Chapter 14 CS-1

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%