Water and Forced-Air Cooling of Vacuum Tubes Nonelectronic Problems in Electronic Tubes

Concentric Line with Circular Inner Conductor andSquare Outer Conductor

In Fig. 9(a), a vertical plane that divides the config-uration into its two mirror-image halves is an equipo-tential of potential zero. Accordingly, if we replacehalf the configuration by a plane conductor in thisneutral plane and set D = h, w= 2h, we get the con-figuration of Fig. 9(c), for which we have, from (17),

Zo= 138

7r2 h tanh-

2loglo

ira

If d is the inner diameter of the outer conductorand a is the outer radius of the inner conductor of theconventional concentric circular transmission line,

dZo = 138 logio-2a

Equations (23 and (24) are equal if

d = 1.079h.

I

E logiom=l

Kinsinh-2

1+ 1ycosh mir

sinh-12

1

sinh m7ri

(24)

(25)

(23)

For the same characteristic impedance the diameterof the circular conductor exceeds the side of the squareconductor by only 7.9 per cent.

Water and Forced-Air Cooling of Vacuum Tubes*Nonelectronic Problems in Electronic Tubes

I. E. MOUROMTSEFFt, ASSOCIATE, I.R.E.

Summary General laws of heat transfer from a hot wall to amoving fluid are applied to water and forced-air cooling of vacuumtubes. The calculated data are compared with experimental results.The practical importance of various factors constituting the mecha-nism of heat transfer is analyzed; the role of the internal structure ofthe tube on the dissipation limits is discussed generally. Rules fordesigning finned air coolers are outlined, and the 'optimum" designis discussed. Numerical examples are given. Some limiting factors incooler design are analyzed.

ITHOUT exaggeration one may state that indesigning electronic tubes there are many moremechanical, metallurgical, and heat engineer-

ing problems than those of pure electronic character.One may also admit that quite frequently nonelectronicproblems are solved by the cut-and-try method ratherthan by calculation. One of such questions is the anodecooling; in spite of the fact that water cooling ofvacuum tubes has found enormous application since1923, even now there is no complete agreement amongindividual experimenters regarding diverse factors in-volved in efficient cooling of the tubes. In this paperan attempt is made to give a common basis for analyz-ing and comparing practical results obtained by variousexperimenters. This is done by simply applying to ourspecific case data long since known in heat engineering.

* Decimal classification: R139. Original manuscript received bythe Institute, January 24, 1941; revised manuscript received,November 6, 1941. Presented, Fifteenth Annual Convention, Bos-ton, Massachusetts, June 28, 1940.

t Westinghouse Electric and Manufacturing Company, Bloom-field, New Jersey.

PART I-WATER COOLINGGenerally speaking, the efficiency of cooling of a hot

wall by moving fluid depends on the physical constantsof the fluid, its velocity, and the dimensions of thecooling arrangement. In vacuum tubes with externalanodes, cooling also depends,-in a degree not to beneglected,-on the anode wall thickness and the in-ternal tube structure, as these parameters determinepatterns of heat distribution throughout the anode.Heat generated on the inside surface of the anode flowsthrough its walls, is transferred to the moving liquidand carried by it away from the tube. From heat engi-neering it is known that all factors constituting themechanism of heat transfer from the wall to the liquidare connected by the following relation :1

(D) = o.24(DpV)08 (1)If the individual factors are measured in any con-

sistent system of units, the three parenthetic expres-sions in this equation are dimensionless. In our discus-sion centimeter-gram-second units are used, thereforethe meaning of the symbols is as follows:

h=rate of heat transfer in calories per second persquare centimeter per degree centigrade

1 William H. Adams, "Heat Engineering," McGraw-Hill BookCompany, Inc., NewYork, N.Y., and London, England, 1933, p. 165.

Proceedings of the I.R.E.190 April, 1942

Mouromtseff: Water and Forced-Air Cooling of Vacuum Tubes

k = thermal conductivity of fluid in calories persecond per centimeter per degree centigrade

p =density of fluid in grams per cubic centimeter,u= viscosity of fluid in grams per centimeter second

(poises)Cp=specific heat of fluid in calories per gram per de-

gree centigradeV=average velocity of the fluid in centimeters per

secondD =equivalent diameter of the cross section of the

fluid channel in centimetersIf h is known, the total anode dissipation can be

represented as the product of three quantities.Pi = S T. h calories per second = 4.2 S T. h watts (2,where S is the total heated area of the anode and T. i.the anode temperature above that of the cooling water

It is important to point out that (1) is applicableonly to cases of "turbulent" flow in contradistinction

Fig. 1-Type 891 water-cooled tube.

to the "parallel," "laminar," or "viscous" flow. In eachindividual case one can decide whether the flow isturbulent or viscous either by direct observation onthe transparent model of the jacket, or by using thewell-known criterion, the so-called, Reynolds numbergiven by

DpVRe =

/A(3)

One can notice at once that Re is nothing else but thesecond dimensionless member in (1). About 60 yearsago Osborn Reynolds established the fact that with

AUOO DIA3irTrII I.SSVt4 me)GROSS SrCTIOK OF ANNULARJACXeT OLSArIANCeS.9.0CMt

Fig. 2-Anode dimensions.

Re smaller than 2100 the flow is always viscous; withRe greater than 4000 it is turbulent. With intermediatevalues of Re the flow is unstable and may change fromturbulent to viscous and vice versa. This law holds forany fluid and it has been repeatedly verified by manyexperimenters.

In order to obtain a clearer picture of the amount ofcooling that can be achieved on vacuum tubes, andalso in order to study the role of the individual factorsinvolved in the cooling mechanism, we shall apply (1)to the well-known 891 or similar types of water-cooledtubes regarding which numerous experimental dataare available.The general view of this tube is represented in Fig. 1;

the principal dimensions of its anode and standardjacket are given in Fig. 2. The recommended rate ofwater flow is from 3 to 8 gallons per minute, and thesafe plate dissipation with 4 gallons per minute, and auniform heat distribution (such as is approximatelyrealized in class C operation) is 10 kilowatts; to this,1320 watts necessary for lighting the filament must beadded. Experimentally, it has been found that underthe specified conditions the water just begins to"hiss."'2' Direct observation through a glass waterjacket shows that at this point minute steam bubblesare formed at the anode surface and are carried awayby the water.

2 I. E. Mouromtseff and H. N. Kozanowski, 'Comparativeanalysis of water-cooled tubes as class B audio amplifiers,' PROC.1. R. E., vol. 23, pp. 1246-1248; October, 1935.

' I. E. Mouromtseff and H. N. Kozanowski, "Analysis of opera-tion of vacuum tubes as class C amplifiers," PROC. I. R. E., vol. 23,pp. 769-771; July, 1935.

191

Proceedings of the I.R.E.

In the following, by using (1) on the chosen example,we shall calculate h, the rate of heat transfer from theanode wall to the cooling water. However, in order tojustify the application of (1) we must first checkwhether the flow under the specified conditions isturbulent. The two physical constants affecting theReynolds number are water density p and viscosity,;both vary with temperature. We assume that the ini-tial water temperature is, ti=25 degrees centigrade;then, from the rate of water flow, 4 gallons per minute(254 cubic centimeters per second), and the figure ofdissipated power, 11.3 kilowatts, we shall determinethat the outgoing water will have temperature te = 37.5degrees centigrade. Hence, the average water tempera-ture tave= 31 degrees centigrade. From physical tablesit is found that at this temperature

p = 0.996 gram per cubic centimeter, = 0.0078 poise (gram per centimeter second). (4)

Water velocity calculated by division of the rate offlow, 254 cubic centimeters per second over the cross-section area of the flow, 2.98 square centimeters, is

V = 85 centimeters per second. (5)Finally, the equivalent diameter of the annular clear-ance between the anode and the jacket (Fig. 2), calcu-lated as explained below, is

D= 0.45 centimeter. (6)Using the numerical data one arrives at the Reynolds

numberRe = 0.45 X 85/0.778 X 10-2 = 4940. (7)

Hence, the flow is turbulent, and therefore (1) is ap-plicable to the case under investigation.As to the diameter De, by definition4, it is equal to

four times the hydraulic radius of the cross-sectionalarea of water flow, the hydraulic radius being definedas the ratio between the area and the wetted perimeterof the cross section. In the particular case of the an-nular cross section, the equivalent diameter is equal tothe difference of the jacket and the anode diameters, orto twice the spacing between the jacket and the anode,2 = 0.45 centimeter.

In order to find the numerical value of the rate ofheat transfer, in addition to the already determinedquantities, one must know heat conductivity k andspecific heat of water C,. From physical tables we have

k 0.00145 calorie per centimeter second perdegree centigrade

Cp 1 calorie per gram per degree centigrade. (8)Inserting (4) to (8) into (1) we finally obtain

h = 0. 1355 calorie per second per squarecentimeter per degree centigrade. (9)

4William H. Adams, see pp. 117 and 235 of footnote 1.

This is the rate of heat transfer from anode to waterin the case under discussion; more generally, the calcu-lated figure also shows the order of magnitude of heattransfer per unit area (1 square centimeter) of water-cooled anodes under more or less conventional condi-tions. When h and the total heated area of the anodesare known, one can calculate the anode temperaturesimply by using (2). In discussions similar to ours, it isgenerally more convenient for various theoretical der-ivations to consider a 1-centimeter long anode zone.From the total power of anode dissipation and thelength of its "hot" portion we find that dissipation perunit length of anode is6

m = 11,320/4.2 X (11.5 + 1.24)= 212 calories per second (10)

where 0.62 centimeter is added on each end of thefilament length to take care of the cooling effect of theanode ends; this will be explained in some later partof the paper. On the other hand, the amount of heattransferred to water within the same zone is

e = h X ird = 0.1355 X 47r= 1. 7 calories per second per degree centigrade. (11)

In the state of equilibrium the relation existsm = eTa. (12)

Therefore,Ta = 212/1.7 = 124.5 degrees centigrade. (13)

and the actual anode temperature above freezing pointis

Toa = 124.5 + 31 = 155 degrees centigrade. (14)This is considerably in excess of the boiling tempera-

ture of water. Some direct measurements made byvarious experimenters6 confirm that the anode tem-perature at sufficiently high dissipation is above theboiling point. Still, on first thought it seems to be some-what puzzling how water can be in contact with sucha hot surface and not boil. The explanation is to belooked for in the following facts: First, practicalmeasurements of permissible tube dissipation areusually carried out with the anode insulated from theground by "water coils" both on the inlet and outletsides. These coils are made of rather long pieces of rub-ber hose and incur considerable pressure drop whenwater is flowing; actually from 20 to 60 pounds waterpressure is usually required to pass water at the recom-mended rate through vacuum tubes. Thus, the pressureinside the jacket, which is in series with the two coilsand between them, may be as high as 30 pounds andmore. The boiling point at this condition reaches the

5 In this expression, the factor 1/4.2 converts watts into caloriesper second.

6 R. LeRossignal and E. W. Hall, "The development of largeradio transmitting valves," Gen. Elec. Co. Jour. (British), vol. 7,p. 185; August,_1936.

192 A pril


value of 135 degrees centigrade or more. Second,"spheroidal" state of water probably precedes the out-right boiling. Cold water coming in contact with a hotsurface forms an extremely thin film of superheatedsteam at the surface so that there may be no directcontact between the water and the anode wall; thisfilm is continuously destroyed and renewed due toturbulency of flow. An increase in heat flow above the"safe limit" results in more energetic formation ofsteam bubbles, that is, in boiling, which frequently canbe accompanied by mechanical vibration of the hoseand even of the tube itself. It is considered undesirableto permit water to boil as this usually results in scaleformation and finally anode puncture due to its localoverheating.Now, we shall more closely discuss the effect of

variation of the individual factors appearing in (1) andthe limitation imposed by them.

WATER VELOCITYThe most important factor in (1) is water velocity V,

because in practice it is the only factor which can bevaried by the operator at will and varied within ratherwide limits. In our example the recommended water-flow limits are from 3 to 8 gallons per minute. Accord-ingly, the heat-transfer figure theoretically can bevaried from

= h4(3/4)08 = 0. 108 calorie per second perto square centimeter per degree centigrade (15)

8= h4(8/4) 8 = 0.235 calorie per second persquare centimeter per degree centigrade (16)

hence, permissible dissipation should vary in the sameratio as heat-transfer figures, that is, proportionatelyto h3:h4 and h8:h4. So, for 3- and 8-gallon-per-minuterate of water flow the dissipation limit will be 8 and 16kilowatts, respectively. However, if one looks intothe published technical data of the tube, one will findthat there is no practical need in dissipation higherthan approximately 10 kilowatts because other limita-tions (in voltage, electronic emission) make it useless.Yet higher rates of water flow are necesssary becausein a great majority of cases of operation heat distribu-tion in the anode is nonuniform. Then, as is demon-strated elsewhere,7 dissipation limit may vary in pro-portion only to 0.4th power of water velocity insteadof 0.8th as implied by (1). This will be discussed lateron in connection with the effect of tube structuralparameters.

In this place, it may be repeated that the immediateeffect of an increase of water velocity is the intensifica-tion of turbulency of flow; a more turbulent flow makesa greater inroad into the thin stationary film of water(or steam) ever present at the anode wall, and thus

I I. E. Mouromtseff, "The influence of grid focusing effect onplate dissipation limit of a vacuum tube," Communications, vol.20, p. 11; December, 1938.

increases heat exchange between the wall and thewater. The existence of slowly moving or stationarylayers of liquid in the vicinity to the wall even in aturbulent flow has been established by direct measure-ments by many experimenters.8

It is obvious that turbulency of the flow can increasenot only due to increased water velocity but also dueto the formation of steam bubbles. Therefore, thehigher the anode temperature, the greater is the rateof heat transfer. However, for the reasons given beforeit is not desirable to carry this too far and to permitwater to boil. It is also logical to admit that the turbu-lency of flow can be also intensified by properly de-signed baffling surfaces built in the path of the water,or by other "mechanical" means.

CLEARANCE BETWEEN ANODE AND JACKETWith a fixed rate of water flow Qa cubic centimeters

per second, variation of the annular clearance e be-tween the anode and the jacket affects two factors:velocity of water V and the equivalent diameter De.Each of these factors independently affects turbulencyas may be seen from (3). However, for a given anodesize the Reynolds number is practically independent ofthe width of clearance (if clearance is generally smallcompared to the anode diameter da). Indeed, watervelocity increases with decreasing clearance practicallyin inverse proportion, while the equivalent diameter ofthe annular cross section decreases in direct proportionto it; hence, the net effect is almost nil. In our numericalexample E= 0.0885 inch and the Reynolds number,Re = 4940. One can show that even with an infinitesimalspacing (e =nearly 0) the Reynolds number will in-crease only slightly. In this limiting case

2 X 254Reo = 2eQW/crda-=r = 5200.

,r X 4 X 0.0078(17)

In other words, with a large or a small clearance,turbulency stays the same if the number of gallons perminute is maintained the same. This, however, doesnot mean that the heat transfer h remains the same.By combining similar factors of different dimensionlessgroups in (1) one will find that h is proportional to the0.8th power of velocity and inversely proportional toD02. Hence, for smaller clearance heat transfer be-comes larger, mainly, due to an increased velocity.Experiments show that improvement in cooling doesnot quite follow the expected law, when the clearancebecomes very small. In fact, it has been found that foreach particular set of operating conditions there is anoptimum value of clearance e, such that either forlarger or smaller clearances the efficiency of coolingdrops. The experiments on a standard tube with a 12-inch anode showed that the optimum clearance variedfrom 0.020 to 0.015 inch depending on the rate of waterflow. Of course, in ordinary designs of water-cooled

8 William H. Adams, see p. 106 of footnote 1.

1942 193


tubes these extremely small clearances can be con-sidered as impracticable. The reason for existence of theoptimum clearance must be looked for in the fact, men-tioned previously, of existence of almost stationaryfilms of liquid at the walls. When the thickness of thewater wall becomes comparable with the thickness ofthese stagnant film,s the turbulency decreases and theflow may even become laminar, in spite of a highervelocity.

Fig. 3-Factor A versus fluid temperature.Factor A shows combined effect of physical constants

on rate of heat transferv0.8 p"0CO.B 4 VO.8h= 0.024-d - 0-=0.024- A cal/sec. cm2/0cd5.2 JA. d

PHYSICAL CONSTANTS OF LIQUIDFrom the four constants appearing in (1), water

density p and specific heat Cp can be consideredpractically independent of water temperature; eitherof these constants is equal to unity. The other twoconstants, heat conductivity k and viscosity,I, varywith temperature. Their over-all effect on rate of heattransfer is shown in Fig. 3, upper curve.

In order to illustrate the discussed relation better, weshall assume that with the same power dissipation,11,320 watts, the temperature of the incoming water is7 degrees centigrade. The values of k and Iu to be usedin (1) are

k = 0.00136 calorie per centimeter per second perdegree centigrade.

,u = 0.0124 poise. (18)

The calculated heat-transfer figure is, then

h = 0. 1075 calorie per second per square

centimeter per degree centigrade (19)

and the anode temperature

Ta= 169 degrees centigrade. (20)

Thus, we arrive at an apparent paradox: in spite ofmuch colder water (7 instead of 25 degrees centigrade)the anode temperature is higher by 14 degrees centi-grade. Moreover, due to an increased viscosity, theReynolds number in the case of the 7-degree-centigradewater becomes practically nonturbulent. This willcause a further reduction in the rate of heat transfer. Nowonder that in this condition water at the anode sur-face may start boiling, and one will be forced to limitthe permissible dissipation.

In addition to the temperature sensitivity, viscosityof water may also vary with pressure; and since thepressure in the water jacket may vary from case tocase to a considerable degree, the cooling effect maybe somewhat different in spite of the apparentsimilarity of other conditions in various experiments.The role of the physical constants becomes particu-

larly important if one decides to use liquids other thanwater, for example, transformer oil, kerosene, or one ofthe known antifreeze solutions. This may be war-ranted by superior insulating properties of oils, or bythe necessity to install a transmitter in a room withoutheating. Table I gives constants for the two mostpopular antifreeze liquids, prestone and alcohol. Fromthe table one can see that except for density of pres-tone the constants of both solutions differ from thoseof water so that all of them tend to reduce the rate ofheat transfer. Particularly important is viscosity whichwith solutions with medium content of antifreeze isfrom 2 to 3 times higher than with water. This meansthat in order to have the same turbulence as withwater the rate of the flow must be doubled or trebled.

TABLE IPHYSICAL CONSTANTS OF COOLING LIQUIDS

Mixture Boiling Freezing Density Viscosity Specific Heat, Cp Heat Conductivity, k% POint POint 300C 10c OOC 300C 10C 0C 300C 10C OOC 300C 10C OOCgm/Cm1 centipoises cal/gm/1C cal/sec/cm/10CWater 100 100 0 1 1 1 0.8 1.35 1.8 1.0 1.0 1.0 0.00145 0.00138 0.00132Prestone 10 101 -4 1.010 1.014 1.016 1.0 1.75 2.5 0.99 0.99 0.99

30 102 -16 1.034 1.042 1.046 1.75 3.1 4.5 0.90 0.88 0.87 (0.001)50 106 -37 1.058 1.07 1.075 3.0 5.5 8.0 0.83 0.79 0.79 0.000635

Alcohol 10 93.5 -5 0.979 0.984 1.16 2.18 3.31 1.0 1.0 1.0 0.00130 0.00122 0.0011640 83.5 -29 0.928 0.942 2.02 4.39 7.14 0.92 0.91 0.90 0.00093 0.00088 0.000.85100 80 0.78 0.80 0.81 1.00 1.47 1.77 0.64 0.60 0.54 0.000433 0.000433 0.00433

AwWATE

'5

AW1400TOI 00.10 A0

0 < 401A ALCOHOL

.04 l AIR

.0e

194 A pril

w4004L..


If necessary to choose from the two solutions, prestonemust be preferred as alcohol easily evaporates at com-paratively low temperatures, and its constants are sum-marily less favorable for heat transfer than those ofprestone. The over-all effect of prestone and alcoholfor low temperatures is plotted in Fig. 3.The physical constants of oils, generally speaking,

are also such that the heat-transfer figure for oils is lessthan for water. Therefore with the same dissipationthe anode temperature is always higher for oil than forwater. With the temperature approaching 200 degreescentigrade oil begins to "sludge"; this contaminates theanode surface and consequently the rate of heat trans-fer drops markedly from its initial value. Nevertheless,oils possess one indispensable advantage, high in-sulating property which makes long insulating watercoils unnecessary. Therefore, in cases when dissipationis not too high, or where perfect insulation is of primeimportance (as for example in X-ray tubes), oil coolingmay become justifiable. One of the best liquids of thiskind seems to be kerosene.

THE ROLE OF THE VACUUM-TUBE STRUCTUREIf the anode of a vacuum tube is heated perfectly

uniformly, the anode diameter d and the cathode lengthL are the only structural parameters determining thedissipation limit. If the permissible maximum tempera-ture is Tinax, the dissipation limit is

Po = rd L h Tmax (21)where h is heat-transfer figure to be calculated from (1)and Tmax, is the anode temperature above that of thecooling water. However, it is known that due to thefocusing effect of the grid in many cases of operationheat is generated in the anode nonuniformly; heat pat-terns depend on grid meshes and on the location of in-dividual filament strands with respect to the gridstays. As a result, the average anode temperatureTae, cannot be allowed to reach the permissible maxi-mum; hence, the anode dissipation limit becomes lessthan with a uniform heat distribution. Usually closelyspaced grid windings of thin wire affects but very littlethe temperature distribution; the entire blame fornonuniformity is to be put on the heavy longitudinalgrid stays. It has been shown9 that in the typicalcase of electron beams concentrated along the mid-linebetween the shadows of adjacent grid stays (points A,B, C, D in Fig. 4) the average anode temperature isgiven by the expression

Tave = Tmax(tanh Q)/Q. (22)The temperature Tma. prevails at the points, A, B, C,and D, where heat is generated by impinging electrons;quantity Q depends on heat-transfer figure h and thetube structural parameters in the following way:

Q = I/5kb X7rd/2n. (23)9 I. E. Mouromtseff, see p. 12 of footnote 7.

Here d =anode diameter (4 centimeters)= anode wall thickness (0.159 centimeter= 1/16 inch)

n =number of grid stays (4)k=heat conductivity of copper (0.9 calorie

per centimeter per second per degree centi-grade)

h=heat-transfer figure (0.1355 calorie persquare centimeter per second per degreecentigrade)

Fig. 4-Focusing effect of grid stays in the 891 tube.Accordingly, if there is grid focusing effect, the anode

dissipation limit will be reduced in proportion toTave/Tmax = (tanh Q)/Q. (24)

Using numerical values in parenthesis, which are thefigures of our practical example of the 891 tube, onewill find that

Q = 1.53 and tanh Q=0.91here

(25)

tanh Q/Q = 0. 595. (26)This means that because of the grid focusing effect thepermissible plate dissipation is to be reduced by 40.5per cent. The reduction can be even greater if the fila-ment strands are not symmetrically located with re-spect to the grid "windows."

Close examination of (22) and (23) reveals thatvariation of any factor in (23) should effect a simul-taneous increase or decrease of both the numerator anddenominator of (22). However, the hyperbolic tangentof a function changes more slowly than the functionitself. Therefore, the net change of TfI/Tav is governedby the variation of quantity Q in the denominator. Thisis especially true if the value of tanh Q is nearly unity(which is the case in our example). Thus, one may rulethat in order to decrease the nonuniformity of tem-perature distribution resulting from the grid focusingeffect one has to increase either the anode wall thick-ness a or the number of grid stays n. The latter state-ment must be supplemented by the consideration ofthe cathode structure; it must be disposed of in sucha manner that all grid "windows" are uniformly utilizedby the electrons. In addition, one may also notethat the larger the anode diameter, the greater is

1942 195


temperature nonuniformity with the same number ofgrid stays. As to the effect of heat-transfer figure h, onearrives at the conclusion that its increase amplifies thenonuniformity of temperature distribution; and, sincein a given arrangement one can change h by changingwater velocity, one must realize that the faster thewater is passed through the tube the more pronouncedare heat spots on the anode. Nevertheless, in spite ofthis the total power dissipation increases generally withthe rate of water flow. This follows from the expressionfor total heat dissipation

Pf total = rd I h Tave (27)or

Pf total = 2n Izbk tanh (7hS/kX rd/2n). (27a)With water-cooled tubes of conventional designs the

hyperbolic tangent in this expression, as already men-tioned, is usually very close to unity; therefore the dis-sipation limit in operation with pronounced gridfocusing effect increases proportionally to the squareroot of the heat-transfer figure, or proportionally to the0.4th power of water velocity V.

In connection with (21) and (10) the question maybe asked: "Is the length of the anode hot portionexactly equal to that of the filament structure, or is itlonger due to the heat spreading throughout the 'coldends' ?" A theoretical treatment of this problem isgiven elsewhere ;10 in application to our numericalexample it can be shown that the contribution of thecold ends is equivalent to an increase of the heatedportion by approximately 0.6 centimeter at each end.

PART II-FORCED-AIR COOLINGINTRODUCTORY NOTES

In recent years, approximately since 1935, water-cooled tubes in a number of new transmitters built inthis country and, later on, in Europe were supplantedby tubes with forced-air cooling. Air-cooled tubes canbe advantageously used in all cases when there is adanger of ambient temperature dropping below thefreezing point. In some other cases they can be fur-nished with inexpensive individual blowers, thus elimi-nating the necessity of water piping and of the use ofwater coils insulating the high-voltage water-cooledanodes from the ground.

Air cooling, the same as water cooling, is not a newengineering problem; yet its application to high-powervacuum tubes is a new and sufficiently specific problemto warrant its careful study both theoretically andexperimentally in order to establish general rules forthe systematic design of air coolers. The problem of de-signing an efficient air cooler for a given type of vacuumtube with an external anode may be formulated inseveral different ways. Thus, one may wish to design

10 I. E. Mouromtseff, "Temperature distribution in vacuumtube coolers with forced air cooling," Jour. Appl. Phys., vol. 12,pp. 491-497; June, 1941.

a cooler of the smallest mechanical dimensions for adefinite maximum power dissipation. Or vice versa, onemay start with the largest permissible diameter of acooler and calculate the feasible maximum dissipation.One may also need a cooler representing not morethan a certain specified resistance pressure, in order toemploy a distinct type of air blower available on themarket. In the case of a portable air cooler its weightbecomes an important factor. Finally, the cost of theindividual coolers may influence the choice of the de-sign. One should note that in all cases the maximumpermissible anode temperature is a critical factor, as itdirectly affects the dissipation ratings. However, thistemperature cannot be specified once and forever andfor all designs, because it depends on physical proper-ties of materials involved in the design and on thedegree of outgassing vacuum-tube parts. Obviously,there is no single solution of the postulated problem ingeneral, but for each particular set of requirements onecan find the best answer or an optimum design for thecooler.

It may be of interest to mention that as far back as1931 an air-cooling jacket for the 863 tube (closely re-sembling the 891), was designed and tested at EastPittsburgh; it was intended for a 5-kilowatt trans-mitter. This project is interesting because the air coolerformed a part of the transmitter equipment, and thetube could be replaced in it as in an ordinary waterjacket. In a few words, this air jacket can be describedas follows: It was machined from a 32-inch aluminumbar; its total weight was less than 4 pounds. The cen-tral bore was drilled to fit the 1-inch tube anode. Anumber of longitudinal slots was milled on the outsideof the cylindrical bar. Due to a longitudinal cut, thejacket could be easily brought about and clampedtightly around the anode. Cooling air was supplied ata rate of 90 cubic feet per minute from the source ofabout 0.4 pound pressure. The described jacket per-mitted approximately 2.5 kilowatts total anode dis-sipation with the anode temperature at about 160degrees centigrade. For protection from oxidation theanode of the tube was gold-plated. The device did notfind much practical application as the transmitter de-signers needed greater anode dissipation if forced-aircooling had to compete with water cooling.

In this part of the paper general principles of de-signing air coolers are, first, outlined; then, the princi-ples are applied to the calculation of a cooler for one ofthe popular types of tubes; finally, various factors in-fluencing cooler designs are discussed.

GENERAL PRINCIPLESA conventional air cooler consists of a core with a

central bore for the tube and of a set of vertical finsextending in radial direction and secured to the outersurfaces of the core. All parts are usually made ofcopper because of its good thermal conductivity. Ageneral view of an 891 tube soldered in an air cooler is

196 A pril


shown in Fig. 5; its horizontal and vertical cross sec-tions are schematically represented in Fig. 6.

Generally speaking, the amount of power dissipationby a cooler Ph is proportional to each of the followingthree factors:

1. Average temperature difference between fins andair, T'ave.

2. Rate of heat transfer from copper to air, h caloriesper second per square centimeter per degree centigrade.

3. Total area of the cooling surface, S square centi-meters.

In short, the power dissipated isPh = Tf ave X h X S calories per second. (28)

It can be computed by summing up individual tem-perature differences in the path of heat flow from thetube to the cooling air; to this sum the ambient tem-perature must, of course, be added. Obviously, at notime and at no place can the actual anode temperaturebe permitted to exceed the safe maximum anodetemperature Ta max. As such, one can designate 20 or30 degrees centigrade below the dangerous temperatureat which either the solder bond between the tube and

/@ \ FOR OPTIMUM DIAMETERSO,_ 2CC511k

Fig. 6o{ptimum design of 891R and similar air coolers.the cooler can be weakened, or the vacuum inside thetube becomes affected by gas liberated from overheatedtube parts. Thus for example, with pure tin solder, 160degrees centigrade can be specified as T. max, because at180 degrees centigrade tin begins to soften; if heated tothis level even occasionally, the solder finally melts atsome points and causes a puncture of the anode wall.With normally exhausted tubes it appears safe toadopt 230 degrees centigrade as the anode temperaturelimit; in such cases, of course, a solder with a highermelting point has to be used; cadmium proved to bevery satisfactory for this purpose.The second factor in (28), the rate of heat transfer

h can be calculated from the expression similar to (1),in which physical constants for water are substitutedfor by similar constants for air

Fig. S-Type 891 tube adapted for air cooling.

For a given ambient temperature ta, the largest pos-sible value of fin temperature Tf,n is restricted by themaximum permissible anode temperature T. ma., themajor part of which is constituted by fin temperature.The actual anode temperature in operation establishesitself as function of the power dissipated in the cooler.

(29)k D V)O__ Cp)0 4

By inspection of this expression one may concludethat the only factor in the designer's hands for control-ling the rate of heat transfer is air velocity. The higherthe velocity the more efficient is the cooling. Its upperlimit may be set either by an objectionable whistlingnoise produced by air escaping from the ducts, or, more

1971942


likely, by a rapidly increasing pressure required toforce the air through the cooler; this pressure increasesas the square of velocity. An excessive pressure maypreclude the use of certain types of good commercialblowers available on the market. In addition, powerrequired for driving the blower increases proportionallyto the cube of the air velocity. One may assume that3000 centimeters per second (6000 feet per minute) isthe highest velocity to be recommended for practicalapplications.

Equivalent diameter of the ducts D in (29) is calcu-lated as four times the ratio of the duct cross-sectionalarea to its perimeter. Its variation from cooler to coolerhas not much influence on the rate of heat transfer as itenters in (29) only with an exponent of 0.2.

Individual physical constants appearing in the sameequation vary with air temperature. However, ifgrouped together, they form a factor whose value re-mains almost constant within a wide range of tempera-tures. This factor Aa is plotted in Fig. 3 as a functionof air temperature; it rises slightly toward the freezingpoint; otherwise for all practical cases it can be as-sumed approximately 0.0215 X 10-2, so that (29) inapplications to air can be modified as follows:

VO.8 V0.8h-0.024 A=5.15X10-6 (30)

DO.2 DO.2

Note: The dimension of the factor A is colories persecond0 2 per centimeter26 per degree centigrade.As in most of the practical designs, D0O2 is not too

much different from unity; the later formula can besimplified still further to read

h = 5.15 X 10-6 V0.8. (31)As to the third factor in (28), the total cooling area

of fins S is one of the important factors determiningthe merits of a cooler design. The larger it is the better.However, once the size of the cooler is chosen, there isa definite maximum cooling area inherent in the design:it depends on mechanical limitations of the coolerstructure. These are the minimum fin thickness a andthe minimum spacing t between the fins at their root,that is, at the core surface. Thus, independent of thegeneral design of the cooler, the smallest fin pitcharound the periphery of the core is (6+t); therefore,with the core diameter D, the largest feasible numberof fins is

nmax 7rDc/(b + t). (32)If, then, Df is the cooler diameter, the total cooling

area of the fins isS = 2(Df- De) X 7rlD,/(b + t).

Since the fin length is usually fixed by the tube le1, the feasible maximum of S can be found from

ais-

= 0.aDC

Solution of this equation givesD= Df/2. (35)

This holds for any fixed figure of fin pitch. As to therole of the over-all cooler diameter Df it may be con-cluded a priori that the larger the diameter, the morepower can be taken care of by the cooler, all otherconditions being equal. However, at the end of thissection we shall more closely enter upon the optimumradial extension of fins. One may generally note thatdue to a great difference in values of factor A shown inFig. 3, the total cooling area of an air cooler must bemuch larger than in a water jacket, if anode dissipationof the same order of magnitude is to be taken care of inboth cases. In a practical design the size of the cooleris limited by the general design of the transmitter andits radio-frequency circuit.

NUMERICAL EXAMPLE OF COOLER DESIGNIn this section a method of calculation of an air

cooler is outlined on the example of the 891-R andsimilar tubes.

For convenience of numerical calculation we firstassume that power dissipation per centimeter lengthof the "hot" portion of the anode (Fig. 6) is constantand equal to 1000 watts, which corresponds to thetotal anode dissipation 11,500 watts. With this as-sumption we calculate maximum anode temperaturefor air flow rate of 1500 centimeters per second. (3000feet per minute). Then, in view of proportionalitybetween power dissipation and anode temperatureabove the ambient, the permissible dissipation for eachindividual rate of air flow is obtained by setting theanode temperature limit for example at 230 degreescentigrade and by reducing or increasing the initiallyassumed power dissipation in proportion to the newanode temperature. It is convenient to refer all calcu-lations to the cooler zone of unit length.Taking into consideration the discussion of the previ-

ous section and with reference to Fig. 6, we shall adoptthe following structural data in Table II for the cooler:

TABLE II

Inches Centimeters

Df =outer fin diameter 7.5 19.1Dc =outer core diameter 3.75 9.53d, =inner core diameter 1.63 4. 146 =fin thickness 0.0625 0.159t=fin spacing at root 0.075 0.191T =fin spacing at free end 0.150 0.381

tave =fin spacing, average 0.1125 0.286w =fin width (in radial direction) 1.884.781, =fin length (in axial direction) 9.022.8n =fin number 86

L' length of lower cold end 2.70 6.85L" =length of upper cold end 1.72 4.38

L +L' =length of heated portion 4.53 11.50

From these data one can compute several quantitiesnecessary for further discussion. These areTotal "heated" perimeter of air ducts:

p = 2nw = 820 centimeters.(34) Total cross-sectional area of air ducts:

(36)

A pril198


A -D D2(l - 1/4) - nbw4

= 150 square centimeters.Hydraulic radius of individual ducts:

rh = A/p = 0.18 centimeter.Equivalent diameter of individual ducts:

D = 4rh = 0. 72 centimeter.Total cooling area of fins in a unit zone:

S = 2nw = 820 square centimeters.Rate of air flow though the cooler:

qa= vA = 1500 X 150= 225,000 cubic centimeters per second= 475 cubic feet per minute.

one can determine the average fin temperature Tf aboveair from the equilibrium condition:

(37) m = ef X Tf ave (46)Tf ave = 153 degrees centigrade.

(38) Due to the radial heat flow through the fins, tem-perature is maximum at the fin roots and graduallyfalls toward their free ends (see Fig. 7). The tempera-

(39) ture distribution in the fin is given in the law derivedelsewhere 11

(40)

(41)The physical constants for air at 60 degrees centigradeare:

p = 1.06 X 10-3 gram per cubic centimeter is the airdensity.

k = 6.25 X 10-5 calorie per second per centimeterper degree centigrade is the thermal conduc-tivity of air.

C,,=0.24 calorie per gram per degree centigrade isthe specific heat of air.

=1.95 X 10-4 gram per second per centimeter isthe air viscosity.

As indicated in the previous section, instead of usingindividual physical constants, one can convenientlyoperate with the factor A a plotted in Fig. 3. The esti-mated average air temperature will be 60 degreescentigrade if we assume an ambient temperature of45 degrees centigrade.

Before proceeding with further calculation we shallcheck whether under the specified conditions the airflow is turbulent. The Reynold's number is

Re = Dpv/,= 5860. (42)Hence, the flow is turbulent, and for the calculation ofthe rate of heat transfer from fins to air, we can applyeither of the equations (29), (30), or (31). We obtain,first

h = 0.00193 calorie per second per squarecentimeter per degree centigrade. (43)

From this figure and from the total cooling area onewill find the total rate of heat transfer for the entireunit zone:

ef =-S = 1. 57 calories per second perdegree centigrade. (44)

Remembering that heat generated within the samezone is

m = 1000 watts -240 calories per second (45)

tz = Tf ave cosh (Qf- qf1,)/cosh Qf. (47)The relation between the average fin temperature,Tf ave and Tf max (which at the same time is the tem-perature of the core, T,), according to the same deriva-tion (7), is:

Tf ave= Tfn.ax(tanh Qf/QQf (48)or

Tf ave = T,(tanh Qf)/Qf. (49)In the last three expressions, Qf, as shown below, de-pends on rate of heat transfer h; fin thickness 5; theirlength w; and heat conductivity of copper kc= 0.9calorie per second per centimeter per degree centigrade.

Qf = qw = w /2h/8kc. (50)Using the numerical values from the previous dis-

cussion one obtainsqf= 0. 164 centimeter-'Qf = 0.784

tanh Qf= 0.653(tanh Qf)/Qf = 0. 833

(51)

(52)

and finally,Tfmax= 153/0.833 = 184 degrees centigrade. (53)

In order to determine the anode temperature Ta, wemust yet find the temperature drop in the core AT,,, andin the layer of solder AT8. The thickness of solder weassume to be 1 millimeter (0.040 inch). Using dimen-sions given in the beginning of this section we find

Tc = m X 1/27rkc X ln(Dc/dc)= 35 degrees centigrade

T= n X 0. 1/7rdck8 = 12.3 degrees centigrade.

(54)(55)

Here k8-=0.225 designates thermal conductivity ofcadmium solder.

In addition, we must check the average air tempera-ture At aave, above ambient. This will be found fromthe total dissipation Ph=2760 calories per second andthe rate of air flow qa= 225,000 cubic centimeters persecond.

Atave = Ph/2qapCp = 24 degrees centigrade. (56)11 See equation (1) on p. 491 of footnote 10.

1942 199


Finally, the anode temperature will be determinedby simple addition of all calculated partial temperaturedrops in various parts of the cooling system:Ta = 45 degrees centigrade + Atave + Tf max + ATC + AT,,

= 45 + 24 + 184 + 35 + 12.3= 300 degrees centigrade. (57)This is in excess of Ta max = 230 degrees centigrade

which we adopted as the permissible maximum anodetemperature at any point. However, the calculatedtemperature, 300 degrees centigrade, corresponds to asituation seldom encountered in practice and certainlynot in our concrete example. Indeed, it is calculated insupposition that all heat generated in any unit zone ofthe cooler is transferred to the cooling air within thesame zone; in other words, thus far in the calculationwe neglected longitudinal heat flow through the corefrom the directly heated portion to its cold ends and,hence, we neglected the contribution to the heat dis-sipation by the cooler's "cold ends." Logically, one canpredict that the heavier the core the more substantialis this help. Mathematical treatment of this problemis given elsewhere.10 Briefly, it can be summarized asfollows: The effect of "cold ends" is manifested, first,in a general reduction of temperature along the entireheated portion of the cooler. In addition, from themiddle of the heated portion the temperature de-creases toward the cold ends. In a symmetrical struc-ture the maximum temperature occurs exactly at thegeometrical middle. With unequal lengths of ends themaximum moves toward the shorter end; its exactlocation can be established by calculat on. Withoutreproducing the theoretical derivation we shall usehere only its final results. The important maximumtemperature at about the middle of the directly heatedportion of the core (omitting temporarily the consid-eration of the radial temperature drop in the core) isgenerally given by:

with ac designating the annular cross section of thecore; L, the length of the heated portion between thepoints with temperature maximum and tb, and kc, heatconductivity of copper.The boundary temperature for each end is given by

tb' = m/e, X 1/(1 + tanh Q'/tanh Q) (62)and

tb' = m/ec X 1/(1 + tanh Q"/tanh Q)Here,

QI = qL' and Q" = aL" (63)with q the same as per (61) and L' and L" representingthe lengths of the two cold ends. Obviously, maximumtemperature Tc max calculated for both ends of thecore must be the same.By a trial calculation one can easily establish that

the maximum core temperature lies at a distance of 6.2centimeters from the longer and 5.3 centimeters fromthe shorter cold end. Then, inserting the available datainto (62) and (58) we obtain

tb = 91.3 degrees centigradetb = 99.8 degrees centigrade (64)

andTcmax= 124. 25 degrees centigrade.

This temperature is to be compared to 184 degreescentigrade of (52). Obviously, the actual heat flow ina radial direction through the unit zone comprisingTC max is less than the initially assumed figure, 240calories per second, in proportion to the calculatedtemperatures:

fmax = 240 X 124.5/184 = 162.5 calories per second. (65)This permits us to compute the actual maximum

radial temperature drop across the core at the hottestcross section; it will be reduced from the former figure(equation (53)) in the same proportion:

Tc max = m/e, - (m/e,- tb)/cosh Q.Here, quantity m/ec is the core temperature cal-

cated in the earlier assumption that there is no longi-tudinal temperature drop in the core (no cold ends).m, as before, is heat generated per unit length.ec is the rate of heat transfer for the entire unit

zone, referred to the outer surface of the core perunit zone. It can be determined from (44) and(46) and the condition of continuity of heat flow:

e,T, = efTf (59)tb is the temperature at the boundary between the

"heated" and "cold" portions of the core.Q is the quantity similar to Qf of (50) pertaining

however, to the heated portion of the core; viz,Q = qL (60)

andq = Ve,/clack (61)

(58) Tc max = 35 X 124.5/184 = 23.6 degrees centigrade. (66)As to the temperature drop through the solder wallone must assume that it stays unaffected by the de-rived distribution of heat flow; indeed, heat generationin the anode is unaffected by it, and one may assumethat longitudinal heat equalization takes place entirelyin the core, as the anode wall is relatively thin. Also,unaffected by the longitudinal temperature distribution,air temperature above the ambient remains as it de-pends only on the total heat dissipation. Thus, the cor-rect maximum anode temperature is

Tamax =45 + 24 + 124.25 + 23.6 + 12.3= 230 degrees centigrade. (67)

This temperature is exactly equal to the establishedpermissible Ta max. Hence, the assumed total dissipa-tion=11.5 kilowatts is the correct maximum permis-sible dissipation for the designed cooler under the

200 A pril


assumed operating conditions. One may note that thecalculated power-dissipation figure equals the limitusually specified for the 891 tube with water cooling.This indicates that, if necessary, air coolers can be de-signed so that practically the same dissipation ratingscan be applied to a tube as with water cooling. Thecalculated radial temperature distribution within thehottest section of the cooler is represented graphicallyin Fig. 7.

It may be of interest to note that in a cooler withno cold ends, the permissible dissipation should havebeen reduced from the originally assumed total dissi-pation, 11,500watts,in proportion to the maximum per-missible anode temperature and the calculated anodetemperature above ambient, that is,

Ph max = 11.5 X (230 - 45)/(295 - 45)= 8.3 kilowatts. (68)

The corresponding cooler would have a length exactlyequal to that of the heated portion of the anode. Ob-viously, for efficient cooling such a structure normallyshould be avoided; however, one may be forced into itif the filament structure occupies practically the entirelength of the anode.

Fig. 7-Radial temperature distribution in the891R tube cooler.

AERODYNAMICAL PROPERTIES OF THIE AIR COOLERIn the design of a water jacket for vacuum tubes one

does not give much consideration to the pressure dropin the jacket as this normally is only a small portion ofthe total pressure generally available at the stationand necessary for forcing water through long insulating

water coils. For example, with 3 gallons per minuterecommended for the 891 and similar tubes, the pres-sure drop across the jacket is only 2 pounds per squareinch, while the available pressure is seldom less than25 pounds per square inch. Things are quite differentwith air coolers. The resistance pressure of a cooler andthe rate of air flow inherently determine the kind of

Fig. 8-Fanning friction factor versus Reynolds number.

blower and power necessary for forcing the air throughthe cooler. For a given cooler design as mentioned be-fore, air pressure required from a blower increases inproportion to the square and power to the cube of airvelocity.

Resistance pressure of the jacket proper consists offriction loss in the ducts Hf and of contraction loss Hcdue to a sudden change of the cross section of the airflow at the entrance into the cooler; adding to thisthe velocity head HI, one obtains the total pressureHtotai necessary to propel air through the jacket withthe desired velocity, v centimeters per second.

Friction loss is given by the expression"2Hf = flpv2/2gri.

Contraction loss is 13H =kpv2/2g

(69)

(70)and velocity head

Iv = pv2/2g. (71)The meaning of the symbols is as follows:v =air velocity in centimeters per second; v = 1500

centimeters per second.g = gravitational acceleration; g = 981 centimeters

per second squared.p = air density; at 40 degrees centigrade,

p = 1.2 X 10-3 gram per cubic centimeter.IC,=total length of the cooler; l, = 22.75 centimeters.f friction coefficient; for the calculated Re = 5860,

f=0.0093, (see Fig. 8).12 William H. Adams, see p. 109 of footnote 1.13 William H. Adams, see p. 121 of footnote 1.

1942 201


k =a factor depending on A /A1, if A is the totalcross-sectional area of the air ducts, and A, thetotal area of the cooler cross section. (See Fig. 9.)In our case A/A1=0.52 and k=0.285.

rh = hydraulic radius of individual ducts of thecooler; rh-=0.8 centimeter.

A-AREA OF THI MAIN DUCTArAREA OF C NTRAC TED ECTIONHeKpV 2/9. 981lC /SEC.__He LOST l1EADEAD RE SED lN CMOF WATEP AIR D NSITY -1.2x' 3 GM, IM.3V*LINEAR VEL OCITY OF F UID IN

SMAL ER SE:CTION IN C /SEC.

-4 -__

_.2 4 6 .8 .10A,/A-_____ -

_____I iCADAMl .HEAT Tr 8I4S11"MI6~fItFig. 9 Plot of K versus A1/A for calculation

of contraction loss.

Substituting the indicated numerical values just givenfor the symbols in the last three expressions, we obtain:Hf = 1.55 centimeters H, = 0. 392 centimeters

H, = 1. 36 centimeters.and(72)

0 __ a4 X 0 G 7 I

400~~~~1

O- Lr + -

_> 7 M^~~~~~~~~~X. AMOI PC '191 PI tf^t#'UI tE

b - -~ ~~~~~~~~~C:r- ooL is*05tl .-OINST4,l14D%sipxagrto9 i|sts-._

- -

~~~~~lV-IOOL .I O.5

.~~~~ ~ ~ ~ ~ ~~~~C_ _irV 3 3 ;0%-KR 75I 010D, .- fA^A.F liOZ rfL ! 3_ CU ZVI! 4 -4 C OLER 10.51H Dl k.- MlY^V 0"1 St

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