Physics Chapter 1 Solutions

1 The filament of a 150 W incandescent lamp is 5 cm long and has a diameter of 0.5 mm. The heat flux on the surface of the filament, the heat flux on the surface of the glass bulb, and the annual electricity cost of the bulb are to be determined. Assumptions Heat transfer from the surface of the filament and the bulb of the lamp is uniform . Analysis (a) The heat transfer surface area and the heat flux on the surface of the filament are As = πDL = π (0.05 cm )(5 cm ) = 0.785 cm 2 q s = Q 150 W = = 191 W/cm 2 = 1.91× 10 6 W/m 2 As 0.785 cm 2 Q Lamp 150 W (b) The heat flux on the surface of glass bulb is As = πD 2 = π (8 cm) 2 = 201.1 cm 2 q s = Q 150 W = = 0.75 W/cm 2 = 7500 W/m 2 As 201.1 cm 2 (c) The amount and cost of electrical energy consumed during a one-year period is Electricity Consumption = Q ∆t = (015 . kW)(365 × 8 h / yr) = 438 kWh / yr Annual Cost = (438 kWh / yr)($0.08 / kWh) = $35.04 / yr 2 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of heat transfer through the wall is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the wall are constant. Properties The thermal conductivity of the wall is given to be k = 0.69 W/m⋅°C. Brick Analysis Under steady conditions, the rate of heat transfer through the wall is wall (20 − 5)°C ∆T = (0.69W/m ⋅ °C)(5 × 6m 2 ) = 1035W Q cond = kA 0.3m L 0.3 m 30 cm 20°C 5°C 3 A transistor mounted on a circuit board is cooled by air flowing over it. The transistor case temperature is not to exceed 70°C when the air temperature is 55°C. The amount of power this transistor can dissipate safely is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is disregarded. 3 The convection heat transfer coefficient is constant and uniform over the surface. 4 Heat transfer from the base of the transistor is negligible. Analysis Disregarding the base area, the total heat transfer area of the transistor is Air, As = πDL + πD 2 / 4 = π (0.6 cm)(0.4 cm) + π (0.6 cm ) 2 / 4 = 1.037 cm 2 = 1.037 × 10 −4 m 2 55°C Then the rate of heat transfer from the power transistor at specified conditions is Power Q = hAs (Ts − T∞ ) = (30 W/m 2 .°C)(1.037 × 10 -4 m 2 )(70 − 55)°C = 0.047 W transist Therefore, the amount of power this transistor can dissipate safely is 0.047 W. 4 A 1000-W iron is left on the iron board with its base exposed to the air at 20°C. The temperature of the base of the iron is to be determined in steady operation. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the iron base and the convection heat transfer coefficient are constant and uniform. 3 The temperature of the surrounding surfaces is the same as the temperature of the surrounding air. Properties The emissivity of the base surface is given to be ε = 0.6. Analysis At steady conditions, the 1000 W energy supplied to the iron will be dissipated to the surroundings by convection and radiation heat transfer. Therefore, Q total = Q conv + Q rad = 1000 W where Q conv = hAs ∆T = (35 W/m 2 ⋅ K)(0.02 m 2 )(Ts − 293 K) = 0.7(Ts − 293 K) W and Q rad = εσAs (Ts4 − To4 ) = 0.6(0.02m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[Ts4 − (293K) 4 ] = 0.06804 × 10 −8 [Ts4 − (293K) 4 ] W Substituting, 1000 W = 0.7(Ts − 293 K ) + 0.06804 × 10 −8 [Ts4 − (293 K) 4 ] Solving by trial and error gives Ts = 947 K = 674  C Discussion We note that the iron will dissipate all the energy it receives by convection and radiation when its surface temperature reaches 947 K. Iron 1000 W