Question

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) \(13.9 \mathrm{~kW} / \mathrm{m}^{2}\)

Short answer

Question: Calculate the convective heat transfer rate between the air and the surface of the 2-cm-diameter tubes given that the temperature of the air is 50°C, the temperature of the tubes is 200°C, the properties of air are \(k = 0.026\,\mathrm{W}/\mathrm{m}\cdot\mathrm{K}\) and \(Pr = 0.7\), the Reynolds number is 8000, and the Churchill and Bernstein convective heat transfer correlation is given. Choose the correct answer from the following options: (a) 10.2 \(kW/m^2\) (b) 11.7 \(kW/m^2\) (c) 12.3 \(kW/m^2\) (d) 13.4 \(kW/m^2\) (e) 13.9 \(kW/m^2\). Answer: (e) 13.9 \(kW/m^2\)

Step-by-step solution

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: $$ \mathrm{Nu} \approx 71.033 $$

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 \(D = 0.02\,\mathrm{m}\): $$ h \approx \frac{71.033 \cdot 0.026}{0.02} \\ h \approx 92.5 \frac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{K}} $$

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) \(13.9\,\mathrm{kW}/\mathrm{m}^2\).

Key concepts

Nusselt Number

The Nusselt number is a dimensionless parameter that provides insight into the convection heat transfer process. It is used to compare the relative significance of conductive and convective heat transfer across a fluid boundary layer.
The Nusselt number (\(Nu\)) is calculated with equations that often depend on the Reynolds number and the Prandtl number. For instance, in the provided exercise, the Churchill and Bernstein correlation formula is used:\[ \mathrm{Nu} = 0.3 + \frac{0.62 \mathrm{Re}^{0.5} \mathrm{Pr}^{0.33}}{\left[1+(0.4/\mathrm{Pr})^{0.67}\right]^{0.25}} \]Here, substituting known values such as Reynolds Number (\(\mathrm{Re} = 8000\)) and Prandtl Number (\(\mathrm{Pr} = 0.7\)), we derive the Nusselt number to understand the heat transfer magnitude across the tube surface. A higher Nusselt number indicates more effective convection compared to conduction.
Knowing how to determine the Nusselt number is essential for analyzing thermal performance in practical applications, such as tube heat exchangers or air conditioning systems.

Reynolds Number

Reynolds number is another key dimensionless quantity in fluid mechanics and heat transfer, which helps predict flow patterns in different fluid flow situations. The Reynolds number (\(\mathrm{Re}\)) provides a measure of the ratio between inertial forces and viscous forces present in a fluid flow.
It is computed as:\[ \mathrm{Re} = \frac{\rho v D}{\mu} \]where:

• \(\rho\) is the density of the fluid
• \(v\) is the velocity of the fluid
• \(D\) is the characteristic length (such as the diameter of a tube)
• \(\mu\) is the dynamic viscosity

A low Reynolds number typically indicates laminar flow, where the fluid moves in smooth layers. On the other hand, a high Reynolds number indicates turbulent flow, which enhances mixing and therefore heat transfer. In the example, a Reynolds number of 8000 signifies turbulent flow across the tube, improving heat exchange efficiency.

Prandtl Number

The Prandtl number is a dimensionless number, crucial in the analysis of heat transfer. It signifies the ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity.The Prandtl number (\(\mathrm{Pr}\)) is defined as:\[ \mathrm{Pr} = \frac{\mu c_p}{k} \]where:

• \(\mu\) is the dynamic viscosity
• \(c_p\) is the specific heat of the fluid at constant pressure
• \(k\) is the thermal conductivity

In simpler terms, it gives insight into the thickness of the velocity boundary layer compared to the thermal boundary layer. A typical Prandtl number for gases like air is around 0.7, as in the given problem.
Prandtl numbers influence the convection equations, like the Nusselt number formula used earlier, helping dictate how thermal and momentum boundary layers interact. This can affect the design and efficiency of thermal systems.

Heat Transfer Coefficient

The heat transfer coefficient (\(h\)) represents the convective heat transfer per unit area and temperature difference between a solid surface and the fluid surrounding it. It is critical for designing efficient heat exchanger systems.To calculate it, we use the equation:\[ h = \frac{\mathrm{Nu} \cdot k}{D} \]where:

• \(\mathrm{Nu}\) is the Nusselt number
• \(k\) is the thermal conductivity of the fluid
• \(D\) is the diameter of the tube or the characteristic length

In the given problem, using a Nusselt number of approximately 71.033, we find that the heat transfer coefficient is about 92.5 \(\frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}}\).
This coefficient is essential in calculating how effectively heat is transferred during convection processes and directly affects the rate of heat transfer, influencing the thermal performance of equipment.