An electric heater is used in a room of total wall area, 137 sq. m., to maintain a temperature of 293 K inside it. The walls of the room have 3 different layers of different materials. Innermost layer is of wood of thickness 2.5 cm, middle layer is of cement of thickness 1 cm and outermost layer is of brick of thickness 25 cm. Find the power of the heater, when the outer temperature is 263 K. Assume no heat loss occurs through the floor and ceiling. The thermal conductivity of wood, cement and brick are 0.125, 1.5 and 1 S.I. unit respectively.
2006-12-16 14:40:53 · 2 answers · asked by Rishu 2 in Science & Mathematics ➔ Physics
Firstly, SI unit has no meaning whatsoever relative to thermal conductivity. The correct unit is W/(m-K). In this case, it's more useful to use thermal resistance, i.e., (m-K)/W, which is then multiplied by the thickness/area for each layer separately. Add the three thermal resistances together and divide the temperature difference by the thermal resistance to get the power conducted through the walls.
2006-12-16 16:09:40 · answer #1 · answered by arbiter007 6 · 1⤊ 0⤋
One more Newton’s law: p=k*(S/L)*dt, where p is leaking power, dt is temperature difference on both sides of isolator, S is area of power leakage, L is thickness of power isolator, while k is constant for given isolating material - conductivity.
Thus p = 0.125*(137/0.025)*dt1 for wood, the same p = 1.5*(137/0.01)*dt2 for cement, and p = 1*(137/0.25)*dt3 for brick, while dt1+dt2+dt3 = 293-263 = 20°K;
Thus 20 = p/185 + p/20550 + p/548, hence p=6.00kW
2006-12-17 11:17:01 · answer #2 · answered by Anonymous · 0⤊ 0⤋