ENGINEERING, DESIGN, AND EQUIPMENT a transform of a solid temperature or a dimensionless space variable a transform of a fluid temperature reactor axial variable dimensionless time function defined by Equation 139 S’c,psR2 dimensionless constant for heat generation or lk, particle porosity (adsorption problem) bed porosity root of transcendental equation Bessel transform of CY Gscs k/sPz G c , Or k.
-
vc,p, U24? CY
b R2
ka
k, b
root of transcendental equation corrected dimensionless time variable
- a’)Pl k,
3(1
R k/r reactor radius variable fluid density
p, u =
‘
particle density CY‘ ” ” p 2 = modified Peclet number = . _1 -k/r
=
a’ kl 1
02
=
I
= dimensionless time variable = time variable =1/A--p = root of Equation 136
w Wi
cm
=
a c,p= ~
e + I.bB(dX)
Literature cited (1) Ayelius, A., Z. angew. math. u.Mech. 6,291 (1926). (2) Brmkley, S.R., Jr., J. Appl. Phys. 18,582 (1946). (3) Brinkley, S. R., Jr., U. S. Bur. Mines, Explosives and Phys. Sci. Division,Rept. 3172 (1951). (4) Carslaw, H.S.,and Jaeger, J. C., “Conduction of Heat in Solids,” Oxford Univ. Press, London, 1948. (5) Danckwerts, P. V., Chem. Eng. Sci.2, 1 (1953). (6) Furnas, C. C., U. S. Bur. Mines, Bull. 361 (1932). and Alberda, G., Chem. Eng. Sci. 2,173 (1953). (7) Kramers, H., (8) Schumann, T.E. W., J . Franklin Inst. 208,405 (1929). (9) Sneddon, I., “Fourier Transforms,” McGraw-Hill, New York, 1952. (IO) Thomas, H. C., J . Am. Chem. SOC.66,1664 (1944). (11) Wilhelm, R. H., and Singer, E., Chem. Eng. Progr. 46, 343 (1950).
(Solid-Fluid Inferactions in Fixed a n d Moving Beds)
Fixed Beds with large Particles NEAL R. AMUNDSON Dcparfmenf of Chemical Engineering, Universify of Minnesota, Minneapolis 14, Minn.
I
N T H E previous discussion the problem of heat release and heat generation from a fixed bed of particles to a moving fluid was
considered. The bed was assumed to be nonadiabatic, so that radial transfer of heat had to be taken into consideration. Axial mixing was also allowed for, but the particles were assumed to be so small that the major resistance to heat transfer was a t the particle surface. This is, in general, valid for smallparticles. As the particle size increases, however, radial temperature gradients will arise in the particles, and intraparticle conduction or diffusion may become the limiting factor in the transfer of heat or mass. With this in mind, then, this paper reconsiders those problems of the previous discussion but with the added complication of another &ace variable, the particle radius, which is introduced as a result of the particle conductivity. The problem in which longitudinal mixing is pertinent is again considered. Axial mixing i s neglected in considering flxed beds with large particles
Consider a fixed bed of radius P packed with particles whose radius is R. Fluid is introduced into the bed a t z = 0 and heat is transferred from the solid t o the fluid. Heat is transferred to the reactor wall by the turbulent mixing process in the fluid and is conducted through the reactor wall by ordinary conductive meana. Particle to particle transfer of heat is neglected, and it is assumed that the temperatures are low enough so that radiative transfer in the bed may be neglected. Heat may be gpnerated inside the particles. If the particles are assumed to be spheres, the equation for conduction with radial symmetry is January 1956
where Q is the rate of heat generation per unit volume of sphere. Q is of exponential form for heat generation arising from a chemical reaction. It will be approximated to here by a linear function a bt, so that
+
The particles are large enough to make necessary consideration of intraparticle conduction but small enough to be considered to be in an environment a t constant temperature, thus ensuring radial symmetry. At the particle surface the heat transfer coefficient, hf enters into the boundary condition
where T i s the temperature of the fluid in the neighborhood of the particle. If the heat transfer coefficient becomes very large, or alternatively, if the radius of the particle is large, the boundary condition may be
t=T
r = R
(42‘)
The heat balance on the fluid is identical with that given in the previous discussion and is
INDUSTRIAL AND ENGINEERING CHEMISTRY
35
ENGINEERING, DESIGN, AND EQUIPMENT
-vfc,p/org+ kf,
(e+ -) -1 bT
3(1 -
P bP
or’)iks
R
The Bessel transform will be applied with E J o ( E m ) , and
-
(gt) br
bT ? = R = a’cfPlbT
J&)
(43)
J 0 ( f )= L
(44)
Here it is assumed that the ambient temperature is zero, so that all temperatures are relative to this temperature. The heat transfer coefficient, U,combines the coefficients inside the reactor wall, the reactor wall itself, and the coefficient exterior to the reactor. The influent temperature must be specified = To(p),
2
=
J o ( T : )= Yb
J o ( T ) = Y , J o ( t ; ) = y:,
while again a t the reactor wall the condition is
T
= y,
Em, a root of tmJ1(En>) =
0
In this section it will be supposed that T o and although this is not essential, so that
ti
Jl(E?n) v ,, = t-
y ; - ToJ1(5m) Em
are constants,
f
Wd Em))
= __
Em
Em
Equations 47 to 52 then become
(53)
(45)
where it is assumed that To(p ) may be a symmetric function of p. It is necessary also to specify the temperature of the solid. This, as discussed above, is given by
t =
t l ( T , p,
6 =
z),
(46)
V
Equations 41 to 46 make up a complete mathematical description of the problem to be solved. Special cases of this problem although not stated in this form, have been obtained by Rosen (S), Thomas ( I O ) , and others (4). The effect of intraparticle diffusion was investigated for sinusoidal inputs by Deisler and Wilhelm (3’). Of course, solutions to problems in which the particle size is small may also be considered to be special cases. The general plan of solution is to reduce the problem to dimensionless form, apply the finite Bessel transform to remove the reactor radius as a variable, and apply the iterated Laplace transform to the axial and time variables. This gives a differential equation in the particle radius variable. The successive transforms, after solution of the differential equation, may then be inverted to give the final solution. This solution is not as complex as might be anticipated, as it can be expressed in terms of a function which has been partially tabulated by Rosen ( 7 ) . If the substitutions 2
T
= =
=
UP
12
p
sR
@ =VrL
If the double Laplace transform is applied in the form e
L(W) =
Som
ydq=w
e
Wdx = V
then the system of Equations 53 to 57 reduces to the three equations
.T
u =
E = -PU k ir 3(1 - a’)P2k, !J= R krr
Vf or ‘ C I P fP2 1 k/,
- u [p2V
-
I)*
P1 Em
- E4V - p
($)s=l
=0
(60)
Equations 58 and 59 may be combined to give
are made, the problem is put into dimensionless form. If the corrected time variable t)=r-x
Equation 58 has the solution
is also used, the set of Equations 41 to 46 becomes (47) = t -
--e
bT
b2T
ax + -
u bu
+ ET T
36
(48)
+ -1 bT --
au2
t =
TI s = 1
where A is an arbitrary constant to be determined by use of Equation 61. Substitution of v into Equation 61 gives an equation in A which may be solved. Insertion of this A into Equation 62 gives finally
=
T,(uP) p,
2)
=
=
0, u = 1
TL(u),
2 =
(51)
0
= t,’(S, u , x), 17 =
0
(52)
INDUSTRIAL AND ENGINEERING CHEMISTRY
(Continued)
Vol. 48,No. 1
ENGINEERING, DESIGN, AND EQUIPMENT manipulation, since the transform variable occurs as both p1 and crp~ A. The terms which cause difficulty are those in which p1 occurs alone in the denominator. This difficulty may be met by writing
-
where w = l / ~apl. The expression for V may be obtained from Equation 59.
-1= - -
PI
A
a:
~ P -I A
Pi(api
- A)
and the inversion follows directly. Thus
where B(w) =
cot w - 1 cot w - 1) 1 w
e(w
+
Now Equation 64 is the triple transform of the sought solution. To find its inverse transform it is convenient t o write it in another form
T t is easy to see that the inverse transform with respect to p2
The function y is difficult to find and will not be given here. The function Y should really be written Y,, so that if
may be found without difficulty
Ym =
x m
Ji(Em)
Em
m
then
T = 2 E X (E XmJo(u#m)
+
m= 1
t
+
E)
e '
e apl - A
-
u ( a p f ? A) ("li -k
i)
4
(€,
2,
4% and p corresponds to
$)
7
is
=
(68)
m
m-1
,where (j =
(Em)
and the average fluid temperature a t any level is T = 4 E z z
'The inverse with respect to pl is somewhat more difficult and can %e found by making use of the previously mentioned function of 'Rosen ( 7 ) . Rosen showed that the inverse transform
m J 0
(E
+ 4:) :4 xm
Equations 67, 68, and 69 give the solution to the problem posed in the early part of the paper. Such a solution may, indeed, seem complicated, but the tabulation of the function @ has already been started. Once complete tables are available, the evaluation of the integrals involving @ will not be a difficult task by modern computing methods. It might be possible to manipulate these integrals as Thomas (9) has done for the integrals appearing in the problem of small particles, but, so far as the writer is aware, this has not been done. The calculation of the heat lost through the reactor walls is not difficult and proceeds from the formula
Q
= -27rk,?K
(F)
u u = l dz
Special Cases. A. NEGLIGIBLE RESISTANCE TO HEATTRANSREACTOR WALL. If there is no resistance at the reactor wall, U is sensibly infinite and therefore E is infinite. This does not change the character of the solution appreciably. If E becomes infinite, &, must be chosen as a root of J,(gm)= 0. Equation 67 is unchanged except for the change in e, and Equations 68 and 69 may then be written
.where
FER AT
land
+ sin 2x1 - cos 2X - 1 = HD,(X) X(sinh 2X - sin 21) = cosh 2X - cos 2X = HD,(X) X(sinh 2X
jHD1
= cosh 2X HD2
,with X = 4~. Rosen tabulated the function @ for use in connection with-the problem of solid diffusion during ion exchange in a fixed bed of exchanger. To find the inverse transform of Equation 65 requires a little 'January 1956
and
B. NEGLIGIBLE RESISTANCE TO HEATTRANSFER AT PARTICLE SURFACE.I n this case h, must be large and hence E small.
INDUSTRIAL AND ENGINEERING CHEMISTRY
31
ENGINEERING, DESIGN, AND EQUIPMENT If one lets B approach zero, the solution may be written in exactly the same form as Equations 67, 68, and 69 or 70 and 71 with
condition on the solid temperature. The initial solid temperature is a function of all three space variables, r, p, and z . The system of equations will then consist of Equations 41, 42, 43, and 44, the condition
T=Tj
-
since H, H D I ( ~and ~ H )z H o 2 ( d j )as =-f 0. Thus the basic form of the solution is unchanged. C. STEADY-STATE SOLUTION.The steady-state solution may be found from Equation 63 by evaluating lim(p,v) and then -+
z = O
(45')
and tho inverse Bessel transform of Equation 72. The details will not be presented, as they are largely manipulative. The solution for the temperature of the fluid is
PI-0
inverting the transform with respect to pz. =1 similar operation may be performed on Equation 64. If ya8and Y,,denote the steady-state Bessel transforms of the solution, there results Y.8
=
Em
AS
~ ~ ( sin4 V'~ZS) [el/;? cos di + (1 - E ) sin 4x1
X
where
and
The average temperature or cup mixing temperature may be Ea obtained by replacing 2E J,(uEm) by 4 1in the summations.
where 9, =
5: 4- P B(t/X)
JdEm)
Em
D. No HEATGENERATION.This case is obtained trivially from Equation 67 by allowing 't- 0, A 0.
F. TRANSIENT RESPONSE FROM A STEADY STATE,WITHOUT HEATGENERATION.This case reduces to the simple formula
If in addition t i is zero-i.e.,
and the average temperature can be obtained in the usual way. This formula reduces to Equation 73 if T I = To,as it should. Rosen has shown also that a good approximation to the function a, accurate to within 1%, is given by
-+
the bed and surroundings are a t a
uniform temperature-then
The final solution may be written in a simple form for this case, since 9 does not depend upon m. Therefore when E
< 0.01 and
7>
50.
(73)
Another problem which has arisen and which can be treated by similar methods is that in which the heat generation term is a function of the space variables only, say, S(T, p , z ) . The heat generation for one such problem occurs as a function of the form
All these non-steadystate problems are interesting since one would not expect a priori that the function Q would be independent of the radius variable, although the method of product solutions might supply some hint of this. E. TRAXSIENT RESPONSE FROM A P R I O R STEADY STATE. It develops that the solution of the problem of the response to a step in the input is not so difficult ns might originally be supposed. Let the fixed bed be operated until a steady state is attained with influent temperature To. At e = 0 suppose the influent temperature jumps to a new steady value, T i . It is then required to find the function which describes the approach to the new steady state. I t is clear that the equations nhich need to be solved are identical with those previously used, with the exception of the initial
This may be treated without difficulty, but will not be done here. EXAMPLE. A fixed bed, 1 foot in diameter, packed with particles 0.5 inch in diameter has a length of 5 feet, The temperature exterior to the bed is zero. The bed is operated at the steady state with a gaseous influent at 200" F. The temperature of the influent suddenly jumps to 500' F. The temperature transient of the effluent is required. The physical parameters and operating variables are as follom:
and
38
R = 0.0208 f t . P = 0.50 ft. G = 1000 lb./hr./sq. ft. C / = 0.28 B.t.u./lb./' F. k. = 0.194 B.t.u./hr./sq. ft./(" F./ft.)
INDUSTRIAL AND ENGINEERING CHEMISTRY
Vol. 48, No. 1
ENGINEERING, DESIGN, AND EQUIPMENT boundary condition suggested by Danckqrerts ( a ) and Kramers and Alberda (6) will be used. The system of equations is then
1 = 5 ft. 01’ = 0.50 U = 12.7 B.t.u./hr./sq. ft./’ F. ca = 0.25 B.t.u./lb./” F. ps = 100 lb./cu. ft.
k,
In order to make use of Rosen’s tables of Q it is necessary to calculate
Ep~
G C/ 31 h / ( l
(g+ g) +
- k,
(E)
=
a
+ bt
= caps at
h/(t - T ) ,
3(1
(1000) (0.5) (0.28) = 0.20 3(5) (93.3) (0.5)
CY’
- 01’)
=
T
R
(77)
- a’)k , ( 3 ) R
bY
p.=R
Also if one may assume that (79) then
k/,
=
2(0.0208) (1000) (0.28) = 2,12 (11) (0.50)
bT &=O,z
and
= I
e
t = ti(r, z), T = T,(Y,p, z), G 0r’cfP2 - (1000) (0.50) (0.28) (0.25) g=-1 kf, (5) (2.12) Now the values of the Ern are needed. values are ( I ) , €1
€2
€8
1. If the roots of the cubic are positive
or complex, Laguerre’s theorem does not apply. However, it may be shown that if an associated cubic 8EU n3
+ 4u
(56
+ 1) n2 +
01
ac
-v, “’Z + k
= 0, z = 1
b’C f q g
c =
+
(&)
3(1 - CY’) Dr R br ? = R
ci(r, z ) ,
e
=
o
c
=
ci(z), e
bC -
ae
=
o
where C, is the concentration of the influent brought to the bed. Some assumption must be made regarding the local mechanism. If it is assumed that n = kc k‘ with k and k’ as constants, then
+
42
has real roots in n, then g ( z ) has the same number of positive roots, all the remaining roots being negative. It is apparent, however, that this cubic has no positive roots in n. This equation thus has three negative roots or one negative and two complex roots. The extension of Laguerre’s theorem allows us to say only that g(z) will have a t most two complex roots and it may have none, all other roots being negative. To find these complex roots, if any, may be a difficult task. Since z is related to p through the formula p
* +z 01
there may be positive roots in p if A/.. is large. In this case
INDUSTRIAL AND ENGINEERING CHEMISTRY
Vol. 48, No. 1
ENGINEERING, DESIGN, AND EQUIPMENT there would be instabilities in the system and temperatures would grow to very large values. Conclusions
A new problem in heat transfer and heat release in fixed beds has been solved. It would seem that this problem would follow as a special case from that considered previously, but the detailed passing of the limit from one problem to the other is tedious to carry out. Thomas (10)has obtained a solution of the problem of intraparticle diffusion in fixed beds, neglecting the resistance a t the particle surface for a problem in ion exchange. His solution is expressed in series form and is probably identical with Rosen’s integral. The integral is used here, as explicit directions for its calculation are now available. The magnitudes of the various effects such as intraparticle conduction, radial mixing, axial mixing, particle surface transfer resistance, reactor wall resistance, etc., can be determined only from extensive numerical calculations and experimental work. The problems considered in this paper have had distributed constants, as opposed to the method of lumped constants used by engineers in their first approximations to physical problems. It remains to show in these problems when the constants must
be distributed and when they may be lumped. This problem is a difficult one mathematically, for it asks for an estimate on the difference of the two solutions when a term has been and has not been omitted from the differential equation. Naturally, the latter result could be obtained by extensive calculation, but it is not only too time-consuming but inelegant and cannot lead to general results, literature cited (1) Carslaw, H. S., and Jaeger, J. C., “Conduction of Heat in Solids,” Oxford Univ. Press, London, 1948. (2) Danckwerts, P. V., Chem. Eng. Sci. 2, 1 (1953). (3) Deisler, P. F., Jr., and Wilhelm, R. H., IND. ENG.CHEM.45, 1219 (1953). (4) Kasten, P. R., and Amundson, N. R., J . Phya. Chem. 56, 683 (1952). ( 5 ) Kramers, H., and Alberda, G., Chem. Eng. Sci. 2, 173 (1953). (6) Munro, W. D., private communication. (7) Rosen, J. B., IND.ENG.CHEM.46, 1590 (1954). (8) Rosen, J. B., J . Chem. Phys. 20, 387 (1952). (9) Thomas, H. C., J. Am. Chem. SOC.66, 1664 (1944). (10) Thomas, H. C., J . Chem. Phys. 19, 1213 (1951). (11) Titchmarsh, E. C., “Theory of Functions,” Oxford Univ. Press, London, 1939.
(Solid-fluid Interactions in Fixed a n d M o v i n g Beds)
Two Problems on Moving Beds CHARLES W. SIEGMUND, WM. D. MUNRO, AND NEAL R. AMUNDSON Deparfmenf of Chemical Engineering, Univerdfy of Minnesofa, Minneapolis 74, Minn.
I
N T H E two previous discussions problems related to heat and mass transfer in fixed bed systems were considered. Particles were assumed to be either large or small, so that intraparticle conduction or diffusion was or was not a factor. In an earlier paper (a) the problem of the moving bed reactor was analvzed in a similar way. and this aaaer shows how the nonadiabatic moving bed reactor may be treated. In another paper ( a ) the problem of the moving- bed adsorber was considered, but only linear equilibrium isotherms were allowed. As a second Droblem in this aaaer a numerical scheme will be develoaed. so that an arbitrary isotherm may be used. “
I
1
(g+ ;dt%) + a + 2
bt
bt = cape
where is the radius variable in the sphere and where a + bt is the linear approximation to the usual exponential expression. At the sphere surface the condition is
1
.-
Treatment of nonadiabatic countercurrent reactor i s considered
Previously (3) the adiabatic moving bed reactor was analyzed from the standpoint of heat transfer and heat release. The same problem will now be attacked, but with the adiabaticity restriction removed. This would seem to change the details of the solution only superficially; however, this is not the case. A reactor whose radius is P and whose length is 1 is fed at the top with spheres of radius R a t a rate of G, pounds per hour per square foot and temperature to. At the bottom of the reactor fluid is admitted at temperature 2‘1. There is an interchange of heat between fluid and solid, and because the reactor is nonadiabatic there is a flow of heat in the radial direction toward the reactor wall caused by the turbulent mixing process in the fluid. Further, there is a conductive process taking place inside the particles and a transfer process a t the particle surface. If e is the time a given particle has been in the reactor and if the particles are small with respect to reactor dimensions, radial symmetry in each sphere may be assumed and the temperature in the sphere must satisfy the equation January 1956
k,
-h ($)T=a = hf(t - T ) , r = R The spheres at the inlet to the reactor have a temperature t = t o ,
e=o
(117)
One can also derive a relation between the fluid and solid temperatures, t and T, by a heat balance over an elemental annular ring. This balance gives
The reactor wall condition may be expressed by
-kl
(E) bP P’P
=
UT,
p =
P
The temperature of the entering fluid is
T=Tt,
2 3 1
(120)
Equations 115 through 120 make up a complete description of the problem. The method of solution follows that used before. The Bessel transform is applied to remove the reactor radius varizb€e. The Laplace transform is then used to obtain an ordinary differential equation which may be solved. A difficulty arises here, as the fluid temperature is not specified a t x = 0 but rather a t x = 1. In the adiabatic case this caused no difficulty, since a simple heat balance over the whole reactor
INDUSTRIAL AND ENGINEERING CHEMISTRY
43
