Heat and mass transfer to cooler coils
Heat transfer to a cooler coil involves three stages: heat flows from the airstream to the outer surface of the fins and pipes, it is then transferred through the metal of the fins and the wall of the piping and, finally, it passes from the inner walls of the tubes through the surface film of the cooling fluid to the main stream of the coolant.
In general, dehumidification occurs as well as cooling so that the behaviour of cooler coils cannot be described in simple terms. A very approximate approach, adopted by some manufacturers, is to establish a [/value for the coil that is given a bias to account for the extra heat flow by virtue of condensation. The duty is then described by
0, = UMLMTDw (10.9)
Where Ut is the biased [/value, At the total external surface area and (LM7X>)aw is the logarithmic temperature difference between the airstream and the water flowing, defined by
(^1 ^wb) (^2 ^wa)
(LM7P)aw = 
Ln[(ri — tVJh)/(t2 — rwj]
Using the notation of Figure 10.5. The biased thermal transmission coefficient is expressed by
(10.11) 
L/[/t — R — SRa + Rm + /?w
Where R. d is the thermal resistance of the air film when the external surface is dry, S is the sensibletotal heat transfer ratio (see equation (10.1)), Rm is the thermal resistance of the tubes (Rt) plus the fins (R{), and Rw is the resistance of the water film within the tubes. Rw must be multiplied by the ratio At/Ai( where At is the total external surface area and A; is the internal surface area, so that it refers to the total external surface area. R is the total thermal resistance, air to water.
Ra is the reciprocal of the sensible heat transfer coefficient on the air side, h. d, which depends principally on the mass flow rate of the airstream. For standard air, a staggered arrangement of tubes, and 316 fins per metre (8 per inch), ha is given by McAdams (1954) as:
(10.12) 
Ha = 27.42vf0’8
Wherein vf is the face velocity entering the cooler coil in Ms1 If the coil is only partially wet then there are two [/values: one using SRa in equation (10.11) and referring to the wetted part of the external surface, and the other using Ra in the equation and referring to the dry part of the surface. Determining the boundary between the wet and dry areas is not straightforward but methods are given in by ARI (1991) and ASHRAE (1996).
The resistance of the fins when dry is described by ARI (1991) as:
(10.13)
If the fins are wet then SRa replaces R, d. The term r is the total surface effectiveness and is defined by
(10.14) 
R = (<)Afi + Ap)/At
Where Afl is the surface area of the fins, Ap the external area of the tubes, At the total external surface area and <) is the fin efficiency. The latter is an involved function depending on the practical features of coil construction. For plate fins (at least) its value seems to be virtually independent of face velocity or fin spacing. On the other hand, the number of rows, fin thickness and tube spacing are all significant, as Table 10.2 shows.
Using copper as a material instead of aluminium improves efficiency but the use of larger diameter tubing reduces it.
The resistance of the metal of the tubes, Rt, referred to the total external surface area, is given by McAdams (1954) as
(10.15)
Wherein At/A; is the ratio of the total external surface to the total internal surface area, A, t is the thermal conductivity of the metal of the tubes, d0 is the outer tube diameter and its inner diameter, both in metres.
Table 10.2 Approximate fin efficiencies, based on ARI (1991), for flat plate fins, 2.5 m s’1 face velocity and tubes of 15 mm nominal outside diameter. It is to be noted that McQuiston (1975) claims a reduction in fin efficiency occurs with increased latent load

For water at temperatures between 4.4°C and 100°C the thermal resistance of the water film inside the tubes, /?w, referred to their inner surface area, is given by the following simplified equation from ARI (1991):
Rw = 4a2/[(1429 + 20.9fwm)v°’8] (10.16)
In which fwm is the mean water temperature, v is the mean water velocity and dt is the inside diameter of the tube in metres.
We can now calculate a (7value for a cooler coil and make an approximate estimate of a chilled water flow temperature to achieve a given duty.
EXAMPLE 10.2
A cooler coil is capable of cooling and dehumidifying air from 28°C drybulb, 19.5°C wet — bulb (sling), 55.36 kJ kg’1, 0.8674 m3 kg“1 and 10.65 g per kg to 12°C drybulb, 11.3°C
Wetbulb (sling), 32.41 kJ kg’1, 0.8178 m“3 kg’1 and 8.062 g per kg. 4.75 m3 s“1 of air
Enters the coil and chilled water is used with a temperature rise of 5.5°C. The coil has the following construction: 6 rows, 2.64 m s_1 face velocity, 316 fins per metre, aluminium flat plate fins of thickness 0.42 mm, tubes spaced at 37.5 mm, face dimensions of 1.5 m width and 1.2 m height, tubes of 15 mm outer diameter and 13.6 mm inner diameter. Determine the following: (a) total external surface area, (b) total internal surface area, (c) ratio of external to internal areas, (d) {/value, assuming the coil is entirely wet with condensate, (e) logarithmic mean temperature difference and (/) chilled water flow temperature. Take the thermal conductivity of copper as 390 W m’1 K’1.
Answer
(a) Number of fins = 1.5 x316 = 474 over a width of 1.5 m.
Length of one fin in direction of airflow = 6 x 0.0375 = 0.225 m Gross surface area of one fin
= (1.2 x 0.000 42 + 0.225 x 0.000 42 + 1.2 x 0.225)2 = 0.5412 m2
Number of tubes per row = 1.2/0.0375 = 32.
Total number of tubes passing through fin plates = 6 x 32 = 192.
External crosssectional area of one tube = n x 0.0152/4 = 0.000 177 m2
Since each fin is perforated on each of its sides by 192 tubes the net surface area of one fin
Is
0.5412 — 0.000 177 x 192 x 2 = 0.4732 m2
Net surface area of all fins = 474 x 0.4732 = 224.3 m2.
External perimeter of one tube = n x 0.015 = 0.0471 m.
Gross external surface area of all tubes = 192 x 0.0471 x 1.5 = 13.56 m2 Area of one tube obscured by one fin edge
= 0.000 42 x 0.0471 = 0.000 019 78 m2
Area of all tubes obscured by all fin edges
= 0.000 019 78 x 192 x 474 = 1.8 m2
Net area of all tubes = 13.56 — 1.8 = 11.76 m2.
Total external surface area = At = 224.3 +11.76 = 236.1 m2.
(b) Internal perimeter of one tube = n X 0.0136 = 0.0427 m.
Total internal surface area of all tubes = A, = 192 x 1.5 x 0.0427
= 12.3 m2
(c) Ratio of external to internal surface areas
= At/At = 236.1/12.3 = 19.2
(d) Plotting the entering and leaving air states on a psychrometric chart, as a check that the coil performance is possible, establishes a contact factor of 0.906 and a mean coil surface temperature of 10.35°C. The total cooling load is then calculated as
Qt = 4.75(55.36 — 32.41)/0.8674 = 125.7 kW By equation (6.6) the sensible component of the load is Qs = 4.75(28 — 12)358/(273 + 28) = 90.4 kW Hence S = Qs/Qt = 0.72.
By equation (10.12)
Ra = /ha = 1/(27.42 x 2.640’8) = 0.016 77 m2 K W~’ when dry Ra = 0.016 77 x 0.72 = 0.012 07 m2 K W“1 when wet
From Table 10.2 the fin efficiency is 0.98 and by equations (10.14) the effectiveness of the surface is
T] = (0.98 x 224.3 + 11.76)/236.1
= 0.98
By equation (10.13) the resistance of the wet fins is
Rt = (—~j^x 0.012 07 = 0.000 246 m2 K W“1
P — 1^.. 0015 .00150 * “ 2 x 390 0.0136
= 0.000 036 2 m2 K W“1 Hence the resistance of the metal fins and tubes is + Rt
= 0.000 246 + 0.000 036 2 = 0.000 282 m2 K W’1 when wet
By equation (10.16) we can calculate Rw if the mean velocity of waterflow through the 32 tubes passing in series through the six rows is known.
Mass flow of water = 125.7/(4.19 x 5.5)
= 5.454 kg s“1
Assuming a mean water temperature of 10°C the density is 999.7 kg m3 and the total volumetric flow rate of water is (5.454/999.7)1000 = 5.456 litres s1.
The total crosssectional area of the tubes through which this flows is
32 x n x 0.01362/4 = 0.004 649 m2
Hence the mean velocity of water flow is
5.456/(0.004 649 x 1000) = 1.174 m s"1
By equation (10.16)
/?w = 0.01360 2/[( 1429 + 20.9 x 10) x 1.1740 8] = 0.000 227 m2 K W“1
Referred to the internal tube surface. Hence, referred to the external surface area
Rw = 0.000 227 x 19.2 = 0.004 358 m2 K W“1
Thus
Ux = 1/(0.004 358 + 0.000 282 + 0.012 07) = 1/0.0167 = 59.8 W m“2 K“1
(e) LMTD = Qt/(Ut x At) = 125 700/(59.8 x 236.1) = 8.9°C
(/) Using the notation of Figure 10.5 and equation (10.10) we have
89 = (28 — twb) — (12 — tW3)
Ln[(28 — rwb)/(12 — fwa)] Since rwb = twa + 5.5
(22.5 — fwb) — (12 — fwa)
1. 9 =
Ln[(22.5 — rwb)/(12 — ?wa)l
Ln[(22.5 — twa)/( 12 — fwa)] = 10.5/8.9 = 1.18 (22.5 — ?wa)/(12 — tm) = exp(1.18) = 3.25
Rwa = (3.25 x 12 — 22.5)/2.25 = 7.3°C
The method of enhancing the [/value of a cooler coil in order to account for the dehumidification occurring, outlined above and used in example 10.2, cannot be regarded as anything other than very approximate. Some of the reasons are:
(a) The heat transfer is crossflow for any particular row, even though it is contraflow
From row to row. Hence the logarithmic mean temperature difference used in the foregoing should be modified to take account of this.
(b) The airflow distribution over the face of the coil is not uniform, being concentrated
In the centre with a tendency to stagnation at the corners. Hence the value of the heat transfer coefficient through the air film, ha, calculated by equation (10.12), is not uniform over the coil face.
(c) The use of the sensibletotal heat transfer ratio, S, must be regarded as an approximation to account for dehumidification. Wellverified techniques for predicting the value of ha are not available according to ASHRAE (1996) and it is best to use experimentally determined data, obtained from manufacturers.
(d) If the bond between the root of the fin and the tube is not good, there is likely to be
A significant loss of heat transfer. This may be true of some plate fins if the plates are
Too thin: there is then insufficient plate material to flow properly when punched and form a good, cylindrical collar (to separate the fins) in the process of manufacture. Furthermore, the collars may be cracked.
(e) It is probable that not all the external surface of the fins and tubes is wet with condensate. In this case the method outlined will give an optimistic assessment of the performance.
(/) The (/value is not constant throughout the cooler coil.
(g) The psychrometric state of the air leaving the coil is not uniform across the face and
Hence calculations based on a single state will not be precise.
Note that the face velocity of 2.64 m s1, used in the example, would result in the carryover of condensate from the fins, making the use of downstream eliminator plates essential.
Most manufacturers determine the performance of their cooler coils themselves but ASHRAE Standard 33 (1978) gives details of acceptable laboratory methods. ARI Standard 410 (1991) then offers procedures for predicting the results of such laboratory tests for other working conditions and coil sizes.
A more involved method of determining the performance of cooling and dehumidifying coils is given by ASHRAE (1991) and ASHRAE (1996) see also BS 5141: (1983).
It is worth noting that there is evidence from Rich (1973) that the heat transfer coefficient through the air film is essentially independent of the fin spacing. An unusual relationship between heat transfer coefficient and the number of rows has also been shown by Tuve (1936) and Rich (1975) when Reynolds numbers for the airflow are low (<10 000), the coefficient for deep coils being much less than that for shallow coils. A tentative explanation offered is that vortices shed by the tubes give nonuniform temperature distribution in the airstream for downstream rows.
Posted in Engineering Fifth Edition