FIN EFFICIENCY

There are several types of fin types such as helical solid fins, serrated helical fins, circular fins, rectangular (H type ) fins, peg pins, longitudinal fins are available due to development in heat transfer study. Out of the various fin configuration longitudinal fin configuration is the simplest fin configuration for estimating the fin efficiency, from which correlation can be made to get other fin type fin efficiency.

a

hf

hb

hi l=b l l=0 hfo

At any cross section as in figure let Tc be the constant temperature of the hot fluid everywhere surrounding the fin and let t be the temperature at any point in the fin and variable. Let be the temperature difference driving heat from the fluid to the fin at any point in its cross section. Then

= Tc – t

If l is the height of the fin varying from 0 to b

‘d /dl = -dt/dl

The heat within the fin which process through its cross section is Q = ka d /dl……….A

Differentiating the equation

‘dQ/dl = ka d² /dl²………..1

Where a is the cross sectional area of the fin. This is equal to the heat which passed into the fin through its sides from l=0 down to the darkened cross section.

If P is the perimeter of the fin, the area of the sides is Pdl and the film coefficient from liquid to fin side, whether on fin surface or tube surface, is hf.

‘dQ = hf Pdl or dQ/dl = hf ………..2

Equation 1 and 2 are same and hence 1-2 =0 kad² /dl² -- hf = 0

Rearranging, d² /dl² - hf /ka =0

The direct solution of this equation is,

0.5 0.5

= C1

e

^l(hfP/ka)

+

C2

e

^-l(hfP/ka)

………..

3 _____

‘m = (hfP/ka)

The general soution is = C1e^ml + C2e^-ml Applying boundary layer concept,

At the outer edge of the fin l =0,

e = c1 + c2

Assume no heat enters from outside edge of the fin, then

‘d /dl =0 when =0 and C1 –C2 =0 C1 = C2 = e/2

Equation 3 becomes,

e = (e^ml+ e^-ml)/2 = e cosh(ml)

Thus an expression has been obtained for the temperature difference between constant fluid temperature and variable fin temperature in terms of the length of the fin.

It is now necessary to obtain an expression for Q in terms of l Differentiating the equation 2, we get

d²Q/dl² = hfPd /dl……….4 /dl = ‘d²Q/dl² hfP

Substituting in equation A Q = kad²Q/hfPdl²

d²Q/dl² = hfPQ/ka………..5 d²Q/dl² - hfPQ/ka = 0

As before the solution is

0.5 0.5

Q = C3

e

^l(hfP/ka)

+

^C4

e

^-l(hfP/ka)

………..

⁶

Q = C3e^ml + C4e^-ml At I =0,

C3 + C4 =0 , C3 = -C4 and dQ/dl =0

‘dQ/dl =hf e = mC3 –mC4 =0 C3 = hf e/2m and C4 = -hf e/2m Q = hf e e^ml /2m - hf e e^ml /2m Q = hf e /m(e^ml - e^ml )/2

Q = hf e /m sinh(ml)

The ratio of heat load Q to the temperature difference at the fin base is Q/ = hf e sinh(ml)

m e cosh(ml)

Q/ = hfP tanh(ml) ………7 ‘m

hf = Qm/ ptanh(ml)

Let hf is the heat transfer coefficient of fin and the bare tube and the heat absorbed by fin through the heat transfer coefficient hf is getting transmitted into the tube by means of base heat transfer coefficient hb. The ratio of heat transfer coefficient hb to the heat transfer coefficient hf is termed as fin efficiency.

According to Fourier’s law of conduction, hb = Q/ lP

η’ = hb/hf = Q P tanh(ml) Qm lP

= tanh(ml) ml

This equation applies only to the fin and not to the bare portion of the tube between fins. The total heat removed from the annulus liquid and arriving at the tube inside diameter is a composite of the heat transferred by the fins to the tube outside diameter and that transferred directly to the bare tube surface. These may be combined by means of the weighted efficiency.

If the heat tranferred through the bare tube surface at the tube outside diameter is designated by Q0, then

Qo = hf Ao where A0 is the bare tube surface at the outside diameter exclusive of the area beneath the bases of the fins. If there are N number of fins on the tube l.P.Nis all of the fins surface. The total heat transfer at the outside diameter is given by

Q = Qb + Qo = hb.l.P.N. + hf.Ao

=(hb.l.P.N.Ao/Ao + hf.l.P.N.Ao/P.l.N)

=(hb/Ao + hf/P.l.N)l.P.N.Ao.

=(l.P.N.tanh(ml) + Ao ) hf ………..8 ml

Calling hfo the composite value of hf to both the fin and bare tube surfaces when referred to the outside diameter of the tube, the fin effectiveness or weighted fin efficiency is by definitionη =hfo/hf. Combining equation 7 and 8

η =hfo/hf = l.P.Nη’ + Ao l.P.N +Ao

= (η’.At + Ao /(At + Ao)) inserting At –At,

=( At –At + Ao + Atη’) /(At+ Ao) = (At+Ao -(1-η’)At)/(At+Ao)

Fin effectivenessη = 1- (1-η’)At/A where At is Area of fin surface

Ex.01. Design a feed water heater for a 10 tph boiler whose exhaust gas flow is 21355 kg/hr at an outlet temperature is 185°C and the desirable outlet temperature is 140°C. The feed water is available at 60°C.

I. Heat duty (gas side)

= 21355 ( 48.26 – 36.09) =259890 kcal/hr

Where 48.26 kcal/kg enthalpy of gas at 185°C and 36.09 kcal/kg enthalpy of gas at 140°C.

II. Water outlet temperature to = 259890/10000 + 60 =85.98°C.

III. For the heat exchanger, gas is flowing inside the tube internal heat transfer coefficient governs the overall heat transfer coefficient. The overall heat transfer coefficient is around 85% of inside heat transfer coefficient.

Assume gas velocity inside the flue tubes 15 m/sec.

Flow area required to maintain the velocity in tubes

= 21355 (162.5 +273)/(3600x273x1.295x15) =0.48716m²

Tube selected 50.8 x 3.25

IV. Number of tubes = 0.48716 /( (0.0508 – 2x0.00325)² = 316 tubes.

V. Heat transfer coefficient

Gas properties at average temperature 162.5°C.

Thermal conductivity = 31.225 x 10^-3 kcal/mhr°C Kinematic viscosity = 28.577 x 10^-6 m²/sec Prandtl number = 0.6775

Reynolds number = vd/γ

= 15 x 0.0443 /28.577 x 10-6 = 23252.559

HTC = 0.023 Re^0.8 Pr^0.3 K/d

= 0.023 x (23252.559)^0.8 x (0.6775)^0.3 x (31.225 x10^-3/0.0443) = 39.2 kcal/m²hr°C

The overall heat transfer coefficient = 0.85 x39.2 =33.32 kcal/hrm²°C.

V Heat transfer area required = Q/(Ux Lmtd) = 259890/(33.32 x89.15) = 87.49m²

VI Heat transfer length

= 87.49/( x 0.0443) = 628 m Number of tube required = 316

Tube length = 628/316 = 1.98 m

Considering tube length for welding to the tube sheet clearance 10 mm on both sides Tube size chosen 2.0 m.

Ex.02. Calculate gas outlet temperature for a evaporator intended to generate steam flow 15.28 kg/sec, pressure 20 bar(a), tube size 38.1 x 3.25 mm thk and thermal conductivity 49.844 W/mK. The gas flow 19.03 kg/sec, inlet temperature is 1180°C and the pattern of tube arrangement furnace width is 3.06 meter and length 2.7meters, tube pitching width side 80mm and depth side 78mm, inline arrangement, number of tubes in width side 36 and depth side 20. Heat transfer area of evaporator 246 sq.m. the evaportor was enclosed in a water wall having EPRS area of 26 sq.m and gas is flowing 90° to evaporator tubes. Consider heat transfer effectiveness of 82% and 71% for enclosure. Partial pressure of water =0.1158 bar and carbon di oxide = 0.1249bar.

Assume gas outlet temperature as 658°C

Heat loss by the gas = 19.03 *( 1418.1 – 745.94) = 12791.2 KW Inside heat transfer coefficient = 14000 W/m²°C (assume) Gas properties at average temperature = 919°C

Density kg/cu.m = 0.299 Dynamic viscosity kg/ms = 4.573 x 10^-5 Prandtl number = 0.71 Thermal conductivity W/mK = 0.083 For inline arrangement

ST/d = 80/38.1 = 2.099; SL/d = 78/38.1 = 2.047 Arrangement factor fe = 0.98 (from tables)

Gas flow area = (3.06 x 2.7 – 36 x0.0381x2.7) = 4.558sq.m Gas velocity = 19.03/(0.299 x 4.558) = 13.96 m/s

Reynolds number = 13.96 x0.299 x 0.0381/4.573x10^-5 = 3478.5

hc = 0.287 x fex Re^0.6 xPr^0.364 x k/d

= 0.287 x0.98x 3478.5^0.6x 0.71^0.364x0.083/0.0381 = 72.1 W/sq.mK

Radiation heat transfer coefficient

Beam length = 3.4 x(0.08 x0.078x1-0.00114)/( x0.0381x1) = 0.144876m

Attenuation factor = (0.8 +1.6x0.1158)(1-0.38x(1192/1000))(0.1158 + 0.1249) ((0.1158+0.1249)x0.14486)

= 0.6947

emissivity of gas = 0.9 x(1-e-0.69470.14486

) = 0.08617

hr = 5.67 x 10^-8 x 0.85 x 0.08617x(1192⁴ –485⁴) (1192 –485) = 11.53 W/sqmK

total gas side heat transfer coefficient = hc + hr =72.1+11.53 = 83.63 W/sqmK 1 do + d0 ln(d0/di) + 1

=

U hidi 2k ho

= 0.0381/(14000x0.0317) + 0.0381 ln(0.0381/0.0317) + 1/83.63 2 x 49.844

U = 82.55 W/sq.mK

Heat transferred to evaporator tubes, Q = U x effectiveness x Ax lmtd

= 82.55 x 0.82 x 246 x 673 = 11206 KW.

Heat transferred to water wall encloser,

Neglecting metal resistance and internal conductance the convection heat transfer coefficient be 82.55 w/sq.mK

Q = 82.55 x 0.71 x 26 x 673 = 1025 KW.

Total heat gained by evaporator and encloser is = 11206 + 1025 = 12231 KW.

7.0 AIRHEATER

7.1 INTRODUCTION

Air heater is a heat recovery unit, which is employed to recover heat from lower temperature levels usually 240°C and below. Air heater is used in most of the industrial and utility boilers. The construction of this equipment being a simple and also a non pressure part. Gas temperature drop 240°c to 170°C can be effectively achieved by means of air heater there by 4% boiler efficiency can be improved. In cases of low capacity , low pressure boilers manufactures prefers to have an air heater instead of economiser. This is to avoid problems arised due to economiser like oxygen pitting, corrosion higher feed water temperature, necessity of having deaerator more over economiser is a pressure part. These are avoided by means of air pre heater

In document Boiler Book (Page 80-87)