HEAT TRASNFER IN BUILDINGS HEAT TRASNFER IN BUILDINGS

HEAT TRASNFER IN BUILDINGS

THERMAL RESISTANCE Q = k. A(T 1 -T 2)/x Q = (T 1 -T 2)/R Where R = thermal resistance = x/k. A or L/k. A 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.

CONVECTION Q = h. A(T 1 -T 2) (W) Q = (T 1 -T 2)/R (W) R = 1/h. A (C/W)

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

RADIATION Q = εσA(T 4 -Tc 4) Q = hrad. A(T 4 -Tc 4) Q = (T 4 -Tc 4)/Rrad = 1/hrad. A

Conduction

EXAMPLE Consider a 0. 8 -m-high and 1. 5 -m-wide glass window with a thickness of 8 mm and a thermal conductivity of k = 0. 78 W/m · °C. Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 20°C while the temperature of the outdoors is -10°C. Take the heat transfer coefficients on the inner and outer surfaces of the window to be h 1 =10 W/m 2 C and h 2=40 W/m 2 C, which includes the effects of radiation.

Example Consider a 0. 8 -m-high and 1. 5 -m-wide double-pane window consisting of two 4 -mmthick layers of glass (k = 0. 78 W/m · °C) separated by a 10 -mm-wide stagnant air space (k = 0. 026 W/m · °C). Determine the steady rate of heat transfer through this double-pane window and the temperature of its inner surface for a day during which the room is maintained at 20°C while the temperature of the outdoors is -10°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h 1 = 10 W/m 2 · °C and h 2 = 40 W/m 2 · °C, which includes the effects of radiation.

Example-Composite Wall A 3 -m-high and 5 -m-wide wall consists of long 16 -cm x 22 -cm cross section horizontal bricks (k = 0. 72 W/m · °C) separated by 3 -cm-thick plaster layers (k = 0. 22 W/m · °C). There also 2 -cm-thick plaster layers on each side of the brick and a 3 -cm-thick rigid foam (k = 0. 026 W/m · °C) on the inner side of the wall, as shown in Fig. The indoor and the outdoor temperatures are 20°C and -10°C, and the convection heat transfer coefficients on the inner and the outer sides are h 1=10 W/m 2 · °C and h 2=25 W/m 2 · °C, respectively. Assuming onedimensional heat transfer and disregarding radiation, determine the rate of heat transfer through the wall.

Solution - I

Solution - II

HEAT TRANSFER THROUGH CONDUCTION – CYLINDER Qcyl = k. A(d. T/dr) A = 2πr. L Q=(T 1 -T 2)/Rcyl = ln(r 2/r 1)/2 π l k = [ln (outer radius/inner radius)]/[2 π x (length) x (thermal conductivity)]

HEAT TRANSFER THROUGH CONDUCTION - Sphere Qcyl = k. A(d. T/dr) A = 4 π r 2 Q=(T 1 -T 2)/Rsph = (r 2 - r 1)/4 π r 2 r 1 k = [ (outer radius - inner radius)]/[4 π x outer radius x inner radius x thermal conductivity] Q=(T∞ 1 -T ∞ 2)/Rsph

MULTI-LAYER CYLINDERS

MULTI-LAYER CYLINDERS Where A 1= 2πr 1 L and A 4= 2πr 4 L

MULTI-LAYER CYLINDERS The ratio d. T/R across any layer is equal to Q· , which remains constant in one-dimensional steady conduction.

Example Steam at T 1 = 320°C flows in a cast iron pipe (k =80 W/m · °C) whose inner and outer diameters are D 1 =5 cm and D 2 =5. 5 cm, respectively. The pipe is covered with 3 -cmthick glass wool insulation with k =0. 05 W/m · °C. Heat is lost to the surroundings at T 2 =5°C by natural convection and radiation, with a combined heat transfer coefficient of h 2 =18 W/m 2 · °C. Taking the heat transfer coefficient inside the pipe to be h 1 =60 W/m 2 · °C, determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation.

Solution