HEAT TRASNFER IN BUILDINGS

HEAT TRASNFER IN BUILDINGS

THERMAL RESISTANCE Q = kA(T1-T2)/x Q = (T1-T2)/R Where R = thermal resistance = x/kA or L/kA The equation for thermal resistance is analogous to the relation for the flow of electric current I I = (V1 – V2)/Re Where is Re = L/σA is the electric resistance across the voltage difference (V1 – V2) and is the electrical conductivity.

CONVECTION Q = hA(T1-T2) (W) Q = (T1-T2)/R (W) R = 1/hA (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(T4-Tc4) Where Q = heat transfer through radiation ε = emissivity (=1 for ideal radiator) σ =Stefan-Boltz-maann constant (= 5.67 x 10-8 Watt/m2K4) A = area of radiator T = temperature of radiator Tc= temperature of surroundings

RADIATION Q = εσA(T4-Tc4) Q = hradA(T4-Tc4) Q = (T4-Tc4)/Rrad Rrad = 1/hradA

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 h1 =10W/m2 C and h2=40W/m2C, 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-mm-thick 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 h1 = 10 W/m2 · °C and h2 = 40 W/m2 · °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 are 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 h1=10 W/m2 · °C and h2=25 W/m2 · °C, respectively. Assuming one-dimensional heat transfer and disregarding radiation, determine the rate of heat transfer through the wall.

Solution - I

Solution - II

HEAT TRANSFER THROUGH CONDUCTION – CYLINDER Qcyl = kA(dT/dr) A = 2πrL Q=(T1-T2)/Rcyl Rcyl = ln(r2/r1)/2 π l k = [ln (outer radius/inner radius)]/[2 π x (length) x (thermal conductivity)]

HEAT TRANSFER THROUGH CONDUCTION - Sphere Qcyl = kA(dT/dr) A = 4πr2 Q=(T1-T2)/Rsph Rsph = (r2 - r1)/4 π r2 r1 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 A1= 2πr1L and A4= 2πr4L

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

Example Steam at T1 = 320°C flows in a cast iron pipe (k =80 W/m · °C) whose inner and outer diameters are D1 =5 cm and D2 =5.5 cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation with k =0.05 W/m · °C. Heat is lost to the surroundings at T2 =5°C by natural convection and radiation, with a combined heat transfer coefficient of h2 =18 W/m2 · °C. Taking the heat transfer coefficient inside the pipe to be h1 =60 W/m2 · °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