Heating and Cooling Times
pitrticles art’ :it, uniform t,wnperature. ( c ) KOheat, is lost, to the surroundings. ( d ) N o heat is supplied from any sources other than those specifically mentioned. ( e ) Heitt supplied to or removed from the bath fluid involves no change of state-Le., vaporization, condensation, etc., are excluded. Provided the system under consideration is equipped with adequate circulation and insulation, these assumptions represent in many instances reasonably close approximittions for practical purposes. Analyzing the performance of a surface countercurrent heat exchanger of given dimensions and wsumed constant over-all heat transfer coefficient when rates of circulation and inlet, temperatures of the t,wo fluids are varied, Figure 1 illustrates the temperature conditions. The heat transfer through a surface element of the exchanger must equal the heat loss of the hotter fluid when passing over this element: - R c ~ d T j = (Ti - fi)UdS (1)
in
CIRCULATING SYSTEMS R. C. FISH,ER Head, Wrightson and Company, Ltd., London, England
-
Heat transmission problems in general engineering are most frequently con nected with steady-state conditions, in which temperature changes take place relative to position only, and not relative to time. There are, however, case8 where the computation of temperature variations over a given time is important. Formulas dealing with such conditions appear infrequently in technical literature, and designers often resort to approximations based on average conditions prevailing during the processing time. Examination of the heat transfer mechanism based on first principles has showii that, once adequate formulas have been developed which include a relatively small number of simplifying assumptions, it is possible to carry out concise numeric*al(.ompiitations which readily lead to practical remlts.
The heat lost by the hotter fluid, when passing between any point inside the heat exchanger and the hot-fluid outlet, must be equal to the heat absorbedby the colder fluid, when passing between the cold-fluid inlet and the same point inside the exchangw:
T
H E methods of calculation most readily available and commonly used in heat transfer problems are based on the assumption that a constant mean-temperature difference exists between the heat receiver and heat supplier. This is the case in most continuous operations. However, in a number of important industrial applications this assumption is not, justified; indeed the mean temperature difference is constantly changing from the beginning to the end of the operation. Exltmples of such applications are the heating or cooling of batches of liquids or solids by a heat supplier or remover. Particular cases selected at random from the petroleum and steel industries, respectively, are the heating of tanks containing fluids which are highly viscous at low temperatures such m lubricating oils, asphalts, etc., and the quenching of metal billets in liquid baths. For such conditions practical methods of computation are not readily available, and the formulas developed below are intended to meet this defidency. Although the numerical examples set out in this paper have been selected from cases arising in the petroleum and steel industries, the basic formulas are quite general and may be found useful, possibly with suitable variations, in similar types of heat transmission problems in other industries. I n the following ., I study it is assumed that: (a)The thermal conductivity and the shape of the solid b o d y u n d e r consideration are such that all its particles are at a uniform temperature at a given time. (b) The thermal conductivity of t h e f l u i d a n d arrangement of the Figure 1. Temperature Conditions bath are such that in a Surface Countemurrent Heat Exchanger at any given time all
(Ti - T’)RCR = ( t i - t )rC,
(2)
Combination of these two equations and integration gives the well-known relation for heat exchange with countercurrent flow. The amount of heat exchanged is:
H
=
RCR ( T
- t)$
(3)
This amount of heat is defined by the dimensions of the exchanger, the over-all heat transfer coefficient, and the rates of circulation of the fluids of given characteristics. From Equation 4 it will be noted that $ assumes an indeterminate value for RCR = rC,. However, mathematical analysis of Equation 4 shows that the true value of (b for this case is: $3-
1 RCR
lfm
(5)
The foregoing derivations are well known but have been included to facilitate the study of the subsequent extensions and developments. The arrangement of the system under consideration is shown in Figure 2. Assuming no vaporization of the fluids, t)hesensible heat content of the entire system must be constant; in other words, the sum of all variations in heat contents must be equal to zero:
+
BCedr f ACRdT BPR(T - t)$ de 0 (6) The first and second terms in Equation 6 represent the change of heat content of the body and of the bath; the third term defines the heat removed from the system by the heat exchanger. Furthermore, the heat lost or gained by the submerged body must be equal to the heat transferred through its surface: 939
Vol. 36, No. 10
INDUSTRIAL AND ENGINEERING CHEMISTRY
940
T=u+t . = , i t
Equation 7 can be rewritten: we obtain:
Substituting this value of ($3 in Equation ti and rearranging, u
dT Til -- = dT
where
+ 8) - r
- Bt
T-T
(9)
a = dt'E/RC'B
(10)
6 = (RCR/SshB)@
(11)
Equation 9 represents the connection between the temperature of the body, 7, and the temperature of the bath, T . In order to solve this linear first-order differential Equation 9, the following groups are introduced:
y = (1
from Equation 18 and intro-
dw/w
This standard form is integrated in the conventional way. For brevity, the substitutions according to Equations 14, 15, and 16 are introduced to give the final solution as:
(12)
w = T - f
v = (T
substituting in the above for ducing into Equation 8,
- Q / ( T - t) + B - a)/2
(13)
4 +$
(14)
Again, in the special case For which Equation 20 has been developed U"
1
P = +1+
=
0
To = t we obtain:
+ r/a -
1
(17)
P-q
Equation 9 is integrated by the following substitutions:
T = p + t
dT = d p
7 = w + t
d~ = dw
r - T
= w - p
This yields the homogeneous differential equation:
+ P)
-w dlL = Pi(1 .---___ dw w - p
Solving in the conventional way by writing p
.*. dp
= uw = vdw
+ wdv
we obtain:
For brevity, the Substitutions according to Equations 14, 15, 16, and 17 are introduced, and allocation of terms with the subscript zero to the beginning of the process results in the final equation: / u o - q / 8 + 0 . 6 /v - p/6-Q.b (19) w = wg /uQ - p/6-Q.K / u - q / 6 f Q . 6
It is of intere'st to consider under what conditions the temperature of the body will become equal to the inlet temperature of the cooling water, t. Obviously this can occur only when the bath temperature is also equal to t. Equation 13, v = ( T - t ) / ( r - t ) , assumes an indeterminate form for T = r = t, but the true value u for the equalization of body and bath temperatures can be obtained from Equation 20 which shows that T - t will become zero; Le., T = t for u = p . Since, however, the body can reach cooling medium temperature only if the bath has also reached the same lowest temperature in the system (i.e., when T = t ) , it follows that = t when both are equal to t. On the other hand, Equation 21 or 22 shows that equality between v and p , which is the condition for r = T = t, will occur at e = m. Consequently, complete equalization of all temperatures in the system takes pl+ce only after infinite time; that this does not coincide with practical experience is due to assumptions made to simplify the mathematical procedure. It is felt, however, that this discrepancy will not lead to any practical difficulties. Two specialized cases of practical interest will be considered. CASE1 is a cooling or heating system with no solid submerged in the bath-for example, any arrangement for heating 'or cooling the contents of a tank by external circulation through a heat exchanger. I n these circumstances B = 0, Equation 6 reduces to the form,
The vertical lines in Equation 19 represent the absolute or numerical value of the quantity enclosed-that is, the modulus of the quantity. I n the special case where To = t, according to Equation 13, uo = 0 and Equation 19 takes the form:
-
The groups ( T t ) and ( T O - t ) have been substituted for w and wg, respectively, according to Equation 12. To calculate the time element in the problem, Equation 8 must be integrated. Using the same substitutions as for integration of Equation 9;
BATH
HEAT EXCHANGER
Figure 2. Diagram of Arrangement of Surface Countercurrent Heat Exchanger
October, 1944 *
941
INDUSTRIAL AND ENGINEERING CHEMISTRY
ACE dT
+ RCR ( T - t)+dO = 0
(23)
=
OB=o
A R4
-q
p wo =
To-t /T --t\
--ln
+ riff -
a=--1
and Equation 7 disappears altogether. Integration of this simple expression yields, after some transformation:
70
-t
=
+
=
0
= -0.94
(17)
- 60 = 200
(12)
1560 - 60 = 1500 w =
CASE 2 is a cooling or heating system with no circulation through a heat exchanger-for example, in a quenching plant comprising merely a heat-insulated quenching tank. I n these circumstances R = 0, and Equation 6 reduces to the form:
B c ~ d r ACRdT
(-0.059) - (+0.677)
v o = - To 70
-t -t
7
-t
=
260
150 - 60 - o,06 1560 - 60-
(25)
Equation 7 remains unaltered, and the general correlation in this case is:
2.316 =
/V /V
- 0 . 677/Q.44
+ 0.059/'.44
This equation is probably best solved by trial and error, giving u = 0.341. This correlation is well known and appears as a special case of the general relation derived here.
y=-
T-t
= 0.341 =
7 - t
The use of the foregoing correlations can best be demonstrated by numerical examples. The constants and coefficients are afbitrary, since the assessment of their precise values (for example, that of the average heat transfer coefficients) is not included in this study.
TANK
BODY
A = 10,000 Ib. To = 150' F.
COOLINC SYSTBM
B = 20001b. TO = 1560O F.
R T
= 75001bJbr. = 3750Ib./hr.
0.1 1 50 X 5 '(-0.059)-0.677 0.06 - (-0.059) 0.341 - 0.677 0.06 - 0.677 0.341 - (-0.059)
I
6 = 1.98hr. = 1 hr. 59 min. (approx.) EXAMPLE I1
Conditions are identical to those stated in example I, except that there is no circulation and consequently no heat removal through an external cooler. When will the body temperature become 260' F.? The solution is as follows:
B .t.u. (Ib.) (" F.)
t = 60' F . Btu SB = 50 sq. ft. C E = 0.5 (Ib.) (" F.) B.t.u. ___ hB = 5 So = 100 sq. ft. (hr.) (aq. ft.) (" F.) u = 50 B.t.u. r = 260' F. (hr.) ( 8 9 . ft.) (' F.) Btu Cr = 1.0(lb.) (" F.)
C B = 0.1
Substituting values as defined in example I,
e = 10,000 X 0.5 50x5
1
mx In 260 - (25
The solution follows. Since in this case RCR = rC,: $=-=-
1 RCR
'+% a =
6
1 = 0.571 7500 X 0.5 1f50X100
A C R - 10,000X 0.5 = 25 - - BCB 2000 x 0.1
0.5717500 X 0.5 = 8,56 50 X 5
4- RCR SBhB
1 -
+0 -a -
.
~
E
T = 128' F.
- 2000 x
EXAMPLE I
A body with an initial temperature of 1560" F. is introduced into a quenching tank of given dirnensions. The cooling liquid in the tank is circulated through an external cooler of known capacity. What time will it take for the body to reach a temperature of 260' F? What, then, will the bath temperature be? The following data are given:
-. --6o 60'
260
_
1
_
+ 8.56 - 25 2 -
_
_
=
-7,72
p-T(-1+++;)= a
-1 + dr&:) s 25 2 (
=
-
+
&)
EXAMPLE 111
The liquid in a storage tank is to be heated with low-pressure steam by circulation through an external heat exchanger. What time will it take to heat the tank from 100' to 150" F.? The following data are given: ' TANX = 50 0001b. To lobo F. T = 150' F. B.t.u. CR 0.5Ob.) (O F.)
-
A
HEATINQ SYSTEM R = 80 000 lb./hr. s = 2ob sq. ft. B.t.u. loo(hr.) (sq. ft.) (' F.) t 25OOF.
'
-
-0.059
SOLUTION. As steam condenses at constant temperature, the result in the heat exchanger is the same as if the heating medium hadC, = a,
. .... -
-7.72
6 = 2.45hr. = 2 hr. 27 min. (approx.)
Latent heat = 946 B.t.u./lb.
a
25 ( - 1
(1560 - 150) + 25 1) - [1560 + (25 X150)l
=
+ 0.677
4
1
-e
loo x 200 - --RusC R = 1 - e - 80,000 X 0.5
= 0.393
(4)
942
INDUSTRIAL AND ENGINEERING CHEMISTRY =
A
-11~
RP
/To - t/ ___
IT - t /
H
=
90,000 /loo - 250/ X 0.393 In 1150 - 250/
s6,OOO = 0.644
Vol. 36, No. 10
(24)
hr. = 39 min. (approx.)
The foregoing leads to the conclusion that the problems covered by this study are rather complex because of the great number of variables and constants involved. Ext,ernal circumstances may impose considerable limitat,ions on the assumptions which the designer is free to make. Some points which can be decided upon only iii a particular problem are tmherelativib sizes of the cooling system, the volume of the bat'h, and the circulating rat,e. I n this connection, economic considerations will generally dictate the best design. It doe3 not seem possible to suggest a straightforward method of procedurr. which will lead t o an optimum solution; unless data from similar cases are available, it will be necessary for the designer to assume arbitrarily the free variables and compute several arrangements from which to choose the best. Fort,unately, however, the numerical computations, as set out in the examplw, iirc neithw involvtd tior t irnP consuming.
= heat transferred in exchanger, B.t.u./hr. hg = heat transfer coefficient from body to bath, B.t.u./(hq. ft.) (hr.) (', F,.) p = characteristic group defined by Equation 15 p = characteristic group defined by E uation 16 R = circulating rate of bath fluid, 1b.qhr. r = circulating rate of cooling fluid, Ib./hr. S = heat transfer surface in exchanger, sq. ft. SS = external surface of body, sq. ft. S , = heat transfer surface in cooler, sq. ft. T = inlet temperature of warmer fluid to exchanger, O F'. 2" = outlet temperature of warmer fluid from exchanger, O E'. t = inlet temperature of colder fluid to exchanger, ' F. t' = outlet temperature of colder fluid from exchanger, O P. T, = int:rmediate temperature of warmer fluid in exchanger, = T I = ti
v = a = fi = y
6
e
7
6 w
= = = = = =
F. intermediate temperature of colder fluid in exchanger, F'. over-all heat transfer coefficient in exchanger, B.t.u./ (sq. ft.) (hr.) (" F.) characteristic group defined by Equation 13 characteristic group defined by Equation 10 characteristic group defined by Equation 11 characteristic group defined by Equation 14 characteristic group defined by Equation 1 7 time, hr. temperature of body, F. characteristic group defined by Equation 1 characteristic group defined by Equation 12 O
.iCKNOWLEL)GMENT
In conclusion, the author wishes t o thank E'. ( h r r for useful suggestions on certain technical aspects of the problem, to H. Harper for his helo in the presentation of t,he material, and.to A. C. ~Iuellerfor checking the galley proofs. NOMRVCI, 4TI!RE = weight of h i t 1 1 fluid, Ih. R = weight of body, Ih. C B = specific heat of body, B.t.u., ( 1 1 ~ () " F.1 C R = specific hi5:it 01 h t h Huiti, H.t.u./[lh,) ( " b'.) C, = specific h c i t ot cooling fluid, B.t.ii,/(lb,) iob',)
-4
BIBLlOGRAPHY (1)
Howinan, iMueller, and Nagle, Trurts. Am. SOC..We&.
Enyrv.,
62, 283 (1940).
(2) Fishenden, Margaret, and Saunders, 0. A., "Calculation of Heat Transmission", London, H. Id. Stationery Office, 1932. (3) Grober, Heinrich, "Einfuhrung in die Lehre von der Warmeuber-
tragung", Berlin, Verlag von Juliuv Springer, 1926. (4) McAdams, W. H., "Heat Transmission", New York, MCGrawHill Book Co., 1933. (5) Schack, A. (tr. by Goldschmidt and Partridge), "Industrial Heat Transfer", New York. John Wiley & Sons. 1934.
Viscosities of Molten
COUMARONE-INDENE RESINS OMPARED to most of' the cununercially important poly-
C
meric substances, c'oumaroii[.-ind[~nercsins used in mastic flooring tile, in varnishes, and in rJxtending rubbers have low average molecular weights. Various investigators (6) have reported molecular weights of almost 4000, but most of the commercial products are included within the brief range from 550 to 800. Furthermore, it is well known (10, I d , 19) that many of the physics1 properties of these resins change sharply within t,his molecular weight range. The recent investigations of Flory' on the viscosities of molten linear polyesters ( 7 ) and of Kauzrnann and Eyring on the viscosities of linear hydrocarbons (9) have focused attention on the determination of molecular Ivtights by viscosit,y measurement,s made directly on the material without the use of a solvent,. Although there have been few other studies of the variation of the viscosity of molten synthetic- polyniers with niolecdar weight, Dunstan ( 5 ) proposed the formula: log 7 = .1.11 where 11 = Coefficient of viscosity M = molecular weight A , B = constant,s
+H
This equation was proved iiiralid by Albert for honiologous esters (8) over a narrow molecular weight range and by Flory ( 7 )
A. C. ZElTLEMOYER AND STEPHEN KUTOSH Lehigh University, Bethlehem, Pa.
for it wide molecular weight rang
