化工程序設計─儀器裝置設計與選擇

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化工程序設計

儀器裝置設計與選擇

台大化工系台大化工系陳延平陳延平

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化工程序裝置的設計與選擇

 選擇可解決問題之技術與設備選擇可解決問題之技術與設備  經濟上可行及省錢經濟上可行及省錢  合乎環保要求合乎環保要求  易於操作及合乎安全規範易於操作及合乎安全規範  易於購置且有可信的供應廠商易於購置且有可信的供應廠商

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Examples and solutions

 Removal of acid Removal of acid

gases (CO

gases (CO2₂ and H and H22S)S)

 May choose a gas May choose a gas scrubber

scrubber

 What is the proper What is the proper solvent?

solvent?

 Henry’s law is Henry’s law is applicable for applicable for

solvent selection solvent selection

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Examples and solutions

 Consider a large-size compressorConsider a large-size compressor  Should we used a steam turbine-Should we used a steam

turbine-driven or motor-turbine-driven compressor? driven or motor-driven compressor?

 The answer is a steam turbine-driven The answer is a steam turbine-driven

compressor compressor

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Rules of Thumb

（經驗規則）

 Fans are used to raise pressure about Fans are used to raise pressure about

3%. Blowers raise to less than 40 psig. 3%. Blowers raise to less than 40 psig.

Compressors to higher pressures. Compressors to higher pressures.

 Heat transfer coefficients (Btu/(hr)(sqfHeat transfer coefficients (Btu/(hr)(sqf

t)(F) for estimating purposes: water to t)(F) for estimating purposes: water to

liquid, 150. liquid to liquid, 50. liquid t liquid, 150. liquid to liquid, 50. liquid t o gas, 5. condenser, 150. reboiler, 20 o gas, 5. condenser, 150. reboiler, 20

0. 0.

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Rules of Thumb

 Batch reactions are conducted in Batch reactions are conducted in

stirred tanks for small daily production stirred tanks for small daily production

rates. rates.

 Tubular reactors are suitable to high Tubular reactors are suitable to high

production rates at short residence production rates at short residence

times and when substantial heat times and when substantial heat

transfer is needed. transfer is needed.

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Equipments for

Transportation

 To transport To transport

materials from one

equipment to the

the form of

pressure

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Transportation of gases (Woods,

1995)

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Select a device for air flow rate

of 10

33

dm

33

/s and a pressure rais

e of 1000 kPa?

 Check the previous Figure for the Check the previous Figure for the

general selection purpose general selection purpose

 A centrifugal compressor, or a 2- or 3-A centrifugal compressor, or a 2- or

3-stage reciprocating compressor is stage reciprocating compressor is

appropriate appropriate

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Selection of a pump

-First consideration is the viscosity

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Selection of a pump for a flui

d with viscosity of 100 mPa

s, flow rate of 20 L/s (or 0.0

2 m

33

/s), and a head of 30 m

 From the above Figure, a single-stage From the above Figure, a single-stage

centrifugal radial flow pump is centrifugal radial flow pump is

appropriate appropriate

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The Net Positive Suction

Head (NPSH)

 The NPSH is the pressure or head, The NPSH is the pressure or head,

required at the inlet or suction flange required at the inlet or suction flange

in excess of the vapor pressure of the in excess of the vapor pressure of the

liquid at the temperature of the liquid at the temperature of the

pumping condition. pumping condition.

 NPSH = (The pressure supplied at the NPSH = (The pressure supplied at the

flange) – (The vapor pressure of the flange) – (The vapor pressure of the

liquid at the pumping temperature) liquid at the pumping temperature)

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Example and solution

For pumping a liquid, we have the following co For pumping a liquid, we have the following co

nditions nditions

Vapor pressure of liquid at operating temperat Vapor pressure of liquid at operating temperat

ure is 5 kPa ure is 5 kPa

Atmospheric pressure is 90 kPa Atmospheric pressure is 90 kPa Fluid density is 1 Mg/m

Fluid density is 1 Mg/m33

The pump is located 5 m above the liquid The pump is located 5 m above the liquid The friction loss in the suction line is 1.2 m The friction loss in the suction line is 1.2 m

If the required NPSH of the pump is 4 m, will t If the required NPSH of the pump is 4 m, will t

he pump work without cavitating? he pump work without cavitating?

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Solution to the problem

 To avoid cavitation, we need to supplied To avoid cavitation, we need to supplied a NPSH at least 0.5 m greater than the r

a NPSH at least 0.5 m greater than the r

equired NPSH

 Since the supplied NPSH is only 2.47 m, Since the supplied NPSH is only 2.47 m, the pump cavitates

the pump cavitates

m NPSH loss friction z p p NPSH 47 . 2 2 . 1 ) 5 0 ( ) 8 . 9 0 . 1 5 90 ( 21 2 1           

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Transportation of solids

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Problem and solution

 We need to We need to transport 15.5 transport 15.5 dm dm33/s of wheat /s of wheat horizontally a horizontally a distance of 15.2 m. distance of 15.2 m. What is the What is the appropriate appropriate equipment? equipment?

 Check with the Check with the design data book

design data book

 A crew conveyor is A crew conveyor is appropriate

appropriate

 For a trough For a trough

loading of 45%,

find the screw

diameter as 0.32 m

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Heat Exchange Equipments:

General Classification

 FurnacesFurnaces  ExchangersExchangers  BoilersBoilers  CondensersCondensers

 Cryogenic unitsCryogenic units

 General applications General applications for heating and cooling

for heating and cooling

(see the next figure)

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Rough Sizing of Furnace

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Example and solution

 Select a furnace to heat up an oilSelect a furnace to heat up an oil  The heating load is 2 x 10The heating load is 2 x 1077 kJ/h kJ/h

 Consider a wider heat flux rangeConsider a wider heat flux range

 What will be the total area of tubes?What will be the total area of tubes?  Select a A-furnaceSelect a A-furnace

 From the previous figure, estimate the total From the previous figure, estimate the total tube area as 300 m

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Heat Exchanger Design:

General Steps

 Determine the heat loadDetermine the heat load

 Select the heating/cooling mediumSelect the heating/cooling medium  Make an overall selection Make an overall selection

 Decide the inlet and outlet Decide the inlet and outlet

temperatures temperatures

 Estimate the heat transfer coefficientEstimate the heat transfer coefficient  Determine the heat transfer areaDetermine the heat transfer area

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Circulation Systems for

Heat Exchange

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Determination of the Heat

Load

 Sensible heatSensible heat  Latent heatLatent heat

 Heat of reactionHeat of reaction

 The total heat load is the summation oThe total heat load is the summation o

f all above terms f all above terms

 It is easily obtained from the energy bIt is easily obtained from the energy b

alance of process flowsheet alance of process flowsheet

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Selection of the Medium

 High TemperatureHigh Temperature

– Combustion gasesCombustion gases

– Hot mineral oilHot mineral oil

 Usual TemperatureUsual Temperature

– Condensing steamCondensing steam

– Air or waterAir or water

 Very Low TemperatureVery Low Temperature

– AmmoniaAmmonia

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Example and solution

 We wish to heat a system to 200We wish to heat a system to 20000C, what mC, what m

edium may be used?

 From the last figure, we can choose DowtheFrom the last figure, we can choose Dowthe rn J (alkyllated aromatic) if the pressure is a

rn J (alkyllated aromatic) if the pressure is a

bout 2 - 10 atm

 Use steam at higher pressure (20 atm)Use steam at higher pressure (20 atm)  Choose Dowthern A at a lower pressureChoose Dowthern A at a lower pressure

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Selection of Equipment Based on

the General Range (Some figures

are shown below)

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Selection of the Inlet and

Outlet Temperatures

 Normally, use countercurrent configurationsNormally, use countercurrent configurations

 If water is used as the cooling medium, its outlet If water is used as the cooling medium, its outlet

temperature rarely exceeds 50

temperature rarely exceeds 5000C (to prevent erosion C (to prevent erosion

and corrosion) and corrosion)

 For air as the cooling medium, its exit temperature is For air as the cooling medium, its exit temperature is

usually less than 50 usually less than 5000CC

 The “approach temperature” is usually 10 (heat The “approach temperature” is usually 10 (heat

exchange, condensation) or 20

exchange, condensation) or 20 00C (boiling)C (boiling)

 Use the log mean temperature difference (LMTD) as Use the log mean temperature difference (LMTD) as

the driving force in the countercurrent heat exchanger the driving force in the countercurrent heat exchanger

 For many cases, the mean temperature difference is For many cases, the mean temperature difference is

0.75(LMTD) for practical design purpose 0.75(LMTD) for practical design purpose

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Estimation of the overall

heat transfer coefficient

 The overall heat transfer coefficient U is The overall heat transfer coefficient U is determined from the heat transfer

determined from the heat transfer

coefficients for the fluid on the inside and

outside of the tubes (resistance effects are

added to h)

 Approximate values for U are available in Approximate values for U are available in many literature

many literature

 The total area is finally calculatedThe total area is finally calculated  Q (heat duty) = U x Area x Q (heat duty) = U x Area x __TT

od id h h U 1 1 1 _ _

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Direct contact Heat

Exchange Equipments

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Example and solution

 Fatty acid drops are cooled down from 255 to 140 Fatty acid drops are cooled down from 255 to 140

0

0C, at a flow rate of 0.8 kg/s and heat capacity of C, at a flow rate of 0.8 kg/s and heat capacity of

2.8 kJ/kg

2.8 kJ/kg00CC

 A direct contact countercurrent heat exchanger A direct contact countercurrent heat exchanger (spray tower) is chosen. Water inlet is at 60

(spray tower) is chosen. Water inlet is at 60 00C, C,

heat capacity is assumed as 4.2 kJ/kg

heat capacity is assumed as 4.2 kJ/kg00CC

 The volumetric heat transfer coefficient for a spray The volumetric heat transfer coefficient for a spray tower is from 2 to 70 kW/m

tower is from 2 to 70 kW/m33K. Take it as 20 K. Take it as 20

kW/m

kW/m33K in this problemK in this problem

 If the diameter is 0.9 m, estimate the height of If the diameter is 0.9 m, estimate the height of heat transfer region

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Example and solution

 The cooling load is:The cooling load is:

– Q = 0.8 x 2.8 x (255-140) = 257 kWQ = 0.8 x 2.8 x (255-140) = 257 kW

 The outlet water temperature is:The outlet water temperature is: – 257 = 0.35 x 4.2 x (T-60) 257 = 0.35 x 4.2 x (T-60)

– Hence T = 235 Hence T = 235 00CC

 The LMTD is:The LMTD is:

– LMTD = [(140-60) – (255-235)]/[ln(80/20)] = 43 LMTD = [(140-60) – (255-235)]/[ln(80/20)] = 43 00CC

 The volume of tower is:The volume of tower is:

– Q(257 kW) = U (20 kW/mQ(257 kW) = U (20 kW/m33K) x V x 43 K) x V x 43 00CC

– V = 0.3 mV = 0.3 m33

 The height of heat transfer region is found as 0.5 mThe height of heat transfer region is found as 0.5 m  Literature has shown a result of 0.9 to 1.2 m Literature has shown a result of 0.9 to 1.2 m

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Cooling Tower

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Rough Sizing of Cooling

Water

 Liquid loading: 1.7 L/mLiquid loading: 1.7 L/m22 for fully packed cross section, and 2.4 L/m for fully packed cross section, and 2.4 L/m2 2 for for

cross flow configuration

 Volumetric gas flow rate (per cross sectional area) is 1800 dmVolumetric gas flow rate (per cross sectional area) is 1800 dm33/m/m22 s s

 Ratio of liquid and gas flow rate is around 1.0Ratio of liquid and gas flow rate is around 1.0  Height of cooling tower: Height of cooling tower:

– H = 4.6 to 6 m, if approach T is 8.3 to 11 H = 4.6 to 6 m, if approach T is 8.3 to 11 00CC

– H = 7 to 9 m, if approach T is 4.5 to 8.3 H = 7 to 9 m, if approach T is 4.5 to 8.3 00CC

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Example and solution

 To cool down 312 kg/s of water from To cool down 312 kg/s of water from

43 to 29.3

43 to 29.3 00C, a cooling tower is C, a cooling tower is

selected selected

 The wet bulb temperature of air is The wet bulb temperature of air is

23.4 23.4 00CC

 Estimate the size of a fully packed Estimate the size of a fully packed

tower tower

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Example and solution

 The cross sectional are is:The cross sectional are is:

– A = [312 kg/s]/[(1 kg/L)(1.7 L/mA = [312 kg/s]/[(1 kg/L)(1.7 L/m22s)]s)]

– A = 183 mA = 183 m22

 The gas flow rate is:The gas flow rate is:

– V = 183 mV = 183 m22 x 1800 dm x 1800 dm33/ m/ m22s = 330,000 dms = 330,000 dm33/ s / s

– This is about 330 mThis is about 330 m33/s x 1.2 kg/m/s x 1.2 kg/m3 3 =396 kg/s=396 kg/s

 The liquid to gas flow rate is:The liquid to gas flow rate is: – 312/396 = 0.79312/396 = 0.79

– Less air is needed. Let it be 312 kg/s, or 260 mLess air is needed. Let it be 312 kg/s, or 260 m33/s/s

 The cooling range is:The cooling range is:

– Approach T = 29.3 – 23.4 = 5.9 Approach T = 29.3 – 23.4 = 5.9 00CC

 The height of tower is:The height of tower is: – H = 8 mH = 8 m

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Problem to design a heat

exchanger

 Methanol at a flow rate of 100,000 Methanol at a flow rate of 100,000

kg/h is cooled from 95 to 40

kg/h is cooled from 95 to 40 00C.C.

 Water is used as the coolant with a Water is used as the coolant with a

temperature rise of 25 to 40 temperature rise of 25 to 40 00CC

 Design a shell and tube heat Design a shell and tube heat

exchanger: Number of tubes, Shell exchanger: Number of tubes, Shell

diameter, Pressure drop, etc. diameter, Pressure drop, etc.

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Reactor design

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PFTR configurations

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STR configurations

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Problem 1

 Consider the dehydration of toluene to Consider the dehydration of toluene to make benzene

make benzene

 Gas phase reaction plus solid catalystGas phase reaction plus solid catalyst  Temperature is 590 to 650 Temperature is 590 to 650 ooCC

 Pressure is 6 to 11 MPaPressure is 6 to 11 MPa

 Reaction is first order with respect to toReaction is first order with respect to to luene and 0.5 order to hydrogen

luene and 0.5 order to hydrogen  Exothermic reaction: 50 MJ/kmol Exothermic reaction: 50 MJ/kmol

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Solution to Problem 1

 Relatively high pressure and Relatively high pressure and

temperature temperature

 Select a PFTR reactor with catalyst Select a PFTR reactor with catalyst

packed in tubes packed in tubes

 May operate adiabatically or remove May operate adiabatically or remove

some of the heat some of the heat

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Problem 2

 Consider the alkylation of isobutane to Consider the alkylation of isobutane to

produce alkyllate produce alkyllate

 Liquid phase reaction with immiscible sLiquid phase reaction with immiscible s

ulfuric acid catalyst ulfuric acid catalyst

 Temperature 7 to 13 Temperature 7 to 13 ooCC

 Pressure 580 to 650 kPaPressure 580 to 650 kPa

 Very rapid reaction, very complex chaiVery rapid reaction, very complex chai

n mechanism, slightly exothermic n mechanism, slightly exothermic

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Solution to Problem 2

 This is liquid reaction at room This is liquid reaction at room

temperature and high pressure temperature and high pressure

 Use a CFSTR reactorUse a CFSTR reactor

 This is a relatively clean reaction. Use This is a relatively clean reaction. Use

the internal heat exchanger the internal heat exchanger

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Problem 3

 Consider the suspension polymerization of viConsider the suspension polymerization of vi nyl chloride to make PVC

nyl chloride to make PVC

 Liquid monomer drops suspended in water Liquid monomer drops suspended in water with water soluble initiator

with water soluble initiator

 Temperature is 25 Temperature is 25 ooC, pressure is 0.85 MPaC, pressure is 0.85 MPa

 Exothermic reaction: 115 MJ/kmol monomer Exothermic reaction: 115 MJ/kmol monomer converted

converted

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Solution to Problem 3

 Use a batch STR, or a series of CFSTRUse a batch STR, or a series of CFSTR  Water is added to control the removal Water is added to control the removal

of heat, and thus control the of heat, and thus control the

temperature temperature

 Temperature variation must be Temperature variation must be

minimized to keep a good molar mass minimized to keep a good molar mass

distribution of the product distribution of the product