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

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

Step by step solution

01

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 \((0.025\mathrm{m})\).
02

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 \(53.5^{\circ} \mathrm{C}\).
03

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.
04

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 \(47^{\circ} \mathrm{C}\).

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Key Concepts

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

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.

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Most popular questions from this chapter

Consider a 2-m-high electric hot-water heater that has a diameter of \(40 \mathrm{~cm}\) and maintains the hot water at \(60^{\circ} \mathrm{C}\). The tank is located in a small room at \(20^{\circ} \mathrm{C}\) whose walls and ceiling are at about the same temperature. The tank is placed in a 44-cm-diameter sheet metal shell of negligible thickness, and the space between the tank and the shell is filled with foam insulation. The average temperature and emissivity of the outer surface of the shell are \(40^{\circ} \mathrm{C}\) and \(0.7\), respectively. The price of electricity is \(\$ 0.08 / \mathrm{kWh}\). Hot-water tank insulation kits large enough to wrap the entire tank are available on the market for about \(\$ 60\). If such an insulation is installed on this water tank by the home owner himself, how long will it take for this additional insulation to pay for itself? Disregard any heat loss from the top and bottom surfaces, and assume the insulation to reduce the heat losses by 80 percent.

A solar collector consists of a horizontal aluminum tube of outer diameter \(5 \mathrm{~cm}\) enclosed in a concentric thin glass tube of \(7 \mathrm{~cm}\) diameter. Water is heated as it flows through the aluminum tube, and the annular space between the aluminum and glass tubes is filled with air at \(1 \mathrm{~atm}\) pressure. The pump circulating the water fails during a clear day, and the water temperature in the tube starts rising. The aluminum tube absorbs solar radiation at a rate of \(20 \mathrm{~W}\) per meter length, and the temperature of the ambient air outside is \(30^{\circ} \mathrm{C}\). Approximating the surfaces of the tube and the glass cover as being black (emissivity \(\varepsilon=1\) ) in radiation calculations and taking the effective sky temperature to be \(20^{\circ} \mathrm{C}\), determine the temperature of the aluminum tube when equilibrium is established (i.e., when the net heat loss from the tube by convection and radiation equals the amount of solar energy absorbed by the tube). For evaluation of air properties at \(1 \mathrm{~atm}\) pressure, assume \(33^{\circ} \mathrm{C}\) for the surface temperature of the glass cover and \(45^{\circ} \mathrm{C}\) for the aluminum tube temperature. Are these good assumptions?

Consider an industrial furnace that resembles a 13 -ft-long horizontal cylindrical enclosure \(8 \mathrm{ft}\) in diameter whose end surfaces are well insulated. The furnace burns natural gas at a rate of 48 therms/h. The combustion efficiency of the furnace is 82 percent (i.e., 18 percent of the chemical energy of the fuel is lost through the flue gases as a result of incomplete combustion and the flue gases leaving the furnace at high temperature). If the heat loss from the outer surfaces of the furnace by natural convection and radiation is not to exceed 1 percent of the heat generated inside, determine the highest allowable surface temperature of the furnace. Assume the air and wall surface temperature of the room to be \(75^{\circ} \mathrm{F}\), and take the emissivity of the outer surface of the furnace to be \(0.85\). If the cost of natural gas is \(\$ 1.15 /\) therm and the furnace operates \(2800 \mathrm{~h}\) per year, determine the annual cost of this heat loss to the plant. Evaluate properties of air at a film temperature of \(107.5^{\circ} \mathrm{F}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

A vertical 4-ft-high and 6-ft-wide double-pane window consists of two sheets of glass separated by a 1-in air gap at atmospheric pressure. If the glass surface temperatures across the air gap are measured to be \(65^{\circ} \mathrm{F}\) and \(40^{\circ} \mathrm{F}\), determine the rate of heat transfer through the window by \((a)\) natural convection and (b) radiation. Also, determine the \(R\)-value of insulation of this window such that multiplying the inverse of the \(R\)-value by the surface area and the temperature difference gives the total rate of heat transfer through the window. The effective emissivity for use in radiation calculations between two large parallel glass plates can be taken to be \(0.82\).

Is the heat transfer rate through the glazing of a double-pane window higher at the center or edge section of the glass area? Explain.

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