CHBI 201 1 Billion: The British say that a billion is a million million (1,000,000,000,000). American say that a billion is a thousand million (1,000,000,000) and insist that a million million is actually a trillion. The Canadian Press agrees with the Americans http://www.scit.wlv.ac.uk/~jphb/american.html Please note that "tonne" is not a British spelling of "ton" but a quite separate metric unit equal to 1000 kg as distinct from the British ton of 2240 lbs (= 1016.96 kg). Billion: thousand million The old British usage in which a billion was a million 2 is now largely obsolete and most British speakers would assume the American meaning. Careful users avoid the words altogether and use exponent notation. The usage continued trillion = tri+(m)illion = million 3 = 10 18 quadrillion = quad+(m)illion = million 4 = 10 24 centillion = cent+(m)illion = million 100 = 10 600
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CHBI 2011 Billion: The British say that a billion is a million million (1,000,000,000,000). American say that a billion is a thousand million (1,000,000,000)
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CHBI 201 1
Billion: The British say that a billion is a million million (1,000,000,000,000). American say that a billion is a thousand million (1,000,000,000) and insist that a million million is actually a trillion. The Canadian Press agrees with the Americans
http://www.scit.wlv.ac.uk/~jphb/american.html
Please note that "tonne" is not a British spelling of "ton" but a quite separate metric unit equal to 1000 kg as distinct from the British ton of 2240 lbs (= 1016.96 kg).
Billion: thousand million The old British usage in which a billion was a million2 is now largely obsolete and most British speakers would assume the American meaning. Careful users avoid the words altogether and use exponent notation. The usage continued
trillion = tri+(m)illion = million3 = 1018
quadrillion = quad+(m)illion = million4 = 1024
centillion = cent+(m)illion = million100 = 10600
The American naming seems to work on the principle 103+(number×3)
Continuous Process The input and outputs flow continuously throughout
the duration of proces
Semibatch Process Any process neither batch nor continuous
CHBI 201 7
Balances on Continuous Steady-state Processes
Input + Generation = Output + Consumption If the balance is on a nonreactive species, the generation
and consumption will be 0. Thus, Input = Output
Example
Distillation
1000 kg /h
Benzene + Toluene
%50 Benzene by mass
475 kg Toluene/h
M2 kg Benzene/h
m1 kg Toluene/h
450 kg Benzene/h
Input of 1000 kg/h of benzene+toluene containing 50% B by mass is separated by distillation column into two fractions. B: the mass flow rate of top stream=450 kg/hT: the mass flow rate of bottom stream=475 kg/h
CHBI 201 8
Solution of the exampleInput = Output Benzene balance
1000 kg/h · 0.5 = 450 kg/h + m2
m2 = 50 kg/h Benzene
Toluene balance
1000 kg/h · 0.5 = 475 kg/h + m1
m1 = 25 kg/h Toluene
Balances on Continuous Steady-state Processes
..
..
CHBI 201 9
BALANCES ON BATCH PROCESSES
Initial Input + Generation = Final Output + Consumption Objective: generate as many independent equations as
the number of unknowns in the problem
F
(W+A)
B
D F = B + DF.xF = D.xD + B.xB
F.yF = D.yD + B.xB
x: mole fraction of W
y: mole fraction of A
CHBI 201 10
EXAMPLE (Batch Process) Centrifuges are used to seperate particles in the range of 0.1 to 100
µm in diameter from a liquid using centrifugal force. Yeast cells are recovered from a broth ( a mix with cells) using tubular centrifuge. Determine the amount of the cell-free discharge per hour if 1000 L/hr is fed to the centrifuge, the feed contains 500 mg cells/L, and the product stream contains 50 wt% cells. Assume that the feed has
a density of 1 g/cm3.
CentrifugeFeed (broth) 1000 L/hr
500 mg cells/L feed
( d= 1 g/cm3)
Concantrated cells P(g/hr)
50 % by weight cells
Cell-free discahrge D(g/hr)
CHBI 201 11
EXAMPLE (Batch Process) Centrifuges are used to seperate particles in the range of 0.1 to 100 µm in
diameter from a liquid using centrifugal force. Yeast cells are recovered from a broth ( a mix with cells) using tubular centrifuge. Determine the amount of the cell-free discharge per hour if 1000 L/hr is fed to the centrifuge, the feed contains 500 mg cells/L, and the product stream contains 50 wt% cells. Assume that the feed has a density of 1 g/cm3.
Boxes and other symbols are used to represent process units. Write the values and units of all known streams Assign algebraic symbols to unknown stream variables
CombustionChamber
Condenser
100 mol C3H8
1000 mol O2
3760 mol N2
200 mol H2O
50 mol C3H8
750 mol O2
3760 mol N2
150 mol CO2
CHBI 201 13
EXAMPLE (Flow charts) Humidification and Oxygenation Process in the Body: An
exp. on the growth rate of certain organisms requires an environemnt of humid air enriched in oxygen. Three input streams are fed into an evaporator to produce an output stream with the desired composition. A: liquid water, fed at a rate of 20 cm3/min, B: Air, C: Pure oxygen (with a molar flow rate one-fifth of the molar flow rate of stream B)
One more equation is needed Volume is not conserved! Use consistent units (mole, kg) Do not make mole balances in reactive processes.
Humid air
(n0) O2
(n1) N2
(n2) H2O
(n3) H2O
225 L/h
Condenser Dry air(n4) O2
(n5) N2
(n6) H2O
ρH20 is given
In the condenser, 95% of H2O in the inlet air is condensed.
CHBI 201 18
EXAMPLE
A continuous mixer mixes NaOH with H2O to produce an aqueous solution of NaOH. Determine the composition and flow rate of the product, if the flow rate of NaOH is 1000 kg/hr and the ratio of the flow rate of H2O to the product solution is 0.9.
Nsp = number of species Ns = number of streams Nu = total number of variables
EXAMPLE 2
MNaOHH2O
Product
System boundary
CHBI 201 19
EXAMPLE 2 - continue
Streams
Species
FEED WATER PRODUCT
NaOH FNaOH WNaOH PNaOH
H2O FH2O WH2O PH2O
Total F W P
Nu = 3(2+1) = 9
Specifications: ratio of two streams
the % conversion in a reaction
the value of each concentration, flow rate, T, P, ρ, V, etc.
a variable is not present in a stream, hence ,it is 0
Last row in the table
CHBI 201 20
EXAMPLE
A cylinder containing CH4, C2H6, and N2 has to be prepared containing a CH4 to C2H6 mole ratio of 1.5 to 1. Avaliable to prepare the mixture are1) a cylinder containing a mixture of 80% N2 and 20% CH4
2) a cylinder containing a mixture of 90% N2 and 10% C2H6
3) a cylinder containing a mixture of pure N2
What is the number of degrees of freedom?
EXAMPLE 3
CHBI 201 21
Unknowns: 3 xi and 4 Fi
EXAMPLE 3 - continue
F1
CH4 0.2N2 0.8
F2
C2H6 0.1N2 0.9
F4
CH4 xCH4
N2 xN2
C2H6 xC2H6
F3
N2 1
CHBI 201 22
Equations: Material balance (CH4, C2H6, N2)
One specified ratio xCH4/xC2H6 = 1.5 One summation of mole fractions 5 independent equations
Ndf = 7 – 5 = 2
If you pick a basis as F4=1, one other value has to be specified in order to solve the problem.
EXAMPLE 3 - continue
4
Ffor 1ix
CHBI 201 23
Balances on Multiple-unit Processes
1 3
2
4
100 kg/hr
0.5 kg A/kg
0.5 kg B/kg 30 kg/hr
0.3 kg A/kg
0.7 kg B/kg
40 kg/hr
0.9 kg A/kg
0.1 kg B/kg
30 kg/hr
0.6 kg A/kg
0.4 kg B/kg
Q3
x3
Q2
x2
Q1
x1
CHBI 201 24
Balances on Multiple-unit Processes
Q : mass flow ratexA : mass fraction of A
1-xA : mass fraction of B
Number of unknowns = 6Number of equations = 2+2+2 = 6 Therefore, solution exists
100 = 40 + Q1 Q1 = 60 kg/hr
100.(0.5) = 40.(0.9) + 60.(x1) x1 = 0.233
30 + Q1 = Q2 Q2 = 90 kg/hr
x2 = 0.256
30 + Q3 = Q2 Q3 = 60 kg/hrx3 = 0.083
You should treat any junction as a process unit!
1
3
2
CHBI 201 25
CHBI 201 26
It is rare that a chemical reaction A B proceeds to completion in a reactor. Its efficiency is never 100. Some A in the product !
To find a way to send the “A” back to feed you need a seperation and recycle equipment, this would decrease the cost of purchasing more A.
If a fraction of the feed to a process unit is diverted around the unit and combined with the output stream, this process is called bypass.
RECYCLE & BYPASS STREAM
rxn Sep.Feed Product
Recycle
Process Unit
Feed
Bypass stream
CHBI 201 27
Feed: Fresh air with 4 mole% H2O(v) is “cooled” and “dehumidified” to a water content of 1.7 mole% H2O.
Fresh air is combined with a recycle stream of dehumidified air. The blended stream entering unit contains 2.3 mole% H2O. In the air conditioner some of the water is removed as liquid. Take 100 mole of dehumidified air delivered to the room, calculate moles of feed, water condensed, dehumidified air recycled.
EXAMPLE (pg 110)
CHBI 201 28
EXAMPLE - continue
AIR CONDITIONER
n1 (mol)
0.04 W
0.96 DA
100 mol
0.983 DA
0.017 W(v)
n2 (mol)
0.977 DA
0.023 W(v)
n3 mole W(ℓ)
n5 (mol) 0.983 DA, 0.017 W
n4 (mol)
0.017 W
0.983 DA
CHBI 201 29
Overall system: 2 variables (n1, n3)
2 balance equations (two species) Degree of freedom = 0 (n1, n3) are determined!!!
Mixing point: 2 variables (n2, n5)
2 balance equations (two species) Degree of freedom = 0
Cooler: 2 variables (n2, n4)
2 balance equations (two species) Degree of freedom = 0
Splitting point: 2 variables (n4, n5) Donot use SP in the solution
Water blance on Mixing point:0.04n1 + 0.017n5 = 0.023n2
n2 = 392.5 mol
n5 = 290 mol recycled
EXAMPLE - continue
CHBI 201 31
If there is a chemical reaction in a process More complications
The stoichiometric ratios of the chemical reactions Constraints
The stoichiometric equation 2SO2 + O2 2SO3
2 molecules of SO2 reacts with 1 molecule of O2 and yields 2 molecules of SO3
2, 1 and 2 are stoichiometric coefficients of a reaction
CHEMICAL REACTION STOICHIOMETRY
CHBI 201 32
If the reactants are not in stoichiometric proportion one of them will be excess, the other will be limiting
LIMITING & EXCESS REACTANTS
iν
i0n-
in
)( reaction of Extend
fed molesreacted moles A of conversion Fractional
stoich.)
A(n
]stoich)
A(n -
feed)
A(n [
A of excess Fractional
CHBI 201 33
C3H6 + NH3 + 3/2 O2 C3H3N + 3 H2O
Feed: 10 mol % of C3H6, 12 mole % NH3 and 78 mole % airA fractional converison of limiting reactant = 30%Taking 100 mol of feed as a basis, determine which reactantis limiting, and molar amounts of all product gas constituentsfor a 30% conversion of the limiting reactant.