91Ó°ÊÓ

A plane wall has constant properties, no internal heat generation, and is initially at a uniform temperature \(T_{i \cdot}\) Suddenly, the surface at \(x=L\) is heated by a fluid at \(T_{\infty}\) having a convection coefficient \(h\). At the same instant, the electrical heater is energized, providing a constant heat flux \(q_{o}^{\prime \prime}\) at \(x=0\). (a) On \(T-x\) coordinates, sketch the temperature distributions for the following conditions: initial condition \((t \leq 0)\), steady-state condition \((t \rightarrow \infty)\), and for two intermediate times. (b) On \(q_{x}^{\prime \prime}-x\) coordinates, sketch the heat flux corresponding to the four temperature distributions of part (a). (c) On \(q_{x}^{n}-t\) coordinates, sketch the heat flux at the locations \(x=0\) and \(x=L\). That is, show qualitatively how \(q_{x}^{\prime \prime}(0, t)\) and \(q_{x}^{\prime \prime}(L, t)\) vary with time. (d) Derive an expression for the steady-state temperature at the heater surface, \(T(0, \infty)\), in terms of \(q_{o}^{\prime \prime}\), \(T_{\infty}, k, h\), and \(L\).

Step by step solution

01

Initial condition

 As the entire wall is initially at a uniform temperature, Ti, the temperature distribution on the T-x coordinates will be a horizontal line.

02

Steady-state condition

At steady-state, temperature distribution is linear between the two boundaries. The surface at x=L is heated by the fluid at T∞, and electrical heater at x=0 is providing a constant heat flux q₀′′. Draw a straight line from the heated surface (x=0) to the cooled surface (x=L) on T-x coordinates.

03

Intermediate times

For the intermediate times, the temperature distribution will be an increasingly linear curve as time passes. Sketch two curves in between the initial and steady-state conditions; these will show the temperature distribution for the intermediate times. (b) Sketching Heat Flux distributions:

04

Heat flux for the initial condition

The heat flux for the initial condition will be zero as there is no heat transfer initially, so this will be a horizontal line at qx'' = 0.

05

Heat flux for the steady-state condition

At steady-state, the heat flux corresponding to the temperature distributions of part (a) will be constant along the x-direction since qx'' is a constant given by Fourier's Law. So, draw a horizontal line at the constant heat flux value.

06

Heat flux for intermediate times

For the intermediate times, the heat flux will gradually reach the steady-state value from an initially zero heat flux. Sketch two curves that show the heat flux distribution for the intermediate times. (c) Sketching Heat Flux at x=0 and x=L with respect to time:

07

Heat flux at x=0

On the qx′′-t coordinates, sketch a horizontal line starting from the value of q₀′′ at t=0, representing the heat flux at x=0, which remains constant with time.

08

Heat flux at x=L

At x=L, the heat flux varies with time and converges to a steady-state value as time goes to infinity. Sketch a curve starting from zero heat flux at t=0 that approaches the steady-state value of qx'' as t→∞. (d) Deriving steady-state temperature at the heater surface, T(0, ∞):

09

Applying Fourier's Law along the x-direction

Apply Fourier's Law to compute the heat transfer across the wall: \(-k \frac{dT}{dx}|_{x=0} = qâ‚€''\)

10

Applying Convection and Fourier's Law at x=L

Apply convection and Fourier's Law at x = L: \(-k \frac{dT}{dx}|_{x=L} = h(T^\infty - T(L))\)

11

Setting up the steady-state temperature equation

We have two equations and two unknowns (T(0) and T(L)). Solve these equations to find T(0, ∞) in terms of the given variables: \(T(0, \infty) = T(L, \infty) - \frac{k}{hL} (-q₀'' - h(T^\infty - T(L)))\)

12

Simplify the expression

Simplify the expression to get the final formula for T(0, ∞): \(T(0, \infty) = T^\infty + \frac{q₀''}{h} + \frac{k}{hL} q₀''\) This expression gives the steady-state temperature at the heater surface in terms of given variables q₀′′, T∞, k, h, and L.

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Key Concepts

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

Convection Coefficient

The convection coefficient, often denoted as \(h\), plays a critical role in problems involving heat transfer between a solid surface and a fluid in motion. It quantifies the rate at which heat is transferred per unit area per unit temperature difference between the surface and the fluid.

For instance, in our textbook exercise, the surface at \(x=L\) is being cooled by a fluid, and the convection coefficient helps determine how efficiently the fluid can remove heat from the surface. In real-life applications, it's a parameter influenced by fluid properties, flow velocity, and the nature of the surface itself.

When dealing with heat transfer problems, understanding how to calculate the convection coefficient—whether through empirical correlations for different flow conditions or from experimental data—is essential for accurate thermal analysis and design.

Heat Flux

Heat flux, represented by \(q''\), is a measure of the rate of heat energy transfer through a given surface area. It is usually expressed in watts per square meter \(\text{W/m}^2\). In our heat transfer scenario, we're looking at a constant heat flux \(q_{o}^{''}\) applied to the surface at \(x=0\), simulating an electrical heater.

Understanding heat flux is important for managing the thermal performance of systems and ensuring materials are not exposed to temperatures that may cause damage. During analysis, you would sketch how heat flux changes with position and time to visualize the process and better design the thermal system.

Fourier's Law

Fourier's Law is the cornerstone of heat conduction analysis and states that the heat transfer rate through a material is proportional to the negative gradient of temperature and the area through which heat is flowing. Mathematically, it's given by:\[q'' = -k \frac{dT}{dx}\]

In the exercise, Fourier's Law is used to connect the steady-state heat flux \(q_{o}^{''}\) with the temperature gradient at the heater surface. This law also informs the slope of the temperature distribution in steady-state conditions, as seen in the exercise's sketches.

Steady-State Temperature

Steady-state temperature refers to a condition where the temperature in a system does not change with time. In other words, the temperature distribution becomes time-independent. For the given exercise, we derive the steady-state temperature at the heater surface in part (d), which is the temperature after the system has stabilized and there are no further changes in temperature over time.

Reaching a steady-state condition is critical in many engineering applications for performance analysis and system design, ensuring that operational temperatures do not exceed safe or functional limits for the materials involved.

Thermal Conductivity

Thermal conductivity, denoted by \(k\), is a material property that indicates how well a material conducts heat. It is central to the exercise as it directly affects the rate at which heat is transferred through the wall by conduction.

In the provided solution, \(k\) is used to relate the heat flux \(q_{o}^{''}\) to the temperature gradient at the surface of the heater. Thermal conductivity varies between materials and is critical when selecting materials for their thermal management properties in designing and analyzing heat transfer systems.