predict storage-tank heat transfer precisely

5/26/2018 Predict Storage-tank Heat Transfer Precisely

Predict storage-tankheat transfer precisely

Use this procedure to determine the rate of heat transfer fro1nvertical storage tank when shortcut methods are inadequate. a

Jimmy D. Kumana and Samir P. K othari, }/e1mingson, Durham and Richardson, /11c

O Heating or cooling storage tanks can be a majorenergy expense at plants and tankfarms. Though manyprocedures for calculating such heat-transfer requireroents have been published [ 1 35 7 8 JO ] the simplifying assumptions that they use can lead to significanterrors in co1nputed hea t-t ransfer rates. This is of concern bec use effic ierl t sizing of L nks irlsulation heatersand coolers depends on accura te estimates of heat trans..fer to and from the various tank surfaces. And the ulti.mate value of being accura te increases as energy costs~ i n u e to rise.t t The procedure presented here de ter mines the hea tft'ransfer to or from a vertical-cylind rical s.torage tank

s e a e d on the ground- li ke the one in Fig. I . I t includest h effects of tank configuration, liquid level, ambientt e m p e r a t u r e and wind speed, as well as temperature

;;\:ariations within the tank and between air and ground.IApartially worked example shows how to use the techque, and how to do the calculations on a computer.

' e theory"storage ta nks come in many di fferent shapes and[ izes. Horizontal-cylindrical and spherical tanks are

s e d for storage of liquids under pressure; atmosphericl a n k s tend to be vertical-cylindrical, with flat bottoms' ';md conical roofs as shown in Fig. I The example prei e n t e d here is for the latter configuration, but the proce:dure applies to a ny tank for which reliable heat-transfer'correlat ions are available.' . For the sake of simplicity, we assume that the tankcontents are warmer than the ambient ai r, and that weare concerned with heat loss from the tank rather than

e a t gain. But the method may, of course, be applied to:either case.( 1, Consider, then, the categories of surfaces from which~ . h e a t may be transferred across the tank boundaries: wetl , dry sidewa lls, tank bot tom, and roof. In the context

~ d here, wet refers to the portion of the wall sub,,J erged under the liquid surface, whereas dry refers to~ e portion of the wall in the vapor space, above the

q u i d surface.~ ~ ~ general, the heating coils would be located near~ l i e bottom of the tank, in the form of Rat pancakes.". , _refore, the temperatu re o f the air (or vapor) space"' '


Individual film heat-transfer coe fficients Table I

ical cond itions or determine them by m easurement.The area values are a lso easy to obta in :

A4 = 'D (L - L,.)A.., = wD L,.A = D2/ 4A= (wD/ 2)(D2/4 + d2)112

)(7)(8)(9)

The com p lications arise when we try to estimate theoverall hea t-tra nsfer coefficients u,, u .. vb and u,, forthe four surfaces of the tank. For the tank geometry

c h o s ~ n . these ca n for tunately be calculated from thein div idual fi lm heat-tr ansfer coe fficients in the conventional m anner, usi ng publ ished correlations.T he overall coefficients

Ta ble I shows the component coefficient for eachs u r f a c ~ . T he overall heat-tra nsfer coefficient for the d rysidew all of the tank (U4 ) is calcula ted as the sum of theresistances of vapor film, fou ling, me ta l wall, insulation( if any), a nd outside air (convection plus rad iation).

The ou tside-air heat-transfer coefficient (I 109For vertical plates ta ller than 3 ft, Stuhlbarecommends:h = 0.451cL-o";(N0 , Np,)o.1;where 104 < .lv0; 1Vp , ) < 109.Hor izontal -sur face hea t-transfer coefficientscoe fficients a pply to the roof and inside-bottom sof the tank. T he bottom is assumed to be flat. Ffaces facing up [8):

N = 0 14 (N V )0.33Ju r rFor surfaces facing down:

N,v,, = 0.27 (Na , Np , )o.zs128 CHtM ICAL t NCINUR INC : IARCH 22 982


Data for ABC Chemical Co . exampleLiquid Air

0.080.250.007

Heat-transfer coe fficients after first iteration

Table tiVap r

Q.080.250.007

0.002

-.'

Table Ill

Table IV

35 f , a wind velocity of 10 mph . a nd a liqu id50 ii,_ The ot her da ta req uired are given in TNote that the liqll id tempera tu re is controlledto provide a sr ma rg in of safety.Si nce the Pra nd tl a nd Grashof nu m be rs occuedly in the fi lm hea1-uansfer cocfl\cicn t equatiremain relatively un cha nged for a ll the condin terest, let us first ca lculate. their values. T huliqu id phase:

Ne , = L3p2g f3 iT / 2 = 97 .5 L3 iTNp, = c. k = ~ 8 4Sim ilarly , for the vapor p hase, Ne , =107L3 iT and Np, = 0.28. We can now ca lcindividua l film hea t-transfer codlicients, using

propriate L and iT va l\les in the Gra shofequalions. This is an iterati\re procc.ss tl1 t reqt ial estimates for wa ll and gr ound temperatu\\all te.1nperatures.

C oefficient for vapor a t wa ll h.,.., . As an inproxi1nation assun e that the '' al l ternpcratuaverage of the va por and ou1side-air tempT,. = (SO + 35)/ 2 = 42.s r. Then find thenun1ber :

" -Gr - 1.90 X IO;(L - L,. )3(7"v - T,.1.90 x 101(24)3(7.5)1.97 x 10

Emp loy ing Eq. 15, find the Nussel1 nu mberthe coeflicient k = 0.0151, L == 48 ft, L' = 2

N,, = 1 3 8 N a , 1 0 ( Y i , 1 1 - 0.55) = 921h = (92 1.l )(k)/ (L - Lw) = 0.581 Bt u/ f1oe fficient for liqu id a t th e wall (h1 ,,, . Her,\lp, nor lv0 , Np, ) fa lls wi th in the range of theble correla tions (Eq. 16. 18). Let us try b o th , ag

an average for Tic:T , = TL + T,) / 2 = 45 fN ;, == 97 .47L3 (TL - T.,,) = 1.35 x 10

U sing Eq . 16 a nd 18, we get two es1 ima 1es fo rtransfer coefficient (k = 0.1 2, Np, = 484):hL,,, = (0.495k/ L,,, )Nc, p , ) o = o. 04 Bt u/hL.,, = 0 . 4 5 k l ~ N a , N p , 2 ~= 1.415 B1u / ft1n-r

To be conservative, we use 1he highehL,,, = 1.41 5 Bt u/ ft2 h-F.

Coefiicicnl for vapor at roof h,,.). \Ve co nsiAa t plate, wi th a di a1n eter of 20 ft, and usag a in with an average T.,, of 42.s f (k = 0.0

N0 , = 1.9 X 101D 3 Tv - T,,,) == 1.14 x 10hw = (0.27k/ D )(Nc, Np, )0.2; = 0. 154 B1u/

Coefficient for liquid at tank bouom h i ~that the ground temperature ( Tc ) is ; F aboent, and use an average of liquid a nd g ro\lnd tu res as a first approximacion for the tank-botp eratul e:

T = TL T0)/ 2 ==< Ti + T< + 5)/ 2 =130 ~ c - u - - 1 1 - c A _ L _ E ~ N ~ c - 1 . - N E . E ~ R N M - A - c - 2 ~ - . - 9 - , , ~


Rate of heat transfer during unit period Table VI

heat losses can be tabulated and added to get the cum ulative he at loss for a month or year. Of course, this requires climatic data and tank-level estimates for theoverall time-period.Comparison with other methods

Aerstin and Street (1 Joffer a very simple method forcalculating heat loss from tanks. For a tank with l.5 in.of sidewall insula tion, and a wind speed of l 0 mph, therecommended overall U (based on k = 0.019 for theinsulation) is 0.14 for AT = soF and 0.14 forAT = 100. F. Adjusting these values fo r k = 0.028 andAT= 11F, as in our example, yields a n overall U of0.206 Btu/ ft2h-F. T he total exposed sur face is 3,331 ft2(tank bottom not ind uded), and thus the ove rall rate ofheat transfer by their method is:

Q= 0.206 x 3,331 x 17 = 11,666 Btu/ hThis compares with a beat loss of 8,9 13 Btu/ h (forthe exposed surface) calcula ted by the procedure of th isarticle-see T able VI. Thus their method y ields a result313 too high in this case.S tuhlbarg [ ] takes an approach similar to that proP?sed here, but his method differs in how the ou tsideta nkwall film coefficient is computed. Stuhlbarg r'* 'mm cnds the use of a manufacturer's ~ t a table, and doesnot explicitly distinguish be tween the bulk liquid temperature and the outside-wall surface temperature 1ncalculating' the proper heat-transfer coe fficient. -. .

The algebra ic me thod of Hughes and Deuresembles the one presented in rh is aniclewa s Bu t it does not recognize differe nces be tL1id a r1d ' ' a por lc1111Je ra tures inside the la11k 1 naccou nt fo r the in teraction between J Tand "in calcL1 lating a \vi11dcnhat'l

predict storage-tank heat transfer precisely

Documents