Question

A \(0.2-\mathrm{m}\)-long and \(25-\mathrm{mm}\)-thick vertical plate \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) separates the hot water from the cold water. The plate surface exposed to the hot water has a temperature of \(100^{\circ} \mathrm{C}\), and the temperature of the cold water is \(7^{\circ} \mathrm{C}\). Determine the temperature of the plate surface exposed to the cold water \(\left(T_{s, c}\right)\). Hint: The \(T_{s, c}\) has to be found iteratively. Start the iteration process with an initial guess of \(53.5^{\circ} \mathrm{C}\) for the \(T_{s, c}\).

Short answer

Answer: The temperature of the plate surface exposed to the cold water, \(T_{s, c}\), is \(47^{\circ} \mathrm{C}\).

Step-by-step solution

Define the heat transfer equation

To determine the temperature of the plate surface exposed to the cold water, we will use Fourier's law of heat transfer, which is given by: \(q = k \cdot A \cdot \frac{(T_{s, h} - T_{s, c})}{L}\) where: - \(q\) is the heat transfer rate; - \(k\) is the thermal conductivity of the plate \((15 \mathrm{W} / \mathrm{m} \cdot \mathrm{K})\); - \(A\) is the area through which heat transfer occurs and can be found by multiplying the plate's length by its thickness \((0.2\mathrm{m} \times 0.025\mathrm{m})\); - \(T_{s, h}\) is the temperature at the hot surface, \((100^{\circ} \mathrm{C})\); - \(T_{s, c}\) is the temperature at the cold surface, which we need to find; and - \(L\) is the thickness of the plate \((0.025\mathrm{m})\).

Set up the initial guess of \(T_{s, c}\)

As suggested in the problem, we will start our iteration with an initial guess for \(T_{s, c}\) of \(53.5^{\circ} \mathrm{C}\).

Calculate the heat transfer rate

Using the initial guess of \(53.5^{\circ} \mathrm{C}\) for \(T_{s, c}\) and the given values, we can calculate the heat transfer rate \((q)\). The area \(A\) can be calculated as \(0.2\mathrm{m} \times 0.025\mathrm{m} = 0.005\mathrm{m^2}\). \(q = 15 \mathrm{W} / \mathrm{m} \cdot \mathrm{K} \cdot 0.005 \mathrm{m}^{2} \cdot \frac{(100 - 53.5)}{0.025\mathrm{m}} = 73.5\mathrm{W}\) Now that we have the heat transfer rate, we can go through several iterations to find the temperature \(T_{s, c}\) accurately.

Iterate to find the accurate value for \(T_{s, c}\)

In each iteration, we will calculate a new value for \(T_{s, c}\) by using the previously calculated value and substituting it back into the heat transfer equation. Repeat this process until the difference between consecutive temperatures is minimal (e.g., less than \(0.01^{\circ} \mathrm{C}\)). New equation for calculating temperature in each iteration is: \(T_{s, c} = T_{s, h} - \frac{q \cdot L}{k \cdot A} \) Initial guess: \(T_{s, c} = 53.5^{\circ} \mathrm{C}\) Iteration 1: \(T_{s, c} = 100 - \frac{73.5 \times 0.025}{15 \times 0.005} = 47^{\circ} \mathrm{C}\) Iteration 2: \(T_{s, c} = 100 - \frac{73.5 \times 0.025}{15 \times 0.005} = 47^{\circ} \mathrm{C}\) Since there is no change in temperature after the second iteration, we can conclude that the temperature of the plate surface exposed to cold water, \(T_{s, c}\), is \(47^{\circ} \mathrm{C}\).

Key concepts

Thermal Conductivity

Thermal conductivity, represented by the symbol k, is a measure of a material's ability to conduct heat. It indicates how easily heat passes through a material due to a temperature difference. The higher the thermal conductivity of a material, the more efficient it is at transferring heat.

As applied in the exercise, a metal plate with thermal conductivity of 15 W/m·K separates hot and cold water, which means the metal plate is quite efficient at conducting heat. If the plate had a lower thermal conductivity, it would act as a better insulator, and the rate at which heat transfers from the hot water side to the cold water side would be slower. Materials with low thermal conductivity are typically used for insulation purposes, while those with high conductivity are used to enhance heat transfer in applications like heat sinks and cooking utensils.

Heat Transfer Rate

The heat transfer rate, denoted as q, reflects the quantity of heat transferred per unit time. It is measured in watts (W) and is derived using Fourier's law of heat transfer which states that heat transfer through a solid is directly proportional to the temperature difference across the solid, the cross-sectional area perpendicular to the heat flow, and the thermal conductivity of the solid, and inversely proportional to the thickness of the material.

In the exercise provided, we used the temperatures of the hot and cold water surfaces, the thermal conductivity of the plate (15 W/m·K), and the dimensions of the plate to calculate the rate at which heat is conducted from the hot side to the cold side. Sufficiently understanding how heat transfer rate is influenced by these factors is pivotal for students to grasp thermal processes in devices like radiators, heat exchangers, and even in natural phenomena like the cooling of hot objects in a room.

Iterative Calculation

Iterative calculation is a mathematical process used to find an accurate answer through a series of successive approximations. In each iteration, the previous approximation is used to calculate a new, and typically more accurate, estimate.

In the context of our exercise, the temperature of the cold surface of the plate, T_{s, c}, was unknown. The initial guess was provided as 53.5°C, starting the iterative process. After calculating the heat transfer rate q, this value was then used to make a new estimate of T_{s, c} through the simplified heat transfer equation. The iterative process continued until the new estimate of T_{s, c} stopped changing significantly, indicating that the solution had been adequately refined.

This type of calculation is useful for complex problems where an explicit solution is not possible or practical to determine. Iterative methods are widely used in various scientific and engineering computations, including numerical analysis, computer science, and physics.