Air \((\operatorname{Pr}=0.7, k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) at \(200^{\circ} \mathrm{C}\) flows across 2-cm-diameter tubes whose surface temperature is \(50^{\circ} \mathrm{C}\) with a Reynolds number of 8000 . The Churchill and Bernstein convective heat transfer correlation for the average Nusselt number in this situation is $$ \mathrm{Nu}=0.3+\frac{0.62 \mathrm{Re}^{0.5} \mathrm{Pr}^{0.33}}{\left[1+(0.4 / \mathrm{Pr})^{0.67}\right]^{0.25}} $$ (a) \(8.5 \mathrm{~kW} / \mathrm{m}^{2}\) (b) \(9.7 \mathrm{~kW} / \mathrm{m}^{2}\) (c) \(10.5 \mathrm{~kW} / \mathrm{m}^{2}\) (d) \(12.2 \mathrm{~kW} / \mathrm{m}^{2}\) (e) \(13.9 \mathrm{~kW} / \mathrm{m}^{2}\)

Step by step solution

01

Calculate the average Nusselt number using the Churchill and Bernstein correlation

First, input the known values into the Churchill and Bernstein correlation for average Nusselt number given below: $$ \mathrm{Nu} = 0.3 + \frac{0.62 \cdot \mathrm{Re}^{0.5} \cdot \mathrm{Pr}^{0.33}}{\left[1 + (0.4/\mathrm{Pr})^{0.67}\right]^{0.25}} $$ Substitute \(\mathrm{Re} = 8000\) and \(\mathrm{Pr} = 0.7\): $$ \mathrm{Nu} = 0.3 + \frac{0.62 \cdot 8000^{0.5} \cdot 0.7^{0.33}}{\left[1 + (0.4/0.7)^{0.67}\right]^{0.25}} $$ Then, calculate the Nusselt number: $$ \mathrm{Nu} \approx 71.033 $$

02

Calculate the heat transfer coefficient, h

Now that we have the Nusselt number, we can find the heat transfer coefficient (\(h\)) using the following formula: $$ h = \frac{\mathrm{Nu} \cdot k}{D} $$ Substitute \(\mathrm{Nu} \approx 71.033\), \(k = 0.026 \frac{\mathrm{W}}{\mathrm{m}\cdot\mathrm{K}}\), and \(D = 0.02\,\mathrm{m}\): $$ h \approx \frac{71.033 \cdot 0.026}{0.02} \\ h \approx 92.5 \frac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{K}} $$

03

Calculate the heat transfer rate, q"

Finally, we can find the heat transfer rate (\(q''\)) using the following formula: $$ q'' = h \cdot (\Delta T) $$ Substitute \(h \approx 92.5 \frac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{K}}\) and \(\Delta T = 200^{\circ}\mathrm{C} - 50^{\circ}\mathrm{C} = 150^{\circ}\mathrm{C}\): $$ q'' \approx 92.5 \cdot 150 $$ Now, convert the heat transfer rate to \(\mathrm{kW}/\mathrm{m}^2\): $$ q'' \approx \frac{92.5 \cdot 150}{1000} \approx 13.875 \,\mathrm{kW}/\mathrm{m}^2 $$ Comparing the calculated heat transfer rate with the given options, we find that the closest answer is (e) \(13.9\,\mathrm{kW}/\mathrm{m}^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!