DEPARTMENT OF CHEMICAL ENGINEERING
B.Tech. - CHEMICAL ENGINEERING 2012-2013 EVEN
CH2254 CHEMICAL PROCESS CALCULATIONS
UNIT-I
PART-A- QUESTION & ANSWERS
1. Define ppm.
ppm (or) parts per million is used for expressing the concentration of extremely dilute
solution (i.e.) solution containing trace impurities. ppm is the weight fraction for solids and
liquids and a mole fraction for gases
1 ppm =1 mg/1000000 mg=1 mg/l (Density of very dilute solution ~1g/m 3=100mg/10-3l
which is approximately mg/l)
2. Define mole.
Mole is the amount of substance of a system which contains as many elementary entities as
there are atoms in 0.012 kg of carbon -12 and it is abbreviated as mol.
3. Define weight.
The force becomes weight, when the body acts under gravitational acceleration i.e., when
a=g, weight G= (1/gc) mg. Since g and gc are assumed equal for all practical purpose, G= m
i.e., the values of weight and mass become practically equal.
4. Define pressure head.
Often the pressure head is expressed in pressure head
Pressure head= pressure (absolute)/ density
Absolute pressure= gauge pressure + atmospheric pressure.
The more commonly used pressure heads are in terms of mercury and water columns
1 atm = 10.33 m of water.
5. Define intensive property and extensive property.
Intensive property: This state of a system is independent of mass eg. Temaperature.
Extensive property: This state of a system is dependent on mass eg. Volume.
6. Define Weight Per cent.
The weight percentage of each component is found by dividing its perfective weight
WA
100
by the total weight of the system Weight % of A =
W
7. Define Volumetric per cent.
The percent by volume of each component is found by dividing its pure – component
volume by the total volume of the system Volumetric % of A = VA 100
V
WA
MA
8. Define Mole Fraction. Mole Fractions of A =
WA w B
MA MB
9. Define Mole per cent. Mole % of A = Mole fraction 100
10. Distinguish between normal atmospheric pressure, absolute pressure, gauge
pressure and vacuum.
Atmospheric pressure + gauge pressure=absolute pressure
Atmospheric pressure –vacuum pressure =absolute pressure
Atmospheric Pressure is the pressure of the air and the atmosphere. Barometer is used to
measure the atmospheric pressure. Usually atmospheric pressure is t5he zero point for a
relative pressure scale and it is not a constant.
Absolute pressure is the pressure corresponding to the complete vacuum .It is the zero point
for an absolute pressure scale.
Gauge pressure is the pressure measured downward from the atmospheric pressure to zero
absolute pressure (perfect vacuum).
11. Define Specific gravity of a gas
Sp. Gr. of gas = Density of the gas
_________________________
Density of air in the same condts.
12. Define Specific gravity of a Liquid
Sp. Gr. of Liquid = Density of the liquid
_________________
Density of H2O at the same Conditions.
13. Define the following for a binary mixture a) mol fraction b) vol %
Mole fraction: For a mixture of two components
Mole fraction A =
mass of A / mol. Wt. of A
-----------------------------------------------Mass of A/ mol.wt of A + mass of B / mol. Wt of B
Vol% = Vol. occupied by A / vol. Occupied by A and B * 100
14. Distinguish between molecular weight and equivalent wt.?
Molecular weight: the combined weight of all the atoms in a molecular put together
constitutes molecular weight.
Equivalent weight: Molecular weight divided by valency of atoms is called equivalent weight
[Molecular weight / valency]
15. Explain the difference between absolute temperature and relative temperature.
The Kelvin (K) and the Rankine (0R) scales are the absolute temperature scales whose zero
points correspond to the lowest possible temperature whereas the Celsius ( 0 C) and Fahrenheit
(0F) scales are the relative temperature scales whose zero points are arbitrarily fixed by their
inventors ( 0C*1.8)+32=0F;0C + 273 =K; 0F+460=0R.
16. The film heat transfer coefficient ’h’ has the dimension of Btu/(hrs.ft 2.0F) in the FPS
system of units. Convert this into kcal/(s.m 2.0C)
Solution:
h= Btu/(hrs.ft2.0 F)
1Btu=0.252 kcal
1 h =3600 s
1 ft =0.3048 m
0
0
0
0
1 C=1.8 F (or)1 F =1/1.8 C
Film heat transfer coefficient h = 0.252kcal/(3600s)(0.3048m2)(1/1.80C)
=1.356 * 10-3 kcal /(s.m2.0C)
17. Convert the Diffusion co-efficient of 0.87 ft2/h into MKS unit.
Solution:
Diffusion coefficient =0.87 ft2/h 1 ft=0.3048m 1 hour =3600 seconds
In MKS units, diffusion coefficient = 0.87*(0.3048 m) 2/(3600sec) = 2.245*10 -5m2/s
18. Calculate the volume occupied by 130 Kg of CO 2at 2.5 atm and 35 0C.
Basis: 130 Kg of CO2
nCO2=Kg mol of Co2=130/44=2.9545 Kg Mol
V = nRT/P=2.9545*0.082058 (273+35)/2.5=29.869m3
19. In a double effect evaporates plant, the second effect is maintained under a vacuum
of 450 mm Hg. Find the absolute pressure in KgF/cm2 and psia.
Solution:
Since [Atm.pr. – vacuum pr. =Absolute pr ]
(1) 760- 450 = 310 mm Hg absolute
310 mm Hg (abs)= 310mm Hg/760mm Hg*1.0332 kgF/cm2 = 0.421 kgF/m2(abs)
(2) 310 mm Hg (abs) = 310mm Hg/760mm Hg *14.7 psia=5.996 psia(abs)
20. An aqueous solution contains 40% Na2 63 by weight. Express the composition in
mole percent.
Solution:Basis of calculation: 100 gm of solution
Molecular weight:Na2 63 = 106
H2O = 18.02
Na2 63 present = 40gm (or) 40/106=0.377g.moles
3.33
g. moles
H2O present = 60 gm (or) 60/18.02 =
3.71
Mole % Na2 63 = (0.377/3.71) 100 = 10.16
89.8
Mole % H2O = (3.33/3.71) 100 =
100.0
21. A solution of Naphthalene C10H8 , in benzene, C4H6, contains 25 mole percent of
Naphthalene express the composition of the solution in weight percent.
Solution:
Basis of calculation: 100g – moles of solution
Molecular weight:C10H8 = 128.1
C6H6 = 78.1
C10H8 present = 25 g. moles or 25 128.1 = 3200 gms.
C6H6 present = 75 g. moles or 75 78.1 = 5800 gms.
--------9060 gms.
3200
Weight % of C10H8 =
100 35.3
9060
5860
Weight of C6H6
=
100 64.7
9060
---------100.0
----------
22. A solution of Sodium Chloride in water contains 230 gm of NaCl per litre at 20C.
The density of the solution at this temperature is 1.148 gm per cubic centimeter.
Calculate Composition of weight percent.
Solution:Compositions in weight per cent
Weight % NaCl
= 230
100 20.0
1148
Weight % H2O
= 918
100 80.0
1148
--------100.0
--------23. The density of the atmosphere decrease with increasing altitudes. When pressure is
300mm Hg, How many inches of water is it?
Solution: Given P=300 mm Hg
760 mm Hg =406.782 inch H2O=33.8985ft H2O
H=300 mm Hg/760 mm Hg *406.782 inch H 2O =160.57 inch H2O
24. What is Average molecular weight?
When a sample under some process is a mixture of 2 or more gases, the average mol weight
is calculated by adopting a unit molal quantity of the mixture as basis of calculation. The
weight of this molar quantity is then calculated and represents average molecular weight.
25. Calculate the average Molecular weight of a five gas having the following
composition by volume.
CO3 = 13.1%, O2 = 7.7, N2 = 79.2
Solution:CO2 = 0.131 g. moles =
5.76 gm
O2 = 0.077 g. moles =
2.46 gm.
N2 = 0.792 g. moles =
22.18 gm
------------Weight of 1 g. mole =
30.40 gm
------------26. What is the mass flow rate of 300 m3/min of methane at 250 0C and 730 mm Hg
Pressure?
Solution:
Basis: 300m3/min of methane
Q=300m3/min
Assuming ideal gas, Pv =nRT =(weight / mol.wt)RT
PQ=(m / M) RT Therefore m=PQM/RT=730 * 300 * 16 / 62.3549 (273 + 250) = 107.439
Kg / min.
27. An aqueous solution of soda ash contains 20% Na 2CO3 by weight. Express the
concentration as weight % Na2O.
Solution:
Basis: !00 Kg of Na2CO3 solution.
Soda Ash content=20%=20 kg
H2O Content= 80%=80 Kg
Na2CO3
Na2O
+
Co2
106 Kg
62
44
20 Kg
20*62/106
20*44/106
=11.6298
=8.302
Na2O content =20*62/106 =11.698 kg
Composition of Na2O =11.698*100/100=11.698%(wt%)
28. Find the volume of CO2 at 25o C and 750 mmHg if the volume of CO 2 is 15m3 at 760
mmHg and 200C ?
P1 V1/T1=P2 V2/T2 ; 750*V1/760*298=760*15/760*293
V1=15.459m3
29. What is the Clausius Clapeyron equation?
Clausius Clapeyron equation is given by
dp/dt = Hv/(RT2/p)
Where, Hv (or) is the molal latent heat of vaporization.
P = vapor pressure, T = absolute temperature
ln (p2/p3)= Hv[ 1/t1 -1/ t2]
30. Calculate the weight of 100 m3 of H2 measured as 725 mm Hg and 25 0C
Solution:
Basis : 100 m3 of H2
For an ideal gas ,Pv=nRT N=PV/RT=725*100/62.359*(273+25)=3.9014 kg mol
Weight of H2=Kg mol of H2 *Molecular wt of H2=3.9014*2=7.8028 =7.803Kg
31. Define compressibility factor.
Compressibility factor express the non-0indentity of a gas and is denoted by Z .It is defined
as
Z=PV/nRT
Z is a dimensionless quantity and is equal to one for ideal gas, and not equal to one for real
gas; value of Z is obtained from the Nelson-Obert Generalized compressibility charts if P rand
Tr(orVr ) is known.
Part B
1.
2.
3.
4. Convert 499 gm CuSO4 . 5H2O into moles.
Basis : 499 gm of CuSO4 . 5H2O
Atomic weights : Cu = 63.5, S = 32, O = 16 and H = 1
Molecular weight of CuSO 4 . 5H2O
1 63.5 1 32 4 16 5 2 1 116 249.5
Moles of CuSO4 . 5H2O =
499
249.5
= 2 mol.
The relationship of compound as a whole and its constituents is given as
follows:Each mole of NaOH contains one atom of Na
1 mol of NaOH = 1 atom of Na
1 mol of NaOH = 1 atom of Na
= 1 gm- atom of H.
Each mole of NaOH contains 1 atom of Na. The sign refers to equivalent to and
not ‘equal to’.
Similarly for H 2SO4 and ‘S’
1 mol of H2SO 4 = 1 atom of S
1 kmol of H2SO4 = 1 k atom of S
i.e. each mole of H2So4 contains 1 atom of S.
For each mole of H 2SO4 contains 1 atom of S.
For CuSO4 . 5H2O and CuSO4.
1 mol CuSO4 . 5H 2O = 1 mole CuSO4
1 kmol CuSO 4 . 5H2O = 1 kmol CuSO 4
5. How many moles of K 2 CO3 will contain 117 kg K?
Solution:Basis : 117 Kg. of K
Atomic weight of K = 39
Atoms of K
117
3 katom
39
Each mole of K2CO 3 contains 2 atoms of K.
2 atom of K = 1 mole of K 2CO3
2 katom of K = 1 kmol of K2CO3
1
3
2
= 1.5 kmol.
Moles of K 2CO3 =
6. Find the equivalent weights of (1) HCl (2) NaOH (3) Na 2CO3 and H2SO4.
Solution:(1) HCl:Mol. Weight of HCl = 1 1 + 1 35.5
Valence of HCl
= 36.5
= 1.
Equivalent weight of HCl = 36.5
(2) NaOH:Molecular weight of NaOH = 1 23 + 1 16 + 1 1
= 40
Valence of NaOH = 1
Equivalent weight of NaOH =
40
40
1
(3) Na2CO3:Molecular weight of Na2CO 3 = 2 23 + 1 12 + 3 16
= 106
Valence of Na2 CO3 = 2
Equivalent weight of Na2CO3 =
106
53.
2
(4) H2SO4:Molecular weight of H 2SO4 = 2 1 + 1 32 + 4 16
= 98
Valence of H2SO 4 = 2
Equivalent weight of H2SO4 =
98
49.
2
7. Calculate the equivalent weights of the following compounds.
(1) H3PO4 (2) CaCl 2 (3) FeCl 3 (4) Al2 (SO4) 3 and (5) KMnO4.
At weight K = 39, Ca = 40, P = 31, H = 1, O = 16, Al = 27, S = 32, Cl = 35.5, Fe =
56, Mn = 55.
Solution:
(1) H3PO4:Mol. Weight of H 3PO4 = 3 1 + 1 31 + 4 16 = 98
Valence of H3PO 4 = 3
Equivalent weight of H3PO4 =
(2) CaCl2:-
98
32.67
3
Mol. Weight of CaCl 2 = 1 40 + 2 35.5 = 111
'
'
Valence of CaCl2 2
Equivalent weight of CaCl2 =
111
55.5
2
(3) FeCl3:Mol. Weight of FeCl3 = 1 56 + 3 35.5 = 162.5
Valence of FeCl3 = 3
162.5
33
= 54.17
Equivalent weight of FeCl3 =
(4) Al2 [SO4]3 :Mol. Weight of Al2 [So4]3 = 2 27 + 3 32 + 12 16 = 342
Valence of Al2 [SO 4]3 = 6
Equivalent weight of Al 2 [SO 4]3 =
342
57
6
(5) KMnO4:Mol. Weight of KMnO4 = 1 39 + 1 55 + 4 16 = 158
Valence of KMnO 4 = 5
Equivalent weight of KMnO4 =
158
31.6
5
8. 98 gms of sulphuric acid (H2SO4) are dissolved in water to prepare one litre
of solution. Find normality and molarity of solution.
Solution:Basis : One litre of solution
Amount of H 2SC4 dissolved = 98 gm
Mol. Weight of H 2SO4 = 98
Equivalent weight of H2SO4 =
Cm-equivalent of H 2SO4 =
98
49 .
2
98
2
49
gm equivalent of H2SO4
Volume of solution in litre
2
= 2
1
Normality (N) =
Moles of H 2SO4 =
98
1.0 mol
98
moles of H2SO4
Volume of solution in litre
1
= = 1.
1
9.
20 gms of caustic soda are dissolved in water to prepare 500 ml of
solution. Find the normality and molarity of solution.
Molarity (M) =
Solution:Basis : 500 ml of solution.
Molecular weight of NaOH = 40
Equivalent weight of NaOH =
40
40
1
Volume of solution = 500 cm 3
= 0.5 lit.
Cm-equivalent ofNaOH =
20
0.5
40
gm equivalent of NaOH
Volume of solution in lit.
0.5
1
0.5
Normality (N) =
Moles of NaOH =
20
0.5 mol.
40
moles of NaOH
Volume of solution in lit.
0.5
1.0
0.5
Molarity (M) =
10. A solution of caustic soda contains 20% NaOH by weight. The density of
the solution is 1.196 kg/lit. Find the normality, molarity and molality of the
solution.
Solution:Basis : 100 kg of solution
The solution contains 20 kg NaOH and 80 kg water (solvent)
Density of solution = 1.196 kg/lit.
100
83.62 lit.
1.196
20
0.5 k mol.
Moles of NaOH in solution =
40
= 500 mol.
Volume of Solution =
gram moles of NaOH
Volume of solution in lit.
500
5.98
83.52
Molarity (M) =
For NaOH as valence = 1,
Equivalent weight = Molecular weight.
Therefore,
Normality [N] = Molarity (M) = 5.98
gram moles of NaOH
kg. of solvent
500
=
80
= 6.25 mol/kg.
Molality =
11. A solution of caustic soda contains 20% NaOH by weight. The density of
the solution is 1.196 kg/lit. Find the normality, molarity and molality of the
solution.
Solution:Basis : 100 kg of solution
The solution contains 20 kg NaOH and 80 kg water (solvent)
Density of solution = 1.196 kg/lit.
100
83.62 lit.
1.196
20
0.5 k mol.
Moles of NaOH in solution =
40
Volume of Solution =
= 500 mol.
gram moles of NaOH
Volume of solution in lit.
500
5.98
83.52
Molarity (M) =
For NaOH as valence = 1,
Equivalent weight = Molecular weight.
Therefore,
Normality [N] = Molarity (M) = 5.98
gram moles of NaOH
kg. of solvent
500
=
80
= 6.25 mol/kg.
Molality =
UNIT-II
PART-A- QUESTION & ANSWERS
1. State and explain the Raoult’s law.
Raoult’s law states that “the vapour pressure exerted by a liquid forming mixture P i is
directly proportional to the mole fraction of the liquid in the mixture Xi”
Pi=PiXi
Where Pi=Vapour pressure of pure solvent
Raoult’s law holds good for ideal gas and dilute (very dilute) solution. e.g. ideal liquid-liquid
mixture.
2. What is Ideal gas Law?
A gas is said to be ideal if it follows the relation
PV= nRT
Where P= Absolute pressure. V= Volume of n molecule of gas.
N= number of molecules of gas. R= Universal gas content.
T= Absolute temperature.
The above Equation is called as the ideal gas law
3. Define Partial Pressure: The partial pressure of a component of which is present in the
mixture of gases is the pressure exerted by the component if it were alone taken into a
container having the same volume and at the temperature as that of the gaseous mix.
4. Define Vapor pressure
The Vapor pressure of a component is the ratio of the Partial pressure to the mole fraction
of that component.
5. Define Pure component volume: The pure component volume of a component gas
present in gaseous mixture is the volume occupied the component is it were alone taken into a
container at the same pressure and temperature as that if the gaseous mixture.
6. Define Dalton’s law: It states that the sum of the partial pressure of the entire component
is equal to the total pressure.
7. Define Amagal’s Law or Leduc’s Law :It states that the sum of pure component volumes
at all the components is equal to the total volume of the mixture.
8. Define Henry’s law:
As low concentrations of gas in liquid, partial pressure of solute gas is directly proportional to
mole fraction of solute.
Pi =Hixi
H i = Henry’s constant.
9. State VanderWaal’s Equation of state.
VanderWaal’s equation of state is given by
(P+a/V2)(V-b)=RT
where a= 2TR2Tc2/64Pc
(m3) MPa/(Kg mol)2
b= R Tc/8 Pc
m3/Kg mol
PcTc =critical pressure and critical temperature of gas
10. Show that in the case of ideal gases, the volume%, mole% and pressure% are equal.
Solution:
For an ideal gas mixture,
PiV=niRT-------------------------(1)
PtV=ntRT-------------------------(2)
(1)/(2) gives Pi/Pt =ni/nt
Pi=Pt ni/nt =Pt yi
Pressure% ~ Mole%-------------(3)
PVi=niRT--------------------------(4)
PVt=ntRT0------------------------(5)
(4) / (5) gives Vi/Vt=ni/nt
Vi=Vt*ni /nt = Vt * yi
Volume %=mole% ------------------------(6)
Combining (3) and (6) for ideal gas mixture
Pressure%=volume%=mol%-------------(7)
11. Calculate the density of Cl 2 gas at 2500C and 250 atm absolute pressures using the
ideal gas law.
Solution: Since PV=nRT
P =(Wt/mol.wt) RT/V=RT/ M
=PM/RT Density of chlorine gas=100*35.5/0.082058(273+250)=82.719 Kg/m 3
12. Assuming the applicability of the ideal gas law. Calculate the maximum temperature
to which 10 lb of T2, enclosed in 30 cu.ft. chamber.
Solution:Volume at S.C. = 0.307 359
= 128.1 Cu.ft.
P V
T2 T1 2 2
T1 V1
= 273
150
30
14.7 128.1
Temperature at 30 Cu.ft. = 652K (or) 379C
13. Calculate weight of 100 Cu.ft. of water vapour. Measured at a pressure of 15.5 mm
Hg. and 23C.
Solution:15.5 273
Volume of S.C. = 100
1.88 cu.ft.
760 296
Moles of H2O = 1.88 359 = 0.00523 lb mole
Weight of H2O = 0.00523 18 = 0.0942 lb
14. It is desired to compress 10 lb of carbon dioxide to a volume of 20 cu.ft. Calculate
the pressure in pounds per square inch that is required at temperature of 30C.
Solution:Volume of S.C. = 0.228 359 81.7 cu.ft.
V T
P2 T1 1 2
V2 T1
30C 303K
Pressure at 20 Cu.ft. = 14.7
81.7 303
= 66.6 Tsi
20 273
15. A compound whose molecular weight is 103 analyzes as following.
C = 81.5, H = 4.9, N = 13.6
What is formula?
Solution:Basis : 103 Kg of compound
81.5
103 83.95 Kg
Amt of carbon =
100
83.95
6.995
Kg. atons of carbon =
100
16. Define Molarity:
It is defined as the number of gmmole of solute dissolved in 1it of solution.
17. Define Normality:
It is defined as the number of grams equivalent of solute dissolved in 1lit of solution.
18. Define Molality.
Molality: Number of gram moles of solute dissolved in one kilogram of solvent.
19. 98 gms of sulphuric acid (H 2SO4) are dissolved in water to prepare one litre of
solution. Find normality and molarity of solution.
Solution:Basis : One litre of solution
Amount of H2SC4 dissolved = 98 gm
Mol. Weight of H2SO4 = 98
98
49 .
Equivalent weight of H2 SO4 =
2
98
2
Cm-equivalent of H2SO4 =
49
gm equivalent of H2SO4
2
= 2
1
Volume of solution in litre
98
Moles of H2SO4 =
1.0 mol
98
moles of H2SO4
1
Molarity (M) =
= = 1.
1
Volume of solution in litre
Normality (N) =
20. 20 grams of caustic soda are dissolved in water to prepare 500 ml of solution. Find
the normality and molarity of solution.
Solution:Basis : 500 ml of solution.
Molecular weight of NaOH = 40
40
40
Equivalent weight of NaOH =
1
Volume of solution = 500 cm3 = 0.5 lit.
20
0.5
Cm-equivalent ofNaOH =
40
gm equivalent of NaOH 0.5
1N
Normality (N) =
Volume of solution in lit. 0.5
20
0.5 mol.
Moles of NaOH =
40
moles of NaOH
0.5
1.0 M
Molarity (M) =
Volume of solution in lit. 0.5
21. A solution of caustic soda contains 20% NaOH by weight. The density of the solution
is 1.196 kg/lit. Find the normality, molarity and molality of the solution.
Solution:Basis : 100 kg of solution
The solution contains 20 kg NaOH and 80 kg water (solvent)
Density of solution = 1.196 kg/lit.
100
83.62 lit.
Volume of Solution =
1.196
20
0.5 k mol. = 500 mol.
Moles of NaOH in solution =
40
gram moles of NaOH
500
5.98
Molarity (M) =
Volume of solution in lit.
83.52
For NaOH as valence = 1,
Equivalent weight = Molecular weight.
Therefore,
Normality [N] = Molarity (M) = 5.98
gram moles of NaOH 500
Molality =
=
= 6.25 mol/kg.
80
kg. of solvent
22. Find out the grams of HCl needed to prepare 1 liter = NH 4Cl solution.
Normality (N) = (gram equivalent of HCl)/ (volume of solution in lit)
= weight / (equivalent weight)/ volumes in lit
= (weight / (molecular weight/ valency)) / volume in liter
= weight/36.5)/1
=73 g
23. One hundred gms of chlorine gas is mixed with 10 gm. of hydrogen gas. Calculate
the mole fractions of the two gases in the mixture.
100g of cl 2 (g); 10g of H2 (g) Moles of Cl2 = 1.408 Moles of H 2 = 10/2 = 5
Mole fraction of Cl2 = 1.408/6.408 = 21.97% Mole fraction of H 2 = 78.03%
24. A producer gas has the following composition by volume CO-25.0% CO2 -5.0% O2 4.0% and balance nitrogen. What is its average molecular weight?
Basis: 100 kg mol of gas
Substance
Vol %=mol %
Weight
CO
25
700
CO2
5
O2
N2
220
4
66
128
1848
Total Weight =
Average molecular weight=2896/100=28.965
2896
25. An industrial strength drain cleaner contains9 Kg of water and 5 kg of NaOH. What
rare the mole fraction of each component in the bottle of solution?
Solution:
Basis: Drain cleaner contains 9 Kg of H2O and 5kg of NaOH
Wt (kg)
Molecular Wt (Kg/Kg Kg mol
Mole fraction
mol)
H 2O
9
18
9 / 18 = 0.500
0.5 / 0.625 = 0.8
NaOH
5
40
Mole fraction of water= XH2O =0.8
5 / 40 = 0.125
0.125 / 0.625 = 0.2
= 0.625
Mole fraction of water=XNaOH =0.2
Part B
1. By electrolyzing a mixed brine a mixture of gases is obtained at the cathode
having the following percentage composition by weight.
Cl2-67
Br2-28
O2-5
Using the ideal gas law, calculate:
(a) Composition of the gas by volume.
(b) Density of the mixture in grams per litre at 25 oC and 740mm of Hg
pressure.
(c) Specific gravity of the mixture, (air = 1.0) Atomic weights:
Cl-35.5, Br-80.0, O-16.0
Solution:
Basis : 1gm of the gas mixture.
(a) Composition of gas by volume.
Constituents
Cl2
Br2
Amount
Gm.
Gm.mol.
0.67
0.00943
0.28
0.00175
Mol % (=
Vol.%)
74.02
13.74
O2
Total
0.05
1.00
0.00156
0.01274
12.24
100.00
(b) 1 gm of mixture contains 0.01274 mol.
P=740mm of Hg.
T= 25oC=298oCK
R
mm.Hglit.
700 22.4
62.4
273
gm.mol o K
nRT
[from ideal gas law]
P
0.01274 62.4 298
=
0.32 lit.
740
1
Density =
3.12gm / lit
0.32
Hence, V=
(c) Assuming ideal behaviour for air, volume of 1 gm mol at 25 oC and 1 atmosphere
760 22.4 298
25.11 litres.
740 273
Mol. Wt. of air = 28.84
28.84
gm
1.15
25.11
lit
Specific gravity of the mixture
Density of air =
=
3.12
2.71.
1.15
2. In the Deacon process for the manufacture of chlorine, a dry mixture of
hydrochloric acid gas and air is passed over a heated catalyst which promotes
oxidation of the acid. Air is used in 30% excess of that theoretically required.
(a) Calculate the weight of air supplied per kg. of acid.
(b) Calculate the composition by weight of the gas entering the reaction
chamber.
(c) Assuming that 60% of the acid is oxidized in the process, calculate the
composition by weight of the gases leaving the chamber.
Solution:
Basis
Reaction
:
:
1kg of dry HCI gas.
4 HCI + O2 2Cl2 + 2H2O
(a) 4 36.5 kg HCI require 32 kg O 2
1kg HCI require
32
0.219kg.
146
100
0.953kg
23
=0.953 1.3=1.24kg.
=0.219
Theoretical air
Air supplied (30% excess)
(b) Wt. of O2 supplied = 1.24 0.23 = 0.285 kg.
Wt. of N2 supplied = 1.24 0.77=0.955 kg.
Analysis of gas entering reaction chamber:
Constituents
HCI
O2
N2
Total
Amount, kg
1,000
0.285
0.955
2.24
Wt.%
44.00
12.73
42.61
100.00
(a) Composition of the gases leaving the reaction chamber:
Reaction is 60% completer
HCI converted
=
0.6kg
From the chemical reaction,
142
0.6 0.583kg
146
36
=
0.6 0.148kg
146
32
=
0.6 0.132kg
146
= 0.285-0.132=0.153kg
Chlorine formed =
Water vapour
Oxygen used
Oxygen left
Constituents
HCI
O2
N2
Cl2
H2O
Total
Wt, kg
0.400
0.153
0.955
0.583
0.148
2.239
Wt.%
17.87
6.86
42.62
26.04
6.61
100.00
3. A volume of moist air 30m3 at a total pressure of 101.325 kPa and a temperature of
303K (30C) contains water vapor in such proportions that its partial pressure is 2.933
kPa. Without total pressure being changed the temperature is reduced to 288K (15C)
and some of water vapour is removed by condensation. After cooling it is found that the
partial pressure of water vapour is 1.693 kPa. Calculate (a) volume of air at 288K
(15C) and (b) weight of water condensed.
Solution:Basis: 30m3 of moist air at 303K.
Ideal gas law is :
PV = nRT
PV
RT
where, n = moles of moist air
n=
P = 101.325 kPa
R = 8.31451 m3.kPa/kmol.K.
V = 30m3
T = 303K
n = moles of moist air =
101.325 30
8.31451 303
= 1.2066 kmol.
Let n1 be the kmol of air
n2 be the kmol of moisture/water vapour.
p1 = Partial pressure of air at 303K
= 101.325 – 2.933
= 98.392 kPa
p2 = Partial pressure of moisture at 303K
= 2.933 kPa
For ideal gas,
Pressure % = mole %
Pressure fraction = mole fraction
For air
98.392 n1
101.325 n
n1
98.392
1.2066
101.325
= 1.172 kmol at 303K
For water vapour/moisture.
n
2.933
2
101.325 n
n2 =
2.933
1.2066
101.325
= 0.035 kmol at 303K
At 288K,
Partial pressure of water vapour = 1.693 kPa
Let n3 be the moles of water vapour at 288K.
Moles moist air = 1.172 + n3
n3
1.693
101.325 1.172 n3
n3 = 0.02 kmol
Moles of moist = 1.172 + 0.02
air at 288K = 1.192 kmol
Let n’ = 1.192 kmol
PV = n’RT
where V = Volume of air at 288K
n’ = 1.192 kmol
R = 8.31451 m3.kPa/kmol.K
T = 288K
P = 101.325 kPa
V=
n'RT
P
1.192 8.31451 288
101.325
28.17 m3
Moles water condensed = n2 – n3
= 0.035 – 0.02
= 0.015 kmol.
Amount of water condensed = 0.015 18
= 0.27 kg.
4. The gas acetylene is produced according to the following reaction by
treating calcium carbide with water:
CaC2+2H2O C2H2+Ca (OH)2
Calculate the number of hours of service that can be derived from 1.0 kg
of carbide in an acetylene lamp burning 60 litres of gas per hour at a
temperature of 20oC and a pressure of 740 mm Hg.
Solution:
Basis: 1 kg of carbide
CaC2+2H2O C2H2+C2(OH) 2
64 kg calcium carbide gives 26 kg C 2H2
1 kg calcium carbide gives
26
kg C2H 2
64
26 1
kg mol C2H2 0.0156kg mol.
64 26
Assuming ideal gas behaviour for C2H2
Volume of gas produced at 20oC and 740 mm of Hg is to be calculated.
pV 760 22,400
nT
1 273
mm Hglit.
=62,358.9
kg mol o K
R
nRT
P
0.0156 62,358.9 273+20
Volume of acetylene =
=
740
=385.8 litres.
No.of hours of service =
385.8
6.43
60
5. A natural gas having the composition CH4 – 94%, C2H6 – 3% and N2 – 3% is
piped from the well at 25C and 3.0 atm. pressure. Assuming that the ideal gas
law is obeyed, find out:
(a) Partial pressure of N2.
(b) Volume of N2 per 100 cu. M of gas.
(c) Density of the gas.
Solution:Basis: 100 m3 of gas at 25C and 3 atm.
Volume % = mole %
From ideal gas law and Dalton’s law of partial pressure,
Partial pressure = Mole fraction Total pressure
(a) Partial pressure of N2 = 0.03 3.0 = 0.09 atm.
(b) Volume of N 2 per 100 m3 of gas = 3m3 (measured at 3 atm and 25C).
(c) n = No. of moles =
PV
RT
Atm m3
R 0.082
kg mol K
3 94
11.54 kg mole
0.082 273 25
nCH4
nC2H6
nH2 =
33
0.37 kg mol
0.082 273 25
33
0.37 kg mol
0.082 273 + 25
The analysis is as follows:Constituents
CH4
C2H6
N2
Total
Density of the gas =
Kg mol
11.54
0.37
0.37
12.28
Mol wt
16
30
28
Weight
184.64
11.10
10.36
206.10
Weight
Volume
206.10
2.06 kg/m3 2.06gm / lit.
100
6. For 1000 lit/sec of a gaseous mixture of the following composition:
CH4 – 10%, C 2H6 – 30%, H2 – 60% (all by volume) at 30C and 2000 mm
Hg gauge, calculate,
(a) the mole fraction of each component
gm mol
(b) the concentration of each component,
CC
(c) the partial pressure of each component
(d) the molar density of the mixture
(e) the mass flow rate of the mixture
(f) the average molecular weight of the gas.
Solution:Basis : 1000 lit of the gaseous mixture
Pressure
= 2000 mm Hg gauge
= 2760 mm Hg absolute.
(a) Let nCH 4, nC2H 6, nH2 be the moles of the components
From ideal gas law :-
nCH4
PV
mm Hg lit.
, R = 62.36
RT
gm mol K
=
nC2H6
nH2
=
2760 1000 0.1
62.36 273 30
2760 1000 0.3
62.36 303
2760 1000 0.6
62.36 303
Mol fraction of CH4
=
43.82 gm mol
87.64 gm mol
14.61
14.61 43.82 87.64
14.61
0.10
146.07
Mole fraction of C2H6
Mol fraction of H2
14.61 gm mol
43.82
0.30
146.07
87.64
0.60
146.07
(b) Concentration of CH4
14.61
gm mol
1.46 10 5
1000 1000
cc
Concentration of C2H 6 =
43.82
gm mol
4.38 105
1000 1000
cc
Concentration of H2
87.64
gm mol
8.76 10 5
1000 1000
cc
=
(c) Partial pressure of CH4 = 0.1 2760 = 276 mm Hg
Partial pressure of C2H 6
= 0.3 2760 = 828 mm Hg.
Partial pressure of H2
= 0.6 2760 = 1656 mm Hg.
(d) Molar density of the mixture =
146.07
gm mol
0.146
1000
lit.
(e) Mass flow rate of the mixture.
Constituents
CH4
C2H6
H2
Mol
14.61
43.82
87.64
Mol wt
16
30
2
Total
Weight
233.76
1314.60
175.28
1723.64
Mass flow rate of the gaseous mixture
= 1723.64 gm = 1.724 kg.
(g) Average molecular weight = 16 0.1 + 30 0.3 + 2 0.6 = 11.8
7. A mixture of CH4 and C 2H6 has density 1.0 kg/m 3 at 273K (0C) and 101.325
kPa pressure. Calculate the mole % and weight % CH4 and C2H4 in the mixture.
Solution:Basis : 1 kg/m3 density of gas mixture at 273 K and 101.325 kPa.
P Mave
Density of gas mixture = =
RT
RT
Mave = .
P
where = 1 kg/m3
T = 273 K
P = 101.325 kPa
R = 8.31451 m3. kPa/kmol. K
Mavg. = 1
8.31451 273
101.325
= 22.4
Let x CH4 and xC2H6 be the mole fractions of CH 4 and C2 H6 respectively.
Mavg = Mi xi MCH4 .xCH4 MC2H6 .xC2H6
22.4 = 16 xCH4 30xC2H6
….(1)
xi = 1
xCH4 xC2H6 1
xC2H6
….(2)
1 xCH4
….(3)
Put the value of x c2H6 from equation (3) in equation (1) and solve for xCH4 .
22.4
= 16xCH4 30 1 xCH4
x CH4
= 0.543
xC2H6
= 0.457
Mole % of CH4 = xCH4 100 0.543 100
= 54.3
Mole % of C2H6= 0.457 100
= 45.7
Weight of CH 4 in 1 kmol mixture = 0.543 16
= 8.69 kg
Weight of C2H 6 in 1 kmol mixture = 0.457 30
= 13.71 kg.
Weight of gas mixture
= 22.4 kg
Weight % of CH4 in mixture
=
8.69
100
22.4
= 38.8
Weight % of C2H 6 in mixture
= 100 – 38.8
= 61.2
8. In one case 26.6 litres of NO2 at 80 kPa and 298K (25C) is allowed to stand
until the equilibrium is reached. At equilibrium the pressure is found to be
66.662 kPa. Calculate the partial pressure of N2O4 in the final mixture.
Solution:Basis : 26.6 lit. of NO2 at 80 kPa and 298 K.
Volume of NO 2 = 26.6 lit. = 0.0266m3
P1V1
= n1RT1
P1V1
RT1
where P 1 = 80 kPa, T 1 = 298 K
n1 = Initial moles =
R = 8.31451 m3. kPa/kmol. K
V1 = 0.0266 m3
80 0.0266
n1 =
8.6 10 4 kmol
8.31451 298
n1 = 0.86 mol
2 NO2
= N2O4
Let x be the mol of N2O4 in final gas mixture.
NO2 reacted = 2x mol
NO2 unreacted = 0.86 – 2x mol
Final moles = n2 = 0.86 – 2x + x
= 0.86 – x mol
For initial conditions – P1V1 = n1RT1
For final conditions – P2V2 = n2RT2
But here, V1 = V 2 and T1 = T2
P1 n1
P2 n2
80
0.86
66.662 0.86 x
Solving we get x = 0.1434 mol.
Final moles = 0.86 – x
= 0.86 – 0.1434 = 0.7166 mol
Mole fraction of N 2O4 in final gas mixture
xN2O4 0.1434 / 0.7166
= 0.20
Partial pressure of N2O4 = xN2O4 .P
0.20 66.662 13.33 kPa
9. A closed vessel contain a mixture of 40% NO2 and 60% N2O4 at a
temperature of 311K (38C) and a pressure of 531.96 kPa. When the
temperature is increased to 333K (60C), some of N2O4 dissociates to NO2 and
a pressure rises to 679.95 kPa. Calculate the composition of gases at 60C by
weight.
Solution:Basis : 100 kg of gas mixture at 311 K
NO2 in gas mixture = 40 kg
N2O4 in gas mixture = 60 kg
40
0.87 kmol
46
60
0.652 kmol
Moles of N 2O4 =
92
Initial moles = n 1 = 0.87 + 0.652 = 1.522 kmol
Moles of NO 2 =
N2O4 = 2NO 2
Let x be the kmols of N2O4 dissociated at 333K.
NO2 formed = 2x kmol
N2O4 at 333K = 0.652 – x kmol
N2O4 at 333K = 0.87 + 2x kmol.
Total moles at 333K = (0.652 – x) + (0.87 + 2x)
= 1.522 + x kmol
P1V1 = n1RT1
…(1)
P2V2 = n2RT2
…(2)
But V1 – V2 for it being closed vessel.
Taking ratio of equation (1) and (2) we get,
P1 n1 T1
.
P2 n2 T2
where P 1 = 531.96 kPa, P2 = 679.95 kPa
n1 = 1.522, n2 = 1.522 + x
T1 = 311K, T2 = 333K
531.96
1.522
311
679.95 1522 x 333
Solving we get,
x = 0.295 kmol.
NO2 at 333K = 0.87 + 2x
= 0.87 + 2 0.295 = 1.46 kmol.
N2O4 at 333K = 0.652 – x
= 0.652 – 0.295 = 0.357 kmol.
Amount of NO 2 at 333K = 1.46 46 = 67.16 kg
Amount of N 2O4 at 333K = 0.357 92 = 32.84 kg.
Composition of Gases at 333K:
Component
NO2
N2O4
Total
Quantity in kg.
67.16
32.84
100
Weight %
67.16
32.84
100
10. A solution of NaCl in water contains 230 g of NaCl per litre at 20 0C. The density of
the solution at this temperature is 1.148g/cc. Find the composition in (a) weight% (b)
volume % of water (c) mole% (d) atomic % (e) molality and (f) g Nacl / g water.
UNIT-III
PART-A- QUESTION & ANSWERS
1. Write Principle of stoichiometry:
The mass and gas volumetric relationships in chemical reactions are presented together
with the basic units for expressing charging in mass and composition.
2. What is the law of conservation of mass?
The mass of the system remains constant, regardless of the changes taking place within the
system. This statement is known as the law of conservation of mass and is the basis so called
material balance of a process.
3. What is the use of material and energy balances?
A process design starts with the development of a process flow sheet or process flow
diagram. For the development of such a diagram, material and energy balance calculations
are necessary, and these balances follows the laws of conservation of mass and energy.
4. What is a flow sheet?
A process flow sheet is one in which all incoming and outgoing materials and utilities are
shown. It includes,
a) Flow rate or quantity of each stream.
b) Operating conditions of each stream, such as pressure and temperature.
c) Heat added / removed in particular equipment.
d) Any specific information which is useful in understanding the process. For example,
symbolic representation of a hazard, safety precautions, sequence of flow if it is a batch
process, corrosive nature of materials, etc.
5. Define Percentage Excess:
The percentage excess of any reactant is defined as the percentage of the excess to the amt
theoretically required by the stoichiometric equation for combination with the limiting
reactant.
6. Define Percentage yield:
It is defined as the moles of the desired product formed to the moles that would have been
formed if there were no side reactions and the limiting reactant had reached completely and
multiplying by 100.
7. Define Degree of Completion:
The degree of completion of a reaction is ordinarily expressed as the percentage of the
limiting reaching material, which is converted or decomposed into other products.
8. Define Total Conversion:
It is defined as the total fraction of reactant converted to all products.
9. Define Percentage Conversion:
It is defined as the fraction of reactant converted to all products and multiplying by 100.
10. Define Excess reactant:
It is defined as some of the reaching materials are present in excess of the amts
theoretically required for combination with the others.
11. Define Limiting reactant:
It is defined as some of the defining as the material that is not present in excess if that
required combining with any of the other reacting materials.
12. Explain about Duhring lines:
When the temp of a liquid is plotted against the temp of the known reference liquid (e.g.
H2O) at equal pressure; it’s called a Duhring plot. Also a line plotting the Bp of solution
against B.P of the solvent is called Duhring’s line.
13. Define Selectivity:
The ratio of moles of desired product formed to the moles of undesired product formed is
called Selectivity. Selectivity =moles of the desired product / moles of undesired products
14. Explain COX chart:
The Antoine’s eqn is Log10P= A-B / P + C (OR)
ln P = A-B / P + C
Where P - vapour pressure in Kpa
T –temperature in K A,B,C- are constants
This equation correlates pressure and correlates saturated temperature. A graph between
ln P and 1/t is called COX chart. It’s the graphical representation of Antoine’s equation.
15. Define Recycle and recycle ratio:
Recycle: In the process of recycling a part of the out going stream is again fed as the input by
mixing with the fresh stream. It is done to increase the yield or to enrich the product, to
conserve the heat or to improve the operation.
Recycle Ratio: The ratio of the quantity of a reactant recycled to the quantity of the same
reactant entering a recycling operation as fresh feed. Recycle ratio= R/F
16. Define By-passing:
In the process of by-passing a portion of the inlet stream is diverted from the process
and it is remixed at the end of the process.
17. What are the advantages of recycling operation?
Recycling operations have several advantages given below:
(1) To utilize the valuable reactants to their maximum and avoid wastage.
(2) To utilize the heat being lost in the outgoing stream.
(3) To improve the performance of the equipments.
(4) To control the operating variables such as composition, pressure, temperature, etc. in
a reaction.
(5) To improve the selectivity of a product.
18. Define Purging and Purge ratio.
Purging: A purge stream is one that is bled off to remove an accumulation of inerts or
unwanted impurities that might otherwise build up in the recycle stream. Purging is used in
the production of NH3 .In the synthesis step, some of the gas stream must be purged to prevent
build of argon and methane.
Purge ratio: It is the ratio of quantity of purge stream to the quantity of recycles feed.
Purge ratio = P/R
19. What is the need for withdrawing purge stream in the recycling operation involving
feed mixture containing impurities?
Purge stream helps in the removing the accumulation of inerts or unwanted impurities that
might otherwise build up in the recycle stream.
In the production of NH3, in the synthesis step, some of the gas stream must be purged to
prevent the build of argon and methane.
20. What is the significance of stoichiometric equation in solving the material balance
problems with chemical reaction?
Stoichiometric equation of a chemical reaction gives the knowledge of the stoichiometric
mole ratio or weight ratio of all the reactants and products, which enable us to solve the
material balance/energy balance problems. It is also possible to determine the amount of
unreacted components in the present stream if excess reactant is sent.
21. Define Reactant ratio.
The number of moles of an excess reactant per mole of limiting reactant in the reactor feed.
22. Define yield per process.
The yield of an product is the reactant effluent expressed as a percentage of that portion of
the limiting reactant in the reactor feed which is converted and disappears during the courge
of reaction.
23. Define fresh – feed conversion.
The percentage of the limiting reactant in the combined reactor feed that is converted and
disappears.
24. Define liquid volume velocity.
The liquid volume at 60 0 f at limiting reactant fed per for unit volume of effective reactor
(or) catalyst bed.
25. Define weight hourly space velocity.
The weight of limiting reactant feed per hour per unit weight of catalyst in the reactor.
26. Define catalyst resistance time.
The process period length in a fixed bad operation. In continuous motley – bed operation it is
equal to the at of catalyst.
27. Define space – time yield.
The net yield of a product from the reactor per hour per unit of effective reactor volume
28. Define stoichiometry and stoichiometric ratio.
Stoichiometry is a theory of the proportions in which the chemical species combine with one
another.
Stoichiometric ratio is the ratio of stoichiometric co- efficient of two molecular species/
components in the balanced reaction equation. Consider a chemical reaction.
CO+ 2H2 →CH3OH
For the above equation, the stoichiometric ratio of H2 to CO is 2/1= 2
29. What is a stoichiometric equation?
The stoichiometric equation of a chemical reaction is a statement indicating relative moles of
reactants and products that take part in the reaction. For example, the stoichiometric equation,
CO+ 2H2 →CH3OH indicates that one molecule (mol or Kmol) of CO reacts with two
molecule of hydrogen to produce one molecule of methanol.
30. What is a stoichiometric proportion?
Two reactants A and B are said to be present in stoichiometric proportion, if the rate of moles
of A present to the moles of B present is equal to the stoichiometric ratio obtained for the
balanced reaction equation. Consider a chemical reaction.
CO+ 2H2 →CH3OH
For the reactants in the above reaction to be present in stoichiometric proportion, there must
be 2 moles of H2 for every mole of CO present in the feed to the reactor.
Part B
1. A solution contains 50% benzene, 30% toluene and 20% xylene by weight at
a temperature of 100oC. The vapours are in contact with the solution. Calculate
the total pressure and the molar percentage compositions of the liquid and the
vapour. The vapour pressures and molecular weights are as follows:
Components
Benzene
Toluene
Xylene
Vapour pressure at 100oC
1340mm of Hg
560mm of Hg
210mm of Hg
Mol. Wt.
78
92
106
Solution:
Basis: 100 kg of solution.
Molar percentage composition of liquid:
Constituents
Benzene
Toluene
Xylene
Total
Amount kg
50
30
20
100
Mol. Wt.
78
92
106
Mol.
0.641
0.326
0.189
1.156
Mol%
55.45
28.20
16.35
100.00
Calculation of partial pressure:
By Rault’s law,
P=V.P. Mol. Fraction in liquid
Where, p= partial pressure
Pbenzene = 1340 0.5545=743.0 mm Hg
Ptoluene = 560 0.2820 = 157.9 mm Hg
Pxylene = 210 0.1635 =34.3 mm Hg
Total pressure =p=743.0+157.9+34.3
=935.2 mm of Hg
Vapour composition:
p
Total pressure
where, y=Mol. fraction in vapour.
y
743
0.7945 79.45%
935.2
157.9
Ytoluene
0.1688 16.88%
935.2
34.3
Yxylene
0.0367 3.67%
935.2
Ybenzene
2. The spent acid from a nitrating process contains 33% H 2SO4 36% HNO3 and
31% water by weight. This acid is to be strengthened by the addition of
concentrated sulphuric acid containing 95% H 2SO4and concentrated nitric acid
containing 78% HNO3. The strengthened mixed acid is to contain 40% H2SO4
and 43% HNO3. Calculate the quantities spent and concentrated acids that
should be mixed together to yield 1500 kg of the desired mixed acid. (A.M.I.E.
Exam. Summer 1975, Winter 1978 and summer 1979).
Solution:
Basis: 1500 kg of the desired mixed acid.
Let, x=wt. of waste acid (in kg) required
Y=wt. of conc H 2SO4 (in kg) required
Z= wt. of conc. HNO3 (in kg) required
Overall material balance is,
X+y+z=1500
….(1)
Sulphuric acid balance is,
0.33x+0.95y=0.4 1500=600
Nitric acid balance is,
……(2)
…..(3)
0.36x + 0.78z=0.431500=645
600-0.33x
0.95
645-0.36x
From Eq. (3), z=
0.78
From Eq. (2), y=
.....(4)
.....(5)
Putting (4) and (5) in Eq. (1)
600 0.33x 6.45 0.36x
1500
0.95
0.78
x-(0.347x+0.462x)
x
or,
=1500-(631.58+826.92)
or
0.191x=41.5
Therefore,
From (4),
From(5),
x=
41.5
217.3kg
0.191
y=556.1 kg
z=726.6 kg.
Weights of waste (spent) acid, conc. H 2SO4 and co. HNO3 required are 217.3, 556.1
and 726.6 kg respectively.
3. A solution of potassium dichromate in water contains 13% K 2Cr 2O7 by
weight, 1000 kg of this solution is evaporated to remove some amount of
water. The remaining solution is cooled to 20oC. If the yield of K 2Cr2O7 crystals
is 80%, calculate the amount of water evaporated. Solubility of K 2Cr2O7 is 1.390
kg mole per 1000 kg water (at 20 oC).Atomic weights: K39, Cr52.
Solution:
Basis:1000 kg of 13% K2Cr 2O7 Solution,
Assumption: No water of crystallization with K 2Cr2O 7 crystals.
Let,
F= amount of feed solution, kg.
E= amount of evaporation, kg.
C= amount of crystals formed, kg.
M= amount of mother liquor left after crystallization, kg.
Mol. wt. of K2Cr2O 7 =294.
XF=0.13
XC =1.0
0.39 294
0.103
0.39 294 1000
K2Cr2O7 in original solution = 130 kg.
xM
Yield of crystals = 80%= 130 0.8=104kg
So,
C=104kg
Overall material balance gives,
or
F=E+M+C=E+M+104
…..(1)
M-(F-E-104) =(1000-E-104)=896-E
K2Cr2O7 balance gives,
FxF=MxM+Cxc
….(2)
(since, xE=0, ExE term is not there)
So, 10000.13=(896-E) 0.103+104
0.103 E=92.29 + 104 – 130
i.e.,
E=643.6 kg
Amount of water evaporated = 643.6 kg.
4. 1000 kg of sodium carbonate solution containing 25% Na 2CO3 is subjected
to evaporative cooling, during which process 15% of the water present in the
solution is evaporated. From the concentrated solution Na 2CO3, 10H2O
crystallizes out. Calculate how much crystals would be produced if the
solubility of Na2CO3, 10H2O is 21.5 gm per 100gm of water.
Solution:
Basic: 1000kg of sodium carbonate 25% solution (F).
Let
C=crystals formed.
E=evaporated water
M=mother liquor
Water present in the solution = 750kg.
Water evaporated = 750 0.15=112.5kg
F=C+E+M=C+112.5+M
Or,
i.e.,
1000=112.5=C+M
M=(887.5-C)
Mol. wt. of Na2 CO 2, 10H 2O=286
106
xC
0.371
286
21.5
xM
0.177
121.5
Na2CO3 balance is,
FxF=MxM+CxC
10000.25=(887.5-C)0.177+0.377C
Solving,
C=479.4 kg.
Crystals formed = 479.4 kg.
5. Phosphorous is prepared by heating in the electric furnace a thoroughly
mixed mass of calcium phosphate, sand and charcoal. In a certain charge
silica used is 10% in excess of that theoretically required to combine with all
the calcium to form the silicate and the charcoal used is 40% in excess of that
required to combine as carbon monoxide, with the oxygen that would
accompany all the phosphorous as the pentoxide.
(i)
(ii)
Calculate the composition of the original charge.
If the decomposition of the phosphate is 90% complete and the
reduction of the phentoxide is 70% complete, calculate the amount of
phosphorous produced per 100 kg of charge.
Atomic weights: P-31.0, Si-28.0, Ca-40.0.
Solution:
Basis: 1kg mol of calcium phosphate decomposed.
Reaction: (i) Ca3 (PO 4)2+3SiO2 3CaSiO3+P2O 5
(310)
(360)
9142)
(ii) P2O5 + 5C
(142) (512)
2P +5CO
(231)
…(1)
…(2)
310 kg of Ca3 (PO4) 2 require 180kg of SiO2
Since SiO 2 is 10% excess.
So,
SiO2 supplied = 180 1.1=198kg
142 kg of P2O5 require 60kg carbon.
Since charcoal (carbon) is 40% excess.
So carbon supplied = 60 1.4=84 kg.
(i) Charge analysis:
Constituents
Ca3(PO 4)2
SiO2
Charcoal
Total
Amount kg
310
198
84
592
(ii) 100kg. of charge contains 52.36 of calcium phosphate
Since the phosphate decomposition is 90%
Amount of P 2O5 formed (reaction -1)
Wt%
52.36
33.45
14.19
100.00
52.36 0.9
142
21.58kg.
310
Reduction of P2O5 is 70% complete
P2O5 reduced = 21.58 0.7=15.11kg
From reaction (2),
142kg P 2O5 produces 62kg. P.
62
15.11 P2O5 produces
15.11 6.6kg.
142
Amount of phosphorous produced = 6.6kg.
6. Bottled liquid gas of the following composition is sold for house – hold use:
Component
Composition, Mole%
V.Pr. at 30oC, bar
n-Butane
50
3.4
Propane
45
10.8
Ethane
5
46.6
Determine (i) the pressure of the system and the equilibrium vapour
composition at 30 oC and (ii) if all the ethane be removed from the liquid, the
pressure of the system and the vapour composition at 30 oC. Assume Rault’s
law is applicable.
Solution:
Basis: 1kg mole of the bottled gas.
Components
n-butane
Propane
Ethane
Mol frac. In liquid
0.50
0.45
0.05
Vapour pr., bar
3.4
10.8
46.6
(i) Pressure of the system
= partial pressure = 8.89 bar
Equilibrium vapour composition:
y
Partial pressure
Pr essure of the system
Pbu tan s
Total pressure
1.70
=
100 19.12%m
8.89
4.86
Similarly, y propane
100 54.67%m
8.89
ybu tan s
Partial pr., bar
1.7
4.86
2.33
y ethane
2.33
100 26.21%m
8.89
(ii) If ethane is removed, the liquid left will be
=1.0-0.05=0.95 kg mol.
Components
Amount mol
Mol% in – liquid
p.p., bar
Butane
Propane
Total
0.50
0.45
0.95
52.63
47.37
100.00
1.789
5.116
6.905
Vapour comp.
mol%
25.91
74.09
100.00
Pressure of the system = p.p. = 6.905 bar.
7. Carbon tetrachloride is made as following:
CS2+3cl2 CCl4 + S2Cl2
The product gases are found contain
CCl4
S2Cl2
CS2
Mol%
23.3
23.3
1.4
Cl2
32.0
Calculate (i) The percentage of the excess reactants used.
(ii) The percentage of conversion.
(iii) The kg of CCl 4 produced per 100kg of cl 2 converted.
Solution:
Basis: 100kg mol of product gas.
Product gas analysis:
Components
CCl4
S2Cl2
CS2
Cl2
Mol
23.3
23.3
1.4
32.0
In this case CS2 is the limiting reactant and Cl 2 is the excess reactant. From the
product analysis, it is seen that 23.3 kg mols of CS 2 have been converted.
(i)
(ii)
CS2 unconverted = 1.4 kg mol
Percentage conversion
23.3 100
94.33
23.3 1.4
Chlorine required for complete conversion of
CS2 = 24.7 3.0=74.1 kg mol.
Chlorine used
Chlorine in product
=
=
23.33=69.9 kg mol.
32.0
Total chlorine used
Excess chlorine used
=
=
=
69.9+32.0=101.9 kg mol
101.9-74.1
27.8kg.mol.
(i) Percentage of the excess reactant used
27.8
100 37.52
74.1
(ii) From the reaction,
371.0kg of cl 2 yield 154 kg of CCl4
100kg of Cl2 yield
154
100
3 71.0
=72.3 kg of CCl4
8. SO2 reacts with pure O2 to form SO3. If the reaction is carried out with 100%
excess oxygen as necessary for complete oxidation, but under such low
temperature and pressure that the reaction goes only 60% to completion,
calculate:
i.
ii.
iii.
iv.
v.
The mole fraction of SO2, O2 and SO3 in reactants and products
The weight fraction of the three gases in reactants and products
The average molecular weights of reactants and products.
The partial pressure of the products if the total pressure is 2 atm.
If the products were cooled to 20C and the pressure reduced to 720
mm Hg., what volume of gas would be obtained from 100 kg of SO2.
Solution:Basis : 100 gm. Mol. of SO 2 gas
Reaction : SO2 +
1
O2
SO3
2
Oxygen required = 50 gm mol.
Oxygen supplied (by 100% excess) = 50 2 = 100 gm. mol.
Reaction is 60% complete
SO3 formed
= 60 gm mol.
(a) Oxygen left
= 100 – 30 = 70 gm mol.
Analysis of reactants:Constituents
Mol
Mol fraction
SO2
100
0.500
O2
100
0.500
Constituents
Mol
Mol fraction
SO2
SO3
40
60
0.235
0.353
O2
70
0.412
Analysis of products:-
(b) Weight fraction of reactant gases:Constituents
Mol
Wr. in gm
Wt. fraction
SO2
100
6400
0.667
O2
100
3200
0.333
Weight fraction of product gases:Constituents
Mol
Wr. in gm
Wt. fraction
SO2
40
2560
0.267
SO3
60
4800
0.500
O2
70
2240
0.233
(c) Average molecular weight of reactants
= 0.5 64 + 0.5 32 = 48.0
Average molecular weight of products
= 0.353 80 + 0.235 64 + 0.412 32
= 56.46
(d) Partial pressure = Total pressure Mol fraction
p.p. SO2 = 2 0.235 = 0.470 atm.
p.p. SO3 = 2 0.353 = 0.706 atm
p.p. O2 = 2 0.412 = 0.824 atm
100
1.56 gm mol
64
170
1.56 2.65 gm mol of product
100 gm mol of SO2 yield
100
(e) 100 gm SO2 =
Volume of gas
=
=
nRT
P
2.65 62.36 273 20
720
67.25 litres
So, gas produced for 100 kg of SO 2
= 67.25 1000 67,250 litres
9. Two pure organic chemicals A and B are introduced into the apparatus as
shown in the figure below for the purpose of making the compound AB.
Stream flows are adjusted so that mole ratios of A to B in the reactor is 4:1. At
the temperature and pressure employed, this mole ratio effects complete
consumption of B, 90% of the B forms AB, the desired product. The remainder
forms A2B, a useless by product. A2B is incapable of further reaction. The
effluent stream from the reactor is cooled to 20 C, at which temperature A and
AB are completely immissible. A2B however is soluble in A and AB. The
distribution coefficient is
C
k AB 6
CA
where, C AB = mol A2B / mol AB
C A = mol A2B/mol A.
Calculate the amount of all material flowing in each numbered stream,
per 100 mole of desired product AB in stream 4.
Solution:Basis : 100 mol of B reacted.
AB
A + B
Since, 90% of B forms AB, 90 mol of B reacts with 90 mol of A to give 90 mol AB
A2B
2A + B
10 mol B reacts with 20 mol A to give 10 mol A 2B
Mole of A reacted = 90 + 20 = 110
For 90 mol AB produced, 110 mol A reacted
100 mol AB produced =
110
100 122 mol A reacted
90
Stream (1) = 122 mol.
Mol of B required for 100 mol AB,
100
100 111
90
Stream (2) = 111 mol.
The only place that A2B can leave the process is with AB. After startup, the recycle stream composition will change constantly until a steady state has
been reached. Under steady state conditions,
10
100 11.0
90
So, stream (4) will have 100 + 11 = 111 mol
Mol A 2B with stream (4) =
Calculation for Stream (3):
Stream (3) and (4) are in equilibrium with respect to the concentration of A2B.
CAB
11
100
CAB
6 (given)
CA
11 mol A 2B
600
mol A
mol A 4
; mol A = 4 111 = 444
mol B 1
CA
make-up A is 122 mol.
Hence recycle
A = 444 – 122 = 322 mol
CA = concentration of A2B in stream (3)
= 11/600
Mol. A 2B in stream (3) =
11
322 5.90
600
So, stream (4) will have
(i) 100 mol AB; (ii) 11 mol A 2B
Stream (3) will have
(i) 322 mol A; (ii) 5.9 mol A 2B
10. 1000 kg of an impure limestone which analysis 96% CaCO 3 and 4% inert
material is reacted with a sulphuric acid solution containing 70% sulphuric
acid and 30% water. The reacting mass is heated and all the CO 2 generated is
driven off together with some of the water. The analysis of the final solid ‘cake’
is
CaSO4 – 86.54%
CaCO3 – 3.11%
H2SO4 - 1.35%
H2O
- 6.23%
Inerts - 2.77%
Calculate :(a) the degree of completion of the reaction
(b) mass of acid solution fed
(c) mass of gas driven off
(d) composition of gases driven off
Solution:Basis : 100 kg of cake formed
Reaction : CaCO3 + H 2SO4 CaSO 4 + CO2 + H2O
Amount of CaSO 4 formed = 86.54 kg.
100
86.54 63.63 kg.
136
CaCO3 unconverted = 3.11
CaCO3 consumed =
(a) Degree of completion of the reaction
(b) Acid used
Acid in cake
Acid supplied
Lime stone used
63.63
95.34
63.63 3.11
98
86.54 62.36 kg.
136
= 1.35 kg.
=
= 62.36 + 1.35 = 63.71 kg.
= 63.63 + 3.11 + 2.77 = 69.51 kg.
69.51 kg limestone require 63.71 kg H 2SO4
1000 kg limestone require
=
63.71
1000 916.56 kg H2SO4
69.51
Acid solution fed
=
916.56
1309.37 kg.
0.7
(c) Water in acid
= 1309.37 – 916.56 = 392.81 kg.
18
63.63
100
= 11.45 kg (per 69.51 kg, limestone)
Water formed
=
Total water formed by reaction
=
11.45
1000 164.72 kg.
69.51
Amount of water (acid solution and reaction)
= 392.81 + 164.72 = 557.53 kg.
Water left in cake per 1000 kg limestone
6.23
1000 89.63 kg.
69.51
Water evaporated
= 557.53 – 89.63 = 467.9 kg.
CO2 formed
=
44
63.63 28.00 kg.
100
For 1000 kg of limestone,
CO2 formed
=
28
1000 402.82 kg.
69.51
Mass of gas driven off = 467.90 (H 2O) + 402.82 (CO2)
= 870.72 kg
(d) Composition of gases driven off:
Constituents
CO2
H2O
Weight, kg
402.82
467.90
Total
Kg. mol.
9.155
25.994
35.149
Mol%
26.06
73.95
100.00
11. 4000 kg of KCl are present in a saturated solution at a temperature of
80C. The solution is cooled to 20C in an open tank. The solubilities of KCl at
80C and 20C are 55.0 and 35.0 parts per 100 of water respectively. Calculate:
(a) Assuming water equal to 3% by weight of solution is lost by
evaporation, weight of crystals obtained.
(b) The yield of crystals neglecting loss by evaporation.
KCL crystallizes without any water of crystallization.
Solution:Molecular weight of KCl crystals
= 39 + 35.5 = 74.5
Let F, M, E and C represent the amounts of feed, mother liquor, evaporation
and crystals and xF, xM, xE and xC their solid fractions respectively.
xE 0
55
0.355
100 55
35
xM
0.259
100 35
xF
x C 1.0
Overall material balance is,
F=E+M+C
KCl balance is, FxF = Mxm + Cxc
Amount of KCl in feed = 4000 kg.
So,
FxF = 4000
4000
11,268 kg
i.e.
F=
0.355
….(1)
….(2)
(a) E = 0.03 F = 0.03 11,268 = 338 kg.
From equation (1),
(11,268 – 338 – C) = M
Putting Equation (3) in (2),
4000 = (10,930 – C) 0.259 + C
C = 1578 kg.
(b) When E = 0,M = (11,268 – C)
Equation (2) gives
4000 = (11,268 – C) 0.259 + C
Solving,
C = 1460 kg.
UNIT-IV
PART-A- QUESTION & ANSWERS
1. Define humidity.
….(3)
The humidity (H) of a gas in generally defined as the at of water per unit cut of moisture free
gas.
2. Defined Molar humidity.
The number of molar of water per mole of moisture free gas.
3. Define Relative humidity.
It is defined as the ratio of the partial pressure of condensable vapour in gas phase to the
vapour pressure of liquid.
4. Define saturation humidity.
It is defined as the absolute humidity of vapour gas mixture at 100% saturation.
5. What is Absolute Humidity?
This is defined as the weight of water vapour per unit weight of dry air in the mixture.
H = WA/WB = weight of water vapour / weight of dry air.
6. Define Percentage humidity:
It is the ratio of the molal humidity at unsaturated conditions to the molal humidity at
saturation.
%H = Hm /Hms x100 %
7. What is humidification?
The operation that is carried out increase the amount of vapour in a gas stream.
8. Define dehumidification.
It is carried to reduce the vapour content of gas stream is known as dehumidification.
9. Define Dew point.
If an unsaturated mixture of vapour and gas is cooled the relative amounts of the components
and the percentage composition by volume will at first remain unchanged.
10. Define vaporization processor.
It is the Relationship between the quantity of volumes of gases entering and leaving and the
quantity of material evaporated.
11. Define Condensation.
The relative saturation of a Partially saturated mixture of vapour and gas may be increased.
12. Define Adiabatic saturation temperature.
It is temperature that the vapour gas mixture could reach if its were saturated through an
adiabatic process.
13. What is partial saturation?
If a gas contains a vapour in such properties that is partial pressure in less than the vapour
pressure at the liquid.
14. Define molar saturation.
It is defined as the ratio of moles of vapour (condensable) to the moles at dry gas.
15. Define Relative Saturation:
It is the percentage ratio of the partial pressure of water vapor at the given conditions to the
partial pressure of water vapor at saturation.
%R.S = PA/PASX100
16. Define saturated vapour.
The gas phase in saturated with liquid, it contains all the liquid if can hold at temperature and
pressure of system.
17. At 26.42 in Hg the dry bulb temp of air is 150 of, and 1fg cut bulb temp is 120 of
obtain humidity.
Solution:
Molar humidity
=
0.116
Molar humidity correction =
0.020
-------------Corrected molar humidity
0.136
18. Define humidity chart.
Hmax =
Pg
-----------P-Pg
19. Moist air in found to contains 8.18m of water vapour per cubic root at a
temperature of 30oC. Calculate the temp to which it must be heated.
Solution:
Vapour pressure of Water
=
19.4
-------0.15
=
130mm Hg
20. A mixture of benzene vapour and air contains 10.1% benzene by volume.
Calculate dew point of mixture when at temp of 25 oC and pressure of 750mm Hg.
Solution:
Partial pressure of benzene = 0.101 750
=75.7mmhg.
21. Calculate the dew point when the when the mixture is at a temp of 30 oC and a
pressure of 750mmHg.
Solution:
Partial pressure of benzene =
75.7 min
Dew point
=
20.0oC.
22. What is saturation mixture?
If a gas is holding the vapour to its maximum capacity, then the gas is said to be saturated
with vapors. At saturation,
It is found that P.P. of vapor (A) in the mixture it equal to the vapour pressure of A
23. What is unsaturated mixture?
If the gas is holding the vapor less than its capacity the mixture is said to be unsaturated.
24. Define dry bulb temperature?
This is the thermometer dip. temperature of wet air.
25. Define wet bulb temperature?
When an unsaturated air is passed over water, the water gets cooled and attains a constant
temperature which will be lower than the Dry Bulb Temperature of the air. This is called as
the Wet Bulb Temperature of the air.
If the saturation is less, then the difference (D.B.T. – W.B.T.) is more at the saturation,
D.B.T. = W.B.T.
26. Define humid heat?
It is defined as the specific heat of one kg of dry air and the moisture contained by it.
Cs = 0.24+0.46 H Kcal / kg dry air 0C
27. Define humid volume?
It is the volume occupied by dry air and the moisture contained in it and expressed m 3/kg dry
air.
VH = {MA + 1/MB} 22.414 {D.B.T.+273/273}1/Pt m3 /kg dry air.
28. Define Percentage excess air:
It is the percentage ratio of the amount of a substance present in excess to the amount
theoretically required. %excess air =excess amt. / theoretical amt. x 100
Part B
1. The dry bulb temperature and dew point of ambient air were found to be 302
K (29oC) and 291 K(18 oC) respectively, Barometer reads 100 kPa. Calculate:
(a) The absolute molal humidity,
(b) The absolute humidity,
(c) % RH,
(d) % Saturation,
(e) Humid heat and
(f) Humid volume
Data: Vapour pressure of water at 291 K = 2.0624 kPa vapour pressure of water
at 302 K=4.004 kPa.
Solution:
Air-water vapour mixture with DB = 302 K and DP = 291 K.
At DP, partial pressure of water in air is equal to vapour pressure of water.
PA = partial pressure of water vapour in air = 2.0624 kPa
P=total pressure = 100 kPa
Absolute molal humidity = Hm
PA
P PA
2.0624
100-2.0624
kmol water vapour
=0.02106
kmol dry air
=
.....Ans. (a)
Mol. Wt. H2O=18,Mol. Wt. air =29
Mol.wt.H2O
Mol.wt.air
18
=0.02106
19
kg water vapour
=0.0131
kg dry air
Absolute humidity =H= Hm
.....Ans. (b)
At saturation,
DB=WB=DP=302K
Vapour pressure at saturation i.e. at 302K=P6=4.004kPa.
% RH =
PA
100
P6
2.0624
100
4.004
=51.51
=
.......Ans(c)
P Mol.wt.H2O
Saturation humidity = H8 8
P P6 Mol.wt.air
4.004 18
H8
100 4.004 29
=0.02589=
kg water vapour
kg dry air
% saturation =
H
100
H5
0.0131
100
0.02589
….Ans.(d)
Humid heat = C a = 1.006+1.84H
=1.006+1.884 0.0131
=1.03kJ/kg dry air K
=
….Ans. (e)
H
Humid volume = VH
MA
1
22.4136
M
s
DB 101.325
P
273
0.0131 1
302 101.325
=
22.4136
100
19
18
273
=0.8846m3 / kg dry air
.....Ans.(e)
2. A mixture of benzene and dry air at a temperature of 303 K(30 oC) and a
pressure of 101.325 kPa is found to have a dew point of 288 K(15 oC).
Calculate:
(a) Percentage by volume of benzene.
(b) Moles of benzene per mole dry air.
Data: Vapour pressure of benzene at 288K=7.999 kPa.
Solution:
Basis : A mixture of benzene and dry air at 101.325 kPa.
PA = Partial pressure of benzene in mixture.
= vapour pressure at 288 K.
= 7.999 kPa.
For ideal gas mixture,
Volume % = mole % = pressure %.
PA
100
P
7.999
=
100
101.325
=7.9
Volume % of benzene=
....Ans.(a)
1 mol of mixture
Benzene present = 0.0791=0.079 mol.
Dry air present = 1-0.079=0.921 mol.
0.079
0.921
=0.086
Moles benzene per mole of air=
....Ans(b)
3. A mixture of acetone vapour and nitrogen contains 15.8% acetone by
volume. Calculate the relative and percent saturation of the mixture at a
temperature of 293 K(20oC) and a pressure of 101.325 kPa.
Data: Vapour pressure of acetone at 293 K = 24.638 kPa.
Solution:
Basis: 1 mol of acetone nitrogen mixture.
Mol % acetone = volume % acetone = 15.8
15.8
Molfraction of acetone in mixture = y =
100
P= total pressure = 101.325 kPa.
PA=partial pressure of acetone = y.P
=0.158 101.325
=64.98
…..(Ans)
Hm= Moles acetone per mole nitrogen in gas mixture
0.158
1 0.158
0.1876
Consider one mole of saturated mixture at 293 K(20 oC) and 101.325 kPa.
Volume % acetone Pr essure % acetone
in saturated mixture in saturated mixture
24.638
Volume % acetone
=
100
101.325
=24.31
Mole % acetone in saturated mixture = volume % acetone
= 24.31
24.31
1
100
=0.2431 mol
Acetone in saturated mixture =
Nitrogen in saturated mixture = 1-0.2431
=0.7569 mol
HmS moles acetone per mole nitrogen in saturated mixture
0.2431
0.7569
= 58.40
=
…..Ans.
4. A gas mixture containing benzene vapour is saturated at 101.325kPa and
323 K(50oC). Calculate the absolute humidity if the other component of the
mixture in (a) nitrogen and (b) carbon dioxide.
Data: vapour pressure of benzene at 323 K=36.664 kPa.
Solution:
Basis : A gas mixture at 101.325 kPa and 323 K.
(a) Nitrogen and benzene vapour mixture.
Mol Wt. N2=28, Mol Wt. C6H6=78
As the mixture is saturated at 323K, the partial pressure of benzene is equal to
vapour pressure of benzene at 323 K.
PA =36.664 kPa
P=101.325 kPa
Hm
PA
P PA
36.664
101.325-36.664
kmol benzene vapour
=0.567
kmol nitrogen
=
Absolute humidity = Hm
Mol wt.C6H6
Mol wt. N2
78
28
kg benzene
=1.579
.....Ans.(a)
kg nitrogen
=0.567
(b) Carbon dioxide – benzene vapour mixture.
Mol Wt. CO2 = 44
PA = 36.664 kPa
P =101.325 kPa
Hm
=
PA
P PA
36.664
101.325-36.664
=0.567
Kmol benzene vapour
kmol carbon dioxide
Absolutehumidity Hm
Mol wt. C6H6
Mol wt. CO2
=0.567
78
44
kg benzene
....Ans(b)
kg carbon dioxide
5. The dry bulb temperature and dew point of ambient air were found to be 303
K(30oC) and 289 K(16 oC) respectively. Calculate (i) the absolute molal humidity,
(ii) the absolute humidity, (iii) the % RH, (iv) the % saturation and (v) the humid
heat.
=1.005
Data: Vapour pressure of water at 289 K = 1.818 kPa
Vapour pressure of water at 303 K= 4.243 Kpa
Barometric pressure = 100 kPa
Solution:
Basis : Air – water mixture at DB = 303 K and DP = 289 K.
Partial pressure of water in air = Vapour pressure of water at DP
PA =1.818kPa
=Partial pressure of water vapour in air
P=total pressure
=100 kPa
Absolute molal humidity = H m=
PA
P PA
1.818
100 1.818
kmol water vapour
0.01852
kmol dry air
Mol Wt. H2O=18, Mol Wt. air = 29
.....Ans.(i)
Mol.Wt.H2O
Mol Wt. air
18
Absolute humidity = H 0.01852
29
kg water vapour
0.0115
kg dry air
Hm
....Ans.(ii)
At saturation,
DB=WB=DP=303K
Vapour pressure of water at saturation = 4.243 kPa
i.e. at 303 K
Pa = 4.243 kPa
P
A 100
Pa
1.818
100
4.243
42.85
%RH=
.....Ans(iii)
P Mol wt. H2O
8
P P3 Mol wt.air
4.243 18
Saturation humidity = H 8=
100 4.243 29
0.0275
kg water vapour
kg dry air
H
% Saturation = 100
H8
=
0.0115
100
0.0275
=41.82
.....Ans(iv)
Humid heat = C8 1.006 1.84H
=1.006+1.84 0.0115
=1.0272 kJ/kg dry air. K ......Ans(v)
6. A mixture of benzene vapour and nitrogen gas at 297 K and 100 kPa has a
relative humidity of 60%. It is desired to recover 80% of benzene by cooling a
mixture to 283 K(10oC) and compressing to a suitable pressure. Find out the
pressure required for above duty.
Data: Vapour pressure of benzene at 297 K=12.2kPa
Vapour pressure of benzene at 283 K = 6kPa.
Solution:
Basis: A mixture of benzene and nitrogen with 60% RH at 297 K.
PA
100
P8
where PA = partial pressure of benzene
P8 = vapour pressure of benzene
=12.2 kPa.
p
60 A 100
12.2
pA = 8.32 kPa
%RH
at 297 K
In 1m3 of the benzene – nitrogen mixture
Mass of benzene =
pA .MA
V
RT
where pA = 7.32 kPa
MA = 78
T =297 K
R =8.31451 m3. kPa/kmol. K
Mass of benzene=
7.32 78 1
8.31451 297
=0.31kg
Mass of nitrogen =
100-7.32 28 1
8.31451 297
=1.051 kg
Hence,humidity=H=
0.231
1.051
=0.22kg/kg
Hm
PA
P PA
7.32
100-7.32
kmol benzene
=0.07898
Kmol nitrogen
=
Humidity =H=Hm
Mol wt.C6H6
Mol wt. N2
=0.07898
=0.22
78
28
kg benzene
kg nitrogen
In order to recover 80% of benzene, the humidity must be reduced to 20% of
its initial value. As the vapour will be in contact with liquid benzene, the nitrogen will
be saturated with benzene vapours and hence at 283 K.
Ho 0.22
=0.044
20
100
kg benzene
kg nitrogen
Ho humidity of saturated gas.
p M
Ho s A
P Ps MB
where Ho=0.44
MA=78, MB =28
P= total pressure
O8=6kPa at 283 K
(p.p. of benzene vapour in saturated gas is equal to vapour pressure of benzene at
283 K).
6 78
0.044
P 6 28
0.044(P-6)=16.71
0.044P-0.264=16.71
P=385 kPa
Total pressure to which gas should be compressed = 385.8 kPa
….Ans.
7. A mixture of acetone vapour and nitrogen contains 14.8% acetone by
volume, calculate the following at 293 K (20oC) and a pressure of 99.33 kPa.
(a) Partial pressure of acetone.
(b) Moles of acetone per mole of nitrogen.
(c) Relative saturation of mixture at 293 K(20oC), and
(d) Percentage saturation of mixture at 293 K(20oC).
Data: Vapour pressure of acetone at 293K=24.638 kPa.
Solution:
Basis: 1 kmol of mixture of acetone vapour and nitrogen at 293K.
Mole % acetone in mixture = volume % acetone
= 14.8
Mole fraction of acetone in mixture = 14.8/100
=0.148
P =total pressure = 99.33 kPa.
PA=partial pressure of acetone vapour
Y =molefraction of acetone in mixture
PA=y.P
=0.1488 99.33
=14.7 kPa.
Acetone in mixture =
14.8
1
100
=0.148 kmol
Nitrogen in gas mixture = 1-0.148
=0.852 kmol.
0.148
0.852
=0.174 ....Ans(b)
Hm kmols acetone per kmol nitrogen =
Vapour pressure of acetone at 293K=24.638 kPa
P8= 24.638 kPa
% relative saturation =
pA
100
p8
14.7
100
24.638
=59.66
...Ans(c)
=
Consider one kmol of saturated mixture at 293 K (20 oC) and 99.33 kPa.
Volume % acetone = pressure % acetone
P
3 100
P
24.638
100
99.33
24.80
Mole % acetone = volume % acetone = 24.8
Acetone in saturated mixture = 0.248 1
=0.248 kmol
Nitrogen in saturated mixture = 1-0.248
=0.752 k mol
HmS mole acetone per mole of nitrogen in saturated mixture
0.248
0.752
=0.329
% saturation =
Hm
100
Hms
0.174
100
0.329
=52.9
=
....Ans.(d)
8. An air – water vapour mixture has relative humidity of 80% at 293 K(20 oC)
temperature and 100 kPa pressure. Calculate: (a) molal humidity of air, (b)
molal humidity of air if its temperature is reduced to 283 K(10 oC) and the
pressure is increased to 174.65 Kpa condensing out some of the water, (c)
Calculate the weight of water condensed from 500 kg of original wet air in the
process of part (b), (d) calculate the final volume of the wet air of the part (c).
Data: Vapour pressure of water at 293 K(20oC) = 2.40 kPa
Vapour pressure of water at 283K(10oC) = 1.266 kPa.
Solution:
Basis: Air-water vapour mixture of RH= 80%at 293 K and 100kPa.
Vapour pressure of water at 293 K= 2.40 kPa
P8=2.40 kPa
PA= partial pressure of benzene
P8 = vapour pressure of water in air
P
%RH= A 100
P8
pA
100
2.40
pA = 1.92 kPa
80
Initial molal humidity Hm
PA
P PA
7.32
100-7.32
kmol water vapour
=0.01958
Kmol dry air
Hm =
....Ans(a)
Final partial pressure of water vapour (i.e. at 293 K) is equal to the vapour pressure
of water at 283 K (10oC) as air becomes saturated at that stage.
PA at 283K=1.266 kPa
P at 283K = 174.65 kPa
Final molal humidity (i.e. at 283 K and 174.65 kPa).
PA
P PA
1.266
174.65-1.266
kmol water vapour
Final molal humidity =0.0073
Kmol dry air
=
....Ans(b)
Molal humidity at 293 K=0.01958
Humidity at 293 K (20oC)
18
29
kg water
=0.0122
kg dry air
o
Humidity at 283 K(10 C)
on weight basis
on weight basis
=0.01958
18
29
kg water
=0.00453
kg dry air
=0.0073
At 293 K(20oC), air water vapour mixture contains 0.0122 kg water vapour (moisture)
per kg dry air i.e. per 1.0122 kg of wet air.
Consider 500 kg of wet air,
Water present at 293 K in wet air (i.e. initial)
0.0122
500
1.0122
=6.026 kg
Dry air in wet air = 500-6.026
=493.974 kg
water present at 283K (i.e. final)=
0.00453
493.974
1
=2.238.kg
Water condensed from 500 kg wet air.
=Water present at 293K-Water present at 288K
=6.026 – 2.238
=3.788kg
Total amount of wet air at 283K(10oC)
=493.974 +2.238
=496.212kg
2.238
18
=0.1243 kmol
Water in wet air at 283 K=
Dry air in wet air at 283 K=
493.974
29
=17.033 kmol
Amount of wet air at 283 K=17.033 kmol
=17.1573 kmol
PV=nRT
nRT
V
P
Where n= 17.1573 kmol, T= 283 K
m3 .kPa
P=174.65 kPa, R= 8.13451
kmol.K
17.1573 8.13451 283
174.65
3
=231.2m
…..Ans. (d)
Volume of wet air at 283 K(10oC)=V=
9. Carbon dioxide contains 0.053 kmol of water vapour per kmol of dry CO 2 gas
at temperature of 308 K(35oC) and a total pressure of 100 kPa.
Calculate: (a) Relative saturation of the mixture, (b) Percentage saturation of
the mixture, and (c) The temperature at which the mixture must be heated in
order to achieve the relative saturation of 30%.
Data: Vapour pressure of waer at 308 K= 5.60 kPa
Vapour pressure of water at 330 K(57oC) = 16.745 kPa
Solution:
Basis: 1 kmol of dry CO 2 in the mixture.
Water vapour associated with CO 2 in the mixture.
Amount of CO 2 water mixture = 1+0.053
= 1.053 kmol
0.053
Molfraction of water vapour in gas mixture = y=
1.053
=0.0503
P= total pressure = 100 kPa
Partial pressure of water vapour = pA =y.P
=0.0503 100
=5.03 kPa
P8 = Vapour pressure of water at 308 K=5.60 kPa
PA
100
P8
5.03
100
5.60
=89.82
Relative saturation at 308 K(35oC)=
….Ans(a)
Hm = kmol water vapour per kmol CO 2 at 308 K(35oC)
0.0503
0.0503
1
Consider one kmol of saturated mixture at 308 K(35 oC) and 100 kPa.
Volume % of water = pressure % of water
5.6
100
=
100
=5.6
(Partial pressure = vapour pressure for saturated gas)
5.6
1
100
=0.056 kmol
Water vapour in mixture =
CO2 in gas mixture
=1-0.056
=0.944 kmol
HmS k mol water vapour per kmol CO 2 in saturated mixture
0.056
0.944
=0.0593
% saturation =
Hm
100
Hms
0.053
100
0.0593
=89.37
=
...Ans(b)
Relative saturation 30%
Relative saturation
PA
100
P8
Where PA = partial pressure of water vapour
Ps = Vapour pressure of water
5.03
100
P8
Solving P8 = 16.767 kPa
30
Water exerts a vapour pressure of 16.767 kPa (approximately equal to at 330
K(57oC) as per the data supplied.
Hence, the mixture should be heated to 330 K(57oC) to get the desired relative
saturation as vapour pressure at 330K (57oC).
Given (data) is approximately matching the calculated value.
Temperature two which gas should be heated = 330K(57 oC)
…Ans.
10. A gas mixture containing carbon disulphide vapour [CS2] at a temperature
of 300K (27oC) and a pressure of 100 kPa has the percentage saturation of 70.
Calculate the temperature to which gas must be cooled at constant pressure
so as to condense 40% of CS2 present.
Data: Vapour pressure of CS2 at 300 K(27oC)=52.662 kPa
Vapour pressure of CS2 at 287 K (14oC) = 31.997 kPa
Vapour pressure of CS2 at 288 K(15oC) = 33.33kPa
Solution:
Basis: A gas mixture with percentage saturation of 70 at 300 K(27 oC).
Vapour pressure of CS2 at 300K=52.662 kPa
P8=52.662 kpa
P=100 kpa
% saturation =70
pA / P pA 100
P8 / P Ps
pA /100 pA
70=
100
52.662 /100 52.662
% saturation =
Solving we get,
PA=43.778kPa
Partial pressure of CS2=43.778 kPa
Consider 1 kmol of mixture
43.778
1
100
=0.43778
=0.4378 kmol
CS2 in the mixture
Other gas component in mixture = 1-0.4378
=0.5622
Amount of CS 2 condensed = 0.40 0.4378
=0.1751 kmol
CS2 present finally in the mixture = 0.4378-0.1751
= 0.2627 kmol
Amount of gas mixture (final) = 0.5622+0.2627
=0.8249 kmol
0.2627
100
0.8249
=31.85
Mol % CS 2 =
Pressure % CS2 = Mole % CS2 = 31.85
pA
100 31.85
P
pA
100 31.85
100
p A 31.85kPa.
As the mixture is saturated, the partial pressure of CS 2 is equal to vapour pressure at
final temperature, the vapour pressure of CS 2=31.85 kPa.
From the data cited the temperature at which the vapour pressure of CS2 is
approximately 31.85 kPa is 287 K(14oC).
Therefore, the gas must be cooled to a temperature of 287 K(14 oC).
Temperature to which gas must be cooled = 287 K(14 oC)
UNIT-V
PART-A- QUESTION & ANSWERS
1. Define total heating value.
A fuel is heat evolved in its complete combustion under constant pressure at a temperature of
25C.
2. Define net heating value.
It is defined as final state of the water in the system after combustion is taken as vapour at
25C.
3. Define Coke and Carbon.
The heat of formation of carbon is the form of the diamond in accurately known and is equal
to 0.45 32 Kcal per g-atom.
4. What is coal analysis?
It is addition to organic matter it contain minerals constituent of the plants from which it was
formed and inclusion.
5. Define Combustible.
The sum of the fixed carbon and volatile matter of a coal.
6. Define Molecular at of petroleum fraction.
The average molecular at of petroleum fractions may be average boiling point and gravity.
7. Define Hydrogen content.
The relationship between hydrogen content and characteristic factor fro material of constant
boiling points.
8. Define incomplete combustion of fuels.
The standard heating values at fuel correspond to conditions at complete combustion of all
carbon to carbon dioxide gas.
9. Define enthalpy of water vapour.
The enthalpy of superheated water vapour referred to the liquid at 25C
10. Define enthalpy of vapourisation.
The enthalpy of water is equal to the heat of sum vapourisation at 25C.
11. Define weight of dry gaseous products.
A direct Measurements of the dry gaseous products from a combustion process.
12. Define weight of Dry-Air supplied.
Direct Measurement of the ut (as) volume of air used in combustion to the gaseous products.
13. Define wet of moisture in air.
The wet of moisture per mole of dry air depends on the temperature pressure, relative
humidity of the air.
14. Calculate the total volume of gaseous products.
Solution:
Moles of wt gas = 45.1 +2.892 = 48.01b moles
29.92 948
48.0 359
29.08 492
34.150cuft
15. Calculate the volume of wet air.
Solution:
Moles of air = 4.6.6+0.559 47.21 b –moles
Volume at 73’f, 29.08 in
29.92 533
HS 47.2 359
29.08 492
18,870 cuft
16. Define Heating values of fuel
The total heating value of a fuel is the heat evolved in its complete combustion under
constant pressure at a temperature of 25 0C when all the water initially present as liquid in the
fuel and present in the combustion products are condensed to the liquid state.
The net heating value is defined except that the final state of the water in the system after
combustion is taken as vapor at 25 0C.The total heating value is also termed the `gross`
heating value. The net heating value is also termed as the lower heating value.
17. Define Rank of Coal
Fuel ratio of a coal is defined as the ratio of percentage of fixed carbon to that of volatile
matter. The rank of the coal whether bituminous or anthracite, may be estimated from its fuel
ratio.
18. Define Heating value of coal
The total heating value of a coal may be determined by direct calorimetric measurement and
expressed as Btu/lb. The net heating value is obtained by subtracting from the total heating
value the heat of vaporization at 25 0C of the water present in the coal and that formed by the
oxidation of the available hydrogen.
Net H.V. = total H.V. – 8.94 * H * 1050
Where H.V. = heating value, H wt. Fr. Of total hydrogen including available H 2 , H2 in
moisture and H 2 in combined H2O
19. Define proximate analysis?
It is determination is made of each of the proximate analysis defined groups of
moisture, volatile matter, ash and fixed carbon (obtained by deducting %moisture, %volatile
matter, % ash from 100).
20. Define Ultimate analysis:
It is a method of analysis of a fuel which gives the complete composition of
iii) N2 iv) S
v) Ash
i) C
ii) H2
21. Explain the enthalpy –concentration chart with a diagram.
Enthalpy –concentration diagram is useful for binary solutions. The enthalpy per unit solution
is plotted against concentration for a series of constant temperature and constant pressure
lines. With these charts, calculation effects in involved in changing the concentration and
temperature of the solution become simple and rapid.
22. Define Thermal efficiency
The thermal efficiency of a process may be defined as the percentage of the heat input that is
effectively utilized in a desired manner.
23. What is hot thermal efficiency?
If the gas is cooled before use it is sensible heat is available.
24. Define cold thermal efficiency.
It is heating value can be classed as heat effectively utilized in the producer.
25. What is orsat analysis?
The analysis of flue gases is ordinarily determined by the orsat type of apparatus, yielding the
percentages of carbon dioxide, carbon monoxide, oxygen and nitrogen in the moisture free
gases.
Part B
1. The flue gas from an industrial furnace has the following composition by
volume:
CO2-11.73%, CO – 0.2%, N2 -0.09% O2 – 6.81% and N2 -81.17%
Calculate the percentage excess air employed in the combustion if the loss of
carbon in clinker and ash is 1% of the fuel used and the fuel has the following
composition by weight:
C-74%, H 2 -5%, O2-5% N2-1% S-1%, H2O-9% and ash -5%,
Solution:
Basis : 100 kg of the fuel charged to the industrial furnace.
Reactions:
C+O 2 CO2
… (1)
1
H2+ O2 H2O
…. (2)
2
S+O2 SO2
…. (3)
Oxygen balance:
Oxygen required for complete combustion
74 5
1
0.5 kg.mol (Ref.1,2 and 3)
12 2
32
= 7.447 kg.mol.
Oxygen already present in fuel
5
0.157kg.mol
32
Net oxygen demand from air
7.447-0.157=7.29 kg.mol
Carbon balance:
Carbon lost in clinker and ash = 1 kg
Carbon burnt =74-1.0=73 kg=6.08 kg atom
Let x kg mols of flue gas are formed.
Therefore,
(.1173+.002)x=6.8
x= 50.96 kg.mol.
From flue gas analysis,
N2 in flue gas = 50.96 0.8117=41.36 kg.mol.
N2 from fuel = 1 kg =0.036 kg mol.
N2 from air = 41.360-0.036=41.324 kg.mol.
Oxygen supplied from air
41.324
21
10.98 kg mol
79
Excess oxygen = 10.98 -7.29 =3.69 kg.mol
Percentage excess air used
Percentage excess oxygen used
Excess
3.69
100
100 50.62.
Theoretical
7.29
2. Octane is burnt with 10% excess air. Calculate:
(a) Air/fuel ratio by weight.
(b) Air/fuel ratio by volume.
(c)Weight of dry exhaust gas formed per unit weight of fuel.
(d) Mol of oxygen in the exhaust gas per unit weight o fuel.
(e) Mol of water vapour in exhaust gas per unit weight of fuel
(f) Volume of exhaust gas at 1atmosphere and 260 per unit weight f fuel.
The specific gravity of octane may be taken as 0.7
Solution:
Basis : 1 kg mol of octane burnt.
Reaction: C 8H18 12.5 O28 CO 2 +9 H 2O
(a) Theoretical oxygen demand =12.5 kg. mol
Oxygen supplied by 10% excess air
=12.5 1.1 = 13.75 kg mols =440 kg.
Nitrogen supplied by air
79
13.75 51.73kg mol
21
=1448.4 kg
Amount of air supplied
=13.75+51.73=65.48 kg mol
= 1888.4 kg
Molecular weight of fuel =114
Wt.of air 1888.4
16.56.
Wt.of fuel
114
(b) Sp. Gravity of octane = 0.7
Density of octane = 0.7 gm/cc =700 kg/m3
114
Volume of fuel =
=0.163 m3
700
Assuming ideal gas behaviour, volume of air at N.T.P
=65.48 22.4 =1466.75 m3
…(4)
Volume of air 1466.75
8998.5
Volume of fuel
0.163
© Excess O2
=supplied O 2- used O2
=13.75-12.50=1.25 kg mol.
Dry flue gas analysis
Constituents
CO2
O2
N2
Total
Amount
kg mol
8.00
1.25
51.73
60.98
Amount
kg
352.0
40.0
1448.4
1840.4
Wt. of dry exhaust gas 1840.4
16.4
Wt. of fuel
114
(d )
Mol. of O2 in the exhaust gas 1.25
0.011
Wt. of fuel
144
( e)
Mol. of water vapour in the exhaust gast 9.0
0.079
Wt. of fuel
114
( f ) Mols of exhaust gas (wet)
=60.98+9.0=69.98
Applying ideal gas law, volume at 260C and 1 atmosphere
nRT 69.98 0.08206 (260 273)
P
1.0
3
=3060.8 m
Volume of exhaust gas (wet) 3060.8
26.85.
Wt. of fuel
114
3. A producer gas with the composition by volume, 27.3% CO, 5.4% CO 2, 0.6%O2, 66.7%-N2 is burnt with 20% excess air. If the combustion is 98% complete,
calculate the composition by volume of the fuel gases.
Solution:
Basis: 100 kg mole of producer gas burnt.
Oxygen balance:
O2 required for CO combustion
= 27.3 0.5 = 13.65 kg mole
O2 present in fuel =m 0.6 kg mole
Net O2 required = 13.65 -0.6 =13.05 kg mol
O2 supplied by 20% excess air
= 13.05 1.20 = 15.66 kg mol
O2 required for 98% combustion of CO
=273 0.5 0.98 = 13.38 kg mol
Total available oxygen = 15.66+0.6
= 16.26 kg mol.
O2 in excess
Nitrogen balance:
=16.26 – 13.38 =2.88kg mol.
N2 from producer gas = 66.7 kg mol.
N2 from air
= 15.66
79
=58.91 kg.mol
21
N2 from both the sources, which is in the flue gas
= 66.7+58.91=125.61 kg.mol.
Carbondioxide balance:
CO2 from producer gas = 5.4 kg mol.
CO2 from the combustion of CO
= 27.3 0.98 = 26.75 kg mol
Total CO 2 in flue gas
= 5.4 +26.75 =32.15 kg mol.
Carbon monoxide balance:
CO burnt = 26.75 kg mol.
CO left = 27.3 -26.75 =0.55 kg mol
Constituents
CO2
CO
N2
O2
Total
Amount
Kg.mol
Mol%
32.15
19.44
0.55
0.34
125.1
77.93
2.88
1.79
161.19
100.00
4. A furnace is fired with a natural gas that consists entirely of hydrocarbons
(negligible inert gases and sulphur compounds). The Orsat analysis of the flue
gas gives 9.5% CO2, 2.0% O2 and 1.8% CO.
(a) What is the molar ratio of net hydrogen to carbon in the fuel?
(b) What per cent of excess air is being used?
Solution:
Basis: 100 kg mol of dry flue gas.
From Orsat analysis,
Mole of N2
= 100-(9.5+1.8+2.0)=86.7.
Oxygen balance:
O2 supplied by air = 86.7
21
=23.05 kg mol
79
O2 reported in fuel gas (dry)
1
=9.5+ 1.8 +2.0=12.4 kg mol
2
O2 unaccounted = 23.05 -12.4 =10.65 kg mol
= O2 reacted with H 2.
(a)Mols of hydrogen reacted
= 10.65 2 = 21.3
Amount of carbon = 9.5+1.8 =1.3 kg atom.
Mols of H2 21.3
1.885.
Atoms of C 11.3
(b) Mols of O2 required for complete combustion
=Mol required for H 2+Mol required for C
=10.65+11.3=21.95
Amount of excess O2 = 23.05 -21.95=1.1mol
% excess air % excess O2 =
1.1
100 = 5.0
21.95
5. The exhaust gas from a hydrocarbon fuel oil fired furnace, shows 10.2%
CO2, 7.9% O2 and 81.9% N2 by Orsat analysis. Calculate (i) % excess air used,
and (ii) kg of dry air supplied per kg of oil burnt in the engine.
Solution:
Basis :100 kg mol of dry flue gas.
Oxygen balance,
N2 in flue gas
= 81.9 kg mol
21
21.77 kg mol
O2 supplied from air =81.9
79
O2 reported in flue gas = 10.2 (as CO 2) +7.9 (as O2)
=18.1 kg mol
O2 unaccounted
= 21.77-18.1
=3.67 kg ml =O2 used for H2.
O2used actually
O2 excess in flue
= 10.2+3.67=13.87 kg mol.
= 7.9 kg mol
(a) % excess air % excess O2
7.9
100 56.96.
13.87
(b) O2 used for H 2 = 3.67 kg mol.
Amount of H 2 in fuel = 3.67 2=7.34 kg mol
=14.68 kg
Carbon in fuel = 10.2 kg atom
=122.4 kg.
Total weight of the hydrocarbon fuel oil.
=Wt. of H +Wt. of C
=14.68+122.4=137.08 kg
Weight of air
= Mol of O 2 32 +Mol of N2 28
=21.77 32+81.9 28 = 2989.8 kg
Kg dry air
2989.8
21.81.
Kg oil burnt 137.08
6. Determine the fuel gas analysis and air fuel ratio by weight when a medium
fuel oil with 84.9% carbon,11.4% hydrogen, 3.2% sulphur, 0.4%Oxygen and
0.1% ash by weight is burnt with 20% excess air. Assume complete
combustion.
Solution:
Basis : 100 kg of medium fuel oil burnt
Oxygen balance:
Amounts of O2 used are calculated with the help of Eqns.(1), (2) (3) of Ex.1
O2 required for
C=84.9
O2 required for
S=3.2
O2 required for
Total
32
226.4kg
12
32
3.2kg
32
H =11.4 8 = 91.2 kg
_________
320.8 kg
O2 present in fuel
O2 required from air
O2 supplied
O2 excess
=0.4 kg
=320.8-0.4=320.4 kg,
=320.4 1.2 = 384.48 kg
= 384.48 -320.4 =64.08 kg
Air supplied
= 384.48
100
=1671.65 kg
23
Wt. of air 1671.65
16.716.
Wt. of fuel
100
Flue gas Analysis (Wet):
Amount of N 2 = 1671.65
CO2 produced =
77
1287.17 Kg
100
84.9
44 311.3 7.075 kg.mol.
12
SO2 Produced =3.2
64
=6.4 kg =0.10 kg mol.
32
11.4
18 =102.6 kg =5.70kgmol
2
O2 excess
= 64.08 = 2.003kg mol.
Flue gas analysis is given below:
H2O produced =
Constituents
CO2
SO2
O2
N2+
H2O
Total
Amount
kg mol
7.075
0.100
2.003
45.970
5.700
60.848
Amount
kg
11.63
0.16
3.29
75.55
9.37
100.00
7. A boiler is fired using 200 kg hr, of a pure saturated hydrocarbon gas (C nHm)
at atmospheric pressure and 20C. The dry analysis of the fuel gas which
leaves the boiler at atmospheric pressure and 300 C is CO2-12%, O2 -3% and
N2-85%. Estimate the formula of the fuel and total volumetric flow rate of the
gas.
Solution:
Basis: 100 kg mol of dry flue gas
Oxygen balance:
N2 in flue gas
= 85 kg mol.
O2 supplied by air
=85
21
22.59 kg mol
79
O2 reported in flue gas = O2 as CO2 +O2 as O2
= 12+3.0 =15.0 kg mol.
O2 unaccounted
= O2 reacted with H2
= 22.59-15.00 =7.59 kg mol.
H2 reacted
= 7.59 2 = 15.18 kg mol = 30.36 kg
Formulated of the fuel (hydrocarbon )gas:
Amount of carbon = 12 kg atom
Amount of hydrogen = 30.36 kg atom
Atom of H
30.36
2.53
Atom of C
12
As the ratio is more than 2., hence it is a paraffin of formula C nH 2n+2.
n
1
2n 2 2.53
Solving, n= 3.77 =4.0
Hence the fuel is n-butane (C4H10)
Volumetric flow rate:
Amount of fuel =
200
kg.mol
3.45
58
hr.
Assuming ideal gas law, volume at 20C and 1 atmosphere is,
V
nRT 3.45 0.8206 (273 20)
P
1.0
3
82.95 m / hr.
8. A furnace is fired by a gas having the composition H2-52%, CH4-30%, CO-8%,
CnHm-3.%, CO2-2%, O2-0.4% and rest N2. Using a certain quantity of air excess
over stoichiometric complete combustion of the gas is achieved, giving a dry
water gas of 5 m3 per m3 of fuel burned. Estimate:
(a) composition by volume of dry waste gas formed
(b) Per cent excess air used
(c) Weight of water formed per m 3 of gas burned, neglecting considered as
C3H6.
Solution:
Basis: 100 kg mols of fuel burnt
1
O2 CO2
2
CH4+2O 2 CO2+2H2O
Reactions : CO+
C3 H6 (C nHm)+4.502 3CO2+3H2O
1
H3+ O2 -H 2O
2
Oxygen balance.
1
=4 kg mol
2
= 30 2 =60 kg mol
=8
O2 required for CO
O2 required for CH 4
O2 Required for C 3H6 =3.6 4.5 =16.2 kg mol
1
O2 required for H 2
= 52 =26.0 kg mol
2
__________________
Total O 2 required
= 106.2 kg mol
O2 in fuel
= 0.4 kg mol
Net O2 from air
= 106.2 – 0.4 =105.8 kg mol
Dry flue gas formed (with theoretical amount of air):
CO2 formed
fuel
= 8.0 (from CO) +30.0 (from CH 4)+10.8 (from C 3H6) +2.0 (from
= 50.8 kg mol.
79
= 398.0 kg mol
21
N2 from air
= 105.8
Total N 2
= 398.0 +4.0 = 402.0 kg mol
Amount of dry flue gas = 50.8 +402.0 = 452.8 kg mol
Waste gas 5m3
3
Fue l gas
m
Let it be assumed that the fuel gas the flue gas be reported at the same
conditions. Hence volume ratio is identical to mol ratio.
Mols of waste gas = 5 Fuel gas = 500 kg mol
Dry fuel gas with theoretical amount of air
= 452.8 kg mols
Hence excess air = 500-452.8 = 47.2 kg mol
Theoretical air
= 105.8
100
= 503.81 kg mol
21
(b) % excess air used =
47.2
100 9.39
503.81
(a) Composition of waste gas
Air used
= 503.81 +47.2=551.01 kg mol
79
= 551.01
= 435.29 kg mol
100
= 435.29+4.0 =439.29 kg mol
21
= 47.2
= 9.91 kg mol
100
N2 from air
Total N 2
Excess O2
Waste gas analysis:
Constituents
CO2
O2
N2
Total
Amount
( mol)
50.80
9.91
439.29
500.00
Mol % or vol. %
10.16
1.98
87.86
100.00
(c) amount of water formed
= 52.0(from H2)+60 (from CH 4) +10.8 (from C3H6)
=122.8 kg mols = 2210.4 kg
Amount of gas burned = 100 kg mol = 2240 m3 at NTP
Wt. of water formed 2210.4
0.987.
m3 of gas burned
2240
9. The dry fuel gas from an oil fired furnace has a composition of 11.2% CO 2,
5.8% O2 and 83% N2 when analyzed by an Orsat apparatus. Calculate:
(a) Present excess air, and
(b) Weight of combustion air used per kg. of oil fired.
Assume fuel to have 82% C, 12%H, 3% S and balance impurities. Molecular
weight of dry gas is 30.
Solution:
Basis: 100 kg of oil fired.
Oxygen balance:
O2 rqd. For 82 kg carbon
= 218.67 kg = 6.833 kg mol.
O2 rqd. for 12 kg hydrogen
= 12 8 = 96 kg = 3.0 kg mol.
O2 rqd. for 3 kg sulphur
= 3 kg = 0.094 kg mol.
Total O 2 rqd.
= 9.927 kg mol.
Let x kg mols of dry fuel gases are formed.
Carbon balance:
82
=6.833 kg atom
12
Carbon in fuel
=
Carbon in fuel gases
= 0.112 x (from CO2) kg atom.
So,
0.112 x = 6.833
x= 61.0 kg mol.
Amount of N 2 in fuel gas = 61.0 0.83 = 50.63 kg mol.
This N2 has come from combustion air.
21
13.46 kg mol
79
So, O 2 from air
50.63
O2 used
O2 excess
% excess air
= 9.927 kg mol.
= 13.46 – 9.927 = 3.533 kg mol.
% excess O 2
3.533
100 35.60
9.927
(b) Amount of combustion air
50.63
100
64.09 kg mol
79
Molecular weight of air = 28.84
Weight of air
= 64.09 28.84 = 1848.36 kg
Wt. of combustion air 1848.36
18.48.
Wt. of fuel
100
10. A producer gas contains the following (composition in mole %)
CO2-9.2, C 2H4-0.4, CO-20.9, H2-15.6,
CH4-1.9, and N2 – 52
When it is burnt, the products of combustion are found to contain the
following composition:
CO2 -10.8, CO- 0.4, O2-9.2 and N2 – 79.6
Compute:
(i)
Litres of air sued in the combustion of 1 litre of producer gas, both
being at the same temperature and pressure.
(ii)
The per cent excess air used in combustion.
(iii)
The per cent of nitrogen in the products of combustion which came
from the producer gas.
Solution:
Basis: 100 kg mol of producer gas.
Constituents
CO2
C2H4
CO
H2
CH4
N2
Total
Mol
9.2
0.4
20.9
15.6
1.9
52.0
100.0
Atoms of C
9.2
0.8
20.9
-1.9
-32.8
Basis: 100 kg mol of dry combustion gas
Constituents
CO2
CO
O2
N2
Total
Kg mol
10.8
0.4
9.2
79.6
100.0
Atoms of C
10.8
0.4
--11.2
For 100 kg mol of dry combustion gas
Let
x= kg. mol of producer gas
Y = kg mol of air used.
(a) By carbon balance.
0.328 x= 11.2
x= 34.15 kg mol.
By nitrogen balance,
0.52x+0.79y = 79.6
y
or
79.6 0.52x
78.28 kg mol.
0.79
Mol air
(litre of air) T.P.
Mol producer gas (litre of producer gas) T.P.
78.28
2.29
34.15
(b) O2 supplied by air
= 78.28 0.21 = 16.44 kg mol.
O2 excess
= O2 in combustion gas – O2 required for CO
1
= 9.2-0.4 = 9.0 kg mols.
2
=
% excess air
% excess O2
O2 excess
O2 supplied - O2 excess
9.0
121.0
16.44 9.0
(c) N2 from producer gas = 34.15 0.52 = 17.76 kg mol.
% N2 in the products of combustion which cam from the producer gas,
17.76
100 22.31
79.6
11. Blast furnace gas having an analysis by volume on basis of:
CO2 – 13.0 %
CO – 25.0 %
H2 – 3.5 %
N2 – 58.5 %
In burned in a furnace.
Calculate:
(a) Percentage of excess air when the dry product of combustion contains
3.5 % O2
(b) Percentage of excess air when the dry flue gases contain 19.5 % CO 2
5.8% o 2 AND 74.7 % N2
Solution:
Basis: 100 kg mol of blast furnace gas.
Oxygen demand and flue gas formed are computed as under:
Constituents Amount kg mol
CO2
13.0
Oxygen rqd., kg mol
0
CO
25.0
12.50
H2
3.5
1.75
N3
58.5
0
Total
100.0
14.25
Flue gas obtained
13.0
(CO2)
25.0
(CO2)
3.5
(H2O)
58.5
(N2)
100.0
Let y1 = kg mol of dry flue gas obtained
= kg mol of combustion air used
(a) N2 from combustion air = 0.79 z
O2 from combustion air
= 0.21 z
O2 excess in flue gas
= 0.21 z – 14.25
Dry flue gas
= (100 – water vapour) + 0.79 z + (0.21 z – 14.25)
= (100 – 3.5) + z – 14 .25 = 82.25 + z = y1
Solving,
Oxygen from air
Oxygen
0.21z 14.25 3.5
Dry flue gas
82.25 z
100
z = 97.88 kg mol
= 97.88 0.21 = 20.55
Oxygen excess
= 20.55 – 14.25 = 6.3 kg mol
% excess air
= % excess oxygen =
(b) Let
6.3
100 44.2
14.25
y2 = kg mol of dry flue gas in this case
By a carbon balance for the blast furnace and the flue (dry) gases,
0.195 y2 = 38
so,
y2 = 194.87 kg mol
Oxygen in flue gas is 5.8 %
Amount of oxygen = 194.87 0.058 = 11.3 kg mol
% excess air
= % excess oxygen
=
=
Excess oxygen in flue
100
Oxygen requrired
11.3
100 79.3
14.25
12. Calculate the excess air used in a furnace when the flue gas Orsat analysis
is CO2 – 8.2 , O2 – 9.9 , CO – 0.1 , H2 – 0.4 and N2 – 81.4 (all in per cent).
Solution:
Basis: 100 kg mol of flue gas, (dry)
Oxygen balance:
Constituents
CO2
Q2
CO
H2
N2
Total
Amount kg mol
8.2
9.9
0.1
0.4
81.4
100.0
O2 reported kg mol.
8.20
9.90
0.05
–
–
18.15
Oxygen unaccounted
21
21.64 mlo
79
= 21.64 – 18.15 = 3.49 kg mol (used for H2 to give H2 O)
Oxygen used
= 8.2 (for CO2) + 0.05 (for CO) + 3.49 (for H2)
Oxygen supplied
= 81.4
= 11.74 kg mol
Additional amount of oxygen required for the combustion of CO and H 2 of gas
= 0.05 + 0.2 = 0.25 kg mol
Total amount of oxygen required for complete combustion
= 11.74 + 0.25 = 11.99 kg mol
Oxygen excess
= 21.64 – 11.99 = 9.65 kg mol
% excess air ( oxygen) used
=
9.65
100 80.48
11.9
UNIT-VI
PART-A- QUESTION & ANSWERS
1. Define law of conservation of energy.
According to law of conservation of energy, energy can neither be created nor be destroyed,
but it can be transferred from one form to another form.
2. Define mean gram calorie.
1/100 of the energy required to heat one gram of water from 0 to 100 ºC at a pressure of 1
atm is called mean gram calorie.
3. Define characterization factor.
When a crude oil of supposedly uniform character is fractioned into narrow cuts, the specific
gravity of each cut is approximately proportional to the cube roots of its absolute boiling
point at 1 atm pressure. The proportionality factor is taken as an index of the paraffinicity of
the stock. Thus K=3√TB/G Where K is the characterization factor, T B- average boiling poin,
degress rankine at 1 atm pressure, G-specific gravity at 60ºF
4. What is External potential energy?
It is position relative to the earth.
5. What is internal kinetic energy?
It is associated with molecular and atomic structure.
6. What is External kinetic energy?
It is associated with external motion.
7. Define Kopp’s Rule
Kopp’s rule: “The standard heat of capacity of a solid compound is approximately equal to the sum
of the heat capacities of its constituent elements.” This generalization is termed as Kopp’s rule.
8. State Trouton’s rule.
Trouton’s rule state’s that the ratio of the molal heat of vaporisation bof a substance at its
normal boiling point to the absolute temperature .T b is a constant.b/Tb=K
where k= Trouton’s ratio21 for many substances.
9. What is meant by Heat capacity of a substance and give units?
Heat capacity of a substance is defined as the amount of heat required to raise the temperature
of 1 Kg of a substance by 1 Kelvin.
For pure water, heat capacity at 15 0C =4.1855kJ/Kg.k
Heat capacity can be expressed in kJ/Kg.k (or) cal/g.c
10. State the Hess’s law.
Hess’s law of heat summation states that the net heat evolved or absorbed in the chemical
process is the same whether the reaction takes place in one or more steps. (i.e.)H0533 (or)
(H0 R+H0 298+H0P)
Consider the reaction CH4 (g) + 2O2(g)CO2 (g) + 2H2O(g)
11. What do you understand by exothermic and endothermic reactions? Give one
example in each.
When heat is evolved in a chemical reaction, the reaction s called exothermic reaction.
For e.g.
C(s) + O2(g)CO2 Hf =-94051 cal
When the heat is absorbed in a chemical reaction, the reaction is called endothermic reaction
For e.g.
C2H2OH(l) + CH3COOH(l)C2H5OOCH 3(l) + H2O (l) H0 298 =3720 cal
12. What is the effect of pressure on heat capacity of a gas?
Heat capacities for the ideal gas state are independent of pressure. However, they are
functions of temperature.
13. What is Flame temperature?
When a fuel is burnt, lot of heat is generated since the reaction is exothermic. If the
combustion chamber is sealed from surroundings, the heat produced is utilized to heat the
produce (flue gases). Hence the flue gases comes out at a higher temperature which is called
Flame temperature or T.F.T
Part B
1.
2.
3.
4.
5.
6.
7.
8. Calculate the Enthalpy of Zinc Vapour at 950 oC and atmospheric pressure,
Relative to the solid at 30 oC
Data:
Cp solid = 0.105 cal. /gm. oC.
Melting Point = 419oC
o
Cp liquid = 0.109 cal. /gm. C
Boiling Point = 907 oC
Latent Heat of Vaporization = 26900 cal/gm. A
o
Cp Vapor = 4.97 cal/gm atom C
Latent heat of fusion=1660 Cal/gm.atom
Atomic Weight of Zinc = 65.38
Heat absorbed by solid= 0.105(419-30)=40.85 cal/g
Heat of fusion= 1660/65.4=25.38 cal/g
Heat absorbed by liquid= 0.109(907-419)=53.192 cal/g
Heat of vapourization= 26900/65.4=411.31 cal/g
Heat absorbed by vapour=4.97/65.4 (950-907)=3.267 cal/g
Enthalpy of Zinc Vapor=40.845+ 25.38+ 531.92+ 411.31+3.267= 533.994 cal/g (ANS)
9. A stream containing 10% CH 4 and 90% air by volume is to be heated from 373 K
(100 oC) to 573 K (300 oC) at a rate of 0.05 m3 NTP per sec. Calculate the heat
required to be added using mean molal heat capacity data given below:
Data: Mean molal heat capacity in kJ/kmol. K.
o
o
Cpm
Cpm
Gas
373 298 K
573 298 K
CH4
Air
Solution:
37.5974
29.2908
43.0821
29.6132
Basis: 0.05 m3 (NTP) per see of gas stream.
05
22.4
=2.232 x 10-3 kmol/s
Molal flow rate of gas stream
Cpomix Mean molal heat capacity of gas stream
o
= xiCpm
1
Mol fraction CH4 = xCH4 =
100
100
= 0.10
90
100
= 0.90
xair =
o
Cpm
mix Mean molal heat capacity between 373-298 K
=0.10 37.5974+0.90 29.2908
=30.1215 kJ/kmol/K
o
Cpm
mix Mean molal heat capacity between 573-29 8 K
2
= 0.10 x 43.0821 + 0.90 x 29.6132
= 4.3082 + 26.6519
= 30.9601 kJ/kmol. K
o
o
Q n Cpm
mix 2 T2 T0 Cpmmix 1 T1 T0
where T 0 = 298 K
T1 = 373 K
T2 = 573 K
n = 2.232x10-3 kmol/s
Q 2.232 103 30.9601 573 298 30.1215 373 298
= 13.961 kJ/s
= 13.961 kW
10. Calculate the heat that must be added to 3 kmol air to heat if from 298 K (25 oC) to
473 K (200 oC) using mean molal heat capacity for air. Data given below:
o
Cpm
heat 473 and 298 K for air 29.3955kJ/Kmol.K
Solution:
Basis: 3 kmol of iar.
Heat to added = Q
n Cpo mix T To
wehre n = 3 kmol
Cpo mair 29.3955 kJ/kmol.K
T=473 K
To 298 K
Q=3 29.3955 473-298
15432.64 kJ
UNIT-VII
PART-A- QUESTION & ANSWERS
1. Define Internal energy.
The internal energy of a substances in defined as the total quantity of energy that in passes by
virtue at the presence relative positions and mole marts at its component molecule atoms.
2. Define external energy.
The external energy of a body in dependent on its position and motion relative to the earth.
3. What is BTU?
It is defined as the British thermal unit no longer in bared on the energy required to heat a
pound at water one degree Fahrenheit.
4. Define flow Process.
It is one in which streams at materials continuously enter (or) leave the system.
5. Define non flow process.
There in no continuous streams at material enter (or) leave the system deuce the gorge at
operation.
6. What steady – flow process?
In the steady flow process in constancy at temperature and compositions at any given
locations in the process.
7. Define enthalpy.
In the energy equations for both flow and non flow process. H = U + PV.
8. Define heat balance.
Heat balance in a loose terms retry to a special form of energy balance which has come into
general use in all thermal process.
9. Define Heat capacity.
It in an amount of heat required to increase the temperature of a body by one degree.
10. Define specific heat.
It in ratio at the heat capacity at a body to the heat capacity at an equal nears of water.
11. What is special units for heat capacity?
The molal heat capacity at gaseous values for any other unit of an volume.
12. What do you mean heat capacity?
The mean (or) average heat capacity other that tap range, where the mean molal heat capacity
at const.
13. Define four of thermo chemistry.
A given temp and pressure the quantity of energy required to decompose a chemical
compound into its clement.
14. Define heat of fusion.
The fusion of a crystalline solid at its melting point to form a liquid at the same temperature.
15. Define latent heat of Fusion
The latent heat of fusion is the amount heat required when unit qty of a solid is conducted to
liquid at its melting point.
16. Define heat of vaporization.
The heat required to vaporize a substance consists of the energy absorbed in overcoming the
intermolecular forces at attraction in the liquid.
17. Define latent heat of vaporisation
The latent heat of vaporisation is the amount of heat required per unit quantity of liquid when
it is changed from liquid to vapour at its boiling point.
18. Define heat of combustion
The heat of combustion of a substance is the heat of reaction resulting from the oxidation of
the substance with molecular oxygen.
The assignment of negative values to heat of combustion is consistent with the use of changes
of enthalpy as synonymous with heats of formation and heats of vapour since combustion the
value of SH must be negative and hence the heat of combustion is also negative.
H R = -HC.
19. Define standard heat of combustion
It is that resulting from the combustion of a substance, in the state that is normal at 25 0C and
atm. Pressure with the combustion beginning and ending at a temperature of 250C. It is
dependent on the extent to which oxidation is carried.
20. Define heat of reaction
Heat of reaction is the amount of heat to be supplied when a reaction is occurring at constant
pressure (one atom)
21. Define standard heat of reaction.
The standard heat of reaction is the change in the enthalpy of a system as a result of the
chemical reaction-taking place at 1 atm and starting and ending with all the components
(reactants & products) at constant temperature of 25 0C.If stoichiometric number is doubled,
the standard heat of reaction is doubled.
The standard heat of reaction for any reaction can be calculated if the heat of formation (or
heat of combustion) of the compounds taking part in the reaction is known as follows:
H0298= H f, products -H f reactants
H0298= Hc reactants -Hc products
22. Define heat of absorption
When the solvent and solute form an ideal system, the heat liberated during the absorption
system is equal to the latent heat of condensation of the solute. The mixing of non-ideal
liquids is accompanied by the evolution of heat and the heat of absorption is equal to the
algebraic sum of the heats of condensation and mixing.
23. Define Heat of Sublimation
Heat of sublimation: For solids the heat required to convert 1mole of the vapour is more
aptly called the heat of sublimation.
24. Define heat of Condensation
Heat of condensation: The amount of heat required to convert from vapour to liquid.
25. Define heat of mixing.
HEAT OF MIXING
Heat of solution in a system in which both solute and solvent are liquids termed as heat of
mixing. Heat of mixing are frequently expressed on a unit weight rather than a molal basis.
When two liquids are mixed, the enthalpy change or heat effect is called the heat of mixing
measured at constant temperature. Usually at 18 0C or 250C and at the atmospheric pressure. It
is expressed in kJ/kg mol of solution, kJ/kg mol of solute, kJ/kg of solute, kJ/kg of solution.
When two or more substances are mixed to form a gas or liquid solution, we frequently find
heat is absorbed or evolved from the system upon mixing. Such a solution would be called a
real solution.
The heat of mixing and solution has been determined experimentally for many non-ideal
systems. When the equilibrium or solubility data are available at different temperature; the
heat of mixing can be calculated using the relationship
d ( ln r1 )
- L1
------ = ----dT
RT 2
When L1 is the partial modal enthalpy of compt. In solution minus the enthalpy of the pure
liquid at the same temperature and r1 is the activity coeft. of compt.1 in solution.
26. Define Heat of transition.
The equilibrium temperature of transformation in constant although the actual temp of
transformation.
27. What is heat of formation?
A formation reaction is defined as s reaction, which forms a single compound form the
elements contained in it.
e.g. C + ½ O2 +2H2CH3OH
The heat of formation is based on 1 mol of the compound formed. The equations should
indicate the physical state of each reactant and product whether it is gas, liquid or solid. The
heat of formation of a chemical compound is the standard heat of reaction where the reactants
are the necessary elements for the formation of compound which is the only product formed.
Eg.
C(s) + +O2 (g)CO2 H f =94051 cal/g mol
For all the elements, the heat of formation is zero.
28. What is heat of solution?
The enthalpy change accompanists the dissolution of a substances in termed as heat of
solutions.
29. Write kirchoff’s equation?
The effects of temp on the heat of reaction may be iced for the special care of reactions.
30. Explain the sensible heat and latent heat.
When a liquid is heated from room temperature (say 25 0C) to a higher temperature (say 60 0C)
and it remains as a liquid at 60 0C, then the sensible heat absorbed =q= m Cp (60-25)
Where m = mass of liquid
Cp= specific heat of liquid.
If the liquid is further heated to 100 0C, and it vaporizes, then
Heat absorbed =q= sensible heat + latent heat
= m Cp (100-25) + m Where = latent heat of vaporization.
Part B
1.
2.
3.
4.
5.
6.
7. Air containing 21 mole % O 2 and 79 mole % N2 is to be heated from 303 K (30 oC) to
423 K (150 oC). Calculate the heat required to be added if the air flow rate is 3 m 3
(N.T.P) per minute. Data given below
Cpo a bT cT2 dT3
b x 103
11.7551
-5.141
Gas a
O2
26.0257
N2
29.5909
Solution:
c x 106
-2.3426
13.1829
d x 109
-0.5623
-4.968
Basis: 3 m3 N.T.P./min air flow rate.
At N.T.P. 1 kmol of gas occupies a volume of 22.4 m3.
Molal flow rate of air
Cpo mix xiCpio
X O2
21
0.21
100
XN2
79
0.79
100
3
0.134kmol / min.
22.4
XO2 Cpoo2 0.21 26.0257 11.7551 103 T 2.3426 106 T 2 0.5623 10 9 T3
5.4653 2.4685 103 T 0.4919 106 T2 0.1181 109 T2
o
XN2 CpN
0.79 29.5909 5.141 103 T 13.1829 10 5 T 2 4.968 10 9 T 3
2
23.3768 4.0614 103 T 10.4145 106 T2 3.9247 109 T3
Cpo mix 28.8421 1.5929 103 T 9.9226 106 T2 4.0428 109 T3
T2
Q n Cpo mix dT
T1
T2
n 28.8421 1.5929 103 T 9.9226 10 6 T 2 4.0428 10 9 T 3 dT
T1
1.5929 103 2
n 28.8421 T2 T1
T2 T12
2
9.9226 106 3
4.0428 10 9 4
T2 T13
T2 T14
3
4
where n = 0.134 kmol / min
T1 = 303 K
T2 = 423 K
1.5929 103
Q 0.134 28.8421 423 303
(423)2 (303)2
2
9.9226 105
4.0428 109
(423)3 (303)3
(423)4 (303)4
3
4
0.134 3461.05 69.38 158.33 23.84
0.134 3526.16
472.5kl / min
7.875kJ / g 7.875kW.
8. Flue gas leaving the boiler stack at 523 K (250 oC) have following composition on
mole basis.
CO2 = 11.31%, H2O = 13.04%, O2 = 2.17% and N2 = 73.48%
Calculate the heat lost in 1 kmol of gas mixture above 298 K (25 oC) using heat
capacity data given below:
Cpo a bT cT2 dT3
Gas
CO2
H2O
O2
N2
a
21.3655
32.4921
26.0257
29.5909
b x 103
64.2841
0.0796
11.7551
-5.141
Solution:
Basis: 1 kmol of gas mixture.
c x 106
-41.0506
13.2107
-2.3426
13.1829
d x 109
9.7999
-4.5474
-0.5623
-4.968
T2
Q n Cpo mix dT
T1
Cpo mix xiCpio
x CO2
11.31
0.1131
100
xH2O
13.04
0.1304
100
x O2
2.17
0.0217
100
XN2
73.48
0.7348
100
o
XCO2 CpCO
0.1131 21.3655 64.2841 103 T 41.0506 10 6 T 2 9.7999 109 T3
2
2.4164 7.2705 103 T 4.6428 106 T2 1.1084 109 T3
xH2 oCpoH O 0.1304 32.4921 0.0796 103 T 13.2107 10 6 T 2 4.5474 109 T3
2
4.2370 0.0104 103 T 1.7227 109 T2 0.5930 109 T3
XO2 CpoO 0.0217 26.0257 11.7551 103 T 2.3426 106 T 2 0.5623 109 T3
2
0.5647 0.2551 103 T 0.0508 106 T2 0.01220 109 T3
xN2 CpoN 0.7348 29.5909 5.141 103 T 13.1829 10 6 T 2 4.968 10 9 T3
2
21.7434 3.7776 103 T 9.6868 106 T2 3.6505 109 T3
Cpo mix xiCpio
=28.9615+3.7584 10-3 T 6.7159 106 T2 3.1473 109 T3
T2
Q n 28.9615 3.7584 103 T 6.7159 10 6 T 2 3.1473 10 9 T 3 dT
T1
3.7584 103 2
6.7159 106 3
3.1473 109 4
n 28.9615 T2 T1
T2 T12
T2 T13
T2 T14
2
3
4
where n = 1 kmol
T1 = 298 k
T2 = 523 k
3.7584 103
Q 128.9615 523 298
(523)2 (298)2
2
6.7159 109
3.1473 109
(523)3 (298)3
(523)4 (298)4
3
4
16516.34 34686 261 52.66
7071.54 kJ
9. A stream of nitrogen flowing at a rate of 100 kmol/h is heated from 303 K (30
o
C) to 373 K (100 oC). Calculate the heat that must be transferred.
Data: Cpo for nitrogen = 29.5909-5.141x10-3 + 11.1829 x 10-4 T2 – 4.968 x 10-9 T3.
Solution:
Basis : 100 kmol/h of nitrogen gas
T2
Q n Cpo dT
T1
T2
Q n 29.5909 5.141 10 3 T 11.1829 10 6 T 2 4.968 10 9 T 3 dT
T1
5.141 10 3 2
11.1829 10 6 3
4.968 10 9 4
n 29.5909 T2 T1
T2 T13
T2 T12
T2 T14
2
3
4
where n = 100 kmol/h
T2 = 373 K
T1 = 303 K
5.141 103
11.1829 10 6
2
2
Q 100 29.5909 373 303
(373) (303)
(373)3 (303)3
2
3
9
4.968 10
(373)4 (303)4 ]
4
= 202587.2 kJ/h
= 56.274 kJ/s = 56.274 kW
(As 1 J/s = 1 W)
10. A natural gas has the following composition on mole basis:
CH4 84%,C2H6 13% and N2 3%
Calculate the heat to be added to heat 10 kmol of natural gas rom 298 K
(25 C) to 523 K (250 oC) using heat capacity data given below:
Cpo a bT cT2 dT3
o
Gas
CH4
C2H6
N2
b x 103
52.1135
178.0872
-5.141
a
19.2494
5.4129
29.5909
c x 106
11.973
-67.3749
13.1829
d x 109
-11.3173
8.7147
-4.968
Solution:
Basis: 10 kmol of natural gas.
o
Cpmix
x iCpio
84
0.84
100
13
XC2H6
0.13
100
3
XN2
0.03
100
XC114
XCH4 CpoCH 0.84 19.2494 52.1135 103 T 11.973 106 T 2 11.3173 109 T3
4
16.1695 43.7753 10 3T 10.057 10 6T 2 9.506 10 9T 3
XC2H6 CpoC H 0.13 5.4129 178.0872 103 T 67.3749 106 T 2 8.7147 109 T3
2 6
0.7037 23.1513 10 3T 8.7587 10 6T 2 1.1329 10 9T 3
XN2 CpoN 0.03 29.5909 5.141 103 T 13.1829 106 T 2 4.968 10 9 T 3
2
= 0.8877-0.1542 x 10-3 T – 0.3955 x 10-6 T2 – 0.1490 x 10-9 T3
o
o
Cpmix
XCH4 CpCH
XC2 116 CpoC H6 XN2 CpoN
4
2
2
3
16.1695 43.7753 10 T 10.057 106 T2 9.5065 109 T3
0.7037 23.1513 103 T 8.7587 106 T2 1.1329 109 T3
0.8877 0.1542 103 T 0.3955 106 T2 0.1490 109 T3
17.7609 66.7724 103 T 1.6938 106 T2 8.5226 109 T2
Q = Heat to be added
T2
n Cpo mix dT
T1
T2
n 17.7609 66.7724 10 3 T 1.6938 10 6 T 2 8.5226 10 9 T 3 dT
T1
66.7724 103 2
n 3 17.7609 T2 T1
T2 T12
2
1.6938 106 3
8.5226 10 9 4
T2 T13
T2 T14
3
4
where n = 10 kmol
T1 = 298 K
T2 = 523 K
66.7724 103
Q 10 17.7609 523 298
(523)2 (298)2
2
1.6938 106
8.5226 109
(523)3 (298)3
(523)4 (298)4
3
4
10 3996.20 6167.26 65.83 142.61 100866.8kJ
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