A long 35-cm-diameter cylindrical shaft made of stainless steel \(304\left(k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=477 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(\alpha=3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) comes out of an oven at a uniform temperature of \(400^{\circ} \mathrm{C}\). The shaft is then allowed to cool slowly in a chamber at \(150^{\circ} \mathrm{C}\) with an average convection heat transfer coefficient of \(h=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the shaft \(20 \mathrm{~min}\) after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Step by step solution

01

Determine the initial conditions and conversion of given values

Initially, note down the given values and convert the units provided into SI units where necessary. In our case, the time needs to be converted from minute to seconds. - Initial temperature of the shaft, \(T_i = 400 ^\circ C\) - Cooling chamber temperature, \(T_\infty = 150 ^\circ C\) - Thermal conductivity, \(k = 14.9 W/(m \cdot K)\) - Density, \(\rho = 7900 kg/m^3\) - Specific heat, \(c_p = 477 J/(kg \cdot K)\) - Shaft diameter, \(D = 0.35m\) - Convection heat transfer coefficient, \(h = 60 W/(m^2 \cdot K)\) - Time, \(t = 20 \times 60 = 1200s\)

02

Calculate the Fourier number

Now, let's calculate the Fourier number (\(Fo\)) which is given by the formula, \(Fo = \frac{\alpha t}{(D/2)^2}\), where \(\alpha = k/(\rho c_p)\). First, find the value of \(\alpha\): \(\alpha = \frac{14.9}{7900 \times 477} = 3.95 \times 10^{-6} m^2/s\) Now calculate the Fourier number: \(Fo = \frac{(3.95 \times 10^{-6} \times 1200)}{(0.35/2)^2} = 0.162\)

03

Calculate the Biot number

Next, calculate the Biot number (\(Bi\)) using the following formula, \(Bi = \frac{h(D/2)}{k}\) Now plug in the values to find \(Bi\): \(Bi = \frac{60(0.35/2)}{14.9} = 0.703\)

04

Analyze the lumped capacitance approach's suitability and calculate the temperature

Since \(Bi \times Fo > 0.1\), we cannot use the lumped capacitance approach and have to use the analytical one-term approximation method. With \(Bi = 0.703\) and \(Fo = 0.162\), the dimensionless temperature is found from the analytical one-term approximation as \(\theta = 0.827\). Therefore, \(T(t) = T_\infty + \theta \times (T_i - T_\infty)\) Now substitute the values to find the temperature after 20 minutes: \(T(t) = 150 + 0.827 \times (400 - 150) = 356.9 ^\circ C\)

05

Calculate the heat transfer per unit length of the shaft

To compute the heat transfer per unit length of the shaft, we use the total energy equation \(q_L = \rho \cdot V \cdot c_p \cdot (T_i - T_\infty) = m \cdot c_p \cdot (T_i - T_\infty)\) Where \(m\) is the mass of the steel per unit length given by: \(m = \rho \cdot V = \rho \cdot \pi (\frac{D}{2})^2\) per unit length Now use the values provided to find the mass per unit length of the steel: \(m = 7900 \cdot \pi (\frac{0.35}{2})^2 = 3221.86 kg/m\) Now substitute the values into the total energy equation and find \(q_L\): \(q_L = 3221.86 \cdot 477 \cdot (400 - 356.9) = 6.278 \times 10^6 J/m\) The temperature at the center of the shaft after 20 minutes of cooling is \(356.9 ^\circ C\), and the heat transfer per unit length during this time is \(6.278 \times 10^6 J/m\).

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