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Chapter 6: Problem 14

During air cooling of steel balls, the convection heat transfer coefficient is determined experimentally as a function of air velocity to be \(h=17.9 V^{0.54}\) for \(0.5

Short Answer

Expert verified
Question: Calculate the initial values of heat flux and temperature gradient in the steel ball at the surface given the following parameters: air velocity, diameter, initial temperature, and thermal conductivity of the steel ball; air temperature and air velocity during cooling. Answer: The initial values of heat flux and temperature gradient in the steel ball at the surface are 7167.76 W/m² and -477.85 K/m, respectively.

Step by step solution

01

To determine the convection heat transfer coefficient (h) based on the given formula and air velocity, simply substitute the given air velocity (V) into the formula: \(h = 17.9 V^{0.54}\) From the exercise, we have \(V=1.5\mathrm{~m} / \mathrm{s}\), so: \(h = 17.9 (1.5)^{0.54} = 24.72 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) #Step 2: Calculate the heat flux using Newton's Law of Cooling#

Newton's Law of Cooling states that the heat flux (q) is proportional to the difference in temperature between the surface and the air. Mathematically, this can be represented as: \(q = h (T_s - T_\infty)\) where \(T_s\) is the surface temperature of the steel ball, \(T_\infty\) is the air temperature, and h is the convection heat transfer coefficient. From the exercise, we have: \(T_s = 300 ^\circ \mathrm C\) \(T_\infty = 10 ^\circ \mathrm C\) Substitute these values and the value of h into Newton's Law of Cooling: \(q = 24.72 (300-10) = 7167.76 \mathrm{~W} / \mathrm{m}^2\) #Step 3: Calculate the temperature gradient using Fourier's Law#
02

According to Fourier's Law, the temperature gradient is given by: \(\frac{dT}{dx} = -\frac{q}{k}\) where \(\frac{dT}{dx}\) is the temperature gradient, q is the heat flux, and k is the thermal conductivity of the steel ball. From the exercise, we have: \(k = 15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) Substitute the values of q and k into Fourier's Law: \(\frac{dT}{dx} = -\frac{7167.76}{15} = -477.85 \mathrm K / \mathrm m\) #Step 4: Report the initial values of heat flux and temperature gradient#

Based on our calculations, the initial values of the heat flux and temperature gradient in the steel ball at the surface are as follows: Heat flux (q): \(7167.76 \mathrm{~W} / \mathrm{m}^2\) Temperature gradient (\(\frac{dT}{dx}\)): \(-477.85 \mathrm{~K} / \mathrm{m}\)

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Most popular questions from this chapter

Consider two identical small glass balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the container first? Why?

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Consider an airplane cruising at an altitude of \(10 \mathrm{~km}\) where standard atmospheric conditions are \(-50^{\circ} \mathrm{C}\) and \(26.5 \mathrm{kPa}\) at a speed of \(800 \mathrm{~km} / \mathrm{h}\). Each wing of the airplane can be modeled as a \(25-\mathrm{m} \times 3-\mathrm{m}\) flat plate, and the friction coefficient of the wings is \(0.0016\). Using the momentum-heat transfer analogy, determine the heat transfer coefficient for the wings at cruising conditions. Answer: \(89.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)
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