Question

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) \(18.1 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short answer

Answer: The convection heat transfer coefficient is approximately 6.52 kW/m²·K, which is closest to option (a) 5.7 kW/m²·K.

Step-by-step solution

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.

Substitute the given values

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

Calculate the Nusselt number

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

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: $$ h = \frac{105.31 \times 0.61 \,\mathrm{W}/\mathrm{m} \cdot \mathrm{K}}{0.01 \,\mathrm{m}} $$

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) \(5.7 \,\mathrm{kW}/\mathrm{m}^2 \cdot \mathrm{K}\).

Key concepts

Nusselt Number

The Nusselt number, often represented as "Nu," is a dimensionless number that plays a crucial role in the study of convection heat transfer. It gives us an idea of how efficient heat transfer is from a surface into a fluid that is flowing past it. In essence, the Nusselt number is a ratio:

• The numerator represents convective heat transfer, which happens due to the motion of the fluid.
• The denominator signifies conductive heat transfer, which takes place through the fluid's own material properties.

The formula for the Nusselt number in this particular exercise is equation: \[\mathrm{Nu} = \frac{hL}{k} \]Where:

• \(h\) is the convection heat transfer coefficient.
• \(L\) is the characteristic length, specifically the side of the square tube in this context.
• \(k\) stands for the fluid's thermal conductivity.

In our problem, the Nusselt number is calculated using the correlation: \[\mathrm{Nu} = 0.102 \mathrm{Re}^{0.625} \operatorname{Pr}^{1/3} \]To get a value for the Nusselt number, you use specific fluid flow characteristics like the Reynolds and Prandtl numbers.

Reynolds Number

The Reynolds number, often abbreviated as "Re," is a dimensionless quantity that helps us distinguish between different kinds of flow patterns in fluid dynamics. It reflects whether the flow will be laminar or turbulent:

• Laminar flow: Smooth and orderly, usually characterized by lower Reynolds numbers.
• Turbulent flow: Chaotic and mixed, often observed at higher Reynolds numbers.

The Reynolds number tells us how significant inertia forces are compared to viscous forces in the fluid. It's expressed by the formula:\[\mathrm{Re} = \frac{\rho v L}{\mu} \]Where:

• \(\rho\) is the fluid density.
• \(v\) is the fluid velocity.
• \(L\) stands for the characteristic length, for our exercise, it's the side of the square tube.
• \(\mu\) is the dynamic viscosity of the fluid.

In the provided scenario, a Reynolds number of 10,000 indicates turbulent flow across the square tube. This high Reynolds number showcases increased convective heat transfer due to the mixing action of turbulent flows.

Prandtl Number

The Prandtl number, symbolized as "Pr," is essential in the analysis of heat transfer in fluids. It gives us a way to relate the fluid's viscosity to its thermal conductivity. This number is also dimensionless and is calculated as follows:\[\operatorname{Pr} = \frac{\mu c_p}{k}\]Where:

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

What's fascinating about the Prandtl number is that it helps decide the relative thickness of the velocity boundary layer versus the thermal boundary layer in a fluid. A higher Prandtl number implies that the thermal layer is thinner compared to the momentum layer.
In this problem, the Prandtl number given for water is 6, suggesting dominantly a thicker momentum layer. The Nusselt number equation uses this to calculate efficient heat transfer during convection processes.