Question

Air at \(5^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), is used for cooling metal plates coming out of a heat treatment oven at an initial temperature of \(300^{\circ} \mathrm{C}\). The plates \((k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) have a thickness of \(10 \mathrm{~mm}\). Using EES (or other) software, determine the effect of cooling time on the temperature gradient in the metal plates at the surface. By varying the cooling time from 0 to \(3000 \mathrm{~s}\), plot the temperature gradient in the plates at the surface as a function of cooling time. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

Short answer

Include a brief explanation of the lumped system analysis used to solve the problem. Answer: Cooling time affects the temperature gradient in metal plates at the surface by allowing heat to dissipate, causing the temperature to decrease over time. By using lumped system analysis, we verified its applicability through calculating the Biot number, time constant, and temperature-time equation. This allowed us to calculate the surface temperature gradient for various cooling times and observe how the gradient changes as time progresses. The plot of the temperature gradient as a function of cooling time helps visualize this change in the metal plates' surface temperature gradient over time.

Step-by-step solution

Calculate the Biot number to verify applicability of lumped system analysis

To check if the lumped system analysis can be applied to this problem, we need to calculate the Biot number (Bi). The Biot number is given by the following formula: \(Bi = \frac{hL_{c}}{k}\) where \(h\) is the convection heat transfer coefficient, \(L_{c}\) is the characteristic length, and \(k\) is the thermal conductivity of the material. For a flat plate, we can approximate the characteristic length as \(L_{c} = \frac{thickness}{2}\). Given \(h = 30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(k = 180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and thickness = \(10 \mathrm{~mm}\), we can calculate the Biot number.

Compute the time constant and equation for temperature using lumped system analysis

If the Biot number is less than 0.1, then we can apply the lumped system analysis. The lumped system analysis uses the following equation to calculate the temperature of the object as a function of time: \(T(t) = T_{\infty} + (T_{i} - T_{\infty})e^{-t/\tau}\) where \(T(t)\) is the temperature of the object at time \(t\), \(T_{\infty}\) is the surroundings temperature, \(T_{i}\) is the initial temperature of the object, and \(\tau\) is the time constant. The time constant is given by the following formula: \(\tau = \frac{\rho c_{p} L_{c}}{h}\) Given \(\rho = 2800 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p} = 880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and the values calculated in Step 1, we can compute the time constant.

Calculate surface temperature gradient at varying cooling times

Once we know the time constant, we can compute the temperature of the object at the surface for different times between 0 and 3000 seconds by substituting the values of \(T_{\infty} = 5^{\circ} \mathrm{C}\), \(T_{i} = 300^{\circ} \mathrm{C}\), and the time constant into the temperature-time equation obtained in Step 2. To find the temperature gradient, we differentiate the temperature equation with respect to time: \(\frac{dT(t)}{dt} = - \frac{(T_{i} - T_{\infty})}{\tau} e^{-t/\tau}\) Then, substitute the values of \(T_{\infty}\), \(T_{i}\), and \(\tau\) into the equation to calculate the temperature gradient values at each second between 0 and 3000 seconds.

Plot the temperature gradient as a function of cooling time

With the temperature gradient values obtained in Step 3, we can plot the temperature gradient as a function of cooling time in the range of 0 to 3000 seconds. This plot will help us visualize how the surface temperature gradient in the metal plates changes due to time. By following these steps, we can determine the effect of cooling time on the temperature gradient in the metal plates at the surface using the lumped system analysis and verify its applicability.

Key concepts

Convection Heat Transfer Coefficient

The convection heat transfer coefficient (\( h \)) is a crucial factor in thermal engineering that describes the rate of heat transfer between a surface and a fluid moving past it. In our exercise, air is used as a cooling medium with a given convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Let's break this down: The convection heat transfer coefficient is measured in watts per meter squared per kelvin (\(W/m^2K\)), which signifies the amount of heat transferred per unit area of the surface, per unit temperature difference between the surface and the fluid. A higher \( h \) value means more efficient heat transfer, which is vital for applications like cooling metal plates after heat treatment, as mentioned in the exercise. It represents the convective heat transfer occurring due to the movement of air - in essence, the higher the value, the better the cooling effect. Understanding how to calculate and utilize this coefficient is essential in the design of cooling systems and thermal management scenarios.

Temperature Gradient

The temperature gradient is an expression of how the temperature changes with distance, which in the context of our cooling metal plates, is measured at the surface as a function of cooling time. It represents the rate of temperature decrease at the surface as time progresses.

Why does this matter? The temperature gradient can tell us a lot about the rate at which heat is being removed from the metal plates. In the exercise, users would calculate the temperature gradient by differentiating the temperature equation with respect to time. This gives us \(dT(t)/dt\), which, when applied over the range of cooling times between 0 and 3000 seconds, will show us how quickly the surface temperature drops.

A steeper gradient means a more rapid change in temperature, which could be advantageous for rapid cooling processes. However, extremely fast cooling might result in thermal stresses or other undesirable effects. The plot created in the solution would provide a visual representation and help to understand and predict the cooling behavior of the metal plates.

Biot Number

The Biot number (Bi) is a dimensionless quantity used in heat transfer calculations. It is essential when deciding if the lumped system analysis is suitable for a particular problem. The Biot number gives us a ratio of the internal resistance to heat conduction within a body to the external resistance to heat convection from the surface of the body to the surrounding fluid.

The formula, \(Bi = \frac{hL_c}{k}\), requires the convection heat transfer coefficient \(h\), the characteristic length \(L_c\), and the thermal conductivity \(k\) of the material in question. In the exercise, the characteristic length is approximated as half the thickness of the plate, which is quite common in heat transfer problems involving flat plates.

A low Biot number (Bi < 0.1) suggests that the temperature within the object can be approximated as being uniform - hence the justification for using lumped system analysis in the exercise. This analysis simplifies complex heat transfer scenarios, making it manageable to calculate how an object's temperature changes over time, as demonstrated in the steps of the solution.