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Example 8-4 Production of Propylene Glycol in an Adiabatic CSTR Propylene glycol is produced by the hydrolysis of propylene oxide: C3 H 6O + H 2 O → C3 H 8O 2 A + B → C Over 800 million pounds of propylene glycol was produced in 1997 and the selling price was approximately $0.67 per pound. Propylene glycol makes up about 25% of the major derivatives of propylene oxide. The reaction takes place readily at room temperature when catalyzed by sulfuric acid. You are the engineer in charge of an adiabatic CSTR producing propylene glycol by this method. Unfortunately, the reactor is beginning to leak, and you must replace it. (You told your boss several times that sulfuric acid was corrosive and that mild steel was a poor material for construction.) There is a nice overflow CSTR of 300-gal capacity sitting idle; it is glass-lined and you would like to use it. You are feeding 2500 lb/hr (43.04 lb mol/hr) of propylene oxide (PO) to the reactor. The feed stream consists of (1) an equivolumetric mixture of propylene oxide (46.62 ft 3/hr) and methanol (46.62 ft3/hr), and (2) water containing 0.1 wt% H2SO4. The volumetric flow rate of water is 233.1 ft 3/hr, which is 2.5 times the methanol-PO flow rate. The corresponding molar flow rates of methanol and water are 71.87 and 802.8 lb mol/hr, respectively. The waterpropylene oxide-methanol mixture undergoes a slight decrease in volume upon mixing (approximately 3%), but you neglect this in your calculations. The temperature of both feedstreams is 58°F prior to mixing, but there is an immediate 17°F temperature rise upon mixing of the two feedstreams caused by the heat of mixing. The entering temperature of all feedstreams is thus taken to be 75°F (see Figure E8-4.1). Furusawa et al.1 state that under conditions similar to those at which you are operating, the reaction is first order in propylene oxide concentration and apparent zero order in water due to the excess of water with the specific reaction rate

(

)

k = Ae − E / RT = 16.96 ×1012 e −32400/ RT h −1 The units of E are BTU/lb mol. There is an important constraint on your operation. Propylene oxide is a rather low boiling substance (b.p. at 1 atm, 93.7°F). With the mixture you are using, you feel that you cannot exceed an operating temperature of 125°F, or you will lose too much by vaporization through the vent system. Can you use the idle CSTR as a replacement for the leaking one if it will be operated adiabatically? If so, what will be the conversion of oxide to the glycol?

1

T. Furusawa, H. Nishimura, and T. Miyauchi, J. Chem. Eng. Jpn., 2, 95 (1969).

SOLUTION (All data used in this problem were obtained from the Handbook of Chemistry and Physics unless otherwise noted.) Let the reaction be represented by A+ B→C

in which A is propylene oxide (MW=58.08; ?=0.859 g/cm3; Cp=35 BTU/lb mol- °F2) B is water (MW=18.02; ?=0.9941 g/cm3; Cp=18 BTU/lb mol-°F) C is propylene glycol (MW=76.11; ?=1.036 g/cm3; Cp=46 BTU/lb mol-°F2) M is methanol (MW=32.04; ?=0.7914 g/cm3; Cp=19.5 BTU/lb mol-°F) In this problem neither the exit conversion nor the temperature of the adiabatic reactor is given. By application of the material and energy balances we can solve two equations with two unknowns (FA and T). Solving these coupled equations, we determine the exit flow rates (conversion) and temperature for the glass-lined reactor to see if it can be used to replace the present reactor. 1. Mole balance equation: FAo − FA + rA V = 0 2. Rate law:

rA = − k C A •

•

•

3. Stoichiometry (liquid phase, v in = v out = v ):

FB = FBo − ( FAo − FA ) FC = FCo + (FAo − FA ) Ci =

Fi •

v 4. Combining the results and solving for FA recalling that τ =

V •

gives

v

2

Estimated from the observation that the great majority of low-molecular-weight oxygenE containing organic liquids have a mass heat capacity of 0.6 cal/g@ C"15%.

FAo − FA − k C A V = 0 F FAo − FA − k •A V = 0 v FAo = (1 + k τ ) FAo FAo FAo FA = = (1 + k τ ) 1 + τ A e − E / RT

(

)

This equation relates the temperature and molar flow rate of A. 5. The energy balance for this adiabatic reactor in which there is negligible energy input provided by the stirrer is •

Q = ∑ Fi inlet

0 = ∑ Fi

inlet

∫C

Toutlet

∫C

Tinlet

Toutlet

∫C

Tinlet

pi

dT + ∆ H r (Tout )rV

Tinlet

inlet

∑ Fi

Toutlet

pi

pi

r dT + ∆ H r (Tout ) A V νA

1 dT = − ∆ H r (Tout ) (FA − FAo ) νA

Since the heat capacities are all constants the integral shown on the left hand side of this equation can be computed to give

∑ Fi inlet

Toutlet

∫C

pi

dT = FAo C p A ∆T + FBo C pB ∆T + FCo C pC ∆T + FMo C pM ∆T

Tinlet

This equation can be substituted into the energy balance and, along with the fact that, ν A = − 1 the energy balance can be solved to also give FA as a function of T FA =

(F

Ao

)

C pA + FBo C pB + FCo C pC + FMo C pM ∆T + ∆ H r (Tout ) FAo ∆ H r (Tout )

We also need to have an expression for the heat of reaction at Tout . ∆ H r (Tout ) = ∆ H r (Tref ) + ∑ν i i

= − 36,400 +

To u t

∫C

pi

dT

Tref

To u t

∫ [(− 1)35 + (− 1)18 + (+ 1)46] dT

Tref

(

= − 36,400 − 7 Tout − 528o R

)

6. Solution: We have two simultaneous equations in FA and T which must be solved simultaneously. The following MathCAD program does this both graphically and numerically. Example Problem 8-4, Fogler First define all of the constant values FAo := 43.03⋅ FBo := 802.8⋅ FCo := 0.0⋅

lb ⋅ mole

CpA := 35⋅

hr lb ⋅ mole

CpB := 18⋅

hr

lb ⋅ mole

FM := 71.87⋅

CpC := 46⋅

hr lb ⋅ mole

V := 300⋅ gal 3

ft

vdot = 40.687

gal

τ :=

hr

+ 46.62⋅

lb ⋅ mole ⋅ R BTU lb ⋅ mole ⋅ R BTU lb ⋅ mole ⋅ R

CpM := 19.5⋅

hr

vdot := 46.62⋅

BTU

ft

3

hr

+ 233.1⋅

ft

3

hr

min

V vdot

τ = 0.123 hr Sum := FAo⋅ CpA + FBo⋅ CpB + FCo⋅ CpC + FM⋅ CpM

BTU lb ⋅ mole ⋅ R

Since there are two unknowns (FA and T) express all other variable in terms of these two. FB( FA) := FBo − ( FAo − FA) FC( FA) := FCo + ( FAo − FA) DelHr( T) := −36400⋅

BTU lb ⋅ mole

− 7⋅

BTU lb ⋅ mole ⋅ R

⋅ ( T − 528⋅ R)

 −32400⋅ BTU   lb ⋅ mole  12 −1 k( T) := 16.96⋅ 10 ⋅ hr ⋅ exp     1.987⋅ BTU ⋅ T   lb ⋅ mole ⋅ R    To invoke the solver routine in MathCAD we'll first need to provide initial guesses for both FA and T. If we assume about 50% conversion then FA will be about 20. The initial temperature is 528R so let's guess 600R since the reaction is exothermic.

x := 20⋅

lb ⋅ mole hr

y := 600⋅ R To start the solver we first type the key word "Given" then write the expressions we want solved making sure to use the Boolean equal sign Given FAo

x

1 + k( y ) ⋅ τ

x−

Sum⋅ ( y − 535⋅ R) + DelHr( y ) ⋅ FAo DelHr( y )

 FA    := find( x, y ) T FA = 6.146

lb ⋅ mole hr

T = 613.621 R X :=

FAo − FA FAo

X = 0.857

0⋅

lb ⋅ mole hr

We can also put together a plot like Fogler does in his example. To do this let's first define two functions. One will be FA as determined by the material balance and the other will be FA as determined by the energy balance. FAmb( T) := FAeb( T) :=

FAo 1 + k( T) ⋅ τ Sum⋅ ( T − 535⋅ R) + DelHr( T) ⋅ FAo DelHr( T)

Now define a range of values for T Trange := 535 .. 635 And now plot the results

Adiabatic CSTR Example

Molar Flow Rate of A (kgmole/sec)

0.006

0.004

0.002

0 520

540

560

580 Temperature (R)

600

620

640

FAmb FAeb The common point for these two functions, i.e., the solution, is at about 613 R and FA=8x10-4 kgmole/sec = 6.3 lbmole/hr

Example 8-5 CSTR with a Cooling Coil A cooling coil has been located for use in the hydration of propylene oxide discussed in Example 8-4. The cooling coil has 40 ft2 of cooling surface and the cooling water flow rate inside the coil is sufficiently large that a constant coolant temperature of 85°F is maintained. A typical overall heat transfer coefficient for such a coil is 100 BTU/h-ft2-R. Will the reactor satisfy the previous constraint of 125°F maximum temperature if the cooling coil is used? SOLUTION Nothing changes from the previous solution except for the energy balance which now becomes •

Q = UA(Thx − T ) = ∑ Fi inlet

= ∑ Fi inlet

Toutlet

∫C

pi

dT + ∆ H r (Tout )rV

Tinlet

Toutlet

∫C

Tinlet

pi

1 dT + ∆ H r (Tout ) (FA − FAo ) νA

Solving this equation for FA gives the following result FA = FAo +

(F

Ao

)

C p A + FBo C pB + FCo C pC + FMo C pM ∆T + UA(Thx − T ) ∆ H r (Tout )

Again we have two simultaneous equations in FA and T which can be solved using the following MathCAD program. As in the example in the text we obtain an outlet molar flow rate for the propylene oxide of 27.4 lb mol/hr (36.3% conversion) and an exit stream temperature of 563.7°R (103.7°F). This portion of the MathCAD program is a continuation of the adiabatic case. Thus all of the constants and functions of Fa and T have already been defined. We need to add only the values for the heat transfer coefficient, the cooling coil area and the coolant temperature and then solve the resulting simultaneous equations U := 100⋅

BTU 2

hr⋅ ft ⋅ R

A := 40⋅ ft

2

Thx := 545⋅ R

Now invoke the solver as before lb ⋅ mole x := 20⋅ hr y := 600⋅ R Given FAo

x

1 + k( y ) ⋅ τ U⋅ A ⋅ (Thx − y ) − Sum⋅ ( y − 535⋅ R )

x − FAo +

DelHr( y )

0⋅

lb⋅ mole hr

 FA    := find( x, y ) T FA = 27.422

lb ⋅ mole hr

T = 563.655R We may also construct the plot of each balance independently as we did in the adiabatic case FAmb( T) :=

FAo 1 + k( T) ⋅ τ

FAeb(T) := FAo +

Sum⋅ ( T − 535⋅ R ) − U⋅ A ⋅ ( Thx − T) DelHr( T)

CSTR with Heat Exchange Example 0.0063

Molar Flow Rate of A (kgmole/sec)

0.005

0.0038

0.0025

0.0013

0 520

540

FAmb FAeb

560

580 Temperature (R)

600

620

640