I
G.
D.
GALLETLY and C. R. GARBETT
Shell Development Co., Emeryville, Calif.
Pressure Vessels- Let the Tubes Support the Tube Sheet W H E N
HEAT
EXCHANGER
DESIGNS
with fixed tube sheets are specified to satisfy accepted codes and standards, tube sheets obtained may be thicker than necessary. The commonly used standards of the Tubular Exchanger Manufacturers, TEMA (Q),allow for only small support given by the tube bundle .in carrying the pressure difference across the tube sheet. The savings in metal thickness using the tube support are important for large vessels, and structures are correspondingly cheaper to fabricate. Also, thinner sections may reduce the stresses imposed by transient temperature changes. Tubes give extensive support for a fixed tube sheet design; for floating head arrangements a load distribution is effected. Only for single head-Le., return bend tube construction-is support entirely lacking. Miller (8), in extending the method developed by Gardner (3, 4 ) , gives design curves for finding peak stresses in tubes and tube sheet, allowing for
1. Relative rigidities of tubes, sheet, and shell 2. Clamped or free edge on tube sheet 3. Thermal expansion when tubes are heated or cooled
Using Miller’s curves for the vessel (Figure I), an adequate tube sheet thickness for a stress of 17,500 p.s.i. is 2.3 inches, whereas TEMA standards would require a thickness of 5.9 inches. When comparing the plate material for the two cases (2l/2 inches thick us. 6 inches thick for a IO-foot square mill piece) a saving of 7 tons is realized. Moreover, the total amount of drilling saved is about 350 feet. Many studies and tests have been made (2-8, 70) to approve a method
Elastically Supported
Tube Sheet Analysis Miller (8), following Gardner (3, 4 ) , treated the tube-sheet problem as a plate on an elastic foundation (the tubes being
THE STANDARDS TODAY TEMA TEMA Standards* These standards (9) give formulas for tube sheet thickness based on bending 01: shear stress, whichever is the larger. The thickness required to resist bending is:
d o i s
the elastic support of the tube sheet. The basis fo;’ this formula has not been published but can be derived from M~~~~~~~ equations by assigning conservative to some of the parameters. A 1954 addendum to the standards states (9):
Special consideration may be given to conditions of support or loading where t = thickness of tube sheet in tending to reduce tube sheet thickness inches; p = design pressure, p d . , requirements. tube side or shell side, whichever is higher; S = allowable working stress, ASME p.s.i., at design temperature. Values of F and D are different depending ASME Code for Unfired Pressure upon whether stationary or floating Vessels. This code gives design forheads are used, the type of clamping, mulas for flat, unstayed heads and for and others. For fixed tube sheets, braced and stayed surfaces, but no speD is the gasket diameter or the inside cific design formulas for the design of diameter of the shell, For tube side tube sheets in heat exchangers. pressure, F is given by the formula : 1 =
FD
+ K)A2 + 3 K )
F = d(2
(1)
= E,t,(Do - t,)/E&,(d - tt) E = modulus of elasticity, p.s.i. t = wall thickness, inches DO = outside diameter of shell, inches
IC d n
=
outside diameter of tubes, inches
= number of tubes
subscripts s and t refer to shell and tubes, reFpectively. If the shell side pressure is higher, F is: F = dK/(2
+ 3K)
BRITISH
(2)
where
Figure 1. Savings in metal thickness. Miller used a 2.3-inch thick tube sheet on this sample reactor whereas one 5.9 inch thickness i s used in the TEMA formula
for design of tube sheets, and all questions are not fully resolved; this is perhaps the reason the codes and TEMA standards have not yet been modified to take full advantage of the tube support.
British Standards for Fusion-Welded Pressure Vessels. This code ( 7 ) recommends design formulas for guidance only; formulas do not allow for support from the tubes, and treat the tube sheet as a homogeneous plate with edge support and with a reduced Stress allowance because of the drilling. The thickness of the tube sheet is given by: t =D
where
m
S
(4)
D is either the bolt circle diameter
( 3 ) of a wide flanged joint ( B = 7.25),
or D is the gasket diameter for a narrowBy evaluating F and K , the TEMA faced joint ( B = 4.0). This code is standards apparently allow some for now being revised (2, 8).
VOL. 50, NO. 9
SEPTEMBER 1958
1227
the elastic foundation). To solve the problem these assumptions are made :
1. No slip occurs at the junction between tubes and tube sheet. 2. Tubes are stayed to prevent them from buckling. 3. Deflection of the tube sheet is small. Hence, angular distortion of the tube ends due to bending of the tube plate can be neglected. 4. Tubes are uniformly distributed over the whole tube sheet. 5. The forces exerted by the tubes on the tube sheet are distributed continuously.
6. Deflection of a uniformly perforated thin tube sheet under loading is similar to that of a homogeneous plate of the same radius and thickness, and identically loaded. The bending stiffness of the homogeneous plate is equal to the bending stiffness of the tube sheet without holes, multiplied by p , where is p the deflection efficiency. 7 . The maximum stress in the perforated tube sheet is equal to the maximum stress in the homogeneous plate divided by p, where 1.1 is the ligament efficiency (reciprocal of the stress concentration factor due to perfcrationa in the tube sheet).
8 . In-plane forces (due extension or contraction of the tube sheet) can be neglected.
Miller also assumed that the deflection and ligament efficiencies are equal to each other, and that they are given by the mean width between the rows of tubee divided by the tube pitch, With these assumptions, the stresses in the tubes, tube sheet, and shell due to pressure or steady-state temperature difference can be determined. Equations have been solved for the maximum stresses in the tube sheet and tubes and
TUBE SHEET STRESSES The maximum radial stress in the tube-sheet, S,,,, due to both pressure and steady-state temperature difference is given by :
,,s,
=
fi* ~ P ( Q Gf I Gz)
(9)z
(5)
= coefficient of linear thermal ex-
CY
pansion Subscripts s and t refer to the shell and tubes, respectively. GI and Gs are functions of the dimensionless parameters, kR, where
kR
where D = inside diameter ofshell = 2R t = thickness of tube sheet GI, GZ = are dimensionless (Figures 2 . and 3) 6* = p [ ( A - C)/AI - PzQ = p1 - pz x 0 (7'~%na)/(24 - C ) I 1 nu / ( A - C)l = Pressure on shell side PI = pressure on tube side bz = number of tubes n d = outside diameter of tubes = (n/4)D2 A = (nn/4)# C - Eina (Note: Q is the ratio of Q EJ3 the spring constant of the tube bundle to that of the shell. I t is also very nearly equal to the reciprocal of the quantity K used in the 1952 TEMA code.) z = n(d - t J t , = cross-sectional area of metal in one tube B = T(D tJtS = cross-sectional area of shell plate = thickness of shell IS = thickness of tubes tl E = modulus of elasticity = L Y ~T CYST. ~ Y
+
+
_-
+
I
=
(2)1'4R
k* = spring constant of the tube bundle (per unit area) = E,na/lA D , = flexural rigidity of tube sheets = p [ ~ p t 3 / 1 2 ( 1-
Y2)1
21 = length between tube sheets Y = Poisson's ratio p = ligament efficiency o = deflection efficiencv p = p = ( A - C ) / A , according to Miller (8)
Equation 5 can also be inverted to give the thickness of the tube sheet. We obtain
t
=
D
1.1Smax d m
,/+ QG1
GZ ( 6 )
where S ,, is now the maximum allowable stress of the material in the tubesheet. Equation 6 is also somewhat similar to the TEMA formula for thickness of tube sheet. The TEMA formula may be obtained from Equation 6 by making the following substitutions: 1. Substitute Q = 1 / K 2. Put GI = 2, Gz = 3 3. In the expression forp*
3201
A. Put n a / ( A - C) = 0 B. Put yE,na/(A - C) = 0 C. Put A - C/A = p = 1/2
If these simplications are made then Equation 6 will be found to reduce to the TEMA formulas. In comparing step 2 with Figure 2 full advantage is not being taken of the support given by the tubes. This figure also shows that the function GI and Gz have been calculated for both simply-supported and clamped-end tube sheets. Actual tube sheets have boundary conditions which lie somewhere between these two extremes. Contrary to experience with unsupported plates, it will usually (for kR > 3) be found that the assumption of clamped ends gives larger tube-sheet thicknesses than the assumption of simple supports. If it is desired to take advantage of the actual boundary conditions (which will result in smaller thicknesses than the clamped-end assumption) a procedure similar to that described by Yu (70) should be used. The design will now take longer unless it is possible to arrange the tube-sheet supports so that they are almost simple supports. If this can be done, Figure 2 can be used directly in the design. T AND G
1 FLANGE
I
TUBE SHEET
0
2
4
6
8
10 12
14
1 6 18
20
kR
kR
Figure 2. Values of the functions GI and Gz (8)show an advantage only when with d l / ( Q G l Gz) X d p T p , but this i s not :omparing TEMA's F X D mmediately transparent
db + --- Simply supported; -- Clamped
1228
INDUSTRIAL AND ENGINEERING CHEMISTRY
Figure 3. This is a type of tube sheet support which tends toward a simple support
PRESSURE VESSELS
100
1
-3501
I
I
I
I
I
r (INCHES)
Figure 4. Curves give deflections of 4-inch simply supported plate of an elastic foundation and subjected to uniformly distributed unit edge shears the problem has been reduced to some simple formulas and graphs. As far as the designer is concerned, he has only these latter with which to work.
Miller’s Curves vs. TEMA Standards Consider a problem where tube sheets (Figure 1) were stationary and it is desired to calculate their required thickness. To do this with Figure 2, we must know kR, which is itself a function of the unknown tube-sheet thickness, t. Assuming various values for t (2, 3, and 4 inches for this example), corresponding values of kR are calculated and then the values of GI and Gz from Figure 2 are obtained. Substituting these quantities into Equation 5 (withpl = 0 ink*, as this gives the worst case) then the maximum stress in the tube sheet is determined. The calculations are as follows:
p =p =
AQ/(A
- C)
(A
= 10.32
- C)/A
= 0.47
.*.j*/,u = 2610
When plotted, the intersection of the two curves with the horizontal allowable stress line (S,,, = 17,500 p s i . ) shows these required tube-sheet thicknesses: Ends simply-supported t = 2.3 inches t = 4.7 inches Ends clamped There is a considerable disparity in tube-sheet thickness required for the two types of end condition. The designer should approach the simply-supported end condition as closely as possible. Turning now to the TEMA Standards,
the tube-sheet thickness is given as Equation 1. Substituting the numerical values in Equation 2, K is 0.210 and F is 0.91. Hence, from Equation 1, t is 5.9 inches. The TEMA standards formula gives a tube-sheet thickness which is 25% greater than the maximum thickness required by allowing for the elastic support and 157% greater than the minimum thickness. Thus, the standards give no guidance to the designer who uses his ingenuity and provides end conditions for the tube sheets which are close to simple supports. For instance, if the tube sheets are supported as in Figure 3, then the required thickness is only 2.8 inches-i.e., only 48% as thick as the thickness required by the formula in the TEMA standards. Gardner ( 5 ) suggested that the P value used in the example (0.47) is somewhat high and prefers a value of 0.32. If this latter value is used the thicknesses of the tube sheet becomes 3.0 inches (simple supports) and 6.2 inches (clamped ends). The elastically supported tube-sheet of Figure 3 then is 63% of the TEMA thickness rather than 48%.
10
20
30
50
40
r (INCHES)
Discussion Deflection and Ligament Efficiencies. Miller used “ligament” and “deflection” efficiencies to account for the inhomogeneous nature of the drilled plate, following Gardner’s procedure (3). An “equivalent plate constant” method, used, by Horvay (6) and M a k i n (7),gives fictitious values for the elastic constants of the material such as modulus of elasticity and Poisson’s ratio. Neither approach, however, can be said to be “established” and more study will have to be made to decide between the two. Duncan (2) recently reported the results of a series of experiments on perforated tube sheets. One of the objectives of his tests was to determine experimental values for the deflection and ligament efficiencies and compare them with algebraic expressions proposed by various authors. For continuously drilled tube sheets (the type considered here) the experimental deflection efficiency seems to be lower than the value used in the example. Duncan also has some results on ligament efficiencies
Figure 5. These curves show changes of curvature of 4-inch simply supported plate on an elastic foundation and subjected to uniformly distributed unit edge shears and states that the experimental and theoretical values seem to agree. However, it is not clear which theoretical expression he used in his comparison. For the present, we shall assume that the ligament efficiency expression used is adequate. However, the effect of using a lower deflection efficiency on the stresses and deflections should be investigated. For the tube diameter, tube pitch ratio of the authors’ example, Duncan infers that the deflection efficiency, p, would be about 0.2. This compares to the value of p = 0.47 obtained earlier by Miller’s expression for the deflection efficiency. Also, Yu (72) has re-examined Duncan’s data and concluded that a value of p = 0.3 is reasonable. We shall only investigate one tube-sheet thickness herein-Le., t = 4 inches. The ends of the plate are also assumed to be simply supported. Other conditions are the same as for the
Table 1.
Calculations Using Miller’s Design Curves 3Inches; kR = 8.48 t = 4Inches; k R = 6.84 Simple Clamped Simple Clamped Simple Clamped supports ends supports ends supports ends
t = 2Inches; IcR = 11.5 t =
Boundary Conditions GI
Gz
+
Gz) (D/Oa.
4(&ffi 8-x
P.S.1.
8.0 69 428 3,220 19,650
5.45 24 199 3,220 42,220
5.85 37 259 1,431 14,430
4.0 12 124 1,431 30,100
4.7 25 189 805 11,120
3.25 9 97.6 805 21,550
VOL. 50, NO. 9 e SEPTEMBER 1958
1229
tube sheets, and which are also not covered by the TEMA standards are :
L Figure 6. An initially imperfect tube subjected to compressive forces is assumed to look like this L = length between tube sheets; e = Shortening of tube under end load; 6 = lateral deflection of tube under end load; au‘ = End load applied to tube sheet
numerical example discussed previously. The deflections, w , and curvatures, X,, of the plate on an elastic foundation, and subjected to unit edge shears, have been calculated for both values of p. The results are shown in Figures 4 and 5 . Comparing these curves, the edge deflections and maximum curvature are larger for the more flexible plate ( p = 0.2). Assuming the expression for ligament efficiency remains the same for both cases, and calculating the stresses from Figures 4 and 5 all maximum pressure stresses (in tubes, shell, and tube sheet) are lower for p = 0.2 than for p = 0.47. However, the ligament efficiency is assumed to be the same in both cases. This question of ligament efficiency should be studied further, as Gardner (5) recommends a value of 0.32 for this case. Effects of Bowing of Tubes. Previously, Miller made an assumption that the tubes did not bow or buckle. Now when the tube-side pressure is greater than the shell-side, the majority of tubes are in compression. When the shell-side pressure is greater than the tube-side, tubes adjacent to the shell are in compression. For any practical case, some tubes will be in compression and these tubes should be investigated to see that they do not touch and form hot-spots. In so far as it affects Miller’s results the case where the tube-side pressure is greater than the shell-side pressure is the more serious. This is because tubes are never perfectly straight and begin to bow immediately as a compressive load is applied to them. As most tubes are in compression for this case, a general lowering of the elastic support is provided by the tubes. This effect should be considered in the design. The following approximate solution should be reasonably satisfactory for compressive loads which are not close to the Euler buckling load of the tubes. A direct result of Miller’s assumption is that if the tube is compressed by a n amount e , then the upward force on the tube sheet due to this one tube is F = E(e/L)a (7) where E = modulus of elasticity L = length of tube a = cross-sectional area of metal in tube. ( E e/L is the compressive stress in the tube u t . )
1230
IVhen the tube is imperfect, the upward force on the tube-sheet, for the same value of e , should be somewhat less than this value. The difference between the two values depends on the amount of imperfection and the proximity of the compressive stress in the tube to its critical buckling stress, ucr. The support offered by a n imperfect tube can be obtained from Figure 6 which represents a n initially imperfect tube subject to a compressive force, aut*. We shall assume that the initial imperfections, yo, can be represented by yo = 6
sin T X / L
(8)
where
Conclusions
6 = maximum initial eccentricity x =
axial coordinate
Using this expression, 90,assumes that no moment restraint exists at the ends of the tube. But this is conservative. Then the imperfect tube will shorten by an amount X where :
1 x=
(g)’[ (*J
- 1 1 -I- ut* E, (9)
and uer = (k2EtI/aLZ)-
(wL/2)
1. Forces in the plane of the tube sheet. These forces arise because the shell wants to increase its diameter (due to the pressure) and it is restrained from doing so by the tube sheet. Horvay has discussed this problem for perforated plates and Yu for solid plates ( 7 7 ) . Yu’s preliminary results indicate that the influence of the in-plane forces may not be very large. 2. The rotation-resisting capability of the tube bundle on the stresses in the tube sheet. This effect was also discussed by Yu ( I 7 ) . 3. The triangular tube spacing arrangement gives more nearly isotropic behavior of the tube sheet. Caution should be used in applying efficiency factors to square spacing, for which substantial divergencies from isotropic behavior have been noted (6). 4. Transient thermal problems which occur during startup and shutdown of the reactor. Actually, the TEMA standards do not even consider the stresses caused by a steady state temperature difference between tubes, although this can be handled relatively simply by using Miller’s curves.
(10)
where
I = moment of inertia of tube ze = weight of tube per unit length
I t is possible to effect substantial economies in tube-sheet thickness (and cost) by taking full advantage of the support offered by the tube bundle. Further economies may be gained by considering the actual boundary conditions present a t the periphery of the tube sheet. However, before complete confidence can be placed in the design method, experiments must be made with tubes of different diameter, pitch, and others, to determine the values to be used for the ligament and deflection efficiencies. The effects of bowing of the tubes, in-plane forces and transient thermal effects must also be investigated.
Now from Miller’s curves the maximum
literature Cited
compressive stress in the tubes can be obtained. Let us call this quantity c ~ . The shortening of the tubes, if they were perfect, would then be obtained as
(1) British Standards Association, London, “Fusion Welded Pressure Vessels British Standard Code No. 1500: 1949 (Provisional).” (2) Duncan, J. P., Proc. Inst. Mech. Engrs. (London) 169, No. 39, 789 (1955). (3) Gardner, K. A.: J . A j j l . Mechanics 16,
e
=
(at/Et)L
Assuming the end shortening of the imperfect tubes is the same as that of the perfect tubes, we thus substitute 6 for X in Equation 9 and calculate ut*, the stress in the imperfect tube corresponding to the end shortening of the perfect tube. The ratio u*Jat then serves as an approximate measure of the quantity: support given by imperfect tube/support given by perfect tube. If a tube sheet of thickness has been selected using perfect tubes then, to account for the imperfections, we multiply it by C t / u * t . Miscellaneous Topics. Other factors which should be considered in designing
INDUSTRIALA N D ENGINEERING CHEMISTRY
377 11948).
(4) Gardner, K. A , , Ibid., 19,159 (1952). (5) Gardner, K. A. Griscom-Russell Co., Massillon, Ohio, private communication. (6) Horvay, G., Proc. First U. S. Natl. Congr. Appl. Mechanics, p. 247 (1951). (7) Malkin, I., Trans. Am. Soc. Mech. Engrs. 74,387 (1952). (8) Miller, K. A. G., Inst. Mech. Engrs. (London) IB, No. 6, 215 (1952). (9) Tubular Exchanger Manufacturers Association, “TEMA Standards,” 3r d ed., New York, 1954. (10) Yu, Yi-Yuan, J . A j f l . Mechanic 23, 468 (1956). (11) Zbid., 24, 141 (1957). (12) Ibid., p. 315. RECEIVED for review July 1, 957 ACCEPTED March 31, 958
