The top surface of a metal plate \(\left(k_{\text {plate }}=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) is being cooled by air \(\left(k_{\text {air }}=0.243 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) while the bottom surface is exposed to a hot steam at \(100^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the bottom surface temperature of the plate is \(80^{\circ} \mathrm{C}\), determine the temperature gradient in the air and the temperature gradient in the plate at the top surface of the plate.

Short Answer

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Question: Calculate the temperature gradients in both the metal plate and air at the top surface, given the following values: T_surface = 80°C, T_air = 100°C, thermal conductivity of the metal plate (k_plate) = 237 W/m·K, convection heat transfer coefficient (h) = 30 W/m²·K, and thermal conductivity of the air (k_air) = 0.243 W/m·K.

Step by step solution

01

Fourier's Law of Heat Conduction

Fourier's Law of Heat Conduction relates the rate of heat transfer (q) through a material to its thermal conductivity (k), the thickness of the material (L) and the temperature gradient (dT/dx): $$q = -k \cdot L \frac{dT}{dx}$$
02

Convection Heat Transfer Equation

The Convection Heat Transfer Equation relates the rate of heat transfer (q) to the convection heat transfer coefficient (h), the area through which the heat transfer is occurring (A) and the temperature difference between surface and surrounding fluid (ΔT): $$q = h \cdot A\cdot \Delta T$$
03

Set up Metal Plate Temperature Gradient Equation

We know the thermal conductivity (k) of the metal plate, the material's thickness (x), and the bottom surface temperature (T1). Using the Fourier's Law of Heat Conduction, we can write: $$q = - k_{\text{plate}} \cdot \frac{dT}{dx}$$
04

Setup Air Temperature Gradient Equation

We know the convection heat transfer coefficient (h) for the air, and the difference between surface temperature and surrounding fluid (∆T = T_surface - T_air). Using the Convection Heat Transfer Equation, we can write: $$q = h \cdot (T_{\text{surface}} - T_{\text{air}})$$
05

Set up the Equations and Solve for Temperature Gradient

Since the heat transfer rate (q) at the top surface of the metal plate is the same for both metal plate and air, we can set the two equations equal to each other: $$-k_{\text{plate}} \cdot \frac{dT}{dx} = h \cdot (T_{\text{surface}} - T_{\text{air}})$$ Using the given temperature values T_surface = 80°C and T_air = 100°C and given thermal conductivity (k) and convection heat transfer coefficient (h), we can solve for the temperature gradients. For the metal plate: $$\frac{dT}{dx} = -\frac{h \cdot (T_{\text{surface}} - T_{\text{air}})}{k_{\text{plate}}}$$ For the air: $$\frac{dT}{dx} = \frac{h\cdot(100 - 80)}{-k_{\text{air}}}$$
06

Calculate the Temperature Gradients

Now, calculate the gradient values for metal plate and air using the given values: $$\frac{dT}{dx}_{\text{plate}} = -\frac{(30)(100 - 80)}{237} = -\frac{600}{237} = -2.53~\frac{^\circ \text{C}}{m}$$ $$\frac{dT}{dx}_{\text{air}} = \frac{(30)(100 - 80)}{0.243}=\frac{600}{0.243}=2469.14~\frac{^\circ \text{C}}{m}$$
07

Interpret the Results

The temperature gradient in the metal plate at the top surface is -2.53 °C/m, which means that the temperature decreases 2.53 °C for every meter moving up the plate. The temperature gradient in the air is 2469.14 °C/m, which means that the temperature increases 2469.14 °C for every meter moving up the surface.

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Key Concepts

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

Fourier's Law of Heat Conduction
Understanding how heat moves through materials is crucial in many scientific and engineering contexts. This is where Fourier's Law of Heat Conduction comes into play. It states that the rate at which heat is transferred by conduction is proportional to the negative gradient of temperatures and the area through which the heat is flowing. Mathematically, it's expressed as:

\[q = -k \times A \times \frac{dT}{dx}\]
In this formula, \(q\) represents the heat transfer rate per unit area (in Watts per square meter), \(k\) is the material's thermal conductivity, \(A\) is the cross-sectional area of heat flow, and \(\frac{dT}{dx}\) is the temperature gradient in the direction of the heat flow. The negative sign indicates that heat flows from higher to lower temperatures. When applied to the metal plate from the exercise, we can determine the heat flowing through the plate by knowing its thermal conductivity and the temperature difference across its thickness.
Convection Heat Transfer
Heat doesn't only transfer through solid materials; it also passes between a surface and a fluid (such as air or water) moving over it. This transfer by movement of fluids is known as Convection Heat Transfer. The rate at which heat is transferred by convection can be described by the equation:

\[q = h \times A \times \Delta T\]
Here, \(q\) is the heat transfer rate per unit area (in Watts per square meter), \(h\) is the convective heat transfer coefficient, \(A\) is the surface area, and \(\Delta T\) is the temperature difference between the surface and the fluid. Convection can be either natural (or free) when it's caused by buoyancy effects due to density differences, or forced when an external source like a fan causes the fluid motion. In the exercise, by setting up an equation using the air's convection heat transfer coefficient, we can figure out how effectively the air is removing heat from the metal plate's surface.
Thermal Conductivity
The thermal conductivity of a material, often denoted by \(k\), is a measure of its ability to conduct heat. It's defined as the amount of heat that will transfer through one meter of the material when there's a temperature difference of one degree Celsius across that material. It's measured in watts per meter-kelvin (W/m·K). Materials with high thermal conductivity, like metals, are good conductors of heat, whereas materials with low thermal conductivity, like plastics or wood, are considered insulators.

High thermal conductivity is a key property in applications where rapid heat dissipation is crucial, such as in cooling systems for electronic components. Conversely, low thermal conductivity is important in applications like insulation where the goal is to minimize heat transfer. In our metal plate scenario, the high thermal conductivity allows for quick heat transfer through the plate from the hot steam to the cooler air.
Heat Transfer Coefficient
The heat transfer coefficient, indicated as \(h\) in the context of convection, provides a measure of convection heat transfer from a solid surface to a fluid, or vice versa. It's a crucial factor in distinguishing how well a particular surface will transfer heat to a fluid flowing over it. The value of \(h\) can vary widely depending on several factors, including the type of fluid, its velocity, the surface geometry, and the nature of the flow (laminar or turbulent).

Higher values of \(h\) indicate more efficient convection heat transfer capabilities, which can be desirable or undesirable based on the application. For example, in our exercise, an air stream with a given \(h\) is tasked with cooling a metal plate. The effectiveness of this cooling process can be calculated with the provided value, enabling a deeper understanding of the cooling dynamics at play.

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Most popular questions from this chapter

Air flows over a flat plate at \(40 \mathrm{~m} / \mathrm{s}, 25^{\circ} \mathrm{C}\) and 1 atm pressure. (a) What plate length should be used to achieve a Reynolds number of \(1 \times 10^{8}\) at the end of the plate? ( \(b\) ) If the critical Reynolds number is \(5 \times 10^{5}\), at what distance from the leading edge of the plate would transition occur?

A 5 -mm-thick stainless steel strip \((k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=\) \(8000 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is being heat treated as it moves through a furnace at a speed of \(1 \mathrm{~cm} / \mathrm{s}\). The air temperature in the furnace is maintained at \(900^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the furnace length is \(3 \mathrm{~m}\) and the stainless steel strip enters it at \(20^{\circ} \mathrm{C}\), determine the surface temperature gradient of the strip at mid-length of the furnace. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

What is external forced convection? How does it differ from internal forced convection? Can a heat transfer system involve both internal and external convection at the same time? Give an example.

A flat plate is subject to air flow parallel to its surface. The average friction coefficient over the plate is given as $$ \begin{aligned} &C_{f}=1.33\left(\operatorname{Re}_{L}\right)^{-1 / 2} \text { for } \operatorname{Re}_{L}<5 \times 10^{5} \text { (laminar flow) } \\ &C_{f}=0.074\left(\operatorname{Re}_{L}\right)^{-1 / 5} \text { for } 5 \times 10^{5} \leq \operatorname{Re}_{L} \leq 10^{7} \text { (turbulent flow) } \end{aligned} $$ The plate length parallel to the air flow is \(1 \mathrm{~m}\). Using EES (or other) software, determine the effect of air velocity on the average convection heat transfer coefficient for the plate. By varying the air velocity for \(0

Consider two identical small glass balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the container first? Why?

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