Question

During air cooling of oranges, grapefruit, and tangelos, the heat transfer coefficient for combined convection, radiation, and evaporation for air velocities of \(0.11

Short answer

Answer: The initial rate of heat transfer is approximately 5.03 W, the initial temperature gradient inside the orange at the surface is about 0.186 K/m, and the Nusselt number is approximately 2.7.

Step-by-step solution

Calculate Reynolds Number

First, we need to find the Reynolds number (\(Re\)) for the given air velocity. The formula for Reynolds number is: \(Re = \frac{VD}{\nu}\), where \(V\) is the air velocity, \(D\) is the diameter of the orange, and \(\nu\) is the kinematic viscosity of air. At \(3°C\) and 1 atm, the kinematic viscosity of air is approximately \(1.52 \times 10^{-5} \mathrm{~m^2 / s}\). Given the air velocity \(V = 0.3 \mathrm{~m/s}\) and orange diameter \(D = 0.07 \mathrm{~m}\), we can plug in the values and calculate the Reynolds number: \(Re = \frac{0.3 \times 0.07}{1.52 \times 10^{-5}} = 1378\)

Find Heat Transfer Coefficient (h)

Now, we can determine the heat transfer coefficient (h) using the given formula: \(h = 5.05 k_{\text{air}} Re^{1/3} / D\) and given thermal conductivity of air \(k_{\text{air}} = 0.0262 \mathrm{W/m \cdot K}\). Plugging in the values, we get: \(h = 5.05 \times 0.0262 \times 1378^{1/3} / 0.07 \approx 27 \mathrm{~W/m^2 \cdot K}\)

Calculate Initial Rate of Heat Transfer

We can determine the initial rate of heat transfer (\(q\)) using Newton's Law of Cooling: \(q = hA(T_s - T_\infty)\), where \(A\) is the surface area of the orange, \(T_s\) is the initial temperature of the orange, and \(T_\infty\) is the temperature of the surrounding air. Given that \(T_s = 15°C\) and \(T_\infty = 3°C\), and the surface area of the orange \(A = \pi D^2 = \pi \times (0.07)^2 \approx 0.0154 \mathrm{~m^2}\), we can calculate the initial rate of heat transfer: \(q = 27 \times 0.0154 \times (15 - 3) \approx 5.03 \mathrm{~W}\)

Find Initial Temperature Gradient Inside the Orange

To find the initial temperature gradient at the surface of the orange, we can use the heat transfer coefficient and thermal conductivity of the orange: Temperature gradient = \(\frac{q}{h} = \frac{5.03}{27} \approx 0.186 \mathrm{~K/m}\)

Calculate Nusselt Number

Finally, we can determine the Nusselt number (\(Nu\)) using the formula: \(Nu = \frac{hD}{k}\), where \(k= 0.7 \mathrm{~W/m \cdot K}\) is the thermal conductivity of the orange. Plugging in the known values, we get: \(Nu = \frac{27 \times 0.07}{0.7} \approx 2.7\) The initial rate of heat transfer from the orange is approximately \(5.03 \mathrm{W}\), the initial temperature gradient inside the orange at the surface is about \(0.186 \mathrm{K/m}\), and the Nusselt number is approximately \(2.7\).