91Ó°ÊÓ

The convective heat transfer coefficient around a cylinder held perpendicular to a flow varies in a complicated manner. A test cylinder to investigate this behavior consists of a \(0.001\) in-thick, \(12.7 \mathrm{~mm}\)-wide stainless steel heater ribbon (cut from shim stock) wound around a \(2 \mathrm{~cm}\)-O.D., \(2 \mathrm{~mm}\)-wall-thickness Teflon tube. A single thermocouple is located just underneath the ribbon and measures the local ribbon temperature \(T_{s}(\theta)\). The cylinder is installed in a wind tunnel, and a second thermocouple is used to measure the ambient air temperature \(T_{e}\). The power input to the heater is metered, from which the electrical heat generation per unit area \(\dot{Q} / A\) can be calculated ( \(A\) is the surface area of one side of the ribbon). As a first approximation, the local heat transfer coefficient \(h_{c}(\theta)\) can be obtained from $$ h_{c}(\theta)=\frac{\dot{Q} / A}{T_{s}(\theta)-T_{e}} $$ Hence, by rotating the cylinder with the power held constant, the variation of \(h_{c}\) can be obtained from the variation of \(T_{s}\) : where \(T_{s}(\theta)\) is low, \(h_{c}(\theta)\) is high, and vice versa. A typical variation of \(T_{s}(\theta)\) is shown in the graph. A problem with this technique is that conduction around the circumference of the tube causes the local heat flux \(q_{s}(\theta)\) to not exactly equal \(\dot{Q} / A\). (i) Derive a formula for \(h_{c}(\theta)\) that approximately accounts for circumferential conduction. (ii) The following table gives values of \(T_{s}(\theta)\) in a sector where circumferential conduction effects are expected to be large. Use these values together with \(\dot{Q} / A=5900 \mathrm{~W} / \mathrm{m}^{2}\) and \(T_{e}=25^{\circ} \mathrm{C}\) to estimate the conduction effect at \(\theta=110^{\circ}\). \begin{tabular}{cc} Angle (degrees) & \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) \\ \hline 100 & \(65.9\) \\ 110 & \(65.7\) \\ 120 & \(64.4\) \end{tabular} (iii) Comment on the design of the cylinder. Would a \(3 \mathrm{~mm}\)-thick brass tube, directly heated by an electric current, be a suitable alternative? Use \(k=15 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the stainless steel and \(0.38 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for Teflon.

Step by step solution

01

Derive Formula for Corrected Heat Transfer Coefficient

Because there is conduction around the circumference, the local heat transfer coefficient needs correction. To account for this, consider that the actual heat flow may differ slightly from the heat input due to conduction. A common approach is to modify the original formula: \[ h_{c}( heta) = \frac{\dot{Q}/A - q_{cond}( heta)}{T_{s}( heta) - T_{e}} \]where \( q_{cond}(\theta) \) is the heat conducted around the cylinder's circumference. Assuming small conduction effects, this formula approximates the local heat transfer coefficient.

02

Calculate Conduction Effect at Specific Angle

For the heat conduction term \( q_{cond}(\theta) \), assume a linear approximation between angles 100° and 120°. The temperature gradient is: \[ \frac{\Delta T}{\Delta \theta} = \frac{64.4^{\circ}C - 65.9^{\circ}C}{120^{\circ} - 100^{\circ}} = -0.075^{\circ}C/^{\circ} \]Now approximate conduction using Fourier's Law for conduction in a cylinder, treating the observed strip as linear: \[ q_{cond}(\theta) \approx -k \left( \frac{\Delta T}{\Delta \theta} \right) \approx -0.38 \times -0.075 = 0.0285 \text{ W/m}^{2}\]

03

Estimate Heat Transfer Coefficient at \(\theta=110^{\circ}\) Adjusted for Conduction

With the conduction correction, compute \( h_{c}(110^{\circ}) \):\[ h_{c}(110^{\circ}) \approx \frac{5900 - 0.0285}{65.7^{\circ}C - 25^{\circ}C} \approx \frac{5899.9715}{40.7} = 144.9 \text{ W/m}^{2}\text{K}\]

04

Evaluate the Cylinder Design

The Teflon tube's relatively low thermal conductivity hinders effective conduction, making it suitable for these tests. Switching to a brass tube with higher conductivity could skew results by allowing more conduction, potentially affecting accuracy. Hence, the current design optimally balance between heat retention and conduction accuracy.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circumferential Conduction

Circumferential conduction refers to the phenomenon where heat transmits around the circumference of a cylindrical object. Imagine heat moving not only straight through a surface but also sideways, spreading in every direction along a circle. In this exercise, we consider a cylinder that's affected by such circumferential conduction, complicating the measurement of heat transfer.

The presence of circumferential conduction means that the local heat flux, or heat flow at a particular spot on the cylinder, isn't exactly what's expected just by looking at the heat that's added to the system. This can make calculations a bit tricky since the heat flow is partially getting distributed along the circumference, rather than just through its thickness.

This effect is important in engineering because it influences how we calculate and understand the heat transfer coefficient at any point around the cylinder. That's why corrections must be made to account for this circumferential conduction, ensuring that heat transfer analysis accurately reflects the physical conditions of the system.

Thermal Conductivity

Thermal conductivity is a material property that indicates a material's ability to conduct heat. Think of it as a measure of how fast heat moves through a material. It's like an indicator of how good a material is at transferring thermal energy from one part to another inside the same material.

In this exercise, different materials are used, including stainless steel and Teflon, each having different thermal conductivities. Stainless steel, with a higher thermal conductivity, means it transfers heat more effectively than Teflon, which has a lower thermal conductivity.

Understanding thermal conductivity helps predict how the materials in a system will behave under thermal stress, and thus engineers use this property to decide which materials to use depending on whether they want more or less heat conduction. This property plays a crucial role in deciding the effectiveness of the convection test using cylinders.

Heat Flux

Heat flux is essentially the rate of heat energy transfer through a given surface, per unit surface area. It’s like figuring out how much heat is "flowing" through a surface like our cylindrical tube in this exercise.

In the cylinder, heat is applied at a certain rate, and this is where heat flux comes into play. The heat generation per unit area, \(\dot{Q} / A\), tells us how much energy is being put into the system. However, with circumferential conduction, the actual local heat flux \(q_{s}(\theta)\) can differ from this value because heat might spread along the surface instead of solely entering through it.

Monitoring the heat flux accurately is crucial for determining the efficiency of a heating application and for ensuring that you measure the convection heat transfer coefficient as precisely as possible.

Temperature Gradient

A temperature gradient is how much the temperature changes per unit of distance. Imagine walking along the surface of the cylinder and noticing how the temperature drops or rises slightly with each step. This gradient drives heat flow along the surface in systems like the cylindrical tube from the exercise.

The gradient can be observed in the temperature difference between two points. Here, the example calculates the gradient between 100° and 120°, showing how temperature decreases as you move along the cylinder.

This gradient is essential because it affects how heat flows due to conduction. A steeper gradient typically indicates a stronger heat transfer. Therefore, understanding and computing the temperature gradient helps us make sense of how heat conduction impacts the overall heat transfer in systems influenced by circumferential conduction.