Question

2. A composite cylinder is formed by a long cylindrical rod (A) and two concentric cylinders (tubes B and C). Tube B encloses trod A and its inner and outer radii are 20 mm and 40 mm, respectively. Tube C encloses tube B and its outer radius is 50 mm. A thin electrical heater is inserted between rod A and tube B. The rod (A) has a thermal conductivity of 2.5 w/m K, while the tube thermal conductivities are k,-1.5 w/m K and kc=0.75 W/m K. Tube C outer surface is subjected to convection with a fluid of temperature -15°C and a heat transfer coefficient 50 W/m2K. The thermal contact resistance between the heater and tube B surface is 0.002 m-K/W (heater is in perfect contact with rod A). Tubes B and C are in perfect contact. Steady-state conditions are assumed a) Determine the electrical power per unit length of the cylinders (W/m) that is required to maintain the interface temperature between B and C at 15°C. Then, determine b) the heater temperature, and c) the temperature at the center of rod A d) Determine the maximum heat loss (this is also the maximum allowable heater power generation rate) per cylinder length, by adjusting the thickness of tube C. Rod A and tube B dimensions, and convection conditions are the same as before with the BC interface temperature to be kept at 15°C. Assume steady-state conditions.

Please help me with this problem

Answer 1

Given thermal conductivity of K_A = 2.5 w/m K Thermal conductivities of K_B = 1.5 w/m K K_0 = 0.75 Connection temperature = -15 degree C h = 50 w/m^2 K Thermal resistance = 0.002 m^2 k/w a) interface temperature = 15 degree C Electrical power = sum of the thermal conductivity times temperature = (2.5 + (1.5 + 0.75) times 91.5 + 2.23) = 1368 w/m (b) The heater temperature: convection temperature/(n times resistance) times K_B = 15/50 (0.002) times 15 = 100 (c) The temperature at the centre of rod A = heater temperature - Interface temperature = 100 - 15 = 85 degree