For any balanced process:
• Masses or number of moles of input and output
streams could be multiplied by the same factor and
the process remains balanced
• Stream compositions (mass fractions or mole
fractions) are left unchanged
• Stream masses could be changed to mass flow rates.
Also moles changed to molar rates. The process
would still be balanced
• Mass units could be changed to other mass
units. The process would still be balanced.
• Mole units could be changed to other mole
units. The process would still be balanced.
• The procedure of changing flow rates and
leaving compositions unchanged is called
scaling
Basis of Calculation
• It is an amount or flow rate of one stream
or a component in a process
• If a stream or component amount or flow
rate is given, then take this as a basis of
calculation
• If not, then assume a flow rate of a stream
of known composition
Example 4.3-2
Scale-up of a Separation
Process
Given:
• 60-40 (mole%) A&B
50 mol
100 mol
0.6 mol A / mol
0.95 mol A / mol
0.05 mol B / mol
0.4 mol B / mol
12.5 mol A
37.5 mol B
Required:
Achieve Same separation with a
continuous feed of 1250 lb-moles /h.
Scale the flow chart.
Solution:
• Originally batch process
• Scale factor is 1250 lb-moles/h
100 mol
= 12.5 lb-moles/h
mol
• The masses of all streams of a batch
process are converted to flow rates by
multiplying by the scale factor
• Feed:
100 mol 12.5lb-moles = 1250 lb-moles
mol
h
• Top product stream:
50 X 12.5 = 625 lb-moles /h
• Bottom product stream:
12.5 X 12.5 = 156 lb-moles A/h
37.5 X 12.5 = 469 lb-moles B/h
625 lb-moles/h
0.95 lb-mol A /lb-mol
1250 lb -moles / h
0.05 lb-mol B /lb- mol
0.6 lb-mol A /lb- mol
0.4 lb-mol B /lb- mol
156 lb-moles A/h
469 lb-moles B/h
Rules for Balancing a Nonreactive Process
1. The maximum number of independent
equations that can be derived by
writing balances equals the number of
chemical species in the input and
output streams
2. Write balances first that involve the
fewest unknown variables
Example 4.3-3
Balances on a Mixing Unit
Given:
• Initially: NaOH aqueous solution
• 20% NaOH by mass
• Dilute by pure water
• Product: 8% NaOH solution
100 kg
0.2 kg NaOH / kg
0.8 kg H2 O/ kg
m2 kg
0.08 kg NaOH /kg
0.92 kg H2O / kg
m1 kg H2O
V1 liters H2O
Required:
V1 /100? M2/100?
Solution
1. The process could be considered continuous or
Batch
2. Examine the unknown variables and
available balance equations
i. Unknowns: m1, m2 and V1
ii. Equations: can write 2 balance equations only
because there are 2 species (NaOH, H2O)
iii. Need a third equation to solve for three unknowns
(It is the equation relating volume and mass of
water)
NaOH Balance
(0.2 kg NaOH)(100kg)
= (0.08 kg NaOH /kg) m2
m2 = 250 kg NaOH
Total mass balance
100 kg +m1 = m2 = 250 kg
m1 = 150 kg H2O
Volume of diluent water
V1 = 150 kg 1.00 liter = 150 liters
kg
Required ratios:
1. V1/100 kg = 1.50 liters H2O/ kg feed
solution
2. m2 /100 kg = 2.50 kg product solution/ kg
feed solution
Example 4.3-4
Given:
• Stream of humid air
• Condenser
• 95% of H2O v condenses
• Flow rate of condensate = 225 L /h
mol O2 /h
mol N2 /h
mol dry air
21 mol% O2
79 mol % N2
mol H2O(v)
mol H2O(v) /h
225 liters H2O
mol H2O (l) /h
95% of water in feed
Required
Solution
1. Process:
• Continuous
• Non-reactive
• Single unit
2. Basis of solution:
225 L/h Condensate
3. Degree-of-freedom analysis:
•
Unknown variables:
6 (counted on chart)
n3 is given as the basis of calculation
˙
˙
˙
•
Number of independent equations:
3 (= number of species in the problem)
Need to specify three other variables:
˙
•
n2 is calculated as a process condition
(liquid water condensate is 0.95 of n2)
No other variable could be specified
• The problem is underspecified and another
piece of information must be given
• Assume that the entering air contains 10
mole% of water vapor in air
• Then the number of degrees of freedom is
zero (five unknowns and five equations)
and the problem is solvable.
• Start writing out 3 equations in the order of
solving the equation (s) with one unknown
variable first, then simultaneous equations
• First calculate the molar flow rate of
condensate (mol /h) from the volumetric
flow rate (225 L/h) and the density of water
(1 kg/L)
• Then proceed with the species balances
• n˙3 (mol)/ h =
225 L H2O 1.00 kg 1 mol H2O
h
L
18 X10-3 kg
= 12500 mol / h
˙ = 0.95 n˙ , n˙ = 13158 mol / h
• n
3
2
2
• n˙˙2
= 0.1
n˙ 1+ n˙ 2˙
• 0.9 n˙2 =0.1 n˙ 1
• n˙1 = 118422 mol / h
˙ 4 = 0.21 X n˙1 , ˙n4 = 24869 mol / h
• n
˙ = 0.79 X n˙ , n˙ = 93553 mol / h
• n
5
1
5
˙ 6 = 0.05 X n
˙ 2 , n˙6 = 658 mol / h
• n
• yO2 = 0.209
• yN2 = 0.786
• yH2O = 5.50 X 10-3
Example 4.3-5
Given:
• Distillation column
• Feed: 45 wt % benzene (B), rest toluene (T)
• Bottom: 8% benzene in feed
• Feed volumetric flow rate : 2000L/h
• Feed specific gravity : 0.872
˙m2 (kg/h)
0.95 mol B/mol
˙
0.05 mol T/mol
xB2 kg B / kg
Feed
1-xB2 kg T/kg
2000 L/h
˙m1 (kg/h)
Bottom
0.45 kg B/kg
˙ B3 (kg B/h)
m
˙
0.55 kg T/kg
8 % benzene in
˙
feed
m
˙ T3 (kg T/h)
Required
˙
1. m2
˙
˙
˙
2. m3 = mB3 +mT3
xB = mB3 / m3
˙
˙
3.
Solution
1. Process:
•
•
•
•
Continuous
Non-reactive
Single unit
2. Basis of solution:
2000 L/h Feed
• Feed composition is given in terms of mass
fraction, and it is also required to
determine the mass composition of the
bottom product
• First step in the solution is to convert mole
fractions in the top product to mass
fractions
Conversion:
Number of moles of toluene = 5 kmol
Number of moles of benzene = 95 kmol
Mass of benzene = 95 X 78.11= 7420 kg B
Mass of Toluene = 5 X 92.13 = 461 kg T
Total mass = 7420 kg B + 461 kg T = 7881 kg
xB2 = 7420 kg B / 7881 kg mixture
= 0.942 kg B / kg
xT2 = 1-0.942 = 0.058 kg T/ kg
3. Degree-of-freedom analysis:
•
Unknown variables:
4 (counted on chart)
•
•
Number of independent equations:
2 (= number of species in the problem)
Need to specify two other variables:
density relation
Process specification:
(Benzene split -bottom product contains 8%
of benzene in feed)
4. Balances:
Mass flow rate of feed calculated from
Volumetric flow rate and Feed S.G.
˙
2000 L 0.872 kg = 1744 kg/h = m
1
h
L
• Benzene split:
˙ B3 =0.08 (0.45 ˙m1) = 62.784 kg /h
m
• Benzene Balance
˙ =m
˙ x +m
˙
0.45 m
1
2 B2
B3
Substituting:
m˙ 1 = 1744 kg/h
˙
˙ B3 = 62.8 kg/h
m
xB2 = 0.942
m˙ 2 = 766 kg/h
• Toluene Balance
˙ +m
˙
0.55 m˙ 1 = (1-xB2)m
2
T3
˙
m
T3 = 915 kg/h
˙ = 62.8 +915 = 978 kg/h
• m˙ 3 = m˙B3 + mT3
˙
˙
• xB3 = m
B3 /m3 = 62.8 / 978
= 0.064 kg B/ kg
• xT3 = 1- xB3 = 0.936 kg T/kg
General Procedure for SingleUnit Material Balance
Calculation (P. 101 &102)
1. Choose a basis of calculation
a. An amount or flow rate of stream or several
streams given in the problem
b. If no information, take as an arbitrary basis,
the amount or the flow rate of a stream with
known composition
2. Draw a completely labeled flowchart
a. Fill in all known variable values
including the basis of calculation
b. Label unknown stream variables on the chart
2.
Write the requirements of the
problem in terms of labeled
variables.
3.
Problems with mixed mass and mole
units must be solved in only one set
of units.
4.
Do the degree-of-freedom analysis.
5.
Solve the equations in an efficient order.
6.
Scale the balanced process if necessary.
Balances on Multiple-Unit
Processes
• Chemical processes usually involve more
than one process unit (chemical
reactors, mixing units, separation
processes…. etc.)
• Material balance might be done on different
subsystems (In a single-unit processes,
there is only one system)
• Systems may involve subsystems :
The entire process
Interconnected combination of some units
A single unit
A mixing point for two or more process streams
A splitting point into branches
• Inputs and outputs to a system or a
subsystem are the process streams
that intersect the system boundary
Procedure for Material Balance
Calculations on Multiple-Unit
Processes
• Do a degree-of-freedom analysis on the
overall system and on each subsystem
• Isolate and write balances on
subsystems of the process as needed to
solve for unknowns
Example 4.4-1
Two-Unit Process
Given
• Two components: A &B
Required
• Mass flow rates of streams 1, 2 and 3?
• x’s or y’s of the same streams?
2. Basis: Given flow rates collectively
3.
•
•
•
Subsystems:
Overall process
Individual process units
Mixing point
4. Degree-of-Freedom Analysis
Do the analysis for the overall system and for each
subsystem
•
Include only in each analysis the variables in the
streams intersecting a system boundary
• Overall system:
.
2 unknowns (m3, x3) -2 balances (2 species)
= 0 degrees of freedom
.
Possible to solve for m3, x3. These two
variables become known
.
• Mixing point:
.
.
4 unknowns (m1, x1 , m2 , x2 ) - 2 balances (2
species) = 2 degrees of freedom
• Unit 1:
.
2 unknowns (m1,x1) -2 balances (2 species) = 0
degrees of freedom
.
Possible to solve for m1,x1. These two variables
become known
• Back again to mixing point:
.
It is possible now to determine m2 , x2
Calculations
Overall Mass Balance:
.
100 + 30 = 40 + 30 +m3
.
m3 = 60 kg / h
Overall Balance on A:
(0.500) (100) + (0.3)(30) =
(0.9) (40.0) +(0.6)(30) +x3(60)
x3 = 0.0833 kg A/kg
• Total mass Balance on Unit 1:
.
100 = 40 + m1
.
m1 = 60 kg/h
• Species A Balance on Unit 1:
(0.5) (100) = (0.9)(40) + x1 (60)
x1 = 0.233 kg A/kg
• Total mass balance on Mixing Point:
.
.
m1 + 30 = m2 = 90 kg/h
• Species A Balance on Mixing point:
.
.
x1 m1 +(0.3)(30.0) = x2 m2
.
.
using the values of m1 , x1 , m2
x2 = 0.255 kg A /kg
Example 4.4-2
Extraction-Distillation Process
Given:
•
•
•
•
Mixture : Acetone (A) & Water (W)
Composition: 50 wt% (A), 50 wt% (W)
2 stage extraction +Distillation
Details given as flow chart
Required:
•
•
•
•
•
•
•
Mass of stage 1 extract
Mass of stage 1 raffinate
Mass of stage 2 extract
Mass of combined extract
Mass of overhead product from distillation
Mass of bottoms product from distillation
Compositions of all above streams
Solution
1. Process: batch and non-reactive
2. Basis: 100 kg feed mixture
3. Degree-of-freedom analysis:
•
Overall process
4 unknown variables
(m5 , mA6 , m m6 , m w6) - 3 equations =
1 degree of freedom
• Extractor 1
5 unknown variables
(mA2 , mM2 , m W2, m1, xm1)
- 3 equations
= 2 degrees of freedom
• Extractor 2
4 unknown variables
(mA2 , mM2 , m W2, m3) - 3 equations = 1
degree of freedom
• Mixing point
6 unknown variables
(m1 , xm1 , m3 , m A4 , mM4 , mW4) - 3 equations =
3 degrees of freedom
• Combined extractors
3 unknown variables
(m1 , xm1 , m3 ) - 3 equations =
0 degrees of freedom
• Distillation column
7 unknown variables
(mA4 , mM4 , mW4 , m5 , mA6 , mM6 , mW6 ) - 3
equations = 4 degrees of freedom
Start balances on combined extraction units
•Total mass balance:
(100 + 100 +75) kg = 43.1 + m1 + m3
• Mass balance on A :
100 (0.500) kg A =
(43.1) (0.053) + m1(0.275) +m3(0.09)
Solve simultaneously
m1 = 145 kg , m3 = 86.8 kg
• Mass balance on M
(100 +75) kg M =
= (43.1) (0.016) + m1 xM1 + m3 (0.88)
Substitute the values of m1 and m3
xm1 = 0.675 kg MIBK / kg
• Balances Around Extract Mixing Point
Mass balance on A:
m1 (0.275) + m3 (0.09) = mA4
Substitute m1, m3
mA4 = 47.7 kg acetone
• Balance on M
m1 xM1 + m3 (0.88) = mM4
Substitute the values of m1 , m3 , xM1
mM4 = 174 kg MIBK
• Balance on W
m1 (0.725 - xM1) + m3 (0.03) = mW4
Substitute the values of m1 , m3 , xM1
mW4 = 9.9 kg water
• Balances Around First Extractor
Balance on A :
100 (0.50) = mA2 + m1 ( 0.275)
Substitute m1
mA2 = 10.1 kg acetone
Balance on M :
100 = mM2 + m1 xM1
Substitute m1 , xM1
mM2 = 2.3 kg MIBK
• Balance on W
100 (0.500) = mW2 + m1 (0.725 – xM1)
Substitute m1 , xM1
mw2 = 42.6 kg water
• The remaining unknowns m5 , mA6 , mM6 , mW6
may be obtained through overall balance on the
whole process or the distillation column. In either
case, we have only three independent equations
and 4 unknowns and hence 1 degree of freedom.
are stopped at that point.
Recycle Streams
• One situation involving recycle streams is
chemical reactions in reactors
• A chemical reaction rarely proceeds to completion.
• It is therefore necessary to separate any unreacted
material and recycle it to the inlet of the reactor
• This way, the cost of fresh reactant is less and the
product being more pure is sold at a higher price.
• Recycling doesn’t disturb the overall process
material balance
Other reasons for using recycle
in chemical Processes
• Recovery of catalyst and recycling it to
reactor
• Increasing rate of filtration by diluting
concentrated slurry with a recycled part of
the filtrate
• Controlling a temperature of a reactor
where a highly exothermic reaction takes
place (see textbook for details P.110)
Example 4.5-1
Given
•
•
•
•
•
Dehumidifcation process
Mole fraction of fresh humid air
Mole fraction of blended stream
Mole fraction of dehumidified air
Basis: 100 moles of dehumidified air
Required
• Moles of fresh feed (n1)?
• Moles of water condensed (n3)?
• Moles of recycled dehumidified
air (n5)?
Solution
1.
Process: Batch, no reaction
2.
Basis: 100 moles of dehumidified air
3. Degree-of-freedom analysis
• Overall System
2 unknowns (n1 , n3) - 2 independent equations
= 0 degrees of freedom
•
Mixing point
3 unknowns (n1, n5 , n2)- 2 independent equations =
1 degree of freedom
•
Process
3 unknowns(n2 ,n3 , n4 ) -2 independent equations = 1
degree of freedom
•Splitting point
2 unknowns (n4, n5 ) - 2 independent equations = 0
degree of freedom
Be Careful now! The splitting point does not
follow the rule as shown above. Why?
The reason is that only one independent balance
equation can be written for any splitting point in any
mass balance problem, because the streams entering
and leaving this point have the same composition.
Therefore, the correct degree of freedom analysis is:
2 unknowns (n4, n5 ) -1 independent equation=1
• Overall balance on dry air
n1 (0.96) = 100 (0.983)
n1 = 102.4 mol fresh feed
• Overall mole balance
n1 = n3 + 100
Substitute n1
n3 = 102.4-100 = 2.4 mol H2O
• Balances on the mixing point (since the problem
Dry air
n1 (0.960) + n5 (0.983) = n2 (0.977)
Total balance
n1 + n5 = n2
Solve simultaneously after substituting n1
n 5 = 290 mol recycled , n2 = 392.5 mol
Example 4.5-2
Given:
•
•
•
•
•
4500 kg K2 Cr2 O4 solution
xK2Cr2O4 = 0.33 (feed to evaporator)
xK2Cr2O4 = 0.494(feed to crystallizer)
Recycle stream (filtrate) :
36.4% K2 Cr2 O4
Filter cake: 95% by mass crystals
+ 36.4% K2Cr2 O4 solution
Required
Case1:
.
•Rate of evaporation (m2)?
.
.
.
•Rate of production of crystals (m4)?
.
•Feed rate to evaporator (m1)?
.
•Feed rate to crystallizer (m3)?
.
•Recycle ratio (m6 / 4500)?
Case 2:
If the filtrate was not recycled, calculate
the production rate of crystals. What
are the benefits and costs of recycling?
Solution
1.
.
Process: Continuous, non-reactive
2.
Basis: 4500 kg/h
3.
Degree-of- freedom Analysis
•
Overall System
.
. .
3 unknowns (m4 , m5 , m2) - 2 independent equations - 1 more
equation
.
. = 0.95 (m. + m
m
4
4
5) =0
•
Mixing point
.
.
Evaporator.
.
3 unknowns (m1 , x1 , m6 ) - 2 independent equations = 1 degree of
freedom
•
.
4 unknowns (m1 , x1 ,, m2 , m3) – 2 independent equations =
= 2 degrees of freedom
• Crystallizer and Filter
.
. .. .
.
4 unknowns (m3, m4, m5, m6) – 2
independent equations -1 more equation
.
.
.
m4 = 0.95 (m4 + m5)=1
 remaining degrees of freedom = 1
Solution
Overall System
• Process specification
.
.
.
m4 = 0.95 (m4 + m5)
.
.
m5 = 0.05263 m4
• Overall chromate balance
.
.
4500 (0.333) = m4 + 0.364 m5
Solving simultaneously both equations:
.
m4 = 1470 kg crystals/h
.
m5 = 77.5 kg entrained solution/h
Overall mass balance
.
.
.
4500 = m2 + m4 + m5
.
.
Substitute m4 , m5
.
m2 = 2950 kg H2O vapor
Crystallizer Total Mass Balance
.
.
m3 = m 4 + m5 + m 6
.
.
.
.
Substitute
m
,
m
.
4
5
.
.
m3 = 1550 + m6
Water Balance Around Crystallizer
.
.
.
0.506 m3 = 0.636 m5 + 0.636 m6
.
Substitute m5
.
.
m3 = 97.4 + 1.257 m6
Solving both equations
.
m3 = 7200 kg/h crystallizer feed
.
m6 = 5650 kg/h
.
Recycle ratio = m6 /4500 = 5650/4500
= 1.26 kg recycle/ kg fresh feed
• Case 2: No recycle stream.
Degree-of-freedom Analysis
• Overall system has 1 degree-of-freedom
• Evaporator has zero degrees-of-freedom
• Crystallizer has 1 degree-of-freedom
Total mass balance and either water or
.
chromate balance will yield the values of m1
.
.
and m2. Once m2 is known, balances on the
.
. .
.
.
crystallizer are solved for m3 , m4 , m5
.
• m3 = 622 kg crystals/h.
.
• With recycle m3 = 1470 kg crystals/h
• Mass flow rate of discarded filtrate
.
(m5) = 2380 kg/h
2380 X 0.364 = 866 kg/h potassium chromate
wasted
• Recycling enables to recover most of the salt and
selling it. This would outweigh the cost of a pump
pumping power and recycle pipe after some time.
Bypass
A fraction of a feed is diverted around a process unit
and combined with the product. By varying the
fraction of feed bypassed, the product properties are
varied. Bypass and recycle streams balance
calculations are exactly similar.
Mass Balances On Reactive
Systems
Basic Concepts
• Stoichiometric equation: is a balanced
chemical reaction equation
• Stoichiometric coefficients: are the numbers
appearing by molecular species in a
balanced chemical reaction equation
• Stoichiometric ratio: is the ratio of the
stoichiometric coefficients of two molecular
species in a balanced chemical reaction
equation
Example
2 SO2 + O2→2SO3
• Stoichiometric coefficient of SO2 and
O2 are 2 and 1 respectively
• Stoichiometric ratio (SO2 : O2) = 2:1
• Limiting Reactant:
Is the reactant present in less than it
stoichiometric ratio in the feed relative to
every other reactant
•It would run out first if a reaction proceeds
to completion
• Excess Reactant :
Other reactants that are not limiting
• If all reactants are present in stoichiometric
proportion, then there is no limiting reactant
• Fractional Conversion:
Applies to a limiting reactant
It is the ratio :
f = (moles reacted)/ moles fed
• I f 100 moles of a reactant are fed and %
conversion is 80% then 80 moles react and
20 moles remain
• Fractional Excess of reactant:
for reactant A : [(nA)feed - (nA) stoich]
(nA) stoich
Examples:
1. For the reaction:
2 SO2 + O2 →2SO3
If 100 mol of O2 and less than 200
moles of SO2 are present in the feed,
then SO2 is the limiting reactant
2. For the reaction
C2 H2 + 2H2→ C2 H6
If the feed consists of 20 kmol/h
C2 H2 and 50 kmol/h H2 then:
A.
B.
C.
D.
Stoichiometric ratio (H2: C2 H2 ) = 2:1
Feed ratio (H2: C2 H2 ) = 2.5:1
H2 is in excess, C2 H2 is limiting
% excess H2 = [(50 -40)/ 40] X 100 = 25
3.
For the same reaction in the past example:
If feed to a batch reactor consists of 20
kmol C2 H2 , 50 kmol H2 , 50 kmol C2 H6
and 30 kmol H2 reacted. Find the
number of moles of each species at the end of the
reaction?
•
Number of reacted moles of C2 H2 = 15 kmol
(from stoichiometric ratio)
•
Number of remaining moles of H2 = 20 kmol
( 50 -30)
•
Number of remaining moles of C2 H2 = 5 kmol
(20 -15)
• Number of moles of C2 H6 formed
= 15 ( from reaction -by stoichiometry)
+ 50 ( in the original feed)
= 65 kmol
Example 4.6-1
Reaction Stoichiometry
Given
For the reaction:
C3 H6 + NH3+ 1.5 O2 →C3 H3N +3H2O
• Feed (100 mol): 10 mole% C3 H6 , 78
mole% Air, 12 mole% NH3
• Fractional conversion of limiting reactant :
30%
Required
• Determine which reactant is limiting
• % excess other reactants?
• Molar amounts of all product gas
components?
Solution
• Basis : 100 mol feed (given)
• (nNH3/ nC3H6)feed = 12/10 = 1.2
• (nNH3/ nC3H6)stoich = 1/1 = 1
• 1.2 > 1  NH3 is in excess
• (nO2/ nC3H6)feed =16.4 / 10 = 1.64
• (nO2/ nC3H6)stoich = 1.5 /1 = 1.5
• 1.64 > 1.5  O2 is in excess
Propylene is limiting
• (nNH3)stoich = 10 mol (according to
stoichiometry)
• (nO2)stoich = 1.5 X 10 = 15 mol
• % excess NH3 = (12-10)/10 X 100 =
= 20 %
• % excess O2
= [(78 X 0.21)- 15]/ 15 X 100 = 9.3%
•
•
•
•
•
•
•
(nNH3)react = 3 mol
(nO2)react = 1.5 X3 = 4.5 mol
(nNH3)remain = 12-3 = 9 mol
(nO2)remain = 16.4 - 4.5 = 11.9 mol
(nN2) = 78 X0.79 = 61.6 mol
(nH2O)formed = 3 X 3 =9 mol
(nC3H3N)formed = 3 mol
Multiple Reaction Systems
• Reactions are not usually single, but
accompanied by side reactions giving
undesired products
• Side reactions represent economic loss
for example
C2 H4 + 0.5 O2 →C2 H4 O
C2 H4 + 3 O2 →2 CO2 + 2 H2 O
• The degree to which a side reaction is
important can be described by yield and
selectivity
• Yield
= moles of desired product formed
moles formed with no side reactions
and if limiting reactant was
completely converted
• Selectivity
= moles of desired product formed
moles of undesired product formed
• High values of yield and selectivity mean
that side reactions are suppressed
Example 4.6-3
Yield and Selectivity in
Dehydrogenation Reactor
Given
• C2 H6 → C2 H4 +H2
• C2 H6 + H2→ 2 CH4
Feed: 85 mole%, ethane, 15 % inerts
Conversion of ethane: 0.501
Fractional yield ethylene : 0.471
Required
• Molar composition of product gas?
• Selectivity of ethylene to methane ?
Solution
• Fractional conversion of ethane = 0.501
n1 (remaining ethane)
= 85 - 0.501X 85 = 42.4 mol
• Ethylene yield = 0.471
Stoichiometric no. of moles of ethylene = 85 mol
n2 ( moles formed of ethylene)
= 0.471 X 85 = 40.0 mol
• Ethane consumed by the second reaction
= 0.501 X 85 - 40 = 2.6 mol
• Hydrogen consumed by the second reaction
= 2.6 mol
• Hydrogen formed by the first reaction = 40 mol
• n3 (Hydrogen leaving in the product gas)
= 40-2.6 = 37.4 mol
• n4 (Moles of methane formed)
= 2 X 2.6 = 5.2 mol
• n5 (Moles of inerts formed )= 15 mol
• Composition of product gas
Total moles of product gas = 140 mol
 30.3 mol% C2 H6
28.6 mol% C2 H4
26.7 mol% H2
3.7 mol% CH4
10.7 mol% I
Independent Species and
Independent Reactions
• If two molecular species occur in the same
ratio to each other everywhere in a process
and this ratio is needed in calculations,
balances on those species will not be
independent equations
• If two atomic species occur in the same ratio
everywhere in a process, balances on those
species will not be independent
• Chemical reactions are independent if the
stoichiometric equation of any one of them
cannot be obtained by adding and
subtracting multiples of the stoichiometric
equations of the others
Example: the following reactions are not all
independent
[1] A→2B
[2] B→C
[3] A→2C
[3] = [1] + 2 X [2]
Possible Balances For A
Reactive System
1. Molecular species balances
• form: Input + generation = output +
consumption
• Degree- of - freedom analysis:
No. of unknown labeled variables
+ No. independent chemical reactions
- No. independent molecular
species balances
- No. other equations relating unknown
variables = No. degrees of freedom
2. Atomic species balances
• form: Input = output
• Degree-of-freedom analysis:
No. unknown labeled variables
- No. independent atomic species balances
- No. molecular balances on independent
nonreactive species
- No. other equations relating unknown
variables =No. degrees of freedom
Approaches to Material
Balances on reactive systems
• Atomic species balances are the most used
especially if more than one reaction is
involved
• Molecular species balances should be used
for simple systems
Example (Text P.125+126+129)
• Consider the dehydrogenation of ethane as
shown in the figure below
Required:
Obtain the values of the molar flow rates of ethane
and ethylene in the product gas.
Solution:
Basis -given 100 kmol C2H6/min
Degree of freedom analysis (molecular species):
No. unknown variables [2]
+ No. independent chemical reactions [1]
-No. independent molecular species balances [3] = 0
Use Stoichiometric equations to find the generation
or consumption
H2 Balance:
generation = output
GenH2 = 40 kmol /min
C2 H6 Balance:
Input = output + consumption
100 = n1+ 40
 n1 = 60 kmol /min
C2 H4 Balance:
generation = output
 n2 = 40 kmol/min
Degree of freedom analysis (Atomic Species
Balances)
No. unknown variables [2]
- No. independent atomic species balances [2]
- No. molecular balances on independent
nonreactive species [0]
-No. other equations relating unknown
variables [0] = 0
No consumption or generation terms appear
in atomic species balances
 Input = Output
C Balance:
100 kmol C2 H6 /min  2 kmol C/kmol C2 H6 =
n1 kmol C2 H6 /min  2 kmol C/kmol C2 H6 +
n2 kmol C2 H4 /min  2 kmol C/kmol C2 H4
 n1 + n2 = 100
(1)
H Balance
100 kmol C2 H6 /min  6 kmol H /kmol C2 H6= 40
kmol H2/min  2 kmol H /kmol H2+ n1 C2 H6 kmol
/min  6 kmol H /kmol C2 H6 + n2 kmol C2 H4 /min 
4kmol H/ kmol C2 H6
600 kmol H/min = 80 kmol H/min + 6 n1 +4 n2
520 = 6 n1 +4 n2
Solving eqns 1&2 simultaneously:
n1= 60 kmol C2 H6 /min
n2 = 40 kmol C2 H4 /min
(2)
Example 4.7-1
Incomplete combustion of
methane
CH4 +3/2 O2 → CO +2H2O
CH4 + 2O2 →CO2 +2H2O
Feed : 7.8 CH4 mol%, 19.4%O2 , 72.8% N2
Conversion methane: 90%
Gas leaving reactor:8 mol CO2 / mol CO
Required
• Degree-of-freedom Analysis
• Calculate Molar Composition of the
product gas
• Use: atomic species balances
Atomic Species Balances:
5 unknown variables - 3 independent atomic
species balances -1 nonreactive molecular
species balance - 1 specified methane
conversion = 0 degrees of freedom
nCH4 = 0.1 X7.80 = 0.78 mol CH4
N2 Balance, input = output
nN2 = 72.8 mol N2
Now determine nCO, nH2O , nO2
Atomic Species Balances
General form : input = output
then followed by oxygen balance
C Balance
7.8 mol CH4 X (1mol C/1mol CH4) =
= 0.78 mol CH4 X (1 mol C/ 1 mol CH4)
+ nCO (mol CO)X (1 mol C/1 mol CO)
+ 8 nCO (mol CO2) X (1 mol C/1 mol CO2)
 nCO = 0.78 mol CO
 nCO2 = 8 nCO = 8 X 0.78 mol CO2 =
= 6.24 mol CO2
H Balance
7.8 mol CH4 (4 / 1) = 0.78 (4/1)
+ nH2O (2/1)
nH2O = 14 mol H2O
O Balance
19.4 mol O2 (2/1) = nO2 (2/1) +0.78 (1/1)
+ 6.24 (2/1) + 14 (1/1)
 nO2 = 5.75 mol O2
•The molar composition of the gas:
0.78 % CH4, 0.78 % CO, 14% H2O,
5.7% O2 , and 72.5 % N2
Product Separation and Recycle
1. Overall Conversion:
(reactant input to process - reactant output
from process) / reactant input to process
2. Single -pass Conversion:
(reactant input reactor -reactant output from
reactor)/ reactant input to reactor
Example
Consider the reaction A →B
• The overall conversion of A is:
[(75 mol A/ min)in - (0 mol/min)out /(75mol A/min)in ] X
100 = 100%
• The single-pass conversion is :
[(100 mol A/ min)in - (25 mol/min)out /(100mol A/min)in
] X 100 = 75%
•
Reason for recycle is evident, the overall
conversion reached 100% because of perfect
separation. If separation was not not perfect,
overall conversion will still be higher than singlepass conversion
Example 4.7-2
Dehydrogenation of Propane
Required
• Composition of product
mole fractions of all components
• Recycle ratio (moles recycled/ moles fresh
feed)
[n9 + n10 ]/100 mol
• Single-pass conversion
[(n1 - n3)/n1] X 100
Solution
1. Process: Reactive, single reaction
2. Basis : 100 mol fresh feed
3. Degree-of- freedom Analysis
Overall system
3 unknown variables(n6, n7, n8) - 2 atomic balances
( C and H) -1 additional relation (95 % overall
propane conversion) = 0 degrees of freedom
Note : Analysis done based on atomic
species balance
Separator: (physical system, do analysis
in the usual way)
5 unknown variables (n3, n4 , n5 , n9 , n10)
(n6 through n8 are known from balances
on the overall system) - 3 balances
- 2 additional relations (n6 = 0.00555n3 , n10 =
0.05 n7) = 0 degrees of freedom
Mixing point: ( physical system, do
analysis in the usual way - be careful!)
• No need to analyze the reactor because no
more unknowns exist on the chart
First obtain n6 (mol C3 H8) from the
Overall conversion:
n6 = 0.05 (100 mol) = 5 mol C3 H8
Second write two balances on carbon and
hydrogen
• Overall C balance (involving only one
unknown)
100 mol C3 H8 (3 mol C/ mol C3 H8) =
n6 mol C3 H8 (3 mol C/ mol C3 H8)
+ n7 mol C3 H6 (3 mol C/ mol C3 H6)
Substitute n6 = 5 mol
n7 = 95 mol C3 H6
• Overall H balance
(100) (8) = n6 (8) + n7 (6) +n8 (2)
• Substitute n6 = 5 mol, n7 = 95 mol
n8 = 95 mol H2
Product composition :
2.6 mol% C3 H8 , 48.7 mol % C3 H6 ,
48.7 mol% H2
Separator
n6 = 0.00555 n3 , n6 = 5 mol,
n3 = 900 mol C3 H8
n10 = 0.0500 n7 , n7 = 95 mol
 n10 = 4.75 mol C2 H6
n3 = n 6 + n 9 ,
Substitute n3 = 900 mol, n6 = 5 mol
n9 = 895 mol
100 mol + n9 = n1
Substitute n9 = 895 mol
n1 = 995 mol C3 H8
Now calculate recycle ratio and
single-pass conversion
• Recycle ratio =(n9 +n10)/ 100 =9.00
• Single-pass conversion =
= (n1 - n3)/n1 X 100% = 9.6 %
Means of Achieving High
Overall Conversions in
Reactors
1. Design the reactor to yield a low-single pass
conversion, followed by separation of
unreacted material and recycling it.
• This scheme leads to small volume of the
reactor and a decrease in cost.
• The savings in reactor cost may be overcome
by the costs of the separation process unit and
the pump, pipes and fittings in the recycle
line.
2. Design the reactor to yield a high singlepass conversion
• This scheme leads to much larger reactor
volume than the previous one
• No separation unit and no recycling
The option that should be taken is
decided by detailed economic analysis
Purging
• If some material entering in the feed or
produced in a reaction, remains fully in a
recycle stream, then it keeps accumulating
and steady state is not achieved.
• To prevent this accumulation, part of the
recycle stream must be removed from the
process as a purge stream
Example 4.7-3
Recycle and Purge in the
synthesis of Methanol
Given
CO2 +3H2→CH3OH +H2O
• Fresh feed to process : H2 , CO2 ,
0.4 mole % inerts (I)
• Condenser : water and methanol
removed as liquids
• Unreacted materials + inerts: recycled
to reactor
• Purge stream taken from recycle
• Feed to reactor: 28 mole% CO2, 70 mole%
H2 , 2 mole% inerts
• Single-pass conversion of hydrogen:
60 %
Required
Molar flow rate and molar composition
of:
• Fresh feed
• Total feed to reactor
• Recycle stream
• Purge stream
Solution
1. Basis: Neglect given basis 155 kmol CH3OH/hTake 100 mol feed to reactor (combined feed)
since composition is known-Scale the answer at
the end (155/n3)
2. Degree-of-freedom Analysis
i.
Overall system:
7 unknowns
(no , xoc , n3, n4, np , x5C , x5H)
+ 1 reaction
- 5 independent balances
(CO2 , H2 , I, CH3OH,H2 O)
= 3 degrees of freedom
ii.
Recycle -fresh feed mixing point
5 unknowns (no , xoc , nT , x5c , x5H)
- 3 independent balances
(CO2 , H2 , I)
= 2 degrees of freedom
iii. Reactor
4 unknowns (n1, n2 , n3 , n4 )
+1 reaction
- 4 independent balances
(CO2 , H2 , CH3 OH, H2O)
- 1 single pass conversion
= 0 degrees of freedom
iv. Condenser (considering n1, n2, n3, n4 are
known from balances on reactor)
3 unknowns (n5 , x5C , x5H)
- 3 independent balances
= 0 degrees of freedom
v.
Purge - recycle splitting point (considering
n5 , x5C , x5H are known from balances on
condenser)
2 unknowns (nr , np)
- 1 independent balance = 1 degree of freedom
1. Reactor Analysis:
Using single-pass conversion of H2: 60%
 n2 = 0.4 (70 mol H2 fed) = 28 mol H2
H2 balance:
Input + generation = output + consumption
generation = 0
consumption = input - output
= 70 - 28 = 42 mol H2 consumed
CO2 Balance:
output = input - consumption
= 28 - 42/3 (from stoichiometry)
= 14 mol CO2
CH3 OH Balance:
output = generation
n3 = 42 /3 = 14 mol CH3 OH
H2 O Balance:
output = generation
n4 = 42/3 = 14 mol H2O
2. Condenser Analysis
input = output
Total Mole Balance:
n1 +n2 +n3 + n4 +2 mol = n3 +n4 +n5
Substitute n2 = 28 mol, n1 = n3= n4 = 14 mol
n5 = 44 mol
CO2 Balance:
n1 = n5 x5C
Substitute n1 = 14 mol, n5 = 44 mol
 x5C = 0.3182 mol
H2 Balance:
n2 = n5 x5C
Substitute n2 = 28 mol, n5 = 44 mol
x5H = 0.6364 mol CO2/mol
xI = 1-x5C -x5H = 0.04545 mol I/mol
3.Mixing point Balances:
Total Mole Balance:
n0 + nr = 100 mol
I balance:
no(0.00400) +nr (0.04545) =2.0
Solving the two equations simultaneously,
then
no = 61.4 mol, nr = 38.6 mol
CO2 Balance:
nO xOC+ nr x5C = 28
Substitute no = 61.4 mol, nr = 38.6 mol,
x5C = 0.3182 mol CO2 /mol
xOC = 0.256 mol CO2 /mol
xOH = (1-xOC-xOI) =0.740 mol H2 /mol
4. Recycle -Purge Splitting Point
Total mole balance:
n5 = n r + n P
Substitute n5 = 44 mol, nr = 38.6 mol
np = 5.4 mol
5. Flowchart Scaling
Scale factor = 155 kmol CH3OH/h
14 mol CH3 OH
= 11.1 kmol /h
mol
 Fresh feed = 61.4 X 11.1 = 681 kmol/h
Feed to reactor = 100 X 11.1 = 1110 kmol/h
Recycle = 38.6 X 11.1 = 428 kmol/h
Purge = 5.4 X 11.1 = 59.9 kmol/h
Combustion (P.142)
• Combustion is rapid reaction of fuel with
oxygen.
• Heat energy resulting from combustion is
used to generate electricity.
• Products of combustion are CO2 only (if
combustion is complete) or CO & CO2 ( for
partial combustion), H2O, SO2 (if any
sulfur exists in the fuel)
• Composition of the product gas is either on
a wet basis (mole % of components
including H2O). Composition could also be
done on dry basis (mole % of components
without H2O)
• Composition on wet basis could be
converted into dry basis or the reverse by
assuming 100 mol for the given composition
and calculating the desired basis
• Read Example 4.8-1 P. 143-144
Theoretical and Excess Air
• In combustion reactions, oxygen or air is
usually the excess reactant because they are
less expensive than fuel
• Theoretical oxygen: the moles or molar flow
rate of O2 needed for complete combustion
of all fuel fed to the reactor assuming all
carbon is oxidised into CO2 and Hydrogen
is oxidised into H2O
• Theoretical air: the quantity of air in moles
that the conatains the theoretical oxygen
• Percent Excess Air:
(moles air)fed - (moles air)theroretical X 100
(moles air)theoretical
• If fuel flow rate and stoichiometric equations
for complete combustion of fuel are known,
then theoretical oxygen and theoretical air
feed rates can be calculated. If actual feed
rate is also known then the percent excess
air can be calculated
• If theoretical oxygen or air feed rate ,
together with the % excess air is given, then
the actual feed rate may be calculated. If
50% excess air is supplied, for example, then
(moles air)fed = 1.5 (moles air)theoretical
Example 4.8-3 Combustion of
Ethane
Given:
• Fuel: Ethane, conversion 90%
• Air : 50% excess
• CO: 25% reacted ethane
• CO2: 75% reacted ethane
C2H6 + (7/2) O2→2CO2 + 3H2O
C2 H6 + (5/2) O2 →2CO +3H2O
Solution
Basis : 100 mol C2 H6
Degree-of-Freedom Analysis
• 7 unknowns
(no, n1, n2, n3 , n4 , n5 , n6 , n7)
• 3 atomic balances (C,H, O)
• 1 N2 balance
• 1 excess air specification
• 1 ethane conversion specification
• 1 CO/CO2 ratio specification
• (nO2)theroretical =
100 mol C2H6 X 3.5 (mol O2 / mol C2 H6)
= 350 mol O2
0.21 no = 1.50(350 mol O2)
no = 2500 mol air fed
• 90 % Ethane conversion
n1 ( unreacted ethane) =
= 0.1 (100 mol C2 H6) = 10 mol C2H6
reacted ethane = 90 mol
• n4 (mol CO formed)
= 0.25 X 90 mol C2 H6 X (2 mol CO/1mol C2 H6)
= 45 mol CO
• Nitrogen balance
input = output
2500 X 0.79 = n3 = 1975 mol
• Atomic carbon balance: input = output
100 mol C2H6 X (2 mol C/ 1 mol C2 H6 )
= n1 mol C2H6 X (2 mol C/ 1 mol C2 H6)
+ n4 mol CO X (1mol C/ 1mol CO)
+ n5 mol CO2 X (1mol C/ 1mol CO2)
Substitute n1 = 10 mol, n4 = 45 mol
 n5 = 135 mol CO2
• Atomic Hydrogen balance:
input = output
100 mol C2H6 X (6 mol H/ 1 mol C2 H6 )
= 10 mol C2H6 X (6 mol H/ 1 mol C2 H6)
+ n6 mol H2O X (2mol H/ 1mol H2O)
n6 = 270 mol H2O
• Atomic Oxygen balance:
1.5 X350 mol O2 X (2 mol O/ 1 mol O2)
= n2 mol O2 X (2 mol O/ 1 mol O2)
+ 45 mol CO X (1mol O/ 1mol CO)
+ 135 mol CO2 X (2mol O/ 1mol CO2)
+ 270 mol H2O X (1mol O/ 1mol H2O)
n2 = 232 mol O2
• The analysis of stack gas is:
n1 = 10 mol C2 H6
n2 = 232 mol O2
n3 = 1974 mol N2
n4 = 45 mol CO
n5 = 135 mol CO2
2396 mol dry gas
+ n6 = 270 mol H2O
2666 mol total
The stack gas composition on a dry
basis is:
y1= 10 mol C2 H6 = 0.00417 mol C2 H6 / mol
2396 mol dry gas
y2 = 232 mol O2 = 0.0970 mol O2 / mol
2396 mol dry gas
y3 = 1974 mol N2 = 0.824 mol N2 / mol
2396 mol dry gas
y4 = 45 mol CO = 0.019 mol CO /mol
2396 mol dry gas
y5= 135 mol CO2 = 0.0563 mol CO2 / mol
2396 mol dry gas
• Mole ratio of water to dry stack gas
270 mol H2O
= 0.113 mol H2O
2396 mol dry stack gas mol dry gas
Energy Balances
• For designing a process, an engineer has to
account for the energy flowing into and out
of process unit and therefore determine the
energy requirement of a process
• Examples: cooling of CSTR with highly
exothermic reaction, pumping, burning of
fuel to produce energy,
Forms of Energy of a System
• Kinetic Energy
• Potential Energy
• Internal Energy: due to motion of
molecules, atoms, electrons, interaction
between molecules, interaction between
atoms, interaction between electrons…….
•
Energy transferred from or to a system:
A. Work done by a system on surroundings
(example: expansion of a gas to move a
piston), or work done by surroundings
on a system ( example: mixing of a liquid
in a tank) or work due to electric energy
or due to magnetic field
B. Heat transferred from a system to
surroundings or from surroundings to a
system (examples: condensers and
evaporators)
Differences Between Energy
and Mass Balances
• Energy balance equation could be written
as:
Energy input = Energy output + Energy
accumulation
• Note: that there is no generation or
consumption terms like total mass balance
since energy can neither be destroyed nor
created
• Note: Unlike material balance, there is no
component energy balances, only one energy
balance equation involving all streams and
all components
Energy Balances On A Closed
System
• A system is closed when mass doesn’t cross
the system boundaries during the period of
time covered by the energy balance
(example: a closed tank, a batch reactor)
• The integral energy balance equation on a
closed system between two instants of time
is:
∆U + ∆Ek + ∆Ep = Q-W
• ∆ signifies change between final and initial
state
• U = Internal Energy
• Ek = kinetic Energy
• Ep = Potential Energy
• W = work of different forms as mentioned
before
• Q = heat transferred from the system to
surroundings or from the surroundings to
the system
Energy Balances On Open
• An open system has mass crossing its
boundaries. (examples: continuous
reactors, distillation units, extractors…..)
• The steady- state energy balance equation
takes the form:
Rate of Energy Input =
Rate of Energy Output
.
.
.
. .
∆ H+ ∆Ek + ∆Ep = Q-Ws
• Ws = shaft work due to the presence of a pump,
compressor or turbine
• ∆H = change in enthalpy between the outlet and
inlet of a system
Note that the work for an open system
has two components :
• Shaft work (Ws) appearing in the final
equation
• Flow work (Wfl) used to push streams in
and out from a system and this is
implicitly contained in the enthalpy
.
•∆H =
. ^
. ^
( ∑mj Hj )output streams - ( ∑mj Hj )input streams
.
•∆Ek =
. 2
2
.
( ∑mj uj /2 )output streams -( ∑mj uj /2)input streams
.
.
•∆Ep =
.
.
( ∑mj g zj )output streams -( ∑mj g zj )input streams
So finally the energy balance equation could be
written as :
.
.
.
. .
∆H + ∆ Ek + ∆Ep = Q-Ws
Two simplifications:
1) If the process has a single input stream and a
single
. output stream having the same mass flow rate
(m), then for example,
.
. ^ ^
∆H = m (Hout-Hin)
2) For the same value of any of the energy forms for
all input and output streams, the corresponding term
to this form drops from the energy balance equation
.
.
^
.
∆H = H (∑ mj output streams - ∑mj input streams)
.
At steady state, ∆H = 0
Example 7.4-2
Given:
Steam turbine, mass flow rate = 500 kg/hr
Inlet conditions
• Steam pressure and temperature : 44 atm,
and 450 oC
• Velocity = 60 m/s
Outlet conditions
Exit point 5 m below inlet
Steam pressure: atmospheric
Velocity = 360 m/s
Shaft work = 70 kW, heat loss = 104 kcal/h
steam
steam
1 atm
500 kg/hr
saturated
44 atm,
360m/s
450 o C
60m/s
Q = -104 kcal/ h
Ws = 70 kW
Required:
Change in specific enthalpy of steam between
inlet and outlet
Solution (P.324&325)
Use the Energy balance equation on steam
turbine
• Convert mass flow rate to kg/s:
500 kg/h * (1/3600 s/h) = 0.139 kg/s
• Convert heat loss to kW
-104 kcal/h*(1J/0.2390*103)*(1h/3600s)*(1W/103kW)
. in kW
• Calculate rate of change of kinetic energy
0.139 kg/s*(3602-602)/2 J/kg *(1kW/103W)
• Calculate rate of change of potential energy in kW
0.139 kg/s*[9.81*(-5)] J/kg*(1kW/103 W)
• Substitute shaft work (positive value, as is) in the
energy balance equation
• Substitute heat loss ( as negative of the given
value) in the energy balance equation
.
. .
.
.
∆H = Q-Ws- ∆Ek- ∆Ep = -90.3 kW
.
..
^
^
• Finally, calculate H2-H1 by dividing ∆H by m
.
.
^ ^
• H2-H1 =∆H/m = -90.3 kW/0.139 kg/s =
-650 kJ/kg
Tables of Thermodynamic Data
State
• The state of any material is characterized
by temperature, pressure, phase
Values of enthalpy and internal energy
• The absolute value of internal energy or
enthalpy is impossible to determine.
• The change in enthalpy or internal energy
corresponding to a specified change of state
could be measured from the energy balance
equation.
^
^
• The values of U or H listed in
thermodynamic tables are changes
measured relative to a reference state
• A value of zero may be assigned to the
^
^
reference state, then any ∆H = H-0=H ^
• Tables are constructed for different materials to
read the enthalpy and the internal energy values.
(for example, steam tables)
• Reference states of a table may be known, or
unknown, but this is not important since we are
always interested in calculating some the changes
in enthalpy or internal energy between tabulated
states.
• Care must be taken to avoid using different
property tables for the same material (because
enthalpy or internal energy might be based on
different reference states)
Example 7.5-1
• Use of Tabulated Enthalpy Data for
saturated methyl chloride (P. 326 & 327)
Steam Tables
• As water (subcooled liquid) is heated at constant
pressure, it starts to evaporate and after a while, it
boils.
• Boiling occurs at different temps. and press. These
temperatures and pressures are termed saturation
conditions.
• At the saturation conditions, water may be a
saturated liquid, saturated vapor, or a mixture of
saturated vapor and a saturated liquid.
• At temperatures above the saturation temperature
but at the same saturation pressure, water exists as a
superheated vapor.
• The properties of water (saturation temp,
saturation press, superheating temp
specific enthalpy, specific internal energy
specific volume of liquid water and water
vapor) are tabulated in saturated or
superheated steam tables
Example 7.5-2
Use of steam tables (P.328 and 329)
Example 7.5-3
Given:
• Steam mass rate = 2000 kg/h
•Inlet
Steam pressure : 10 bar a and superheat
190 oC (steam temperature is above the
saturation temperature at the given pressure
by 190 oC)
• Q= 0 (turbine operation adiabatic)
• Outlet
Saturated steam pressure: 1 bar
• Neglect P.E. & k .E. changes
Required
.
• Ws in kW
Solution
• The
equation simplifies to:
. energy. balance
. ^ -^
Ws = -∆ H = m (H
out Hin)
• First convert mass flow rate to kg/s2000/3600
kg/s
• Second obtain the inlet steam enthalpy
from superheated steam table B.7:
Saturation temp. at 10 bar = 180 oC (from
saturated steam tables)
Superheated temp. at 10 bar = 180 +190
= 370 oC
^
 Hin (10 bar, 370 oC)= 3201 kJ/kg
(by interpolation)
• Details of interpolation at P = 10 bar
^ kJ/ kg
t oC
H
350
3159
370
x?
400
3264
370-350
50
x- 3159
3264-3159
x = (3264-3159)*(370-350)/50 + 3159 = 3201
kJ/kg
• Third obtain the outlet steam enthalpy
from saturated steam tables:
^
Hout (1 bar, saturated)= 2675 kJ/kg
.
.
W= -∆H = 2000 kg/3600 *(2675-3201)=
= 292 kW
Energy Balance Procedure
A. Draw and label a flow chart for the energy balance
problem
B. Include the labels in the chart :
1. Pressure of each stream
2. Temp of each stream
3. Phase of a stream (liquid, gas or solid)
4. Indicate if stream is pure or multicomponent
5. Specific enthalpy
6. Mass flow rates (material balances could be used
to obtain missing m.f.r. of streams)
7. Apply the energy balance equation and find the
unknown
Example 7.6-1
Energy Balance on one- component
Process
Given:
Inlet
• Two water streams entering a boiler:
feed stream1  120 kg/min, Temp.= 30oC
feed stream2  175 kg/min, Temp. = 65oC
Boiler pressure 17 bar (absolute)
Outlet
• Steam flowing in a 6-cm ID pipe
• Neglect kinetic energies of inlet streams
• Neglect potential energy change (a usual
assumption)
120 kg/min H2O (l)
30oC
H^= 125.7 kJ/kg
175 kg/min H2O (l)
65oC
H^= 271.9 kJ/kg
Boiler
295 kg/min H2O (v)
17 bar (saturated)
^H= 2793 kJ/kg
6-cm ID pipe
Required
.
• Q (kJ/min)?
Solution
• Obtain the mass flow rate of steam from a
simple material balance
• Use steam tables to determine the specific
enthalpies of liquid water ( read the
tabulated values neglecting the effect of
pressure), and the specific enthalpy of the
produced saturated steam
• Apply the energy balance equation
. .
.
.
.
Q- Ws = ∆ H + ∆Ek + ∆Ep
.
.
.
Q = ∆ H + ∆Ek
.
∆H = 295 kg/min*(2793 kJ/kg)
- 120 kg/min*(125.7 kJ/kg)
- 175 kg/min *(271.9 kJ/kg)
= 7.61*105 kJ/min
.
∆Ek
Must first obtain the velocity of steam in the
exit line
.
u(m/s)= V( volumetric flow rate of steam,
m3/s) / A (Area of pipe, m2)
Volumetric flow rate is obtained from the
given mass flow rate and density of steam.
Note that the density of steam is the inverse of
steam specific volume that could be found
from steam tables
Area of pipe = πR2
= 3.1416 *32cm2* (1m2/104cm2)
= 2.83*10-3m2
.
V = 295 kg/min *(1min/ 60s)*(0.1166 m3/ kg)
 u (m/s)= 295 kg/min *(1min/ 60s)*(0.1166
m3/ kg)/2.83*10-3 m2 = 202 m/s
.
.
∆Ek = m u2/2
= 295 kg/min*(2022/2) J/kg (1kJ/103J)
= 6.02 *103 kJ/min
.
.
.
 Q =∆ H + ∆Ek = 7.61*105+ 6.02*103kJ/min
= 7.67 *105 kJ/min
It is typical of energy balance problems on processes
with chemical reactions, phase changes or large
temperature changes to neglect kinetic energy
changes as they represent a small fraction of the total
energy requirement for the process.
Example 7.6-2
Two-component Process
Given:
Heating
Inlet
• Liquid:60 wt% ethane, 40%
n-butane
• 150 K, P = 5 bar
Outlet
• Same liquid
• 200 K, P = 5 bar
• Neglect potential and kinetic energy changes
• Use tabulated enthalpy data for ethane & n-butane
from Perry’s Chemical Engineer’s Handbook on P.
223 & P. 234
Required
• Heat input per kg of the
mixture?
1 kg /s (150 K, 5 bar)
1 kg /s (200 K, 5 bar)
0.6 kg /s ethane (l)
0.6 kg /s ethane (l)
0.4 kg /s butane (l)
0.4 kg /s butane (l)
Solution
• No material balance is necessary since
there is only one stream
.
.
.
.
 Q = ∆ H = Hout - Hin
=[ 0.6 kg /s C2 H6* 434.5 kJ /kg
+ 0.4 kg /s C4 H10 * 130. 2 kJ /kg]
-[0.6 * 314.3 + 0.4*30] = 112 kJ/s
^
Q = 112 kJ/s / 1kg /s = 112 kJ /kg
Example 7.6-3
Simultaneous Material and Energy
Balances
Given:
Inlet
Saturated Steam 1 atm, 1150 kg /h
Superheated Steam 400oC, 1atm
Outlet
Superheated 300oC, 1atm
1150 kg /hr H2O (v)
1 atm, 100o C
^
H = 2676 kJ /kg
.
.
m2 kg /hr H2O (v)
m1 kg /hr H2O (v)
1 atm, 300o C
1 atm, 400o C
^ = 3074 kJ /kg
H
^ = 3278 kJ /kg
H
Required
• Mass flow rate of steam at 300oC?
• Volumetric flow rate of steam at
400oC?
Solution
•
•
•
•
Heat transfer, Shaft work = 0
Negligible change of K.E. & P.E.
Two unknowns
Must use both material and energy balance
Mass balance equation
.
.
1150 kg/h + m1 = m2
Energy balance equation
.
∆H=0
.
.
Hout = Hin
(1)
.
1150 kg/h * 2676 kJ /kg + m1*3278 kJ/kg
.
= m2* 3074 kJ /kg
(2)
Solving eqs. 1 & 2 simultaneously
.
m1 = 2240 kg/h
.
m2 = 3390 kg/h
Obtain the specific volume of superheated
steam at 400oC and 1 atm from table B.7
^
V = 3.11 m3/kg
• The volumetric flow rate of this steam is
2240 kg/h *3.11 m3/kg = 6980 m3/h
• If specific volume data were not available,
the ideal gas equation may be used to
calculate the density of steam.
Enthalpy Changes Based on
Inlet or Outlet Conditions
Until now, we considered reading enthalpies
from tables of thermodynamic data (see past
examples on steam turbines, heating ethane
and butane) and the reference state is that
used to generate the table
What if no available thermodynamic data
table exists for a particular species in the
process?
• Choose one of the inlet or outlet states as
^
the reference state for the species so that at
least one H may be set equal to zero.
• Calculate the enthalpy of each component
in each inlet or outlet stream relative to the
chosen reference state by using
hypothetical paths having constant
pressure, constant temperature or constant
pressure and temperature. This will be
explained in the next few slides.
Calculation of a Species
Enthalpy
• To perform the calculation of the enthalpy
of a species at a particular state (inlet or
outlet), choose any convenient path from
the reference state to the required state of
^ or ∆H
^ for that
the species and determine H
i
i
path.
• Paths chosen are either constant pressure
or constant temperature or both.
Constant Temperature Path
• For a solid or liquid species undergoing pressure
changes at constant temperature and volume:
^ = ^V ∆P
∆H
• For an ideal gas species undergoing pressure
changes at constant temperature and volume:
^ =0
∆H
• For a real gas ( at temperatures well below 0oC or
well above 1 atm), either use thermodynamic tables
or thermodynamic correlations.
Constant Pressure
Path:
For ideal gases
ˆH   T2 c (T) dT
T p
1
For real gases at
constant pressure
For solids or liquids
Constant temperature
and pressure (phase
Changes)
ˆ H  ˆ H v (latent heat of vaporizati on)
For example: change of
phase from liquid to
vapor or from vapor to
liquid usually at the
normal boiling point
General Procedure For Energy
Balance Calculations on Closed
and Open Systems (Text, P.361)
1. Draw and label fully a chart for the energy balance
problem as mentioned earlier
2. Write the appropriate form of the energy balance
equation (closed or open system) and delete any of
the terms that are either zero or negligible
3. Check if the values of the specific internal energy
or specific enthalpy at the inlet and the outlet for
all components are given or could be obtained
from information in the problem or could be read
from thermodynamic tables
• If not, choose one of the inlet or outlet conditions
for each species as its reference state
4. For a closed system, construct a table showing the
reference states chosen at the top and with
columns for initial and final masses or number of
moles of each components and with other columns
for the components’ final and initial specific
internal energies
• For an open system, a similar table is constructed
but with columns showing molar or mass flow
rates of components and their inlet and outlet
specific enthalpies
5. Calculate the values of the specific internal
energies or enthalpies corresponding to each
component by using hypothetical paths from the
reference state to a particular initial or final state
if the system is closed or from the reference state to
a particular inlet or outlet condition if the system
is open
6. Insert the calculated values of internal energies or
enthalpies in their appropriate places in the table
7. Calculate the changes in internal energies or the
changes in enthalpy based on the values in the
table
8. Calculate other terms in the energy
. . balance
equation that have not been neglected (i.e. Δ
K.E, Δ P.E, Ws, Q)
9. Solve the energy balance equation for the
required unknown term
Example 8.1-1
Energy Balance on a Condenser
Given
Condenser
Inlet:
Mixture (Ac (v), N2) 100 mol/s, 65oC, 1 atm,
known mol fraction
Outlet:
Mixture (Ac (v), N2) 36.45 mol/s, 20oC, 5
atm, known mol fraction
AC (l)  63.55 mol /s, 20oC, 5 atm
36.45 mol/s
100 mol/s
0.092 mol Ac(v) /mol
0.669 mol Ac(v) /mol
0.908 mol N2 /mol
0.331 mol N2 /mol
20oC, 5 atm
65oC, 1 atm
Condenser
63.55 mol AC(l) /s
20oC, 5 atm
Required:
Cooling rate?
Solution
1. No need for material balance
calculations
2. Application of energy Balance:
   H
Q
3. To calculate the enthalpy change, reference
states must be chosen for both Ac and N2:
• Table B.8 Lists specific enthalpies of N2
relative to N2(g, 25oC, 1 atm) [this the
reference state chosen for nitrogen]
• No tabulated enthalpy data for acetone, so
one of the process stream conditions and let
^
o
it be Ac (l, 20 C, 5 atm) so that HAc at this
state is equal to zero
Construct an inlet -outlet enthalpy table
having references written at the top of the table as
follows:
References: Ac (l, 20oC, 5 atm), N2(g, 25oC,1 atm)
.
Substance nin
^
H
.
^ out
H
kJ/mol
H3
mol/s
kJ/mol
nout
mol/s
Ac (v)
66.9
H1
3.35
Ac (l)
_
_
63.55
0
N2
33.1
H2
33.1
H4
in
5. Calculate all unknown specific enthalpies
• As mentioned before, specific enthalpies
are obtained relative to the reference states.
• For N2, these are directly read from table
B.8
• For Acetone, the specific enthalpies are
calculated as enthalpy changes between the
process state and the reference state.
For example :
H^1 = ∆H^ ( Ac(l), 20oC, 5 atm)
(Ac (v), 65 oC, 1 atm)
The following process path from the
reference state makes the calculation of
the inlet specific enthalpy of Acetone
possible:
Ac(l, 20 o C, 5 atm) ˆ H 1a Ac(l, 20 o C, 1atm )
ˆ H 1b Ac(l, 56 o C, 1atm ) ˆ H 1c Ac(v, 56 o C, 1atm )
ˆ H 1d Ac(v, 65 o C, 1atm )
Ĥ 1  ˆ H path  ˆ H 1a  ˆ H 1b  ˆ H 1c  ˆ H 1d
 V̂Ac ( l ) (1atm  5atm ) 
56 o C
65 o C
20 o C
56 o C
 Cp( Ac (l ) dT  ( Ĥ v ) Ac   Cp( Ac ( v ) dT
C p ( Ac(l ))  0.123  18.6  10  5 T
C p ( Ac( v ))  0.07196  20.10  10  5 T  12.78  10  8 T 2  34.76  10 - 12 T 3
Substituti ng the specific volume and doing the integratio ns
 Ĥ 1  (0.0279  4.68  30.2  0.753)  35.7kJ / mol
Proceeding in a similar manner, values of
all other unknown specific enthalpies of
Ac are calculated
Substance
.n
in
^in
H
.n
out
^out
H
kJ/mol
mol/s
kJ/mol
mol/s
Ac (v)
66.9
35.7
3.35
32.0
Ac (l)
_
_
63.55
0
N2
33.1
1.16
33.1
-0.10

6. Calculate H
  ( 3.35 mol / s)( 32.0 kJ/mol)  (63.55)(0)  ( 33.1)(-0.1 ) - (66.9)(35. 7) - (33.1)(1.1 6)
 H
 - 2320 kJ/s

7. Solve the energy balance for Q
  H
  -2320 kJ/s
Q
• To calculate enthalpy changes at constant
pressure, formulas for heat capacities of
different components are needed.
• In the absence of tabulated formulas or
data, use is made of kopp’s rule to estimate
the at capacity of a solid or liquid
compound at or near 200C.
Cp Ca (OH)2 = (Cp)Ca +2(Cp)O +2(Cp)H = 26 +
2X17 +2X9.6 = 79 J/mol oC
Enthalpy changes for mixtures
• For mixtures of gases or liquids (streams),
calculate the total enthalpy change as the
sum of the enthalpy changes for the pure
components. Neglect enthalpy changes due
to mixing, this is true for gases at 1atm and
for liquids of similar structure.
• Also for highly dilute solutions of solids or
gases in liquids, neglect enthalpy change of
a solute
• Enthalpy changes for heating or cooling a
mixture can be simplified by using a heat
capacity of the mixture given by the
equation:
(Cp)mix (T) = ∑all components yi Cpi(T)
yi could either be the mol fraction or the mass
fraction of a component depending on the
units of Cp
T2
Ĥ   (C p ) mix (T)dT
T1
Example 8.3-4
Heat capacity of a mixture
Given
Heating
Inlet
60%(by volume) C2H6 , 40% C3H8
150 mol /h, 0oC
Outlet
same mixture at 400oC
Required:
?
Q
Use heat capacity of a
mixture?
Solution
•Cp data at 1 atm from table B.2
(Cp)mix = 0.6 (0.04937 + 13.92X10-5 T
- 5.816 X 10-8T2 + 7.28 X10-12T3)
+ 0.4 (0.06803 + 22.59 X 10-5 T
-13.11 X 10-8T2 + 31.71 X 10-12T3)
= 0.05683 +17.39 X10-5 T
- 8.734 X 10-8 T2 + 17.05 X 10-12 T3
T2
Ĥ   (C p ) mix (T)dT  34.89 kJ / mol
T1
  H
  n H
  150 mol / h * 34.89 kJ /mol
Q
 5230 kJ/h
Usually for gas mixtures, It is reasonable to
assume the mixture behaves as an ideal gas if
A. The inlet or outlet pressure or both are not
specified
B. The mixture of gases may be nearly ideal
C. There might be moderately small changes in
pressure (of few atmospheres)
Therefore, for all of these cases, the specific
enthalpy change is zero at constant
temperature
Example 8.3-5
Energy Balance on a Gas
Preheater
Given:
• Heating
Inlet
• Mixture 10% CH4, 90% air (by volume) 
20oC
Outlet
• Same mixture  300oC
• Gas flow rate 2 X 103 L (STP) /min
2000 L (STP)/min,
2000 L (STP)/min,
0.1 mol CH4 /mol
0.1 mol CH4 /mol
0.9 mol mol air /mol
0.9 mol mol air /mol
20oC,
n mol / min
Heater
300oC,
n mol / min
Required
?
Q
Rate of
heating kW?
Solution
• Assume ideal gas behavior (neglecting
effect of pressure because it is not specified
for both streams)
• Obtain molar flow rate:
2000L (STP)/min *(1 mol/22.4 L(STP)) =
89.3 mol/min
• No need to consider mass balances since
there is only one stream
• No Shaft work , No
changes in kinetic or
potential energy
  H
   n Ĥ   n Ĥ
Q
i i
i i
out
in
• Construct enthalpy table
References: CH4 (g, 20oC,1atm), air (g, 25oC, 1 atm)
Substance
CH4
Air
.n
^
H
.n
mol/ min
kJ/mol
mol/min
in
8.93
80.4
in
0
^
H2
out
8.93
80.4
^
H
out
kJ/mol
^
H1
^
H3
• The reference condition of air is chosen
such that both values of specific enthalpy at
the inlet and the outlet are read directly
from Table B.8
• The reference condition of methane was
chosen at the inlet assuming it exists as a
pure component
•Enthalpies of air are read directly from table
B.8,
^
H2 = -0.15 kJ/mol
^
H3 = 8.17 kJ/mol
• Specific enthalpy change of methane in the
outlet gas mixture at 3000C relative to pure
methane at the reference temp. of 20oC
CH4 (g, 20oC, 1 atm) CH4 (g, 300oC, P)
• Effect of pressure on enthalpy is neglected
• Heat of mixing is neglected
 specific enthalpy change of methane is for
heating pure methane from 20oC to 300oC
 
H
300o C
 (Cp )CH 4 dT , Substitute for Cp from table B.2
20 o C

300o C
5
8 2
 12 3
 (0.03431  5.469  10 T  0.3661  10 T  11  10 T ) dT
20 o C
 12.09 kJ / mol
  H
  (8.93 mol / min)( 12.09kJ / mol )
Q
 [( 80.4)( 8.17)  (8.93)( 0)  (80.4)( 0.15)
 776 (kJ / min)( 1 min/ 60s )(1kw / 1kJ / s)  12.9 kW
Example 8.4-2
Vaporization and Heating
Given
Heating and vaporization at constant
pressure
Inlet
100 g-moles / h Liquid n-C6 H14  25oC
and 7bar
Outlet
100 g-moles / h vapor n-C6 H14  300oC
and 7bar
Required
?
Q
Rate of
heating ?
Solution
• No need to consider mass balances since
there is only one stream
• No Shaft work , No changes in
kinetic or potential energy
  H

Q
• Construct enthalpy table
References: C6 H14 (l, 25oC,7atm)
.
Substance nin
C6 H 14
^
H
in
.n
out
mol/ min
kJ/mol
mol/min
100
0
100
^
H
out
kJ/mol
H1
The following process path from the
reference state makes the calculation of
the outlet specific enthalpy of hexane
possible:
hexane(l, 25o C, 7 atm)
ˆ H hexane (l, 25o C, 1atm )
1a
ˆ H 1b hexane (l, 69o C, 1atm )
ˆ H 1c hexane (v, 69o C, 1atm )
ˆ H 1d hexane (v, 300o C, 1atm )
ˆ H 1e hexane (v, 300o C, 7atm )
Ĥ 1  ˆ H
path
Q  n ˆ H
 ˆ H 1b  ˆ H 1c  ˆ H 1d  85.5 kJ / mol
path
 2.38 kW
neglecting pressure effects ˆ H 1a , ˆ H 1e

69 o C

25 o C C p(hexane ( l ))
dT 
300o C
 C p(hexane ( v )) dT  ˆ H v
69 o C
C p (hexane (l ))  0.2163kJ / mol .o C
C p (hexane ( v ))  0.13744  40.85  10  5 T  23.92  10  8 T 2  57.66  10 - 12 T 3
ENERGY BALANCES ON
REACTIVE SYTEMS
Basic Concepts
1. Reactions are either exothermic or
endothermic
2. The heat of reaction, is the enthalpy
change for a process in which
stoichiometric quantities of reactants at
temperature T and pressure P react
completely in a single reaction to form
products at the same temperature and
pressure
For the reaction:
υ A A + υ BB  υ CC
Where υ is the stoichiometric coefficient
of a reactant or a product
   Ĥ (T , P )
H
r o o
Ĥ r (To , Po )  specific enthalpy of reaction
or heat of reaction
n reac tan t out - n reac tan t in n product out - n product in
  extent of reaction (mol /s) 

 reactant
 product
Properties of heat of reaction
• Heat of reaction is negative for exothermic
reactions and positive for endothermic reactions
• Heat of reaction is independent of pressure
• Value of heat of reaction depends on the way the
stoichiometric equation is written and also on the
states of aggregation of both reactants and
products
• Standard heat of reaction is the heat of reaction
when both reactants and products are at 1 atm and
25oC
Heat of Formation
Standard heat of formation is the enthalpy change
associated with the formation of one mol of a
compound at 25oC and 1 atm from its constituent
elements as they normally occur in nature
• Standard heats of formation for many compounds
are listed in table B.1 in your text book. More
information could be found in Perry’s Chemical
Engineers’ Handbook
• Standard heat of formation for an elemental
species is zero
Heat of Combustion
The standard heat of combustion is the
enthalpy change associated with the
combustion of that substance in presence of
oxygen to give CO2 and H2O, both reactants
and products exist at 25oC and 1 atm
• Table B.1 lists standard heats of combustion
for a number of compounds. More
information could be found in Perry’s
Chemical Engineers’ Handbook
Measurement of Heat of
Reaction
• Heat of reaction may be measured in
calorimeter (a closed reactor immersed in a
fluid contained in a well insulated vessel)
• The rise or fall of fluid temperature can be
measured to determine how much is the
heat transfer to of from the reactor
• By knowledge of the heat capacities of
reactants and products and the amount of
heat transfer, the standard enthalpy of the
reaction may be calculated
• The calorimeter technique has limitations,
for example, if the reaction doesn’t proceed
at a high rate, or if a mixture of products
form rather than a pure compund.
Calculation of Heat of Reaction
Hess’s law:
If the stoichiometric equation for some
reaction1 can be obtained by algebraic
multiplication by constants) on the
stoichiometric equations for other reactions 2,3
…….etc., then heat of reaction1 can be
obtained by performing same algebraic
operations on reactions 2,3,………..
Hess’s law can be applied to
obtain heat of reaction by knowledge
of heats of formation of reactants
and products:
Ĥ o r    i Ĥ o f i 
i
o


Ĥ
 i
fi products
o


Ĥ
 i
fi
rectants
When applying the above equation, the
Standard heats of formation of all
elemental species should be set equal to
zero
Hess’s law can also be applied to
obtain standard heats of reactions that
involve only combustible substances and
combustible products :
Ĥ o r    i ( Ĥ o c ) i 
i
  i (Ĥ o c ) i rectants
  i (Ĥ o c )i
products
If any of the reactants or products
are themselves combustion products, then
their heats of combustion should be
set equal to zero in the above equation.
Energy Balances On Reactors
(Text, P. 450 & 451)
• Follow the steps of the general energy
balance procedure explained for a physical
system
• The only differences are:
A) The choice of the reference states for
reactants and products are taken at 25 oC
and 1 atm for both of them and their states
as indicated by the chemical reaction
equation
• Need to calculate the extent of reaction
(Text, P.451)
• Need to calculate the enthalpy change for
the reactor in a different way than a
physical system (Text, P.451)
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