The base plate of a 800 W household iron has a thickness L = 0.6 cm,...

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The base plate of a 800 W household iron has a thickness L = 0.6 cm, a surface area A = 400 cm, a thermal conductivity k = 60 W/m-K and an emissivity & = 0.6. The iron is left on the iron board with its base exposed to the air ( T=20C). The convection heat transfer coefficient between the base surface and the surrounding air is h=35 W/m.K. In this problem, we will assume that the electrical energy provided base plate P heating element air Th room walls (large enclosure) T base plate emissivity: conductivity: k surface area: A -T to the heating element is fully converted into thermal energy, in the form of a uniform heat flux along the inner surface of the base plate. Part 1 Starting with an energy balance for the base plate, demonstrate that the outside surface temperature of the base plate (T) is the solution of a 4th-order polynomial equation, and calculate it numerically. [hint: the surrounding walls in the room are modeled as a large enclosure at temperature T] Part 2 [solution: T = 400C] In this part, we will only consider the conduction heat transfer within the base plate. 1- Express the differential equation and the boundary conditions for steady one- dimensional heat conduction through the plate. 2- Obtain a relation for the variation of temperature in the base plate by solving the differential equation. 3- Using the result from part 1 (T = 400C), evaluate the inner surface temperature T [solution: T = 402C] The base plate of a 800 W household iron has a thickness L = 0.6 cm, a surface area A = 400 cm, a thermal conductivity k = 60 W/m-K and an emissivity & = 0.6. The iron is left on the iron board with its base exposed to the air ( T=20C). The convection heat transfer coefficient between the base surface and the surrounding air is h=35 W/m.K. In this problem, we will assume that the electrical energy provided base plate P heating element air Th room walls (large enclosure) T base plate emissivity: conductivity: k surface area: A -T to the heating element is fully converted into thermal energy, in the form of a uniform heat flux along the inner surface of the base plate. Part 1 Starting with an energy balance for the base plate, demonstrate that the outside surface temperature of the base plate (T) is the solution of a 4th-order polynomial equation, and calculate it numerically. [hint: the surrounding walls in the room are modeled as a large enclosure at temperature T] Part 2 [solution: T = 400C] In this part, we will only consider the conduction heat transfer within the base plate. 1- Express the differential equation and the boundary conditions for steady one- dimensional heat conduction through the plate. 2- Obtain a relation for the variation of temperature in the base plate by solving the differential equation. 3- Using the result from part 1 (T = 400C), evaluate the inner surface temperature T [solution: T = 402C]