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Hailstones are formed in high-altitude clouds at \(253 \mathrm{~K}\). Consider a hailstone with diameter of \(20 \mathrm{~mm}\) that is falling through air at \(15^{\circ} \mathrm{C}\) with convection heat transfer coefficient of $163 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming the hailstone can be modeled as a sphere and has properties of ice at \(253 \mathrm{~K}\), determine how long it takes to reach melting point at the surface of the falling hailstone. Solve this problem using the analytical one-term approximation method.

Short Answer

Expert verified
Answer: It takes approximately 183.60 seconds for the hailstone to reach its melting point.

Step by step solution

01

Calculate surface area and volume of the hailstone

Given diameter of the hailstone is 20 mm. We will first convert it to meters and then calculate the surface area (A) and volume (V) of the hailstone sphere: Diameter (D) = 20 mm = 0.020 m Radius (r) = D / 2 = 0.010 m Surface area (A) = 4 * pi * r^2 Volume (V) = (4/3) * pi * r^3 Calculating A and V: A = 4 * pi * (0.010)^2 = 0.001256 m^2 V = (4/3) * pi * (0.010)^3 ≈ 4.19 × 10^{-6} m^3
02

Determine the Biot number

To determine the Biot number (Bi), first we need to calculating thermal conductivity (k), heat capacity (c), and density (ρ) of ice at 253 K given data. We can consider the properties of ice at 253 K as: k = 2.180 W/m·K c = 2110 J/kg·K ρ = 920 kg/m³ Now we can calculate the Biot number: Bi = (h * r) / k where h = 163 W/m²·K (convection heat transfer coefficient) Bi = (163 * 0.010) / 2.180 ≈ 0.746
03

Calculate Fourier number using one-term approximation

We'll now calculate Fourier number (Fo) - which depends on time (t) - using the one-term approximation method, for half of the Biot number (Bi/2 ≈ 0.373): Fo = 0.373^2 / 3.289 = 0.042
04

Calculate time to reach melting point

We can now calculate the required time (t) to reach the melting point using Fourier number (Fo): t = Fo * (V/A) * c * ρ / k t = 0.042 * (4.19 × 10^{-6}/0.001256) * 2110 * 920 / 2.180 t ≈ 183.60 s So, it takes approximately 183.60 seconds for the hailstone to reach the melting point at the surface while falling through the air at 15°C.

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Most popular questions from this chapter

A semi-infinite aluminum cylinder $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\alpha=9.71 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$ ) of diameter \(D=15 \mathrm{~cm}\) is initially at a uniform temperature of \(T_{i}=115^{\circ} \mathrm{C}\). The cylinder is now placed in water at \(10^{\circ} \mathrm{C}\), where heat transfer takes place by convection with a heat transfer coefficient of $h=140 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. Determine the temperature at the center of the cylinder \)5 \mathrm{~cm}$ from the end surface 8 min after the start of cooling. Solve this problem using the analytical one-term approximation method.

During a picnic on a hot summer day, the only available drinks were those at the ambient temperature of \(90^{\circ} \mathrm{F}\). In an effort to cool a 12 -fluid-oz drink in a can, which is 5 in high and has a diameter of $2.5 \mathrm{in}$, a person grabs the can and starts shaking it in the iced water of the chest at \(32^{\circ} \mathrm{F}\). The temperature of the drink can be assumed to be uniform at all times, and the heat transfer coefficient between the iced water and the aluminum can is $30 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$. Using the properties of water for the drink, estimate how long it will take for the canned drink to cool to \(40^{\circ} \mathrm{F}\). Solve this problem using lumped system analysis. Is the lumped system analysis applicable to this problem? Why?

A potato may be approximated as a \(5.7-\mathrm{cm}\)-diameter solid sphere with the properties $\rho=910 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.25 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\(, \)k=0.68 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and \)\alpha=1.76 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\(. Twelve such potatoes initially at \)25^{\circ} \mathrm{C}$ are to be cooked by placing them in an oven maintained at \(250^{\circ} \mathrm{C}\) with a heat transfer coefficient of $95 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The amount of heat transfer to the potatoes during a 30 -min period is (a) \(77 \mathrm{~kJ}\) (b) \(483 \mathrm{~kJ}\) (c) \(927 \mathrm{~kJ}\) (d) \(970 \mathrm{~kJ}\) (e) \(1012 \mathrm{~kJ}\)

A short brass cylinder $\left(\rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.389 \mathrm{~kJ} /\right.\( \)\mathrm{kg} \cdot \mathrm{K}, k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and \)\left.\alpha=3.39 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\( of diameter \)4 \mathrm{~cm}$ and height \(20 \mathrm{~cm}\) is initially at a uniform temperature of $150^{\circ} \mathrm{C}\(. The cylinder is now placed in atmospheric air at \)20^{\circ} \mathrm{C}$, where heat transfer takes place by convection with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate \((a)\) the center temperature of the cylinder; \((b)\) the center temperature of the top surface of the cylinder; and (c) the total heat transfer from the cylinder \(15 \mathrm{~min}\) after the start of the cooling. Solve this problem using the analytical one-term approximation method.

Water mains must be placed at sufficient depth below the earth's surface to avoid freezing during extended periods of subfreezing temperatures. Determine the minimum depth at which the water main must be placed at a location where the soil is initially at \(15^{\circ} \mathrm{C}\) and the earth's surface temperature under the worst conditions is expected to remain at $-10^{\circ} \mathrm{C}$ for 75 days. Take the properties of soil at that location to be \(k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and $\alpha=1.4 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\(. Answer: \)7.05 \mathrm{~m}$

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