Building Energy Analysis HEAT TRANSFER Conduction n Convection

Building Energy Analysis

HEAT TRANSFER Conduction n Convection n Radiation n

CONDUCTION Conduction is a process of heat transfer by means of molecular agitation within a material without any motion of the material as a whole. If one end of a metal rod is at a higher temperature, then energy will be transferred down the rod toward the colder end because the higher speed particles will collide with the slower ones with a net transfer of energy to the slower ones. For heat transfer between two plane surfaces, such as heat loss through the wall of a house, the rate of conduction heat transfer is:

HEAT TRANSFER THROUGH SINGLE SLAB Q = KA(∆T/∆ x) Where Q = heat transfer k = thermal conductivity A = surface area ∆T = temperature difference ∆ x = distance T 1 K ∆x T 2

Example Determine the heat loss per square meter surface area of a composite wall made of 220 mm thick brick faced on the outside with 40 mm thick concrete and on the inside with a 15 mm thick insulation and 10 mm thick timber as shown. Thermal conductivities for brick, concrete, insulation and timber respectively are 0. 71, 0. 93, 0. 6 and 1. 75 W/m. K. The inside temperature of the wall is 18 C when the outside surface temperature of the concrete is -3 C.

Solution n n n Q = UAΔT 1/U = {x 1/k 1 + x 2/k 2 + x 3/k 3 + x 4/ k 4} x 1= 10 mm, k 1= 1. 75 x 2= 15 mm, k 2 = 0. 6 x 3= 220 mm, k 3= 0. 71 x 4= 40 mm, k 3= 0. 93 n ΔT = 18 – (-3) n Thus Q = {18 – (-3)}/ {(0. 01/1. 75) + (0. 015/0. 6) + (0. 22/0. 71) + (0. 04/0. 93)} = 54. 75 W/m 2 n

CONVECTION n

RADIATION Radiation is process of heat transfer by electromagnetic waves. Unlike conduction and convection, radiation heat transfer can occur in vacuum. Q = εσA(T 4 -Tc 4) Where Q = heat transfer through radiation ε = emissivity (=1 for ideal radiator) σ =Stefan-Boltz-maann constant (= 5. 67 x 10 -8 Watt/m 2 K 4) A = area of radiator T = temperature of radiator Tc= temperature of surroundings

CONDUCTION THROUGH PLANE SLAB Q = k. A(∆T/∆ x)

THERMAL DIFFUSIVITY n The thermal diffusivity indicates the ability of a material to transfer thermal energy relative to its ability to store it. The diffusivity plays an important role in non-steady conduction. n K = thermal conductivity ρ = the density of the material. Cp = specific heat. n n

REPRESENTATIVE CONDUCTIVITY AND DIFFUSIVITY VALUES

CONVECTION n Convection heat transfer occurs both due to molecular motion and bulk fluid motion. Convective heat transfer may be categorised into two forms according to the nature of the flow: Natural or Free Convection. Here the fluid motion is driven by density differences associated with temperature changes generated by heating or possibly cooling. In other words, fluid flow is induced by buoyancy forces. Thus the heat transfer itself generates the flow which conveys energy away from the point at which the transfer occurs. n Forced Convection. Here the fluid motion is driven by some external influence. Examples are the flows of air induced by a fan, by the wind, or by the motion of a vehicle, and the flows of water within heating, cooling, supply and drainage systems. In all of these processes the moving fluid conveys energy, whether by design or inadvertently. n

CONVECTION Q = hc. A(T 1 -T 2) = hc. A∆T Where n hc= convective heat transfer coefficient n

CONVECTION

Example n A fluid flows over a plane surface 1 m by 1 m with a bulk temperature of 50 o. C. The temperature of the surface is 20 o. C. The convective heat transfer coefficient is 2, 000 W/m 2 o. C. n Q = 2, 000 (W/m 2 o. C) (1 (m)) (50 (o. C) - 20 (o. C)) = 60, 000 (W) n

THERMAL RESISTANCE Q = k. A(T 1 -T 2)/x n Q = (T 1 -T 2)/R Where n R = thermal resistance = x/k. A or L/k. A n n The equation for thermal resistance is analogous to the relation for the flow of electric current I I = (V 1 – V 2)/Re Where is Re = L/σA is the electric resistance across the voltage difference (V 1 – V 2) and is the electrical conductivity.

Heat loss through cylindrical surfaces n Q = k. Ad. T/dr n A = 2Πr. L n Q = k(2ΠL)ΔT/ln r 2/r 1

EXAMPLE n Consider a steam pipe of length L 20 m, inner radius r 1 6 cm, outer radius r 2 8 cm, and thermal conductivity k 20 W/m · °C, as shown in Figure 2– 50. The inner and outer surfaces of the pipe are maintained at average temperatures of T 1 150°C and T 2 60°C, respectively. Determine the rate of heat loss from the steam through the pipe. n Q = k(2ΠL)ΔT/ln r 2/r 1 n