Question

A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of 5∘C. The tube wall has inner and outer radii of 25 and 75 mm, respectively, and a thermal conductivity of 10 W/m-K. The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is 0.01 mK/W. The outer surface of the heater is exposed to a fluid with T[infinity]=10∘C and a convection coefficient of h=100W/m2K. The heater power per unit length of the tube (W/m) required to maintain the heater at a uniform temperature of 25∘C is :

A. 2377
B. 706.38
C. 727.67
D. 1434.85

Answer 1

To find the heater power per unit length required to maintain a uniform temperature on the outer surface of the cylindrical tube, we need to consider both heat conduction through the tube wall and heat convection from the outer surface. By calculating the rates of heat transfer due to conduction and convection and summing them up, we can determine the total power required.

Step-by-step explanation:

To determine the heater power required to maintain a uniform temperature of 25°C on the outer surface of the cylindrical tube, we need to consider the various heat transfer mechanisms involved. First, we calculate the rate of heat transfer due to conduction through the tube wall: Q_cond = [(T_i - T_o) / (ln(r_o / r_i)/2πk)] * L. where Q_cond is the rate of heat transfer by conduction, T_i and T_o are the inner and outer temperatures, r_i and r_o are the inner and outer radii, k is the thermal conductivity, and L is the length. Next, we calculate the rate of heat transfer due to convection from the outer surface of the heater: Q_conv = h * A * (T_infinity - T_o). where Q_conv is the rate of heat transfer by convection, h is the convection coefficient, A is the outer surface area of the heater, T_infinity is the fluid temperature, and T_o is the outer surface temperature. The total power required to maintain the heater at a uniform temperature is the sum of the power due to conduction and convection: P_total = Q_cond + Q_conv. Finally, we divide the total power by the length of the tube to get the heater power per unit length: P_per_length = P_total / L. Substituting the given values into the equations yields the answer: Answer: A. 2377