Reactor Design Ii

Reactor Design II-Fourth Year Dr.Ali N.Khalaf

University of Basrah College of Engineering Chemical Engineering Department Fourth Year First Semester Reactor design II

Chapter One Non-Isothermal Reactor Design

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

1- Optimal Temperatures for Isothermal Reactors: Reaction rates almost always increase with temperature. Thus, the best temperature for a single, irreversible reaction, whether elementary or complex, is the highest possible temperature. Practical reactor designs must consider limitations of materials of construction and economic tradeoffs between heating costs and yield, but there is no optimal temperature from a strictly kinetic viewpoint. Reversible reactions, since these are a special form of multiple reactions, usually exhibit an optimal temperature with respect to the yield of a desired product. One must specify the temperature at which to operate. Consider the elementary, reversible reaction

Suppose this reaction is occurring in a CSTR of fixed volume and throughput. It is desired to find the reaction temperature that maximizes the yield of product B. Suppose Ef>Er, as is normally the case when the forward reaction is endothermic. Then the forward reaction is favored by increasing temperature. The equilibrium shifts in the desirable direction and the reaction rate increases. The best temperature is the highest possible temperature and there is no interior optimum. For Ef<Er, increasing the temperature shifts the equilibrium in the wrong direction, but the forward reaction rate still increases with increasing temperature. There is an optimum temperature for this case. A very low reaction temperature gives a low yield of B because the forward rate is low. A very high reaction temperature also gives a low yield of B because the equilibrium is shifted toward the left. The outlet concentration from the stirred tank, assuming constant physical properties and is given by

𝐶𝐵,𝑜𝑢𝑡 =

𝐶𝐴0 𝜏 𝑘𝑓 1+ 𝜏 𝑘𝑓 + 𝜏 𝑘𝑟

…………………….. ……………….. (1)

We assume the forward and reverse reactions have Arrhenius temperature dependences 𝑑 𝐶𝐵,𝑜𝑢𝑡

with Ef<Er . Setting

𝑇𝑜𝑝𝑡𝑖𝑚𝑎𝑙 =

𝑑𝑇

𝑅 𝑙𝑛

= 0 gives

𝐸𝑟 𝑘𝑟 𝜏(𝐸𝑟 −𝐸𝑓……………………..

………………. (2)

𝐸𝑓

2- Influence of Temperature on Reactor Operation: Because most reactions we not carried out isothermally, we now focus our attention on heat effects in chemical reactors. The basic design equations, rate laws, and stoichiometric relationships derived and used in Chapter 4 for isothermal reactor design 2 Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf are still valid for the design of non-isothermal reactors. The major difference lies in the method of evaluating the design equation when temperature varies along the length of a PFR or when heat is removed from a CSTR. Thus, in a batch reactor, the temperature of the reaction mixture may change with time. In a flow reactor, the temperature may change with time and position, or the feed and effluent streams may have different temperatures. To account for these effects, reactor analysis must include the energy balance. The material and energy balance equations are a coupled set of equations that together describe the performance of an ideal reactor. Typically, the energy balance is expressed in terms of the reactor temperature, which may vary with space and time. The nature of thevariations depends on the reactor type and mode of operation. In perfectly mixed batch reactors, the temperature is the same at all locations in the reactor, but changes with time as the reaction proceeds. Sometimes the reactor is operated with heat transfer through the reactor walls, or via heating or cooling coils inserted into the reactor. Adiabatic operation is also possible. In a plug flow reactor, the temperature varies with axial position (i.e., in the direction of flow) in the reactor. There may or may not be heat transfer through the reactor walls. In a CSTR, the temperature is uniform everywhere in the reactor and is equal to the outlet temperature. However, the inlet and outlet temperatures are different. There may ormay not be heat transfer with the surroundings, either through the vessel walls or by heating or cooling coils within the reactor. There are three types of operation: - Isothermal Operation. - Adiabatic Operation. - Non-isothermal Operation (with heat exchange). Let’s calculate the volume necessary to achieve a conversion, X, in a PFR for a firstorder, exothermic and adiabatic reaction.

𝑑𝑉

Material Balance: Rate Law:𝑟𝐴

=𝐴𝑒

𝐹𝐴𝑜 𝐸 − 𝑅𝑇

Stoichiometry: Combine:

𝑑𝑉 𝐹𝐴𝑜

𝑥

= ∫0

𝑥 𝑑𝑥

= ∫0

𝑟𝐴

𝐶𝐴 𝐶𝐴 = 𝐶𝐴𝑜 (1 − 𝑥 ) 𝑑𝑥 𝐸

− 𝐶𝐴𝑜 (1−𝑥) 𝐴 𝑒 𝑅𝑇

We cannot solve this equation because we don’t have x as a function of V or T. We need another equation. That equation is Energy Balance Equation. 3 Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf

3- Energy Balance for Batch Reactors: The batch reactor is internally uniform in both composition and temperature. The flow and mixing patterns that are assumed to eliminate concentration gradients will eliminate temperature gradients as well. Homogeneity on a scale approaching molecular dimensions requires diffusion. Both heat and mass diffuse, but thermal diffusivities tend to be orders-of-magnitude higher than molecular diffusivities. Thus, if one is willing to assume compositional uniformity, it is reasonable to assume thermal uniformity as well. For a perfectly mixed batch reactor, the energy balance is (

𝐻𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝐻𝑒𝑎𝑡 𝑎𝑑𝑑𝑡𝑖𝑜𝑛 𝑜𝑟 𝑅𝑒𝑚𝑜𝑣𝑒𝑑 )+( ) = (𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝐻𝑒𝑎𝑡) 𝑏𝑦 𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝐹𝑙𝑢𝑖𝑑 𝑏𝑦 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 ……. (3)

T = reaction temperature K Ta= External heating or cooling fluid or wall temperature K A = heat transfer area m2 Cpi = specific heat kJ/Kmol U = overall heat transfer kJ/s.m2.K ∆Hr=Heat of reaction per mole of A reacting

……………………… (4)

3.1 Isothermal operation: Isothermal operation means dT/dt=0 and therefore T=T0 and − 𝑉 𝑟𝐴 ∆𝐻𝑟 = 𝑈𝐴 (𝑇 − 𝑇𝑎 )

…………………………………. (5)

Heat generation = Heat removed

3.2 Adiabatic Operation: For constant heat of reaction ∆Hr and constant volume

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

………….…………………... (.6)

………….…………………... (7)

3.3 Non-isothermal batch reactor: The general design material balance equation is still valid and may also be expressed in terms of fractional conversion, x. ………….…………………... (.8)

This equation is solved with the energy balance equation

𝑁𝐶𝑃 = ∑ 𝜃𝑖 𝐶𝑃𝑖 5

Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf Note for constant Volume batch reactor use Cp, and for constant pressure batch reactor use Cv.

Example 1: Consider the reaction A → B, (–rA) = kCA, k350 = 0.65 min–1,(ΔHr) = 30(kcal/mol) in a 10liter batch reactor with initial concentration of A = 2 mol/l and temperature To = 350 K. Determine the heat transfer rate required to maintain the temperature at 350 K at95% conversion Isothermal operation means dT/dt=0T = To=350 K

Then:Qg=Qr

𝑄𝑔 = −𝑉𝑟𝐴 ∆𝐻𝑟 𝑄𝑞 = −𝑉𝑘𝐶𝐴 ∆𝐻𝑟 = −𝑉𝑘∆𝐻𝑟 𝐶𝐴𝑜 𝑒 −𝑘𝑡 Qr = –0.65 × 2.0 e–0.65t × 10 × 30

Qr = –390e–0.65t(kcal/min).

Since Q < 0, it shows that heat is removed from the system, which is undergoing exothermic reaction. The time corresponding to a 95% conversion for the first order reaction is determined by

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

The heat transfer rate (Qt≡95%) at 95% conversion is

Q = –390e–0.65 × 4.608

Q= –19.51 kcal/min Example 2: Consider the oxidation of carbon monoxide in an adiabatic constant-volume batch reactor. If the reaction is assumed to be essentially irreversible; it is described by the 1 overall stoichiometry: 𝐶𝑂 + 𝑂2 → 𝐶𝑂2 2 The rate of reaction in the presence of water is given by a power law rate expression:

A mixture consisting of 1% CO, 1% O2, 1% H2O, and 97% N2 (mole percentages) is placed in a constant-volume batch reactor of 0.1 m3 volume at an initial pressure of 1 bar and an initial temperature of 700 K. Calculate the time required to achieve 99% conversion of the CO, and calculate the pressure and temperature in the reactor at that time. ∆Hr = - 4.513x103 J/molA The table of Cv( J/mol.K) data is presented below

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Cv

Solution The oxidation of CO is a highly exothermic chemical reaction, and, as the reactor is adiabatic, the energy balance will be quite important. It is thus necessary to solve both the mole and energy balance equations. We start the solution by considering the mole balance equation. The mole balance equation for CO can be written in terms of the concentration because the volume of the reactor is constant. The mole balance is:

First note that the number of moles of water, and hence the concentration, is constant during the reaction, as water does not react and the reactor volume is constant. The water concentration is given by

The initial concentrations of CO and oxygen can be easily calculated:

𝑋

𝑂𝑟 ∶ 𝐶𝑂2 = (𝐶𝐶𝑂 ))0 (𝛳𝑂2 − ) 2 With these substitutions and resulting simplification, the rate equation becomes

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf …………….... (A) We first develop a stoichiometric table:

Substituting for XCO = 0.99, we calculate the total number of moles present to be 1.711 mol. The volume is unchanged; therefore, the pressure in the reactor is

 If we want to find the temperature at the end of reaction and the reaction time only.

𝑁𝐶𝑃 = ∑ 𝜃𝑖 𝐶𝑃𝑖 At inlet

𝜃𝐶𝑂

=1 ∶

𝜃𝑂2

=1∶

𝜃𝐶𝑂2

=0 ∶

𝑁𝐶𝑃 = ∑ 𝜃𝑖 𝐶𝑉𝑖 = 19.8 ∗ 1 + 17.13 ∗ 1 = 36.93

𝐽 𝑚𝑜𝑙𝐴. 𝐾

𝟒. 𝟓𝟏𝟑 ∗ 𝟏𝟎𝟑 𝑻 = 𝟕𝟎𝟎 + 𝒙 𝟑𝟔. 𝟗𝟑 𝑻 = 𝟕𝟎𝟎 + 𝟏𝟐𝟐. 𝟐 ∗ 𝒙 𝑻 = 𝟖𝟐𝟏 𝑲

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

0.99 0.172 𝑡= ∫ 3.37x109 0

𝑑𝑥 (

e

−20131 ) 700+122.2∗𝑥

(1 − 𝑥)(1 − 0.5𝑥)0.25

The above equation solved by Simpson Rule to find the reaction time. 𝑡 = 318 𝑠𝑒𝑐  If we want to find the temperature and conversion as a function of time The general energy balance equation is

For adiabatic operation Q = QRemoved=0. Then: .

…………….... (B)

For batch reactor …………….... (C) Substitute Eq. (C) in Eq. (B) gives

𝑑𝑥 𝑁𝐴𝑜 . ∆𝐻𝑟 . 𝑟𝐴 𝑑𝑇 = 𝑑𝑡 𝑑𝑡 (∑ 𝑁𝑖 𝐶𝑃𝑖 ) 𝑟𝐴 =

𝑈𝐴 (𝑇 − 𝑇𝑎 ) − (∑ 𝑁𝑖 𝐶𝑃𝑖 )

𝐶𝐴𝑜 𝑑𝑥 𝑑𝑡

The resulting equations (A, B and C) must be solved simultaneously with the mole balance, to give the fractional conversion and reactor temperature as a function of time. The solution of this system of coupled nonlinear ordinary differential equations is performed numerically, using the Runge–Kutta method. 10

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Or the above equation may written in the finite form as below

∆𝑇 = 𝐶1 ∆𝑥 − 𝐶2 ∆𝑡 ( 𝑇 − 𝑇𝑎 ) Energy Balance …………….. (11) ∆𝑥 = 𝐶3 𝑟̅𝐴 ∆𝑡

𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝐵𝑎𝑙𝑎𝑛𝑐𝑒

𝑟𝐴 = 𝑓 (𝑥, 𝑇)

𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑒 ………….. .. (13)

𝑁𝐴𝑜 . ∆𝐻𝑟 𝐶1 = ∑ 𝑁𝑖 𝐶𝑃𝑖

𝑈𝐴 ∶ 𝐶2 = (∑ 𝑁𝑖 𝐶𝑃𝑖 )

....…..… (12)

1 ∶ 𝐶3 = 𝐶𝐴𝑜

A plot of the fractional conversion of CO as a function of time is shown in Figure 1. The temperature increases as the conversion increases, which are the expected result, because the reaction is exothermic and the reactor is adiabatic. The fractional conversion reaches 99% after 318 s, at which point the temperature is 821 K.

Figure 1: temperature and conversion profile.

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf 4- Energy Balance for Flow Reactors (PFR and CSTR): Since the reaction is non-isothermal, the temperature will vary along the reaction length the temperature is related to conversion through energy balance equation: For a flow reactor:

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

.... (14)

This is the general energy balance equation for flow system.

5- Adiabatic Reactors (PFR and CSTR) with constant CP: The starting equation at steady state is .... (15)

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

….............. (16)

And the conversion is:

…………………….............. (17)

………………………….............. (18)

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

𝑟𝐴 = 𝐴 𝑒

𝐸 𝑅𝑇

−

𝐶𝐴𝑜 ( 1 − 𝑥 )

Or we can find the volume by integrating the design equation numerically 𝑥 𝑉 =∫ 𝐹𝐴𝑜 0

𝑑𝑥 (𝐶𝐴0 (1 − 𝑥 )𝐴𝑒

𝐹𝐴𝑜

𝑉=𝐶

𝐴0

15

𝑥

∫ ( 𝐴 0

𝑑𝑥 −𝐸 ∆𝐻𝑟 𝑅( 𝑇𝑜 + 𝑥 ∑ 𝐶𝑃𝑖 𝜃𝑖 𝑒

−𝐸 ∆𝐻𝑟 𝑅( 𝑇𝑜 + 𝑥 ∑ 𝐶𝑃𝑖 𝜃𝑖

)

)………………..…(19) (1−𝑋)

Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf Example 3: The elementary irreversible organic liquid phase reaction is carried out adiabatically in a flow reactor. An equimolar feed in A and B enters at 27° C, and the volumetric flow rate is 2 dm3/s and CA0=0.1 kmol/m3. a) Calculate the PFR and CSTR volume necessary to achieve 85% conversion. b) What is the maximum inlet temperature one could have so that the boiling point of the liquid (550° K) would not be exceeded even for complete conversion? c) Plot the conversion and temperature as function of PFR volume. d) Calculate the concentration achieved in one 500 dm3 and in two 250-dm3 CSTRs in series. HA°(273)= -20 kcal/mol, HB°(273)= -15 kcal/mol, HC°(273)= -41 kcal/mol CPA=CPB=15 cal/mol.K, CPC=30 cal/mol.K k = 0.01 dm3/mol.s at 300° K, E =10000 cal/mol Solution Part a:

A+B→C

𝑘 =𝐴𝑒 0.01 = 𝐴 𝑒 𝐴 = 4.27 ∗ 105 5

Then 𝑘 = 4.27 ∗ 10 𝑒

−

−

𝐸 𝑅𝑇

10000 1.897∗300

dm3 /mol. s

− 5271.481 𝑇

Volume of CSTR equals the area of rectangle VCSTR = 0.85 x 206 = 175 dm3 Volume of PFR equals the entire Area under the curve VPFR ~ 305 dm3

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

Solution: Part b What is the maximum inlet temperature one could have so that the boiling point of the liquid (550 K) would not be exceeded even for complete conversion?

Solution: Part C Plot the conversion and temperature as function of PFR volume.

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

Solution: Part D

𝑉 𝑥 = 𝐹𝐴𝑜 𝑟𝐴 For one CSTR:

𝑉 𝑥 = − 5271.481 𝐹𝐴𝑜 2 2 5 𝐶𝐴𝑜 (1 − 𝑥) 4.27 ∗ 10 𝑒 300+200𝑥 𝑥 = 0.92

For two CSTR

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𝑉𝐶𝑆𝑇𝑅

Reactor Design II-Fourth Year Dr.Ali N.Khalaf 𝑥1 . 𝐹𝐴𝑜 = 1 − 5271.481 = 500 𝐶𝐴𝑜 2 (1 − 𝑥1 )2 4.27 ∗ 105 𝑒 300+200𝑥1 𝑥1 = 0.881

𝑉𝐶𝑆𝑇𝑅

2

=

(𝑥2 − 𝑥1 ) . 𝐹𝐴𝑜 (1 − 𝑥1 ) 2

𝐶𝐴𝑜 (1 − 𝑥2 )2 4.27 ∗ 105 𝑒

− 5271.481 300+200𝑥2

= 250

𝑥2 = 0.97 Example: 4 The reversible first-order gas reaction A⇋R is to be carried out in a CSTR. For operations at 300 K the volume of reactor required is 100 liters for 60% conversion of A. What should be the volume of the reactor for the same feed rate and conversion but with operations at 400 K? Data: kl= 1x103exp [-2416/T] min -1 ∆Hr= -8000 cal/mol at 300 K K = 10 at 300 K: Feed consists of pure A: Total pressure stays constant Solution:

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

Example 5: The gas-phase reaction A +B→ C is carried out in a pilot plant tubular reactor at about 2 atm and 300oC. The feed rate is 0.120 m3/sec and feed is 40% A, what reactor volume is needed for 95% conversion under adiabatic operation? k=1.4x107exp (-7700/T) m3/mol.sec :∆Hr = -14000 J/mol : CPA = CPB= 15 J/mol.K Solution:

∑ 𝛳𝑖 𝐶𝑃𝑖 =1*15+1.5*15=37.5 J/mol.K 𝑇 = 573 +

14000 𝑥 = 573 + 373.3𝑥 37.5

−7700 𝑟𝐴 = 1.4 ∗ 10 𝑒𝑥𝑝 𝐶𝐴𝑜 ( 1 − 𝑥 ) 𝐶𝐴𝑜 (1.5 − 𝑥 ) 𝑇 7

𝐶𝐴𝑜 =

0.4 ∗ 2 = 0.017𝐾𝑚𝑜𝑙/𝑚3 0.082 ∗ 573

𝐹𝐴𝑜 = 0.017*0.12 =0.00204 Kmol/sec 0.95 𝑉 =∫ 𝐹𝐴𝑜 0

20

𝑑𝑥 2 (𝐶𝐴0 (1 − 𝑥)(1.5 − 𝑥) ∗ 1.4 ∗ 107 𝑒

−𝐸 ∆𝐻𝑟 𝑅( 𝑇𝑜+ 𝑥 ∑ 𝐶𝑃𝑖 𝜃𝑖

)

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

𝑉=

𝐹𝐴𝑜 1.4 ∗

0.95

∫

2 107 𝐶𝐴0 0

𝑑𝑥

( (1 − 𝑥 )(1.5 − 𝑥 ) exp

−7700

)

573+373.3𝑥

After the integration by Simpson Rule we can find the volume. -------------------------------------------------------------

6- Flow Reactors with Heat Exchange: The general energy balance for continuous flow reactors is given by

6.1- PFR with Heat Exchange: A non adiabatic PFR exchanges heat with the surroundings. Several different designs are used industrially. One possibility would be to impose a constant heat flux at the wall of the reactor. Such a reactor may be constructed in different ways but the simplest scenariois illustrated in Figure 2. This design is very similar to a double pipe heat exchanger, consisting of two concentric tubes with the central tube acting as the reactor and the surrounding jacket containing the heat transfer fluid.

Fig.2:Jacket PFR-double pipe heat exchanger.

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf When reactors are operated in the nonisothermal mode, the diameter of the reactor is usually relatively small so that an effective rate of heat transfer can be achieved. Use of a small reactor diameter can lead to a requirement of very long reactor length if a large volume isrequired. Rather than have a single very long tube, a common design uses multiple reactor tubes surrounded by a large shell containing the heat transfer fluid, resembling a shell and tube heat exchanger. This type of arrangement is shown in Figure 3.

Fig.3:Multi-tubular reactor used for non adiabatic operation.

Where (a) is heat exchange area per unit volume of the reactor,

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf 𝐴 𝜋𝐷𝐿 4 𝑎 = = 𝜋𝐷2 = 𝑉 𝐿 𝐷 4

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

……... (20)

……... (21)

- Solving CRE Problem for Reactors with Heat Exchange:

.. (22)

…(23)

If the coolant temperature varies down the reactor we must add coolant balance:

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Heat Exchanger Energy Balance- Variable Ta Co-current Coolant Balance:

….(24)

……………. (25)

….……………. (26)

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

The above Equations (Eq.20 to Eq. 24)are the general form of the energy balance for a PFR with external heat exchange. It may be written explicitly in terms of in terms of the reactor length to give the temperature and conversion profile along reactor length

𝑑𝑉 = 𝐴𝑡𝑢𝑏𝑒 𝑑𝑧 𝐹𝑜𝑟 𝑀𝑢𝑙𝑡𝑖𝑡𝑢𝑏𝑒 𝑅𝑒𝑎𝑐𝑡𝑜𝑟…………………….. (27) 𝑑𝑉 = 𝐴𝑅𝑒𝑎𝑐𝑡𝑜𝑟 𝑑𝑧 𝐹𝑜𝑟 𝐷𝑜𝑢𝑝𝑙𝑒 𝑃𝑖𝑝𝑒 𝑅𝑒𝑎𝑐𝑡𝑜𝑟………………... (28) 𝜋𝐷2 𝑑𝑉 = 𝑑𝑧 4

𝒅𝑻 𝑨 . ∆𝑯𝒓 . 𝒓𝑨 = 𝒅𝒛 𝑭𝑨𝒐 (∑ 𝜽𝒊 𝑪𝑷𝒊 )

𝑼𝒂 (𝑻 − 𝑻𝒂 ) − 𝑭𝑨𝒐 (∑ 𝜽𝒊 𝑪𝑷𝒊 )

𝑭𝑨𝒐 𝒅𝒙 𝑨 𝒅𝒛 𝒅𝑻 ∆𝑯𝒓 𝒅𝒙 𝑨 ∗ 𝑼𝒂 (𝑻 − 𝑻𝒂 ) = − 𝒅𝒛 (∑ 𝜽𝒊 𝑪𝑷𝒊 ) 𝒅𝒛 𝑭𝑨𝒐 (∑ 𝜽𝒊 𝑪𝑷𝒊 ) 𝒓𝑨 =

The above equation may written in the finite form as

∆𝑇 = 𝐶1 ∆𝑥 − 𝐶2 ∆𝑧 ( 𝑇 − 𝑇𝑎 ) Energy Balance

…………….. (29)

∆𝑥 = 𝐶3 𝑟̅𝐴 ∆𝑧

𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝐵𝑎𝑙𝑎𝑛𝑐𝑒

𝑟𝐴 = 𝑓 (𝑥, 𝑇)

𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑒

𝐶1 =

∆𝐻𝑟 ∑ 𝜃𝑖 𝐶𝑃𝑖

𝐶2 =

𝐴 ∗ 𝑈𝑎 𝐹𝐴𝑜 (∑ 𝜃𝑖 𝐶𝑃𝑖 )

....…..… (30)

…………….. (31)

𝐶3 =

𝐴 𝐹𝐴𝑜

The three types of reactor operations yield different temperature profiles within the reactor and are shown in Figure 4 for an exothermic reaction.

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

Figure 4: temperature profiles for PFR in three different mode of operation.

6.2- CSTR with Heat Exchange:

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Figure 5: CSTR with heat exchange.

̇ =0 and 𝑸̇ = 𝑼𝑨(𝑻 − 𝑻𝒂 ) If 𝑾𝒔

Solving for x:

.………… (32)

Solving for T:

.… (33)

Example 6: A second order, liquid phase reaction, A + B → C, is to be performed in a single-stage CSTR. Given the following data: 1. Determine the size of the reactor required to achieve 85% conversion of A. 28 Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf Answer: V = 7.56 m3 2. Estimate the heat-transfer area needed to maintain the reactor temperature at 27oC. Answer: A=3 m2 3. Calculate what the feed temperature must be if the reactor is to be operated adiabatically at 27oC. Answer: To = 8.8°C

Example 7: Consider the reaction A → B, rA = kCA, k300= 0.05 min-1 , ∆Hr = -20 kcal/mole in a 10 liter reactor with CA0 = 2 moles/liter and To =300 K. At what rate must heat be removed to maintain the reactor isothermal at 300 K for (a) a batch reactor at 90% conversion? This is an isothermal reactor with r = 0.05CA. If batch, then

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Note, however, that the significance of heat removal or addition (Q positive or negative) is quite different with the batch reactor, requiring a heat removal that varies in time, the CSTR requiring a constant heat removal rate, and the PFR requiring a heat removal rate that varies with position z in the reactor. These are sketched in Figure 6.

Figure 6:heat removal rate that varies with position z or t.

7- Multiple Steady States in a CSTR: One of the interesting operating features of a nonisothermal CSTR is the existence of multiple steady states under certain operating conditions. In a reactor with multiple steady states, there is more than one set of steady-state operating conditions possible. From a mathematical perspective, the existence of multiple steady states implies that there is more than one solution (temperature and conversion) that satisfies the mole and energy balance equations. In the following, we use some graphical analysis to illustrate how multiple steady states arise and the implications for reactor operation. A simple first order reaction with a constant-density fluid is chosen for simplicity, but the principles may be extended to more complex cases, with a concomitant increase in mathematical complexity. For illustration purposes, consider a first-order reaction occurring in a nonisothermal CSTR. The material balance equation is:

𝑉 𝑥 = 𝐹𝐴𝑜 𝑘 𝐶𝐴𝑜 ( 1 − 𝑥 ) This equation can be rearranged to give an expression for conversion of A:

𝑥𝑀𝐵

𝑘𝜏 = 1 + 𝑘𝜏

The steady-state energy balance written in terms of conversion and constantheat capacity is given by Equation.

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Note that 𝑄̇ = 𝑈𝐴(𝑇 − 𝑇𝑎 )is positive if heat is added to the reactor, that is, when (Ta>T). The above equation can be rearranged as follows:

.… (34)

……………………………………………… (35)

…… (39)

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Equation39 is a nonlinear equation. For a given set of operating conditions, the reactor temperature is the only unknown value in the equation. However, owing to the nature of the equation, it is possible that under some sets of operating conditions there will be more than one solution to Equation 39. The function G(T) represents the rate of heat generation in the reactor. The second function isR(T) represents the rate of heat removal by heat transfer with the surroundings and the rate of fluid removal from the reactor. Note that R(T) is a linear function of temperature in this case because CP and ∆Hr are assumed to be constant. The intercept of the straight line depends on, for example, the value of the reactor inlet temperature. A typical behavior of the function R(T) at different inlet temperatures is shown in Figure 7.

Fig.7: Variation of heat removal line with inlet temperature. An increase in the value of the reactor inlet temperature simply shifts the line to the right, while maintaining the value of the slope. The slope of the curve can be adjusted by changing the value of (CP0(1+ κ). Also the linear energy line depends on the value of κ as shown in Figure 8.

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Fig.8: Variation of heat removal line with κ (κ=UA/CP0FA0 ). A plot of the generation function, G(T), verses T gives a sigmoidal curve because of the exponential temperature dependence of the rate constant. The shape of the curve depends in large part on the value of the activation energy. A typical plot is shown in Figure 9.

. Fig.9: Shape of the heat generation function at different reactor temperatures. The shape of the curve depends also on the value of space time τ. A typical plot is shown in Figure 10. 33

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Fig.10: Variation of heat generation curve with space-time. All possible reactor operating temperatures must satisfy both functions, G(T) and R(T),such that the two functions have equal values. If both of the functions are plotted on the same graph, the points of intersection of the two lines represent permissible solutions to the reactor energy and mole balance equations. Figure 11 illustrates such a graph. In this figure, a single G(T) curve is shown, which would correspond to specific values of the kinetic parameters and enthalpy of reaction. Multiple R(T) curves are shown, each of which corresponds to a different reactor inlet temperature, with all other parameters the same.

Fig.11: Plot of a single heat generation curve and multiple heats removal curves

Consider first the R(T) line that corresponds to the lowest value of the inlet temperature. This line is at the extreme left of Figure 10. This line intersects the G(T) at a single 34

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf point, labeled 1 in Figure 11. This point of intersection corresponds to the only solution to the mole and energy balance equations; thus this point represents the reactor operating temperature. If the reactor inlet temperature is increased, with all other conditions held constant, the R(T) line moves to the right. The sigmoidal G(T) curve is not affected by the change in the inlet temperature because T0 does not appear in the expression for G(T). The second R(T) curve from the left intersects the G(T) curve at two points, marked with points 2 and 6,respectively. Both these points represent possible operating temperatures for the reactor, the low- and high-temperature stable operating conditions for this reactor. The next R(T) line in Figure 10 intersects the G(T) curve at three points, labeled 3, 5, and 7 respectively. Points 3 and 7 represent stable-steady states. Point 5 is an unstable steady state which cannot be achieved in practice. Continuing to increase the inlet temperature, the next R(T) curve illustrated has two stable steady states at 4 and 8, the final curve has a single steady state at point 9. The steady state that is realized in the reactor depends on either the initial conditions or the history of the reactor as shown in Figure 12.

Fig. 12: Plot of CSTR stability diagram.

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Example 8: Consider a first order adiabatic reaction in a CSTR with the following characteristics:

The rate constant k is expressed as k = exp (15.32 – 7550/T), sec–1. Determine the operating points for both the mass and heat balance equations. Solution: The first order reaction is represented by (–rA) = kCA, and applying the material balance and the energy balance Equations gives

𝑥𝑀𝐵

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𝑘𝜏 = 1 + 𝑘𝜏

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf

τ= V/Qo = 15.0/0.05 = 300.0 sec.

𝑥𝑀𝐵

300 𝑒𝑥𝑝(15.32 –

7550

) 300 𝑘 𝑇 = = 7550 1 + 300 𝑘 1 + 300 𝑒𝑥𝑝(15.32 – ) 𝑇

𝑥𝐸𝐵

𝑇 − 298 = 210

The figure shows that the steady state values are (XA, T) = (0.02, 300), (0.5, 362), and (0.95, 410). The middle point is unstable and the last point is the most desirable because of the high conversion

Example 9: Determine the operating conditions in a CSTR for a first order exothermic reaction with heat transfer under the following conditions: Feed temperature, TO, = 350 K Initial concentration, CAO= 1.0 mol/L Temperature of the cooling medium, TC = K 350 K Mean residence time, τ =1.0 min 37

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf kJ

(–ΔHr)= 200kJ/mol.𝐶𝑝𝑜 = 1.0 : UA/FA.Cp= 1.0 mol.K Reaction rate constant, k= exp (25 – 10000/T) min–1 Solution:

𝑮(𝑻) =

𝑥𝑀𝐵 = 𝑥𝑀𝐵

−𝟐𝟎𝟎 ∗ 𝟏 ∗ 𝟏 ∗ 𝐞𝐱𝐩 (𝟐𝟓 – 𝟏 + 𝟏 ∗ 𝟏 ∗ 𝐞𝐱𝐩 (𝟐𝟓 –

𝟏𝟎𝟎𝟎𝟎

𝐓 𝟏𝟎𝟎𝟎𝟎 𝐓

)

)

𝐤𝐉/𝐦𝐨𝐥

𝑘𝜏 1 + 𝑘𝜏 1 ∗ exp (25 –

10000

= 1 + 1 ∗ exp (25 –

)

T 10000 T

)

𝑹(𝑻) = 𝑪𝒑𝒐 (𝟏 + 𝜿)(𝑻 − 𝑻𝟎 )𝜿 =

𝑼𝑨 𝑪𝑷𝒐 𝑭𝑨𝒐

𝑹(𝑻) = 𝟏(𝟏 + 𝟏)(𝑻 − 𝟑𝟓𝟎) = 𝟐(𝑻 − 𝟑𝟓𝟎)𝐤𝐉/𝐦𝐨𝐥 Both G(T) and R(T) at varying effluent temperature. Figure below shows the profiles of the heat generation and heat removal terms, and the intersections give the operating conditions. The values of the fractional conversion XA corresponding to the temperatures at the sepoints are given in the table below.

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Example 10: The first order irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains A and an inert liquid in equimolar amounts, where FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K? UA= 8000 cal/min·K: Ta= 300K ∆HR=-7500 cal/mol:CpA = CpB =20 cal/mol·K Cpi =30 cal/mol·Kτ=100 min : E=40,000 cal/mol: k=6.6 x 10-3 min-1 at 350K Solution:

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Plug rate law into G(T) and simplify

Steady state temp G (T) =R (T) for T0 = 450K

Use design equation to get XA as an explicit equation:

The steady state temperature for T0 = 450K Then x=

41

is 400 K

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf 8-Adiabatic Equilibrium Conversion: In an exothermic reaction, the value of ∆Hr is negative. Therefore, an increase in the temperature decreases the value of K, with a concomitant decrease in the equilibrium yield. A conflict arises between the need for a high temperature to maximize the reaction rate, and a low temperature to maximize the conversion. For endothermic reaction, equilibrium conversion increases with increasing temperature. A typical equilibrium curve and temperature conversion trajectory for the reactor sequence is shown in Figure 13. For exothermic Reactions: As T increases, the equilibrium shifts to the left. i.e. both K and Xe decreases For endothermic Reactions: As T increases, the equilibrium shifts to the right. i.e. both K and Xe increases

Fig.13: Equilibrium conversion and equilibrium constant for exothermic and endothermic reactions.

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Figure 14 shows a typical equilibrium composition line that might be obtained for an exothermic reaction. The line has an equilibrium conversion near one at low temperature but the conversion decreases as the temperature increases. Now consider the performance of two types of reactor, the adiabatic reactor and the isothermal reactor. For a given set of operating conditions, the adiabatic reaction line governs the relationship between the temperature and the fractional conversion. Assuming constant ∆Hr and CP, the energy balance equation is:

Fig.14:The equilibrium line shows the relationship between the temperature and conversion at equilibrium for a specified starting composition. The adiabatic reaction line shows the maximum conversion that can be obtained for adiabatic operation for a given inlet temperature. The isothermal reaction line shows the best conversion for isothermal operation.

A typical adiabatic reaction line is shown in Figure 15 for an arbitrary value of T0. In the reactor, both the temperature and conversion will increase along this line until the equilibrium line is encountered at X2 and T2. At this point, the rate is zero, and the temperature and composition of the process stream remain constant for any additional reactor volume.

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Fig.15: The equilibrium conversion for adiabatic operation. If the same reactor was operated isothermally, the vertical operating line shown in Figure 14 would be obtained. As the temperature is constant, the fractional conversion finally attained, equal to X1, and is higher than that achieved in the adiabatic reactor. Isothermal operation may not be practical, and does not result in a minimum reactor volume. It is possible to determine an optimal temperature progression along the reactor so as to maximize the conversion for the minimum possible reactor volume. Consider the diagram shown in Figure 15. This diagram shows lines of constant reaction rate correspondingto set values of temperature and conversion. Each solid line represents a locus of the set of temperature and conversion that give a specified value of the rate. The top line corresponds to a rate of zero; in other words, the equilibrium line. We have alreadyseen that the equilibrium conversion drops as the temperature rises. As we move from the top to the bottom of the figure, the value of the reaction rate increases. Note, however, that as the value of the reaction rate increases, the maximum conversion achievable at that rate decreases. This observation suggests a route for minimizing the reactor volume. One could start the reaction at a high temperature to take advantage of the high rate at that temperature, and then progressively lower the temperature to increase the equilibrium yield. The dashed line in Figure 16 illustrates the type of temperature and conversion pathway that must be followed to achieve this optimal design. This reactor design would require a complex cooling pattern along the reactor length, which might not be practical.

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Fig 16: Each solid line in this diagram represents a constant value of reaction rate. Higher conversions than those shown in Figure 16 can be achieved for adiabatic operations by connecting reactors in series with interstate cooling. Inter-stage cooling system consists of a series of reactors or catalyst chambers. In each of the reactor stages, the temperature rises from inlet to outlet. Heat-exchangers (inter-stage coolers) are placed between the reactor stages as shown in Figure 17, where the temperature is reduced without affecting the composition of the process stream. The operating line of a single-reactor does not naturally follow a trajectory that shadows the locus of maximum reaction rates. By subdividing the reactor into a number of stages and removing heat from the reaction mixture after each reactor stage, the temperature is forced lower before the stream enters the next vessel. This allows working closer to the locus of maximum rates (see Figure 16). Clearly the larger the number of steps, the closer the trajectory of the system would approach the theoretical optimum. An inordinately large increase in the number of inter-stage coolers, however, would tend to increase construction costs. We would need to determine the optimum number of reactor stages.

Fig.17: Schematic diagrams of two reactors in series.

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf 8-Multibed Adiabatic Reactors for Equilibrium Limited Reactions: For relatively fast reactions that are equilibrium limited, the adiabatic reactor is commonly used in industry. As seen above, the adiabatic reactor experiences a temperature rise that limits the conversion. Therefore, in applications where very high conversions are desired, it is common to use multiple reactors in series, with cooling between each reactor. Consider, for example, an exothermic reaction that occurs in a series of adiabatic fixed bed catalytic reactors. The operating path would be similar to the one shown in Figure 18.This plot shows the equilibrium line as a function of temperature and conversion. For adiabatic operation, the temperature and conversion in the first reactor follow the adiabatic reaction line denoted Bed 1. When the fraction conversion reaches a value of X1, which is approaching the equilibrium line, the process stream is cooled by passing it through a heat exchanger. As this is a catalytic reaction, no further conversion occurs in this stage, as shown by the flat line. When the process stream is cooled by a predetermined amount, the stream enters reactor number two, in which it undergoes a further adiabatic reaction until the conversion is X2. The stream is cooled as before, and then fed into the last bed, in which the conversion is brought to X3. Using this strategy, a high level of conversion can be achieved using adiabatic reactors. A diagram of the reactor that corresponds to Figure 18and Figure 19. The exact number of beds and the degree of cooling between the beds depends on the reaction involved and operating and capital costs.

Fig. 18: Multiple adiabatic reactors in series.

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Fig.19: Multiple bed adiabatic catalytic reactor with interstage cooling. The most classic application of the multibed adiabatic fixed bed reactor is the oxidation 1 of sulfur dioxide 𝑆𝑂2 + 𝑂2 ⇋ 𝑆03 , equilibrium limited reversible reaction. This 2 reaction is used to make sulfur trioxide, which, when added to water, makes sulfuric acid. Very high yields are desired (in excess of 99%), but the reaction is highly exothermic and equilibrium limited. As a result the multibed approach has been adopted, with typical industrial units having three or four beds. Example 11: For the elementary solid-catalyzed liquid-phase reaction: A ⇋ B .Make a plot of equilibrium conversion as a function of temperature. Determine the adiabatic equilibrium temperature and conversion when pure A is fed to the reactor at a temperature of 300 K. Additional information:CPA = CPB = 50 cal/mol.K :∆Hr = -20000cal/molA Keq (300K) =100000 Solution:

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What conversion could be achieved in Example 5.10 if two interstage coolers were available that had the capacity to cool the exit stream to 350 K? Also determine the heat duty of each exchanger for a molar feed rate of A of 40 moles. Assume that 95% of equilibrium conversion is achieved in each reactor. The feed temperature to the first reactor is 300 K. For an entering temperature of 300 K the adiabatic equilibrium conversion was 0.40. For 95% of equilibrium conversion, the conversion exiting the first reactor is 0.38. The first reactor: The exit temperature from the first reactor is found from a rearrangement of Equation (e):

𝑋𝐸𝐵 = 0.0025(𝑇 − 300) 𝑇 = 400 𝑋 + 300 = 400 ∗ 0.38 + 300 = 452 𝐾 50

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf We now cool the gas stream exiting the reactor at 460 K down to 350 K in a heat exchanger. The gas stream is then sent to the second reactor.

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Example 12: The reversible conversion of A+B ⇋ C+D is carried out in liquid phase in a series of staged PFR's with interstage cooling. The lowest temperature to which the stream may be cooled is 300K. The feed is equal molar in A and B and contains no C or D. The PFR's have a sufficient residence time to achieve 99.9% of the equilibrium conversion in each stage. The feed enters at 300K and the reaction is carried out adiabatically. If two reactors and one interstage cooler are used, what is the maximum conversion (±O.05) that may be achieved after the second PFR? Data: ∆Hr = -40000 cal/molA: CPA = CPB = CPC=CPD=53.3cal/mol.K Keq (300K) = 3.22x108 FAo = 5 mol A /min

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The first stage:

∑ 𝛳𝑖 𝐶𝑃𝑖 =1*53.3+1*53.3=106.6 cal/mol.K 𝑋𝐸𝐵 =

106.6(𝑇 − 300) 40000

𝑋𝐸𝐵 = 0.00266(𝑇 − 300) 𝑋𝐸𝐵 = 0.00266𝑇 − 0.798 This equation cuts the equilibrium curve at x=0.36 and T= 436 K

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf The second stage: 𝑋𝐸𝐵 = 0.36 +

106.6(𝑇 − 300) 40000

This equation cuts the equilibrium curve at x=0.66 and T= 415 K So the maximum conversion for this system is 0.66 The heat load for the interstage cooling is: 𝑄1 = 𝐹𝐴𝑂 (1 − 𝑋1 )𝐶𝑃𝐴 (300 − 436) 𝑄1 = 5 mol A /min(1 − 0.36)53.3cal/mol. K(300 − 436)𝐾 𝑄1 = − 23.196 Kcal/min

9-Optimum Feed Temperature: We now consider an adiabatic reactor of fixed size or catalyst weight and investigate what happens as the feed temperature is varied. The reaction is reversible and exothermic. At one extreme, using a very high feed temperature, the specific reaction rate will be large and the reaction will proceed rapidly, but the equilibrium conversion will be close to zero. Consequently, very little product will be formed. At the other extreme of low feed temperatures, little product will be formed because the reaction rate is so low. A plot of the equilibrium conversion and the conversion calculated from the adiabatic energy balance is shown in Figure 20.

Figure 20: Equilibrium conversion for different feed temperatures.

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf We see that for an entering temperature of 600 K the adiabatic equilibrium conversion is 0.15, while for an entering temperature of350K it is 0.75. The corresponding conversion profiles down the length of the reactor for these temperatures are shown in Figure 21.

Figure 21: Adiabatic conversion profiles for different feed temperatures. We notice that the conversion and temperature increase very rapidly over short distance (i.e., a small amount of catalyst). This sharp increase is sometimes referred to as the "point" or temperature at which the reaction "ignites." If the inlet temperature were lowered to 500 K, the corresponding equilibrium conversion would increase to 0.38; however, the reaction rate is slower at this lower temperature so that this conversion is not achieved until closer to the end of the reactor. If the entering temperature were lowered further to 350 K, the corresponding equilibrium conversion would be 0.75, but the rate is so slow that a conversion of 0.05 is achieved for the specified catalyst weight in the reactor. At a very low feed temperature, the specific reaction rate will be so small that virtually all of the reactant will pass through the reactor without reacting. It is apparent that with conversions close to zero for both high and low feed temperatures, there must be an optimum feed temperature that maximizes conversion. As the feed temperature is increased from a very low value, the specific reaction rate will increase, as will the conversion. The conversion will continue to increase with increasing feed temperature until the equilibrium conversion is approached in the reaction. Further increases in feed temperature for this exothermic reaction will only decrease the conversion due to the decreasing equilibrium conversion. This optimum inlet temperature is shown in Figure 22. For reversible, exothermic reactions optimize feed temperature to maximize XA. High T0: moves XA,EB line to the right. Reaction reaches equilibrium fast, but low XA . Low T0 would give high XA,e but the specific reaction rate k is so small that most of the reactant passes through the reactor without reacting (never reach XA,e)

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Figure 22: Finding the optimum feed temperature.

10-Optimal Temperatures for Isothermal Reactors: Reaction rates almost always increase with temperature. Thus, the best temperature for a single, irreversible reaction, whether elementary or complex, is the highest possible temperature. Practical reactor designs must consider limitations of materials of construction and economic tradeoffs between heating costs and yield, but there is no optimal temperature from a strictly kinetic viewpoint. Ofcourse, at sufficiently high temperatures, a competitive reaction or reversibility will emerge. Multiple reactions, and reversible reactions, since these are special form of multiple reactions, usually exhibit an optimal temperature with respect to the yield of a desired product. Example12:

Additional data 56

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(–ΔHr) = 70000 J/mol; CP(stream) = 4.2 J/g.K : ρ(stream) = 1,000 gm/L; Solution The optimum temperature Topt is determined by Equation:

(1) The optimum temperature Topt is:

The minimum volume VR is determined by:

𝑉=

57

𝑋. 𝐹𝐴𝑜 (𝑟𝐴 ) 𝑜𝑝𝑡

Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf

FAO = QoCAOTherefore, the volumetric flowrateQo is

(2) The minimum volume is determined by

𝑋. 𝐹𝐴𝑜 𝑉= (𝑟𝐴 ) 𝑜𝑝𝑡

= 288.0 liter 3) The feed temperature TO for the adiabatic operation at theoptimal temperature Topt is determined from

𝜃𝑖 𝐶𝑝 𝑖 (𝑇𝑜 − 𝑇𝑜𝑝𝑡 ) = − 𝑥. ∆𝐻𝑟 𝑥. ∆𝐻𝑟 𝜃𝑖 𝐶𝑝 𝑖 𝑇𝑜 = 278.3 𝐾 = 5.1 𝐶 𝑂 𝑇𝑜 = 𝑇𝑜𝑝𝑡 −

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Algorithm for isothermal reactors.(Source: Fogler, H. S. Elements of Chemical Reaction Engineering

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Algorithm for nonisothermal CSTR design

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf Homework Problems: 1 - An elementary second-order liquid-phase reaction occurs in a CSTR. The reactor has a volume of 1200 L and is heated by a steam jacket.

The feed temperature is 27°C, and the flow rate is 30 L/min. The inlet concentrations of A and B are 2 mol/L. The desired conversion of A is 60%. a. Determine the reactor temperature required to achieve the desired conversion. b. Determine the temperature of the steam in the jacket required to operate the reactor at the temperature determined in part (a).

2- A tubular flow reactor is to be designed to produce butadiene from butene by dehydrogenation (gas-phase reaction). The first-order reaction rate expression is written in terms of the partial pressure of butene as

The constant k has units of mol (hrL.atm) −1. The reaction is endothermic, the reactor is often operated adiabatically; therefore, steam is often added to the feed to provide thermal energy for the reaction. For this reactor, the feed is a mixture of 10 mol of steam for each mole of butene. The reactor pressure is 2 atm and the feed temperature is 650°C. Values of the rate constant at different temperatures are given in the following table:

Assume that the heat of reaction is a constant 1.1 × 105 J/mol butene. The heat capacity of the feed stream may be considered constant at 2.1 kJ/kg ⋅K. a. Calculate the reactor volume for a 20% conversion of butene if the reactor is operated isothermally at 650°C with a total inlet molar flow rate of11000 mol/h. b. Determine the reactor volume for 20% conversion of butene for a total inlet molar feed rate of 11000 mol/hr in an adiabatic reactor. 61

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Reactor Design II-Fourth Year Dr.Ali N.Khalaf 3- The following reaction is carried out in an adiabatic plug flow reactor: Calculate the maximum conversion that may be achieved if the feed enters at27°C. Only A and B are in the feed. The following data are available:

4- A first-order endothermic reaction is carried out in a constant-volume batch reactor.

The enthalpy of reaction is 50,000 J/mol. The reactor volume is 1 m3 and contains1000 kg of mixture. There are initially 10,000 mol of reactant A in the reactor. The reactor is heated to 400°C, during which time 10% of the initial A reacts. When the temperature reaches 400°C, the reactor operates adiabatically. The heat capacity of the mixture is a constant 2000 J/kg. K. a. After the heating stops, how much time is required to achieve a final conversion of 70% of the A originally present before the heating started? b. What is the final temperature in the reactor? 5- A first-order liquid-phase exothermic reaction A → B is carried out in a batch reactor. There is no heat transfer through the reactor wall. The reaction mixture is to be held at a constant 40°C by the addition of an inert coolant at a temperature of 25°C. The flow rate of coolant varies with time so as to maintain the reactor temperature at 40°C. The following data are available:

At time t = 0, only compound A is present in the tank at a concentration of8000 mol/m 3. The initial volume is 1.5 m3. a. Calculate the conversion of A 2 h after the start of the reaction. 62

Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf b. Calculate the rate of coolant addition 2 h after the start of the reaction. c. Calculate the total amount of coolant added after 2 h.

6- Consider an adiabatic constant pressure batch reactor in which a second-order gasphase reaction occurs A + B → C (−rA ) = kCACB The rate constant is independent of temperature and equals 0.01m3/mol ⋅ s.The initial reactor temperature is 450 K and the pressure is 100 kPa. The initial volume of the reactor is 1.0 m3. The reactor initially contains 30% by volume A,30% B, and 40% inerts.The heat capacity of the mixture is a constant CV = 1000J/kg K. The average molecular mass of the initial material is 30 g/mol. The internal energy change on reaction is a constant ∆Hr= –60 kJ/mol. Calculate a. The time required to reach 80% conversion b. The reactor volume at 80% conversion c. The reactor temperature at 80% conversion 7- Consider an adiabatic plug flow reactor in which a second-order ideal gas-phase reaction occurs A + B → C. The reactor feed is composed of 40 mol% A, 50 mol% B, and 10 mol% inert material. The total inlet molar flow rate is 10 mol/s. The reactor inlet temperature is 300 K and the pressure is 101.325 kPa. The reaction rate is

The concentrations are in mol/m3. The enthalpy of reaction at 298 K is equal to −50000 J/mol. The heat capacities are A: 10 J/mol . K, B: 5 J/mol .K, C: 15 J/mol. K, and inert: 15 J/mol. K The desired fractional conversion of A is 80%. Calculate the reactor volume required. 8- The elementary, reversible gas phase reaction A⇋ B is to be carried out in a CSTR with heat exchange. Pure A is fed to the reactor. The heat exchange coil in the reactor is maintained at 400K. The rate coefficient is known at 400K, but heat of reaction is unknown.

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Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf A) Calculate the conversion (X) if the steady state CSTR is operated at 450K. B) Calculate the heat removal rate R (T) at this temperature. C) Calculate the heat of reaction (∆Hr). 9- A first-order, reversible reaction A⇋ B is taking place adiabatically. The equilibrium constant at 400 K is 3 and the heat of reaction, which can be assumed constant, is –50 kJ/mole. Pure A with a specific heat of 100 J/mole-K (which is the same as that of B) enters the reactor at 300 K. The equilibrium conversion as a function of temperature is plotted in the figure shown below. 1.2

Equilibrium Conversion

1

0.8

0.6

0.4

0.2

0 300

350

400

450

500

550

600

650

Temperature

a. Show the equations that were used for constructing the plot shown. b. What is the maximum conversion that can be achieved, if the reactor is operated adiabatically? c. If two adiabatic reactors are used, with inter-stage cooling, and if the first reactor exit stream is cooled to 350 K, what maximum conversion can be achieved in the second reactor? What would be the exit temperature from this second stage? 10 -A first-order irreversible liquid-phase reaction ( A  B ) is taking place in a CSTR .A enters the reactor at 300 K at a concentration of 3 mol/liter. The reactor volume is 18 liters and flow rate of pure A is 0.06 liters/s. Other parameters are: 0 H Rx  5000 cal/mole, C pA  C pB  30 cal/mole  K

k  exp(15.32 

64

7550 -1 )s T

Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf 0 The heat generated term given by (H Rx  X ) is plotted in the figure shown

6000

Heat Generation Function

5000

4000

3000

2000

1000

0 250

300

350

400

450

Temperature (K)

1- For the conditions given, determine the number of steady states, and the Conversions and temperatures at these steady states. 2- What is the minimum incoming temperature at which multiple steady states are possible? 3- What is the maximum incoming temperature above which only one steady state occurs? 11-Ascertain whether the following exothermic reaction: 2A ⇋2P (k1 and k1 at 80°C) could be carried out in the reactor shown below:

Calculate the volume and heat removed from the CSTR and the PFR. Do the magnitudes of the heat being removed appear feasible? Why or why not? Data:

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Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf

12-The ester of an organic base is hydrolyzed in a CSTR. The rate of this irreversible reaction is first-order in each reactant. The liquid volume in the vessel is 6500 L. A jacket with coolant at 18°C maintains the reactant mixture at30°C. Additional data:

The average heat capacity is approximately constant at 1.0 kcal L-1 °c- 1 . (a) What is the conversion of ester in the reactor? (b) Calculate the rate at which energy must be removed to the jacket to maintain 30°C in the reactor. If the heat transfer coefficient is 15 kcal/sec. m2 K, what is the necessary heat transfer area? (c) If the coolant supply fails, what would be the maximum temperature the reactor could reach? 13-A reaction is carried out in an adiabatic CSTR with a volume of 10000 L.The feed solution with reactant A, at a concentration of 5 M, is supplied at 10L/ s. The reaction is first-order with rate constant:

(a) Calculate the reactor temperature and exit concentration for feed temperatures of 280, 300, and 320 K. (b) To maintain the reactor temperature below 373 K, a well-mixed cooling jacket at 290 K is used. Show that it is possible to get 90 percent conversion in this reactor with a feed temperature of 320 K. 14- The reversible, first-order reaction shown below takes place in a CSTR.

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Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf

(a) For a reactor space time of 10 min, what is the conversion for a 300 K operating temperature? What is the conversion at 500 K? (Remember: the equilibrium constant depends on temperature.) (b) If the feed temperature is 330 K and the feed concentration is 5 M, what is the necessary heat-removal rate per liter of reactor volume to maintain 300 K operating temperature? 15- The reaction A + B ⇋ C + D is carried out adiabatically in a series of tubular reactors with interstage cooling as shown in the figure below. The feed is equimolar in A and B and enters each reactor at 27oC. The heat removed between the reactors is -87.5kcal/min. (a) What is the outlet temperature of the first reactor? (b) What is the conversion of A at the outlet of the first reactor? (c) Is the first reactor close to equilibrium at the exit?

State any assumptions that you make while solving the problem. Data:

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Chemical Engineering Department- University of Basrah

Reactor Design II-Fourth Year Dr.Ali N.Khalaf 16-A liquid phase exothermic reaction A →B is carried out at 358K in a 0.2 m3 CSTR. The coolant temperature is 273K and the heat transfer coefficient (U) is 7200 J/min·m2·K. What is the heat exchange area required for steady state operation? CPA =CPS=20 J/g•K CA0= 180 g/dm3 υ0= 500 dm3/min T0= 313 K ρ= 900 g/dm3 ∆H°RX(T) = -2500 J/g E=94852 J/mol·Kk(313K)= 1.1 min-1 Answer: A=227.4 m2

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Chemical Engineering Department- University of Basrah