Jakob (1949) suggests the following correlation be used for square tubes in a liquid cross-flow situation: $$ \mathrm{Nu}=0.102 \mathrm{Re}^{0.625} \operatorname{Pr}^{1 / 3} $$ Water \((k=0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6)\) flows across a \(1-\mathrm{cm}\) square tube with a Reynolds number of 10,000 . The convection heat transfer coefficient is (a) \(5.7 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(8.3 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(11.2 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(15.6 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(18.1 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Step by step solution

01

Understand the given correlation formula

The given correlation formula is $$ \mathrm{Nu}=0.102 \mathrm{Re}^{0.625} \operatorname{Pr}^{1 / 3} $$ where Nu is the Nusselt number, Re is the Reynolds number and Pr is the Prandtl number. Our goal is to find the convection heat transfer coefficient "h" which can be calculated using the Nusselt number as follows: $$ \mathrm{Nu}=\frac{h L}{k} $$ where L is the characteristic length of the square tube and k is the thermal conductivity.

02

Substitute the given values

We are given the values for Re, Pr and k. We can now substitute them in the correlation equation. $$ \mathrm{Nu}=0.102 (10,000)^{0.625} (6)^{1 / 3} $$

03

Calculate the Nusselt number

We can now calculate the Nusselt number. $$ \mathrm{Nu} \approx 105.31 $$

04

Calculate the convection heat transfer coefficient "h"

Now, we can use the Nusselt number to find the convection heat transfer coefficient "h" using the equation: $$ \mathrm{Nu}=\frac{h L}{k} $$ Rearranging to solve for "h": $$ h = \frac{\mathrm{Nu} \times k}{L} $$ Substitute the values for Nu, k and L (1 cm = 0.01 m) into the equation: $$ h = \frac{105.31 \times 0.61 \,\mathrm{W}/\mathrm{m} \cdot \mathrm{K}}{0.01 \,\mathrm{m}} $$

05

Find the final value and match with the given options

Calculate the final value of the convection heat transfer coefficient "h": $$ h \approx 6516.3 \,\mathrm{W}/\mathrm{m}^2 \cdot \mathrm{K} $$ Converting to kW: $$ h \approx 6.52 \,\mathrm{kW}/\mathrm{m}^2 \cdot \mathrm{K} $$ This is closest to option (a) \(5.7 \,\mathrm{kW}/\mathrm{m}^2 \cdot \mathrm{K}\).

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!