Question

During air cooling of a flat plate \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), the convection heat transfer coefficient is given as a function of air velocity to be \(h=27 V^{0.85}\), where \(h\) and \(V\) are in \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\mathrm{m} / \mathrm{s}\), respectively. At a given moment, the surface temperature of the plate is \(75^{\circ} \mathrm{C}\) and the air \((k=0.266 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) temperature is \(5^{\circ} \mathrm{C}\). Using EES (or other) software, determine the effect of the air velocity \((V)\) on the air temperature gradient at the plate surface. By varying the air velocity from 0 to \(1.2 \mathrm{~m} / \mathrm{s}\) with increments of \(0.1 \mathrm{~m} / \mathrm{s}\), plot the air temperature gradient at the plate surface as a function of air velocity.

Short answer

Question: Calculate and plot the air temperature gradient at the plate surface as a function of different air velocities ranging from 0 to 1.2 m/s.

Step-by-step solution

Calculate the Convection Heat Transfer Coefficient (h)

We are given the formula to calculate the convection heat transfer coefficient as a function of air velocity: \(h = 27V^{0.85}\). So, for each air velocity, we will calculate h.

Calculate the Heat Transfer by Convection (Q_conv)

We can use the convection heat transfer equation to find the heat transfer from the plate to the air: \(Q_{conv} = hA(T_{s} - T_{air})\), where A is the surface area, \(T_{s}\) is the surface temperature of the plate, and \(T_{air}\) is the air temperature. In this exercise, we're given \(T_{s} = 75^{\circ} C\) and \(T_{air} = 5^{\circ} C\). But, since we want to analyze only the effect of air velocity on the air temperature gradient, we can ignore the factor \(A(T_s - T_{air})\) and consider only the h values we calculated in Step 1 for each air velocity.

Calculate the Heat Transfer by Conduction (Q_cond)

Using Fourier's Law of heat conduction, we can calculate the heat transfer by conduction as \(Q_{cond} = k_{air}A\frac{dT_{air}}{dy}\), where \(k_{air}\) is the thermal conductivity of the air (given as \(0.266 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)), \(dT_{air}\) is the temperature difference, and \(dy\) is the distance perpendicular to the surface of the plate. Since we want to analyze only the effect of air velocity on the air temperature gradient, we can ignore the factor \(k_{air}A\) and consider only the part of the equation involving the air temperature gradient: \(\frac{dT_{air}}{dy}\).

Equate Heat Transfer by Conduction and Convection

For steady-state heat transfer conditions, the heat transfer due to conduction and convection are equal: \(Q_{cond} = Q_{conv}\). So, for each air velocity, we can equate the expressions for heat transfer and solve for the air temperature gradient (\(\frac{dT_{air}}{dy}\)). Equating these expressions, we get: \[h = k_{air} \frac{dT_{air}}{dy}\]

Solve for Air Temperature Gradient

Now, we can find the air temperature gradient by solving the equation obtained in Step 4 for each air velocity: \[\frac{dT_{air}}{dy} = \frac{h}{k_{air}}\] Using the h values calculated in Step 1 and the given value of \(k_{air}\), we can find the air temperature gradient for each air velocity.

Plot Air Temperature Gradient vs. Air Velocity

For each air velocity from 0 to 1.2 m/s with increments of 0.1 m/s, calculate the air temperature gradient using the expressions derived in the previous steps, and then plot the air temperature gradient as a function of air velocity. This plot will show us the effect of the air velocity on the air temperature gradient at the plate surface.

Key concepts

Air Velocity

In the world of convection heat transfer, air velocity plays a crucial role. Air velocity essentially refers to how fast the air flows over the surface in question, like in this exercise where the air flows over a flat plate. The speed at which air moves can affect how heat is transferred from the plate to the air.

• Higher air velocity tends to increase the heat transfer coefficient.
• This happens because fast-moving air can carry heat away more efficiently.
• In this exercise, the relationship between the heat transfer coefficient (h) and air velocity (V) is given by the formula: \(h = 27V^{0.85}\).

By knowing the air velocity, we can predict how effective the cooling process will be. So, adjusting air velocity can be a powerful way to control temperatures in various applications.

Temperature Gradient

Understanding the temperature gradient is essential in analyzing heat transfer. The temperature gradient is the rate at which temperature changes in a material. It is particularly significant when thinking about heat moving through substances, like from the flat plate to the air.In this exercise, we are interested in the temperature gradient at the surface of the plate. The temperature gradient is influenced by several factors, including:

• The difference in temperature between the plate surface and the air.
• The thermal properties of the air, like its thermal conductivity.
• The air velocity, which affects how quickly heat is removed from the surface.

This gradient is measured in units of temperature change per distance, often \( \frac{^{\circ}C}{m} \). Understanding this concept allows us to analyze how efficiently the heat is being transferred from the plate into the air.

Fourier's Law

Fourier's Law is a fundamental principle governing conduction heat transfer. It describes how heat moves through materials due to temperature differences. Simply put, it states that the rate of heat transfer by conduction is proportional to the negative gradient of temperatures and to the material's thermal conductivity.In mathematical terms, Fourier's Law is given by:\[ Q_{cond} = -k \frac{dT}{dx}\]Where:

• \(Q_{cond}\) is the heat transfer by conduction.
• \(k\) is the material's thermal conductivity.
• \(\frac{dT}{dx}\) is the temperature gradient.

Applying this to air flow over the plate, Fourier’s Law helps us calculate how quickly heat is conducted from the air near the surface. By understanding this law, we can better control heat flows in engineering designs.

Steady-State Heat Transfer

In steady-state heat transfer, the temperature of the system doesn't change with time. That means the amount of heat entering any part of the system equals the heat leaving.In our exercise, assuming steady-state conditions simplifies our calculations. Since the heat is not accumulating or depleting in the plate, we equate convection and conduction rates:\[ Q_{cond} = Q_{conv} \]Essentially, this means that:

• The heat being conducted into the air equals the heat being convected away into it.
• It helps us determine the temperature gradient when air flows steadily over the surface.

This assumption of steady-state allows for the simplification that helps solve for the air temperature gradient concerning varying air velocities.

Thermal Conductivity

Thermal conductivity is a property that defines a material's ability to conduct heat. It plays an important role in determining how heat is transferred by conduction.In the described exercise, we deal with the thermal conductivity of air, represented as \(k_{air}\). It is given as 0.266 W/m·K in the problem. Higher conductivity means heat moves more easily through the material.

• Knowing conductivity helps calculate conduction rates in materials using Fourier’s Law.
• We see how it affects the temperature distribution at the plate's surface.

By understanding thermal conductivity, engineers can select and design materials that properly manage heat flow in applications like cooling systems and electronics. High or low conductivity materials might be chosen depending on whether you want heat to be retained or removed.