Question

Water is boiling in a 12 -cm-deep pan with an outer diameter of $25 \mathrm{~cm}$ that is placed on top of a stove. The ambient air and the surrounding surfaces are at a temperature of \(25^{\circ} \mathrm{C}\), and the emissivity of the outer surface of the pan is \(0.80\). Assuming the entire pan to be at an average temperature of \(98^{\circ} \mathrm{C}\), determine the rate of heat loss from the cylindrical side surface of the pan to the surroundings by \((a)\) natural convection and \((b)\) radiation. (c) If water is boiling at a rate of \(1.5\) \(\mathrm{kg} / \mathrm{h}\) at \(100^{\circ} \mathrm{C}\), determine the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of water. The enthalpy of vaporization of water at \(100^{\circ} \mathrm{C}\) is 2257 \(\mathrm{kJ} / \mathrm{kg}\). Answers: $46.2 \mathrm{~W}, 47.3 \mathrm{~W}, 0.099$

Short answer

Short Answer: The heat loss from the cylindrical side surface of the boiling pan due to convection is 46.2 W, and due to radiation is 47.3 W. The ratio of the total heat loss from the side surfaces to the heat loss by evaporation of water is 0.099.

Step-by-step solution

(Step 1: Find the convective heat transfer coefficient (hc))

As given, Outer diameter of pan (D) = 25 cm = 0.25 m Height of pan (H) = 12 cm = 0.12 m Pan Temperature (Ts) = 98 °C = 371 K Ambient Temperature (T∞ ) = 25 °C = 298 K The constant properties of air at a film temperature \((T_s+T_\infty) / 2\) = 334.5 K are: k = 0.0262 W/(m⋅K) ν = 15.89 × \(10^{-6} \mathrm{m^2/s}\) Pr = 0.708 First, we find the Grashof (Gr) and Rayleigh (Ra) numbers: \(\mathrm{Gr} = \frac{g \cdot \beta \cdot (T_s - T_\infty) \cdot D^3}{\nu^2} \) \(\mathrm{Ra} = \mathrm{Gr} \cdot \mathrm{Pr}\) By knowing the Rayleigh number, we can find the convective heat transfer coefficient (hc). For laminar flow (Ra\(_{D} \leq 10^9\)), \( h_c = 0.15(\mathrm{Ra}_D)^{1/3} \cdot \frac{k}{D} \) For turbulent flow (Ra\(_{D} > 10^9\)), \( h_c = 0.10(\mathrm{Ra}_D)^{1/3} \cdot \frac{k}{D} \)

(Step 2: Calculate convective heat loss (Qconv))

To calculate convective heat loss (Qconv), we need the convective heat transfer coefficient (hc) and the area of the cylindrical side surface (Ac). \(A_c = \pi D H\) Now apply the formula for Qconv: \(Q_{conv} = h_c A_c (T_s - T_\infty)\)

(Step 3: Calculate radiative heat loss (Qrad))

For the radiative heat loss, we need the emissivity (ε), the Stefan-Boltzmann constant (σ), and the area of the cylindrical side surface (Ar). The emissivity is given as 0.80, and Ar is equal to Ac. The Stefan-Boltzmann constant is \(5.67 \times 10^{-8} \mathrm{W/(m^2 K^4)}\). Now apply the formula for Qrad: \(Q_{rad} = \epsilon \sigma A_r (T_s^4 - T_\infty^4)\)

(Step 4: Find the heat loss due to evaporation of water)

We are given the rate of evaporation of water (1.5 kg/h) and the enthalpy of vaporization (2257 kJ/kg). \(\dot{m} = 1.5\,\text{kg/h} = \frac{1.5}{3600}\,\text{kg/s}\) \(h_v = 2257\,\text{kJ/kg} = 2257\times10^3\,\text{J/kg}\) Now apply the formula for Qevap: \(Q_{evap} = \dot{m}h_v\)

(Step 5: Compute the ratio of heat lost from side surfaces to that by evaporation)

Now that we have calculated Qconv, Qrad, and Qevap, we can calculate the requested ratio using the formula: \(\frac{Q_{conv}+Q_{rad}}{Q_{evap}}\) By performing the calculations, we get: a) Qconv = 46.2 W b) Qrad = 47.3 W c) Ratio of heat loss = 0.099