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A 0.2-m-long and \(25-\mathrm{mm}\)-thick vertical plate \((k=\) \(1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) separates the hot water from the cold air at \(2^{\circ} \mathrm{C}\). The plate surface exposed to the hot water has a temperature of \(100^{\circ} \mathrm{C}\), and the surface exposed to the cold air has an emissivity of \(0.73\). Determine the temperature of the plate surface exposed to the cold air \(\left(T_{s, c}\right)\). Hint: The \(T_{s, c}\) has to be found iteratively. Start the iteration process with an initial guess of \(51^{\circ} \mathrm{C}\) for the \(T_{s, c}\).

Short Answer

Expert verified
1. Calculate the thermal resistance of the plate. 2. Estimate the convective heat transfer coefficient. 3. Initialize an initial guess for the temperature of the surface exposed to cold air, T_{s, c}. 4. Perform an iterative process by calculating heat transfer through the plate and radiation, checking the energy balance, and updating the guessed temperature, T_{s, c}, until the energy balance condition is satisfied.

Step by step solution

01

Calculate Thermal Resistance of the Plate

The thermal resistance of the plate, \(R_{cond}\), can be calculated as: $$ R_{cond} = \frac{L}{kA} = \frac{0.025}{1.5 \cdot (0.2 \cdot 1) }, $$ where \(L\) is the thickness of the plate, \(k\) is the thermal conductivity, and \(A\) is the cross-sectional area.
02

Estimate Convective Heat Transfer Coefficient

We estimate the convective heat transfer coefficient, \(h_{c}\), using an average value of \(10~\mathrm{W/m^{2}K}\). If a more accurate value is needed in a specific problem, you may use empirical correlations for \(h_{c}\).
03

Initialize Guess for Plate Surface Exposed to Cold Air Temperature

We will start with an initial guess for the temperature of the surface exposed to cold air \(T_{s, c} = 51^{\circ} \mathrm{C}\).
04

Calculate Heat Transfer Through the Plate (Iteration Step)

The heat transfer through the plate can be calculated using the guessed temperature and the thermal resistance of the plate: $$ q_{cond} = \frac{T_{s, h} - T_{s, c}}{R_{cond}}, $$ where \(T_{s, h} = 100^{\circ} \mathrm{C}\) is the temperature of the surface exposed to hot water.
05

Calculate Heat Transfer Through Radiation (Iteration Step)

The heat transfer through radiation can be calculated using the guessed temperature and the emissivity of the surface, as well as the Stefan-Boltzmann constant \(\sigma\): $$ q_{rad} = \epsilon \sigma A \left( T_{s, c}^4 - T_{\infty}^4 \right), $$ where \(\epsilon = 0.73\) is the surface emissivity, \(\sigma = 5.67 \times 10^{-8} ~\mathrm{W/m^{2}K^{4}}\) is the Stefan-Boltzmann constant, and \(T_{\infty} = 2^{\circ} \mathrm{C}\) is the temperature of the cold air.
06

Check Energy Balance and Update Guessed Temperature (Iteration Step)

Now we need to check if the energy balance condition is satisfied: $$ q_{cond} = q_{conv} + q_{rad}, $$ where \(q_{conv} = h_{c} A (T_{s, c} - T_{\infty})\). If not, update the guessed temperature \(T_{s, c}\) using the equation $$ T_{s,c,new} = T_{s,h} - R_{cond} \cdot \left(q_{conv} + q_{rad}\right), $$ and repeat steps 3-5 until the energy balance condition is satisfied. In this way, we will iteratively find the correct temperature of the surface exposed to the cold air, \(T_{s, c}\).

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Key Concepts

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

Thermal Resistance
When studying heat transfer, the term thermal resistance is central to understanding how easily heat flows through a material. It's analogous to electrical resistance, but instead of impeding electric current, thermal resistance hinders the flow of heat. Thermal resistance (\(R_{cond}\)) for a conductive medium, like the vertical plate in our exercise, can be quantified using the formula \[ R_{cond} = \frac{L}{kA} \] where \(L\) represents the thickness of the material, \(k\) is the thermal conductivity, indicating the material's ability to conduct heat, and \(A\) is the cross-sectional area through which heat is being transferred. This concept is vital in determining how effectively a material will insulate against heat transfer. Understanding and computing thermal resistance is pertinent in a variety of fields, ranging from building construction to electronic device manufacturing.
Convective Heat Transfer Coefficient
The convective heat transfer coefficient (\(h_c\)) is crucial in the analysis of heat exchange between a solid surface and a fluid, like air or water, that is moving past it. It quantifies the ease with which heat is transferred from the solid surface to the fluid or vice versa. A high coefficient indicates efficient heat transfer, commonly seen with turbulent fluid flow or increased surface area. Engineers often use empirical correlations or experimental data to find \(h_c\), but approximate values may suffice for initial estimations. For our vertical plate, we've used an estimated value of \(10~\mathrm{W/m^2K}\), which is a typical starting point for air. Knowing \(h_c\) allows for the calculation of the heat lost or gained through convection, a key component in maintaining or adjusting the temperatures of systems and environments.
Radiative Heat Transfer
In addition to conduction and convection, radiative heat transfer is another mode by which heat can travel. It's a process by which heat is emitted as electromagnetic waves, and it can occur in a vacuum as well as through transparent media. Critical to the calculation of radiative heat transfer is the emissivity (\(\epsilon\)) of a material, which is a measure of how effectively a surface emits thermal radiation compared to an ideal 'black body.' In our example, the plate has an emissivity of 0.73, meaning it emits 73% of the energy it would if it were a perfect emitter. The Stefan-Boltzmann law helps us quantify this emission, as seen in the equation \[ q_{rad} = \epsilon \sigma A (T_{s, c}^4 - T_{\infty}^4 ) \] where \(\sigma\) is the Stefan-Boltzmann constant and \(T_{\infty}\) is the cold air temperature. Both practical applications like heating and cooling systems and theoretical studies in astrophysics or climate science extensively use radiative heat transfer principles.
Iteration Process
The iteration process is a methodical approach used to incrementally approach a solution when a direct calculation is not feasible. This technique is particularly useful when working with complex systems where variables are interdependent, as is the case in our example where the temperature of the plate surface exposed to cold air isn't known beforehand. To solve for it, we begin with an initial guess based on experience or estimation and then use iterative calculations to refine this guess. With each iteration, we apply the known principles, such as energy balance equations, and use the outcomes to improve the subsequent guess. This cycle repeats until the solution converges to an acceptable level of accuracy. Iterative methods are an indispensable tool in various scientific and engineering fields, as they allow for solving real-world problems that are mathematically intricate or computationally intensive.
Energy Balance
The principle of energy balance is a cornerstone of thermodynamics; it ensures that in a stable system, the amount of energy entering the system is equal to the amount of energy leaving it, after accounting for any storage or energy transformations within the system. When we refer to heat transfer, we apply this balance to ensure the calculated rates of heat transfer by conduction, convection, and radiation are equivalent, considering the governing equation \[ q_{cond} = q_{conv} + q_{rad} \] where all \(q\)'s represent the different modes of heat transfer. It is this energy equilibrium that allows us to set up an iteration that will converge on a correct value for the surface temperature. This concept not only grounds us in practical applications, like designing energy-efficient buildings or industrial processes, but also in understanding natural phenomena, such as Earth's energy budget which drives climate and weather patterns.

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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.

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