# Plate Fin-and-Tube Heat Exchangers

A common type of heat exchanger used in industrial ventilation is the plate fin-and-tube heat exchanger (Fig. 9.7). Liquid or gas flows in the

Tubes, with a gas or a liquid circulating outside the tubes between the

Plates.

The plates can be straight or wavy. Wavy plates, due to their greater sur­face area, enhance the heat transfer between air and the plates but are unsuit­able in dusty environments.

The complete study of annular fins is essential in order to calculate the amount of conduction in the plate fin. A schematic diagram of annular fins is given in Fig. 9.8.

The temperature distribution within the annular fin is given by the differ­ential equation

P — + r^7“I7(T“T“) = 0- (9J2>

In Eq. (9.32), b is the heat transfer coefficient between the air and the surface of the annular fin, A is the thermal conductivity of the fin, and is the temperature

Airflow  Of air far away from the fin. The differential equation (9.32) together with the boundary conditions is difficult to solve and is not considered in this chapter.

Normally, heat exchanger tubes are arranged in a staggered manner. The continuous plate is considered as being built up of regular hexagons, with a hole in the center for the pipes. This arrangement is shown in Fig. 9.9.

If hexagons are replaced with an annular area of the same size, the outer radius of the annular area w’ill be

R0=JZ/= 0.525/, (9.33)

Where / is the distance between the pipe centerlines in the heat exchanger. The inner hole diameter in the annular area is the same as the pipe’s outer diameter.

The fin heat transfer is determined by using fin efficiency. The tin effi­ciency is calculated using a theoretical approach where the whole fin is consid­ered to be at the same temperature as the fin base. The required parameters necessary to determine the fin efficiency are shown in Fig. 9.10.

The heat transfer through the fin is

‘/’theor = bAs(Tb — TJ = bAsQt, (9.34)

Where As is the surface area of the fin.

The fin efficiency is defined by the division of the actual by the theoretical heat transfer, i. e.,

Yt — *ftacrual — ^ — G

«Atheor kAs6t hA, ■ The actual heat flow through the fin is

Factual = G0f = hrAsBt. (9.36)

The actual heat flow is calculated by multiplying the fin outer surface area by the fin efficiency. The outer surface area is easy to determine; hence, if the fin efficiency is known, the heat transfer from the fin is easily calculated.

The efficiency depends on the fin geometrical configuration, the fin ther­mal conductivity, and the heat transfer coefficient at the fin surface.

The use of Eq. (9.36) to determine the heat transfer of annular fins is a practi­cal approach. The fin efficiency curve is constructed by solving Eq. (9.34) for  Surface Fins/cm Surface Fins/cm

 ! 1.15 5 3.61 2 1.74 6 4.6 3 2.62 7 5.71 4 3.02 8 8.11

 St PrM (heat transfer) arid / (friction)
 0.1 0.08 0.06 0.05 0.04 0.03 0.02 0.015 0.01 0.008 0.006 0.005 0.004  T~i m………… rr

 Friction

 —

 J I I i II i I

 0.003 0. 1 0.15 0.2 0.3 0.4 0.50.6 0.8 1 1.5 2 3 4 5 6 8 10 15 20

RE = x io-3 №

FIGURE 9.12 Diagram for determining the heat transfer coefficient for a straight fin plate.

 (9,37) The abscissa in Fig. 9.12 is

Re =

Here

(9.38)

Qm is the mass flow of air through the heat exchanger, and Arain is the total area between the fins perpendicular to the airflow.

D/, is the hydraulic diameter of one cross-section between the plates, calculated from

Ar

 (9.39) Dh = 2

 Px HN

H is the height of the plates in the heat exchanger (assuming the plates are in a vertical position, which is normal), and Npg is the number of gaps between the plates, ix in Eq. (9.37) is the dynamic viscosity of the gas.

The ordinate in Fig. 9.12 is St Pr2/% where Pr is the Prandt! number of air and is defined as

 Pr = (9.40)

For air at 2CM-Q °C, the Prandtl number is Pr = 0.71. St is the Stanton number, defined as

 H St (9.41

Use Fig. 9.12 to calculate the Reynolds number, Re. Then by reading from one of the curves 1 … 8, depending on how many fins there are per centimeter, the value of St Pr2/3 can be obtained.

The convective heat transfer coefficient is then

2/3

 (9.42) „ St ■ Pr

 Pr 2/3

Next consider the wavy plate with herringbone waves. This arrangement with the relevant geometrical parameters is shown in Fig. 9.13.

The heat transfer coefficient for a herringbone plate is calculated from ’

 (9.43) B = A ReD Pr1/j//D

In Eq. (9.43) we have

(9.44)

Where D is the tube outer diameter including the collar of the fin plate, and Ac is the flow area in the heat exchanger where maximum velocity occurs. For a symmetric, staggered configuration, the maximum velocity occurs between the pipes situated vertically on each other.

The factor / is calculated from

 ( Y-0.205 D /
 / *Ј Pd
 ( ‘ Pd S V /
 PV 0.272 Pi)
 -0.133
 -0.558
 -0.357
 / = 0.394 Re (9.45)

The relevant geometrical parameters for Eq. (9.45) are given in Fig. 9.13.

Example 2

The data of a heat exchanger is given as follows:

• Copper pipes have outer diameter of 12 mm.

• Symmetric, staggered pipe configuration; distance between the pipe centerlines is 32 mm.

• Aluminum plates ASTM 1110 of thermal conductivity 200 W m-3 K”1.

• Plate thickness is 0.3 mm; distance between plates is 3 mm.

• Number of plates is 200.

• Height of plate is 600 mm; depth in flow direction is 290 mm.

• Number of pipe rows is 10, 18 pipes in one vertical row. The mass flow of air through the heat exchanger is 0.9 kg s_1 Calculate

A. the fin efficiency,

B. the heat conductance or reciprocal of heat resistance on the air side if the plates are straight, and

C. the air-side conductance if the plates have herringbone waves with Xf = 4.3 mm and pd = 1 mm.

The physical properties of air can be determined at 20 °C.

Solution

A. We calculate first the total heat transfer surface of the air side. The dis­tance between the plates is s = 3 mm, plate thickness is t = 0.3 mm, and outer diameter of the pipe is dQ = 12 mm. The height of plates is H = 600 mm, and depth of the plates is L = 290 mm. The number of gaps between the plates is Npg = 199, and the number of plates is Np! = 200. The total number of pipes is Np = 10 ■ 18 = 180.

Apg = NPg^do + 2 t)sNp = NpgirDsNp

= 199-ir — 0.0126 ■ 0.003 • ISO = 4.3 m2

The area of the plate fins is

A„i = 2 ■ Nt, i H ■ L-N 7tD_

Ipj — z. ■ ipi ■ j n ■ —Jj“

 = 60.6 m~ 77 ■ 0.0126"

0.

 = 2■200 6 0.29- 180 —

As can be seen, the outer surface area of the pipes between the fins contributes in a small heat transfer area, as expected.

The surface area Amin is

Amjn = Npg s ■ H = 199 • 0.003 • 0.6 = 0.358 m"

At 20 °C, the physical properties of air are ju. = 181.1 ■ 10 " N s m ‘ cp = 1.007 kj kg-1 K- A = 0.0257 W nr1 K~ and Pr = 0.709 For qm = 0.9 kg s~1 Eq. (9.38) gives

Q m A™m 0.358 2-51 kg m

From Eq. (9.39) we get

0.358

Dh = 2‘A,, = 0.006 m.

‘0.6 ■ 199

Equation (9.37) gives

Re = 2.51_:.iMQ6 = 832 .

181.1 • Ur7

The plates at 1 cm distance are 10/3.3 = 3.03. Using Fig. 9.12 we can read St • Pr2/3 = 0.0085.

From this we obtain

H = 1007 ■ 2.51 • = 27.0 Wm~2 K"1 .

0.709

For I = 32 mm, we obtain from Eq. (9.33)

R0 = 0.525 ■ 32 mm = 16.8 mm,

R,. = q = LL2±2_M} = 6.3 mm,

And

RQ — rt = 16.8 — 6.3 = 10.5 mm = 0.0105 m.

The thermal conductivity of aluminum is A = 200 W nr1 K-1; thus we ob­

And

 2-27 0.0063 = 0.180. Ut = Jlh/Xt ■ r, = J —

/220 ■ 0.0003

The above two parameters, the fin efficiency is obtained from Fig. 9.11 as 7} = 0.95.

B. Applying Eq. (9.36), we obtain the heat conductance on the air side as

C„ = h(vAp, + APf;) = 27.0(0.95 • 60.6 + 4.3) = 1638 W K 1 .

C. Because Pt — I = 32 mm and the pipe arrangement is staggered and symmetric,

We have

Pi = sin 60° ■ P, = sin 60° • 32 = 27.7 mm.

Ac in Eq. (9.44) is Ac = 199 • 17(32 — 2 ■ 6.3) • 3 = 0.197 m2 .

Then in Eq. (9.44)

0.

 = 3180 ReD — 9 ■ 0.0126

0. 197 181.1 ■ 10

 ( 32 -0.272 " 3 ‘ -0.205 / 4.3 -0.558 ( 1 27.7 V V 12.6 V / 1 V 7 3 V /
 -0.357

 -0.133

 And in Eq. (9.45) / = 0.394-3180 0.014. Finally, from Eq. (9.43) we get

H = 0.0257 • 3180 ■ 0.709l/3 ■ 0.0147/0.0126 = 85 Wm"2K_1 .

We see that the wavy profile of the fin plate increases the heat transfer coefficient greatly.

We now have

 Y = 0.0105- [. V: — [ 2-85 } 7s;5,= 0.53

 And 220 • 0.0003

Ui = j… — 0.0063 = 0.32 .

‘ V220-0.0003

From Fig. 9.11, we then get rj = 0.86 and finally

Ga = 85(0.86 ■ 60.6 + 4.3) = 4795 W K

9.3.2.S Liquid-Side Conductance

And Total Conductance of Heat Exchanger

The liquid-side conductance is calculated from

G, = 0.023 Ren/s-Pr?-3-Ar77-L,

TI — U. UiJ IVeD, — l’rI ■ A( • 77 • U, (9.46)

Where A; is the thermal conductivity of the liquid, Pr; the Prandtl number for the liquid, and L the total length of the pipes in the heat exchanger.

For ReD; we have

4 Qi _ p/WjD Di is the inner diameter of the pipes.

For the total conductance Gtor of the heat exchanger, we have

_!_ = J. + J — + JL. (9.48′-

Gtot C, G, G, ‘

Here 1 /Gc is the heat exchanger contact resistance. The reason for the con­tact resistance is that there exists a resistance to heat flow between the outer surface of the pipe and the collar of the plate fins. Normally, the fins are attached to the pipes by mechanical expansion of the tubes out into the plate-fin collars. Because of this manufacturing method, the contact will not be ideal. Small gaps between the pipe surface and the collar of the tins will occur.

It is very difficult to estimate the magnitude of the contact conductance Gc. Normally the total conductance of the heat exchanger is determined, and Gc is calculated from Eq. (9.48). Only in the case that the plate fins are welded

To the pipes with a metallurgical contact is the contact conductance infinite,

Leading to zero contact resistance, that is 1 /G(. = 0.