Question

An electrical water \((k=0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) heater uses natural convection to transfer heat from a \(1-\mathrm{cm}\)-diameter by \(0.65\)-m-long, \(110 \mathrm{~V}\) electrical resistance heater to the water. During operation, the surface temperature of this heater is \(120^{\circ} \mathrm{C}\) while the temperature of the water is \(35^{\circ} \mathrm{C}\), and the Nusselt number (based on the diameter) is 6 . Considering only the side surface of the heater (and thus \(A=\pi D L\) ), the current passing through the electrical heating element is (a) \(3.2 \mathrm{~A}\) (b) \(3.7 \mathrm{~A}\) (c) \(4.6 \mathrm{~A}\) (d) \(5.8 \mathrm{~A}\) (e) \(6.6 \mathrm{~A}\)

Short answer

Answer: (d) 5.8 A

Step-by-step solution

Find the heat transfer coefficient (h) using Nusselt number

Using the Nusselt number definition, we can find the heat transfer coefficient (h), where Nu = hd/k: $$ Nu = \frac{hD}{k} $$ Rearrange the equation to solve for h: $$ h = \frac{Nu \cdot k}{D} $$ Plug in the given values (Nu = 6, k = 0.61 W/m·K, D = 0.01 m): $$ h = \frac{6 \cdot 0.61}{0.01} = 366\ \mathrm{W/m^{2}K} $$

Find the heat transfer rate (Q) using heat transfer coefficient

Now, we will find the heat transfer rate (Q) using the formula Q = hAΔT, where A is the surface area and ΔT is the temperature difference: $$ Q = hA \Delta T $$ To find A, we will use the equation for the side surface area of a cylinder (A=πDL): $$ A = \pi \cdot 0.01 \cdot 0.65 = 0.0205\ \mathrm{m^{2}} $$ And \(\Delta T = 120 - 35 = 85^{\circ} \) Now plug in the values of h, A, and ΔT: $$ Q = 366 \cdot 0.0205 \cdot 85 = 641.085\ \mathrm{W} $$

Determine the current using Ohm's Law

Now we can use Ohm's Law (P=IV) to find the current (I), where P is power and V is voltage: $$ I = \frac{P}{V} $$ The given voltage is 110 V. Plug in the values of P and V: $$ I = \frac{641.085}{110} = 5.828\ \mathrm{A} $$ Looking at the given choices, the closest value to 5.828 A is 5.8 A. So, the correct answer is (d) 5.8 A.