Question

A 50-cm-long, 0.2-cm-diameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally. The surface temperature of the wire is measured to be 130°C when a wattmeter indicates the electric power consumption to be 4.1 kW. Then the heat transfer coefficient is

a. 43,500 W/m². °C
b. 68,330 W/m². °C
c. 10,038 W/m². °C
d. 37,540 W/m². °C
e. 137 W/m². °C

Answer 1

The boiling heat transfer coefficient for the given conditions is approximately 43,500 W/m².°C, option (a), found by dividing the heat transfer rate (4.1 kW) by the product of the wire's surface area and the temperature difference (30°C).

Step-by-step explanation:

The boiling heat transfer coefficient can be determined using the rate of heat transfer equation, Q = hAΔT, where h is the heat transfer coefficient, A is the surface area, and ΔT is the temperature difference between the surface and the boiling water.

The surface area (A) of the cylindrical wire is calculated using the formula A = πDL, where D is the diameter and L is the length of the wire. Given D = 0.002 m and L = 0.5 m, the surface area is:

A = π(0.002 m)(0.5 m) = 0.00314 m².

The electric power consumption (P) is the rate of heat transfer (Q) to the water. Using the provided power P = 4.1 kW and the surface temperature of the wire at 130°C with boiling water at 100°C, the temperature difference (ΔT) is 30°C.

Calculating the heat transfer coefficient (h):

Q = hAΔT => 4.1 kW = h(0.00314 m²)(30°C)

To find h, rearrange the equation:

h = Q / (AΔT) = 4100 W / (0.00314 m² × 30°C) = 43,312.10 W/m².°C

This result most closely matches option (a) 43,500 W/m².°C. Therefore, this is the heat transfer coefficient for the given conditions, and option (a) can be considered the approximately correct answer within rounding errors.