Question

Consider a \(0.2-\mathrm{m}\)-diameter and \(1.8-\mathrm{m}\)-long horizontal cylinder in a room at \(20^{\circ} \mathrm{C}\). If the outer surface temperature of the cylinder is \(40^{\circ} \mathrm{C}\), the natural convection heat transfer coefficient is (a) \(2.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(4.1 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(5.2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(6.1 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short answer

Question: A horizontal cylinder with a diameter of \(0.20\, \mathrm{m}\) and an outer surface temperature of \(40^{\circ}\mathrm{C}\) is in a room of temperature \(20^{\circ}\mathrm{C}\). Using the following properties of air: kinematic viscosity \((15.89 \times 10^{-6}\,\mathrm{m^2/s})\), thermal diffusivity \((2.22 \times 10^{-5}\,\mathrm{m^2/s})\), and thermal conductivity \((0.0257\, \mathrm{W/m\cdot K})\), calculate the natural convection heat transfer coefficient. Options: A) 1.47 W/m²·K B) 1.82 W/m²·K C) 2.19 W/m²·K D) 2.53 W/m²·K

Step-by-step solution

Calculate the temperature difference and temperature ratio

In order to find the Rayleigh number, we first need the temperature difference between the cylinder and the surrounding room. Since the cylinder's outer surface temperature is given as \(40^{\circ}\mathrm{C}\) and the room's temperature is given as \(20^{\circ}\mathrm{C}\), the temperature difference \(ΔT\) is: \(ΔT = 40 - 20 = 20 ^{\circ}\mathrm{C}\) Next, we need to calculate the average temperature between the cylinder surface and the room, which can be calculated as: \(T_{avg} = (T_{cylinder} + T_{room})/2 = (40 + 20) / 2 = 30^{\circ}\mathrm{C}\)

Calculate the Rayleigh Number, Prandtl Number, and Grashof Number

The Rayleigh Number can be calculated as the product of the Grashof Number and the Prandtl Number. We can find those numbers using the following formulas for a horizontal cylinder: \(Gr = \frac{g \cdot b \cdot D^3 \cdot ΔT}{ν^2}\) Here, \(g\) is the acceleration due to gravity \((9.81\, \mathrm{m/s^2})\), \(b\) is the thermal expansion coefficient \((1/T_{avg})\), \(D\) is the diameter of the cylinder \((0.2\, \mathrm{m})\), \(ΔT\) is the temperature difference \((20^{\circ}\mathrm{C})\), and \(ν\) is the kinematic viscosity \((15.89 \times 10^{-6}\,\mathrm{m^2/s})\). \(Pr = \frac{ν}{α}\) Here, \(α\) is the thermal diffusivity \((2.22 \times 10^{-5}\,\mathrm{m^2/s})\). Using these formulas, we can calculate the two numbers and then find the Rayleigh Number using: \(Ra = Gr \cdot Pr\) After calculating all the numbers, we can proceed to find the Nusselt number.

Determine the Nusselt Number

For a horizontal cylinder, we can use the following correlation to determine the Nusselt number: \(Nu = c_1 \cdot Ra^{c_2}\) \(c_1\) and \(c_2\) are constants depending on the geometry of the object; for a horizontal cylinder with the given conditions, they are \(c_1 = 0.60\) and \(c_2 = 0.25\). We can now use this formula to calculate the Nusselt Number with the Rayleigh Number calculated in Step 2.

Calculate the natural convection heat transfer coefficient

After calculating the Nusselt Number, we can find the natural convection heat transfer coefficient, \(h\), using the following formula: \(h = \frac{k \cdot Nu}{D}\) Here, \(k\) is the thermal conductivity \((0.0257\, \mathrm{W/m\cdot K})\) and \(D\) is the diameter of the cylinder \((0.2\, \mathrm{m})\). Now, we can plug in all the values and calculate the natural convection heat transfer coefficient. Then, compare the result with the given options to find the correct choice.