# HEAT TRANSFER

Heat will move from a hot body to a colder one, and can do so by the following methods:

1. Conduction. Direct from one body touching the other, or through a continuous mass

2. Convection. By means of a heat-carrying fluid moving between one and the other

3. Radiation. Mainly by infrared waves (but also in the visible band, e. g. solar radiation), which are independent of contact or an intermediate fluid.

Conduction through a homogeneous material is expressed directly by its area, thickness and a conduction coefficient. For a large plane surface, ignoring heat transfer near the edges:

Area x thermal conductivity

Conductance =

Thickness

= A x k L

And the heat conducted is

Qf = conductance x (T — T2)

Example 1.8

A brick wall, 225 mm thick and having a thermal conductivity of 0.60 W/(m K), measures 10m long by 3m high, and has a temperature difference between the inside and outside faces of 25 K. What is the rate of heat conduction?

_ 10 x 3 x 0.60 x 25

Uf—————————

F 0.225

= 2000 W (or 2 kW)

Thermal conductivities, in watts per metre Kelvin, for various common materi­als are as in Table 1.2 . Conductivities for other materials can be found from standard reference works.

Table 1.2

 Material Thermal conductivity (W/(m K)) Copper 200 Mild steel 50 Concrete 1.5 Water 0.62 Cork 0.040 Expanded polystyrene 0.034 Polyurethane foam 0.026 Still air 0.026

Convection requires a fluid, either liquid or gaseous, which is free to move between the hot and cold bodies. This mode of heat transfer is complex and depends firstly on whether the flow of fluid is ‘natural’, i. e. caused by thermal currents set up in the fluid as it expands, or ‘forced’ by fans or pumps. Other parameters are the density, specific heat capacity and viscosity of the fluid and the shape of the interacting surface.

With so many variables, expressions for convective heat flow cannot be as simple as those for conduction. The interpretation of observed data has been made possible by the use of a number of dimensionless groups which combine the variables and which can then be used to estimate convective heat flow.

The main groups used in such estimates are as shown in Table 1.3. A typi­cal combination of these numbers is that for turbulent flow in pipes expressing the heat transfer rate in terms of the flow characteristic and fluid properties:

Nu = 0.023 (Re)08 (Pr)04

The calculation of every heat transfer coefficient for a refrigeration or air — conditioning system would be a very time-consuming process, even with modern methods of calculation. Formulas based on these factors will be found in standard reference works, expressed in terms of heat transfer coefficients under different conditions of fluid flow.

Table 1.3

 Number Symbol Group Parameters Typical Relevance Reynolds Re Pvx Я Velocity of fluid, v Density of fluid, p Viscosity of fluid, p Dimension of surface, x Forced flow in pipes Nusselt Nu Hx K Thermal conductivity of fluid, k Dimension of surface, x Heat transfer coefficient, h Convection heat transfer rate Prandtl Pr Cp Я k Specific heat capacity of fluid, Cp Viscosity of fluid, p Thermal conductivity of fluid, k Fluid properties Grashof Gr Яg p2 x 3d Я2 Coefficient of expansion of fluid, p Density of fluid, p Viscosity of fluid, p Force of gravity, g Temperature difference, 0 Dimension of surface, x Natural convection

Where heat is conducted through a plane solid which is between two fluids, there will be the convective resistances at the surfaces. The overall heat trans­fer must take all of these resistances into account, and the unit transmittance, or ‘U value is given by:

Rt = Ri + Rc + Ro

U = 1/Rr

Where Rt = total thermal resistance

Ri = inside convective resistance

Rc = conductive resistacne

RO = outside convective resistance

Example 1.9

A brick wall, plastered on one face, has a thermal conductance of 2.8 W/(m2 K), an inside surface resistance of 0.3 (m2 K)/W, and an outside surface resistance of 0.05 (m2 K)/W. What is the overall transmittance?

Rt = Ri + Rc + Ro

= 0.3 + — + 0.05 2.8

= 0.707 U = 1.414 W/(m2 K)

Typical overall thermal transmittances are:

Insulated cavity brick wall, 260 mm thick, sheltered 0.69 W/(m2K)

Exposure on outside Chilled water inside copper tube, forced draught 15-28 W/(m2 K)

Air flow outside

Condensing ammonia gas inside steel tube, thin 450-470 W/(m2 K)

Film of water outside

Special note should be taken of the influence of geometrical shape, where other than plain surfaces are involved.

The overall thermal transmittance, U, is used to calculate the total heat flow. For a plane surface of area A and a steady temperature difference AT, it is

Qf = A X U X AT

If a non-volatile fluid is being heated or cooled, the sensible heat will change and therefore the temperature, so that the AT across the heat exchanger wall will not be constant. Since the rate of temperature change (heat flow) will be proportional to the AT at any one point, the space-temperature curve will be exponential. In a case where the cooling medium is an evaporating liquid, the temperature of this liquid will remain substantially constant throughout the process, since it is absorbing latent heat, and the cooling curve will be as shown in Figure 1.5 . Figure 1.5 Changing temperature difference of a cooled fluid

Providing that the flow rates are steady, the heat transfer coefficients do not vary and the specific heat capacities are constant throughout the working range, the average temperature difference over the length of the curve is given by:

AT = ATmax — ATmM

Ln(A7max/A7mm)

This is applicable to any heat transfer where either or both the media change in temperature (see Figure 1.6). This derived term is the logarithmic mean tempera­ture difference (LMTD) and can be used as AT in the general equation, providing U is constant throughout the cooling range, or an average figure is known, giving

Qf = A X U X LMTD Figure 1.6 Temperature change. (a) Refrigerant cooling fluid. (b) Fluid cooling refrigerant. (c) Two fluids

Example 1.10

A fluid evaporates at 3°C and cools water from 11.5°C to 6.4°C. What is the logarithmic mean temperature difference and what is the heat transfer if it has a surface area of 420 m2 and the thermal transmittance is 110 W/(m2 K)?

A7"max = 11.5 — 3 = 8.5K A7"min = 6.4 — 3 = 3.4K

LMTD = IL-Ь ln(8.5/3.4)

= 5.566 K

0f = 420 X 110 X 5.566

= 257 000 W or 257kW

In practice, many of these values will vary. A pressure drop along a pipe carrying boiling or condensing fluid will cause a change in the saturation tem­perature. With some liquids, the heat transfer values will change with temper­ature. For these reasons, the LMTD formula does not apply accurately to all heat transfer applications.

If the heat exchanger was of infinite size, the space-temperature curves would eventually meet and no further heat could be transferred. The fluid in Example 1.10 would cool the water down to 3°C. The effectiveness of a heat exchanger can be expressed as the ratio of heat actually transferred to the ideal maximum: Taking the heat exchanger in Example 1.10:

„ = 11.5 — 6.4 = 11.5 — 3.0 = 0.6 or 60%

Radiation of heat was shown by Boltzman and Stefan to be proportional to the fourth power of the absolute temperature and to depend on the colour, material and texture of the surface:

Qf = aeT4

Where a is Stefan’s constant (=5.67 X 10-8 W/(m2 K4)) and e is the surface

Emissivity.

Emissivity figures for common materials have been determined, and are expressed as the ratio to the radiation by a perfectly black body, viz.

Rough surfaces such as brick, concrete, or tile, regardless of colour 0.85-0.95

Metallic paints 0.40-0.60

Unpolished metals 0.20-0.30

Polished metals 0.02-0.28

The metals used in refrigeration and air-conditioning systems, such as steel, copper and aluminium, quickly oxidize or tarnish in air, and the emissivity fig­ure will increase to a value nearer 0.50.

Surfaces will absorb radiant heat and this factor is expressed also as the ratio to the absorptivity of a perfectly black body. Within the range of temperatures in refrigeration systems, i. e. -70°C to + 50°C (203-323 K), the effect of radia­tion is small compared with the conductive and convective heat transfer, and the overall heat transfer factors in use include the radiation component. Within this temperature range, the emissivity and absorptivity factors are about equal.

The exception to this is the effect of solar radiation when considered as a cooling load, such as the air-conditioning of a building which is subject to the sun’s rays. At the wavelength of sunlight the absorptivity figures change and calculations for such loads use tabulated factors for the heating effect of sun­light. Glass, glazed tiles and clean white-painted surfaces have a lower absorp­

Tivity, while the metals are higher.

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