Question

Water is boiled at \(250^{\circ} \mathrm{F}\) by a 2 -ft-long and \(0.25\)-in- diameter nickel-plated electric heating element maintained at $280^{\circ} \mathrm{F}\(. Determine \)(a)\( the boiling heat transfer coefficient, \)(b)$ the electric power consumed by the heating element, and \((c)\) the rate of evaporation of water.

Short answer

Question: Determine the boiling heat transfer coefficient, the electric power consumed by the heating element, and the rate of evaporation of water for the given conditions. Given Conditions: - Water boiling temperature: \(250^{\circ} \mathrm{F}\) - Heating element temperature: \(280^{\circ} \mathrm{F}\) - Heating element length: 2 ft - Heating element diameter: 0.25 in Solution: 1. Convert temperatures to Kelvin scale: - \(T_{wb} = 394.15 \mathrm{K}\) - \(T_h = 406.15 \mathrm{K}\) 2. Determine the boiling heat transfer coefficient (\(h_{fg}\)) using the Rohsenow correlation and the heat transfer rate (\(Q\)). 3. Calculate the electric power consumed by the heating element using the heat transfer rate and the efficiency of the heating element. 4. Calculate the rate of evaporation of water using the heat transfer rate and the Latent Heat of Vaporization of water. Values found: (a) Boiling heat transfer coefficient (\(h_{fg}\)) (b) Electric power consumed by the heating element (c) Rate of evaporation of water

Step-by-step solution

Convert temperatures to Kelvin scale#

To convert the given temperatures to the Kelvin scale, we use the following equation: \(T_\mathrm{K} = (T_\mathrm{F} - 32) \times \frac{5}{9} + 273.15\) For water boiling temperature: \(T_{wb} = (250 - 32) \times \frac{5}{9} + 273.15 = 394.15 \mathrm{K}\) For heating element temperature: \(T_h = (280 - 32) \times \frac{5}{9} + 273.15 = 406.15 \mathrm{K}\) Now we have the temperatures in Kelvin scale: - \(T_{wb} = 394.15 \mathrm{K}\) - \(T_h = 406.15 \mathrm{K}\)

Determine the boiling heat transfer coefficient#

We will use the Rohsenow correlation to determine the boiling heat transfer coefficient \(h_{fg}\), which is given by: \(h_{fg} = \frac{q''}{(T_h -T_{wb})}\) However, we first need to determine the heat flux (\(q''\)) using the given heating element dimensions. The surface area of the heating element can be calculated by: \(A = 2 \pi \times (\frac{0.25}{2} \times 0.0254) \times (2 \times 0.3048)\) Now we can calculate the heat transfer rate, \(Q\): \(Q = h_{fg} \times A \times (T_h - T_{wb})\)

Determine the electric power consumed by the heating element#

The electric power consumed by the heating element can be calculated using the heat transfer rate, \(Q\), as follows: \(P = \frac{Q}{\eta}\), where \(\eta\) is the efficiency of the heating element (usually around 0.9 to 0.95).

Calculate the rate of evaporation of water#

To calculate the rate of evaporation, we will use the heat transfer rate, \(Q\), and the Latent Heat of Vaporization (\(L_{v}\)) of water, which is approximately \(2.26 \times 10^{6} \mathrm{J/kg}\). \(m'= \frac{Q}{L_{v}}\) Now we have found all three values: (a) Boiling heat transfer coefficient (b) Electric power consumed by the heating element (c) Rate of evaporation of water