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General Material Balance
CENG 101
CENG 101
Lecture 2. Chemical Engineering Processes (5 h)
Lecture 2. Chemical Engineering Processes (5 h)
Learning Objectives: Surrounding
Learning Objectives: out1
(1) Material Balances - -Single Phase/No Reaction
(1) Material Balances Single Phase/No Reaction System
Elementary calculation
Elementary calculation
Process flow chart and description
Process flow chart and description
Gas, liquid and solid
Gas, liquid and solid generation or depletion
Equation of States
Equation of States in1 accumulation
(2) Material Balances - -Single Phase/Reaction
(2) Material Balances Single Phase/Reaction
Reaction Stoichiometry
Reaction Stoichiometry
Reaction Balance
Reaction Balance
Reaction Equilibrium
Reaction Equilibrium
Combustion reaction
Combustion reaction
Learning Guides:
Learning Guides: in2
(1) Lecture handouts
(1) Lecture handouts
(2) Chapters 4-5 of Textbook: Elementary principles of
(2) Chapters 4-5 of Textbook: Elementary principles of
chemical processes
chemical processes
(input + generation) - (output + depletion) = Accumulation
Example 10. A government population survey of a new town shows Process Classification
that in a year 15,000 people establish residence in the town but
during the same year 5000 people moves to the surrounding
suburbs. 8000 birth and 2000 death was recorded during the year.
Surrounding
How many people live in the city if the last year population is out1
350,000 people. How many women in the city if the average male- System
female ratio is 0.82
generation or depletion
in1 accumulation
in2
(input + generation) - (output + depletion) = Accumulation
Batch process
dM/dt
Continuous process
Semibatch process
Example 11. A concrete mixer mixes 1 ton of aggregates (pebbles), Example 12. Cement are produced by roasting the raw material in a
0.5 ton of sand, 0.75 ton of cement and 0.5 ton of water loaded at rotating kiln kept at high temperature. 1 ton per minute of raw
the concrete plant at Tsing yi and delivered to the construction site material enters the kiln producing 0.7 ton of cement. It is known the
at HKUST. The trip took about 1 h and 15 min and the mixture was gaseous by-products are produced during the roasting. From the
continuously mixed at a rotation speed of 1 revolution per 5 minutes data please determine their emission rate.
from the plant to the site. Calculate the total mass discharged at the
site and its composition.
Example 13. A car burn 5 L of gasoline during a 40 km trip. Is this
a batch, continuous or semibatch process. Please reason.
Problem solving procedure
1. Read the problem carefully.
Identify the processes, inputs and outputs, given information and unknown.
2. Draw the process flowchart.
Represent each processes by a box and label the box. Use arrows to
represent the input and output.
3. List all the given or known information.
Write the given information next to the input and output streams
4. Label the unknown and classify.
Identify and classify the unknowns: Unknown asked and required by the
problem, unknown needed to calculated to answer the problem, unknown
that are physical properties that can be found in textbooks or handbooks.
5. Clearly label the unknown that you need to FIND and SOLVE.
Box, underline or highlight the unknown that you need to answer.
6. List the assumptions you made in order to make the calculation.
E.g., basis of calculation, steady-state condition?, ideal gas?,
incompressible fluid?
7. Write down the general material balance equation and simply.
Draw a box around the process where material balance has to be made,
identify the inputs and outputs that cross the box.
8. Check the consistency of units.
Check what units should the final answer be. Make sure that conversions
are clearly detailed out.
9. Conduct overall material balance and component balance
A total of N+1 material balance equations can be written for a system that
contains N components. The last balance equation is used to check the
calculation.
Example 14. Different color of paints can be obtained from blending a few
primary colors. A company decided to take advantage of this fact to market a
small computerized paint blending machine to hardware stores. The blender Ideal Gas
consists of four storage tanks containing, red, green, blue and white paints, and
two mixing tanks. The computer controls the amount of each colored paints that (1) An ideal gas has the following behavior:
is added to the first mixing tank before they are added to the second tank where
they are blended with white paint to get the final desired color.
Boyle’s Law PV = k1
0.5 Kg Red paint contains 3000 ppm Ba, 10 ppm Cd, 5 ppm Hg, 15 ppm Pb Charle’s Law V / T = k2
0.2 Kg Blue paint contains 20 ppm Ba, 250 ppm Cd, 2 ppm Hg, 10 ppm Pb Clapeyron Law P / T = k3
0.7 Kg Green paint contains 20 ppm Ba, 150 ppm Cd, 25 ppm Hg, 10 ppm Pb Amagat Model V / n = k4
1.2 Kg White paint contains 20 ppm Ba, 10 ppm Cd, 30 ppm Hg, 100 ppm Pb Avogrado’s Law P / n = k5
1 / nT = 1 / k6
Were blended to produce the paint desired by a customer at the store. Please
determine what the composition of the blended stream leaving each of the
mixing tanks. When you multiply them all together, you get:
P3V3 / n3T3 = k1k2k3k4k5 / k6
PV / nT = R = cube root (k1k2k3k4k5 / k6)
k1 = atm-L k2 = L / K
k3 = atm / K k4 = L / mol
k5 = atm / mol 1 / k6 = 1 / mol-K
(2) Assumptions for ideal gas:
(a) volumeless, point particles
(b) no interaction
(c) elastic collision
http://jersey.uoregon.edu/vlab/Piston/
http://www.utdallas.edu/~parr/chm1341/1341b612.html
Problem 15: The high heat capacity and latent heat associated with
Virial Equations
water and its availability made it an ideal working fluid. Steam
driven piston and cylinder assembly is the corner stone of industrial
revolution. Therefore significant amount of work were done to
Z = PV = 1 + B’P + C’P2 + D’P3 + …. determine the exact behavior of steam.
RT
Determine the final volume of 1 mole of steam compressed
Z = PV = 1 + B + C + D + …. isothermally from 1 bar to 100 bar.
RT V V2 V3
Steam: B = -53.4 cm3mol-1, C = 2,620 cm6mol-2, D = 5,000 cm9mol-3
2
C’ = C-B 2
3
B’ = B D’ = D-3BC +3 2B
RT (RT) (RT)
Compounds B(cm3mol-1) C(cm6mol-2)
Methane -53.4 2620
Ethane -156.7 9650
Ethylene -140 7200
Methyl chloride -242.5 25,200
Sulfur hexaflouride -194 15,300
Steam -152.5 -5800
Virial equation is derived from experimental PVT data and is the next best
thing to a TS-diagram. Use virial equations if P > 50 bar.
Note: The data is only for gas phase unless specified.
Generalized Compressibility Factor Correlation
Z = PV = Z0 + ωZ1
RT
Z0 and Z1 are read from a table as a function of Tr and Pr
Tr = T/Tc Pr = P/Pc
Generalized Virial Coefficient Correlation Virial Equations
BP P
Z = 1 + BP = 1 + RTc Tr
RT c r
Taylor expansion for f(p,T)
BPc = B0 + ωB1
RTc
B0 = 0.083 - 0.422
Tr 1.6
B1 = 0.139 - 0.172
Tr 4.2 where the constant are evaluated experimentally:
Tr = T/Tc Pr = P/Pc
at low pressure:
Virial equation is derived from experimental PVT data and is the next best
thing to a TS-diagram. Use virial equations if P > 50 bar.
Note: The data is only for gas phase unless specified.
Cubic Equation of States Equation of States
Redlich-Kwong Equations
PVT Equation Pressure Accuracy Computation
a
P = RT - 0.5V(V+b)
Range
V-b T
PVT table All Very high Easy
PVT chart All High Easy
0.42748 R2Tc2.5
a= Virial Equation
Pc 4-terms All High Very difficult
0.08664 RTc 3-terms < 100 bar High-Moderate Difficult
b=
Pc 2-terms < 50 bar Moderate Moderate
1-term (i.e., ideal < 10 bar Low Easy
gas equation)
Tr = T/Tc Pr = P/Pc
Z-correlation All Moderate Moderate
Equation
Virial coefficient < 100 bar Moderate Moderate-
Vapor volume Liquid volume correlation equation Difficult
Redlich-Kwong < 50 bar Moderate Moderate-
Equation Difficult
Ideal Gas Equation < 10 bar Low Easy
Homework # 2 Feb 28, 2003 Material Balance with Reaction
Mar 14, 2003
(1) Identify the chemical reaction
Individual Problems a) Reaction equations are explicitly given,
Chapter 4:Problem 4.6
Chapter 5: Problem 5.34 CO(g) + 2 H2 (g) → CH3OH (l)
Chapter 5: Problem 5.37
Team Problem b) Reaction is described in the problem,
Chapter 4: Problem 4.37
carbon monoxide was hydrogenated to
form methanol
- Hydrogenation -> addition of hydrogen
- Dehydrogenation -> removal of hydrogen
- Hydration -> addition of water
- Dehydration -> removal of water
- Condensation reaction -> joining of two
molecules forming water as byproduct
- Oxidation -> addition of oxygen atom
- Combustion -> oxidation forming CO2 and water as
main products
- Isomerization -> rearrangement of molecular structure
(c) Indirectly deduced from the incoming reactants and
outgoing products.
Material Balance with Reaction Material Balance with Reaction
(2) Write the stoichiometric (balanced) reaction equation (6) Conduct material balance on the reaction
CO(g) + 2 H2 (g) → CH3OH (l) (a) best done in term of moles
(3) Identify the limiting and excess reactants (b) conduct elemental (atomic) balance, or
1 mole CO and 2 mole H2 is in stoichiometric proportion (c) conduct molecular balance
1.25 moles CO and 2 mole H2
H2 is the limiting reactant
CO is the excess reactant
CO is 25 percent excess
(4) Determine whether conversion, selectivity or yield are
given.
Conversion (X) = moles reacted/moles fed
Selectivity (S) = np1/Σ npi
Yield (Y) = C*S
(5) Determine whether the reaction is equilibrium limited.
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
Example 24. Butane was dehydrogenated to produce 1,2-
butadiene using Pt-catalyst supported on carbon black. 1- Material Balance with Reaction
butene, cis and trans-2-butenes were also produced in
quantities. 100 moles/h of butene enters a reactor. If the
conversion is 30% and the selectivity is 65%. aA + b B → cC + d D
(a) How much 1,2-butadiene was produced?
(b) How much butenes were generated?
(c) How much hydrogen is leaving the reactor? Conv. A, Sel. C
(d) What are the results if a recycle stream of 30 moles/h is nin, xin nout, xout
used? Reactor
(e) What are the results if a bypass stream of 30 moles/h is
used?
A is limiting reactant
(na)out = (na)in - X (na)in
(nb)out = (nb)in - (b/a)X (na)in
(nc)out = (nc)in - (c/a)X(S)(na)in
(nd)out = (nd)in - (d/a)X(1-S)(na)in
Material Balance with Reaction Material Balance with Reaction
aA + b B → cC + d D aA + b B → cC + d D
n’in, x’in Conv. A, Sel. C n’out, x’out n’in, x’in Conv. A, Sel. C n’out, x’out
nin, xin Reactor nout, xout nin, xin Reactor nout, xout
nbypass, xbypass nrecycle, xrecycle
A is limiting reactant A is limiting reactant
(n’a)out = (n’a)in - X (n’a)in (n’a)out = (n’a)in - X (n’a)in
(n’b)out = (n’b)in - (b/a)X (n’a)in (n’b)out = (n’b)in - (b/a)X (n’a)in
(n’c)out = (n’c)in - (c/a)X(S)(n’a)in (n’c)out = (n’c)in - (c/a)X(S)(n’a)in
(n’d)out = (n’d)in - (d/a)X(1-S)(n’a)in (n’d)out = (n’d)in - (d/a)X(1-S)(n’a)in
Different Representation of VLE Data Different Representation of VLE Data
Ideal Solution: Example 25: Determine the T-x-y relationship for ethanol and water
using Antoine Equation assuming that ethanol-water form
Raoult’s Law: an ideal solution. Plot the resulting data. Please use
ambient pressure.
PA = YAPT = PA*XA
YA= (PA*/PT ) XA
Antoine Equation:
LnPA* (kPa) = A - B/(T(C) +C)
Compound A B C
Benzene 13.8594 2773.78 220.07
Ethanol 16.6758 3674.49 226.45 T(C) Pe* Pw* Xe Xw Ye Yw
n-heptane 13.8587 2991.32 216.64 Tmin 78.29185 101.3 44.19879 1 0 1 0
80.46185 110.315 48.2752 0.85469 0.14531 0.930752 0.069248
82.63185 119.9885 52.66227 0.722419 0.277581 0.855695 0.144305
Toluene 14.0098 3103.01 219.79
84.80185 130.3574 57.37836 0.601839 0.398161 0.774474 0.225526
86.97185 141.4599 62.44257 0.491758 0.508242 0.686714 0.313286
Water 16.262 3799.89 226.35
89.14185 153.3356 67.87472 0.391118 0.608882 0.592027 0.407973
91.31185 166.0253 73.69541 0.298978 0.701022 0.490009 0.509991
93.48185 179.5714 79.92597 0.214501 0.785499 0.380239 0.619761
95.65185 194.0176 86.58853 0.136941 0.863059 0.26228 0.73772
Tmax 99.99131 225.7879 101.3 2.28E-16 1 5.09E-16 1
Vapor-Liquid Equilibria
105
Water-air system
100
95 water vapor
T(C)
90
Evaporation
85
liquid water
80
75
0 0.2 0.4 0.6 0.8 1 PT = Pair + PH2O
X,Y (ethanol)
Humidity
Relative Humidity
What is the boiling point of pure ethanol and water?
What is the bubble point temperature of a mixture containing
HR = PH2O/PV
0.25 mole fraction of ethanol? What is its dew point temperature?
What is the bubble and dew point temperature of a solution Absolute Humidity
containing 30 wt.% water? HSP = mH2O/mair
= (MWH2O* PH2O/PV)/MWairPair
Problem 26. Dehumidifier uses a vapor-compression cycle to condense A natural convection cooling tower is shown below. Please explain in
moisture from indoor air. The dry air leaves the dehumidifier at a detail its operating principle.
higher temperature.
(a) Wet air with a relative humidity of 90% and T = 30°C enters the
dehumidifier and leaves with a relative humidity of 60% at T = 33°C.
Find the amount of moisture removed per kilogram of dry air. The
barometric at that balmy day is 755 mm Hg.
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