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Chapter 14 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
1
Case Study #1 - Aspirin Production
Material Balances to Determine Flows
Reactor: Species Balances:
SA:
�
˙ n SA, feed = ˙ n SA,out + rcons,SA (1) AAH:
�
˙ n AAH , feed = ˙ n AAH ,out + rcons,AAH (2) PA:
�
˙ n PA, feed = ˙ n PA,out (3) ASA:
�
rform,ASA = ˙ n ASA,out (4) AA:
�
rform,AA = ˙ n AA,out (5) Conversion:
�
rcons,SA = 0.995 ˙ n SA, feed (6) Stoichiometry:
�
rcons,AAHrcons,SA
= 11
(7)
�
rform,ASArcons,SA
= 11
(8)
�
rform,AArcons,SA
= 11
(9)
From Equation 6
�
rcons,SA = 0.995 ˙ n SA, feed = 0.995˙ m SA, feed
MWSA= 0.995 2,000,000g /hr
138.1g /gmol= 14,400 gmol
hr
From Equation 7: rcons,AAH = rcons,SA = 14,400 gmol/hr From Equation 8: rform,ASA = rcons,SA = 14,400 gmol/hr From Equation 9: rform,AA = rcons,SA = 14,400 gmol/hr From Equation 1:
�
˙ n SA,out = ˙ n SA, feed − rcons,SA =˙ m SA, feed
MWSA− rcons,SA = 2,000,000g /hr
138.1g /gmol−14,400 gmol
hr= 72.4 gmol
hr (=10.0 kg/hr) From Equation 2:
�
˙ n AAH ,out = ˙ n AAH , feed − rcons,AAH =ρAAH ˙ V AAH , feed
MWAAH− rcons,AAH
SA 2000 kg/hrAAH 5000 L/hrPA 1250 L/hr
coolingwaterReactor
Mixer
ultrapure water22,000 L/hr25°C
70°C
Pump HeatExchanger
FiltrationSequence
ASAproduct
Acidwaste
coolingwater
Chapter 14 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
2
�
=1.08 g
cm3⎛ ⎝ ⎜
⎞ ⎠ ⎟ 5,000,000
cm3
hr
⎛
⎝ ⎜
⎞
⎠ ⎟
102.1g /gmol−14,400 gmol
hr= 38,500 gmol
hr
(= 3,930 kg/hr)
From Equation 3:
�
˙ n PA,out = ˙ n PA, feed = ρPA ˙ V PAMWPA
=1.68 g
cm3⎛ ⎝ ⎜
⎞ ⎠ ⎟ 1,250,000 cm3
hr
⎛
⎝ ⎜
⎞
⎠ ⎟
998.0g /gmol= 21,400 gmol
hr
(= 2,100 kg/hr)
From Equation 4:
�
˙ n ASA,out = rform,ASA = 14,400 gmolhr
(= 2,600 kg/hr)
From Equation 5:
�
˙ n AA,out = rform,AA = 14,400 gmolhr
(= 865 kg/hr) (Total mass flow rate out = 9,500 kg/hr)
Mixer
Total mass balance:
�
˙ m reactor out + ˙ m water in = ˙ m mixer out (10)
Species balance:
Water:
�
˙ n water,in = ˙ n water,out + rcons,water (11) AA:
�
˙ n AA,in + rform,AA = ˙ n AA,out (12)
Conversion:
�
rcons,AAH = 100% ˙ n AAH ,in (13)
Stoichiometry:
�
rform,AArcons,AAH
= 21
(14)
�
rcons,waterrcons,AAH
= 11
(15)
From Equation 13:
�
rcons,AAH = ˙ n AAH ,in = 38,500gmol /hr
From Equation 14:
�
rform,AA = 2rcons,AAH = 77,000gmol /hr
From Equation 12:
�
˙ n AA,out = ˙ n AA,in + rform,AA = 14,400 + 77,000 = 91,400 gmolhr
(= 5,490 kg/hr)
From Equation 15:
�
rcons,water = rcons,AAH = 38,500gmol /hr
From Equation 11:
�
˙ n water,out = ˙ n water,in − rcons,water =ρwater ˙ V water,in
MWwater− rcons,water
Chapter 14 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
3
�
=1 gcm3
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 22,000,000
cm3
hr
⎛
⎝ ⎜
⎞
⎠ ⎟
18g /gmol− 38,500 gmol
hr=1,184,000 gmol
hr
(= 21,300 kg/hr) From Equation 10:
�
˙ m mixer out = ˙ m reactor out + ˙ m water in = 9,500 + 22,000 = 31,500 kghr
Filtration Sequence
Product Stream: This will consist of all of the ASA and SA that enter. These amounts are
ASA: 14,400 gmol/hr SA: 72 gmol/hr (the product stream will be 99.5 mole% ASA)
Acid Waste Stream: This will consist of all the AA, PA, and water that enter. These are:
AA: 91,400 gmol/hr PA: 21,400 gmol/hr Water: 1,184,000 gmol/hr (the acid stream will be 7.0 mole% acetic acid, which is approximately 3.2 molar)
Equipment Sizing
Reactor
rcons,SA = rrxn,SAVreactor = krxn(cSAV)reactor
= krxn cSA,out Vreactor
�
= krxn˙ n SA,out
˙ V outVreactor
So,
�
Vreactor =rcons,SA ˙ V outkrxn ˙ n SA,out
=rcons,SA
krxn ˙ n SA,out
˙ m outρout
�
=14,400 gmol
hr⎛ ⎝ ⎜
⎞ ⎠ ⎟ 9,500,000
ghr
⎛ ⎝ ⎜
⎞ ⎠ ⎟
0.5s−1( ) 72 gmolhr⎛ ⎝ ⎜
⎞ ⎠ ⎟ 1.2
gcm3
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 3600shr
⎛ ⎝ ⎜
⎞ ⎠ ⎟
= 880,000cm3 = 880L
= 233 gal
Pump
The shaft work per mass of fluid is
Chapter 14 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
4
�
ws = P1 − P2ρ∑
where the density of the streams is given as 1.2 g/cm3 (which converts to 74.9 lbm/ft3)
So,
�
ws = 22 + 260psi74.9lbm / ft
3 144in2
ft2⎛
⎝ ⎜
⎞
⎠ ⎟ 1hp550 ft lbf /s
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ = 0.986 hp
lbm /s
Actual power (Equation 7.10) is
�
W = ˙ m ws = 31,500,000 ghr
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 0.986 hp
lbm /s⎛
⎝ ⎜
⎞
⎠ ⎟
1hr3600s
⎛ ⎝ ⎜
⎞ ⎠ ⎟
2.205lbm1,000g
⎛
⎝ ⎜
⎞
⎠ ⎟ = 19.0hp
But this is power delivered to the fluid. The total power required to operate the pump (Equation 7.11) is
�
Power tooperate the pump = Power delivered to the fluidEfficiency
= 19.0hp0.85
= 22.4hp
Energy Balances
Reactor
The heat of reaction is
�
Δ ˆ H rxnrcons,SA = −85,800 Jgmol
⎛
⎝ ⎜
⎞
⎠ ⎟ 14,400 gmol
hr⎛ ⎝ ⎜
⎞ ⎠ ⎟ = −1.24x109 J
hr
Thus, the amount of cooling water needed is that which will provide that much sensible cooling:
�
˙ Q water = ˙ m waterCpwaterTout −Tin( )
so,
�
˙ m water =˙ Q water
CpwaterTout −Tin( )
= 1.24x109 J /hr
4.184 Jg°C
⎛
⎝ ⎜
⎞
⎠ ⎟ 40°C −18°C( )
= 13.4x106 ghr
= 13,400 L/hr
= 3,550 gal/hr
Mixer
As explained in Technical Information item #4,
�
˙ Q reactor outlet = − ˙ Q ultrapurewater
�
˙ m reactor outletCpreactor outletTfinal −Treactor outlet( ) = − ˙ m ultrapurewaterCpwater
Tfinal −Tultrapurewater( )
solving:
�
Tfinal =˙ m reactor outletCpreactor outlet
Treactor outlet + ˙ m ultrapurewaterCpwaterTultrapurewater
˙ m reactor outletCpreactor outlet+ ˙ m ultrapurewaterCpwater
Chapter 14 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
5
�
=9,500,000 g
hr⎛ ⎝ ⎜
⎞ ⎠ ⎟ 1.67
Jg°C
⎛
⎝ ⎜
⎞
⎠ ⎟ 70°C( ) + 22,000,000 g
hr⎛ ⎝ ⎜
⎞ ⎠ ⎟ 4.184
Jg°C
⎛
⎝ ⎜
⎞
⎠ ⎟ 25°C( )
9,500,000 ghr
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 1.67
Jg°C
⎛
⎝ ⎜
⎞
⎠ ⎟ + 22,000,000 g
hr⎛ ⎝ ⎜
⎞ ⎠ ⎟ 4.184
Jg°C
⎛
⎝ ⎜
⎞
⎠ ⎟
= 31.6°C
Heat Exchanger
The heat duty is
�
˙ Q duty = ˙ m mixer outCpmixer outTexchanger out −Texchangerin( )
�
= 31,500,000 ghr
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 3.41
Jg°C
⎛
⎝ ⎜
⎞
⎠ ⎟ 31.6 − 25°C( ) = 709,000,000 J
hr
= 672,000 Btu/hr
Cooling Water Requirement:
Water entering temperature: 18°C Water exiting temperature: 26.6°C (to maintain ∆T ≥ 5°C)
�
˙ m water =˙ Q water
CpwaterTout −Tin( )
= 709x106 J /hr
4.184 Jg°C
⎛
⎝ ⎜
⎞
⎠ ⎟ 26.6°C −18°C( )
= 19.7x106 ghr
= 5,200 gal/hr
Heat Exchanger Size:
∆T1 = 5°C = 9°F ∆T2 = 7°C = 12.6°F
�
ΔTlogmean = 12.6°F − 9°Fln12.6°F
9°F=10.7°F
From Table 10.4,
�
U ≈ 50 Btuhr ft2°F
From Equation 10.30,
�
A =˙ Q duty
UΔTlogmean=
672,000 Btuhr
50 Btuhr ft2°F
⎛
⎝ ⎜
⎞
⎠ ⎟ 10.7°F( )
= 1,260 ft2
Economics
Capital Investment
31.6°C26.6°C 18°C
25°CMixer outletCooling water
HeatExchanger
Chapter 14 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
6
Reactor Cost: M & S814
47.0 V gal( ).61 = 1469814
47.0 233gal( ).61 = $2,300
Cooler Cost: M & S814
398 A ft 2( ).65 = 1469814398 1260 ft 2( ).65 = $74,400
Pump Cost:
M & S814
421 V galmin
⎛⎝⎜
⎞⎠⎟.46
Where V = 3.15x107 cm
3
hr⎛⎝⎜
⎞⎠⎟264.17 gal
m3⎛⎝⎜
⎞⎠⎟
1m3
106cm3
⎛⎝⎜
⎞⎠⎟
1hr60min
⎛⎝⎜
⎞⎠⎟= 139 gal
min
So Pump Cost =1469814
421 139 galmin
⎛⎝⎜
⎞⎠⎟.46
= $7,300
Filtration Sequence Cost (given) $2,900,000 Total Delivered Cost = 110%($2,300 + $74,400 + $7,300) + $2,900,000 = $2,992,000 Capital Investment = Lang Factor • Delivered Cost = 5.69($2,992,000) = $17,030,000 Depreciation = 85% of Cap. Investment/10 years = .85($17,030,000)/10 = $1,447,000/yr
Operating Cost
Direct Operating Costs
Feed:
SA:
�
$1.33lbm
⎛
⎝ ⎜
⎞
⎠ ⎟ 2000kghr
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 2.205lbm
kg⎛
⎝ ⎜
⎞
⎠ ⎟ 24hrday
⎛
⎝ ⎜
⎞
⎠ ⎟ 365daysyear
⎛
⎝ ⎜
⎞
⎠ ⎟ = $51,380,000 / yr
AAH:
�
$0.49lbm
⎛
⎝ ⎜
⎞
⎠ ⎟ 5000Lhr
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 1.08kgL
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 2.205lbm
kg⎛
⎝ ⎜
⎞
⎠ ⎟ 24hrday
⎛
⎝ ⎜
⎞
⎠ ⎟ 365daysyear
⎛
⎝ ⎜
⎞
⎠ ⎟ = $51,110,000 / yr
PA:
�
$0.34lbm
⎛
⎝ ⎜
⎞
⎠ ⎟ 1250Lhr
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 1.68kgL
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 2.205lbm
kg⎛
⎝ ⎜
⎞
⎠ ⎟ 24hrday
⎛
⎝ ⎜
⎞
⎠ ⎟ 365daysyear
⎛
⎝ ⎜
⎞
⎠ ⎟ = $13,790,000 / yr
TOTAL = $116,280,000/yr
Ultrapure Water:
�
$0.03lbm
⎛
⎝ ⎜
⎞
⎠ ⎟ 22,000L
hr⎛ ⎝ ⎜
⎞ ⎠ ⎟ 1.0kgL
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 2.205lbm
kg⎛
⎝ ⎜
⎞
⎠ ⎟ 24hrday
⎛
⎝ ⎜
⎞
⎠ ⎟ 365daysyear
⎛
⎝ ⎜
⎞
⎠ ⎟ = $12,750,000 / yr
Cooling Water:
�
$0.031000gal⎛
⎝ ⎜
⎞
⎠ ⎟ 3,550 + 5,200 gal
hr⎛ ⎝ ⎜
⎞ ⎠ ⎟ 24hrday
⎛
⎝ ⎜
⎞
⎠ ⎟ 365daysyear
⎛
⎝ ⎜
⎞
⎠ ⎟ = $2,300 / yr
Electricity:
�
$0.05kW hr
⎛
⎝ ⎜
⎞
⎠ ⎟ 22.4hp( ) 745.7W
hp⎛
⎝ ⎜
⎞
⎠ ⎟ 1kW1000W⎛ ⎝ ⎜
⎞ ⎠ ⎟ 24hrday
⎛
⎝ ⎜
⎞
⎠ ⎟ 365daysyear
⎛
⎝ ⎜
⎞
⎠ ⎟ = $7,300 / yr
Chapter 14 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
7
Operating Cost = 1.55($116,280,000+$12,750,000+$2,300+$7,300) = $200,000,000/yr
Sales:
Aspirin:
�
$3.97lbm
⎛
⎝ ⎜
⎞
⎠ ⎟ 2,600kghr
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 2.205lbm
kg⎛
⎝ ⎜
⎞
⎠ ⎟ 24hrday
⎛
⎝ ⎜
⎞
⎠ ⎟ 365daysyear
⎛
⎝ ⎜
⎞
⎠ ⎟ = $199,910,000 / yr
Acid Waste:
�
$0.013lbm
⎛
⎝ ⎜
⎞
⎠ ⎟ 28,900kg
hr⎛ ⎝ ⎜
⎞ ⎠ ⎟ 2.205lbm
kg⎛
⎝ ⎜
⎞
⎠ ⎟ 24hrday
⎛
⎝ ⎜
⎞
⎠ ⎟ 365daysyear
⎛
⎝ ⎜
⎞
⎠ ⎟ = $7,260,000 / yr
TOTAL SALES = $207,170,000/yr Gross Profit = s-c-d = $207,170,000 – $200,000,000 – $1,447,000 = $5,723,000/yr NAPAT = $5,723,000/yr (1 - .35) = $3,720,000/yr
ROI = NAPATCapitalInvestment
=$3,720,000 / yr$17,030,000
= 21.8%