Consider a thin 16-cm-long and 20-cm-wide horizontal plate suspended in air at \(20^{\circ} \mathrm{C}\). The plate is equipped with electric resistance heating elements with a rating of \(20 \mathrm{~W}\). Now the heater is turned on and the plate temperature rises. Determine the temperature of the plate when steady operating conditions are reached. The plate has an emissivity of \(0.90\) and the surrounding surfaces are at \(17^{\circ} \mathrm{C}\). As an initial guess, assume a surface temperature of \(50^{\circ} \mathrm{C}\). Is this a good assumption?

Short Answer

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Question: Based on the given information and solution, determine the reasonability of the initial temperature guess of 50°C for the steady-state temperature of the plate. Answer: To evaluate the reasonability of the initial temperature guess of 50°C for the steady-state temperature of the plate, we must first calculate the steady-state temperature (T_plate) using the energy balance equation, considering both radiation and convection heat transfers. After determining T_plate, we can then compare it to the initial guess (T_guess) of 50°C to assess whether the guess is reasonable or not. If T_plate is close to T_guess, then the initial assumption is reasonable. However, if there is a significant difference between the two values, we should refine the initial guess and re-calculate the steady-state temperature.

Step by step solution

01

Write down the known properties and parameters

Here are the known properties and parameters for this problem: -Length of the plate, L = 16 cm -Width of the plate, W = 20 cm -Initial surrounding air temperature, T_inf = 20°C -Heater rating, Q_heater = 20 W -Plate emissivity, ε = 0.90 -Surrounding surface temperature, T_surr = 17°C -Initial temperature guess, T_guess = 50°C
02

Calculate area and heat transfer coefficient

In order to analyze the heat transfer processes, we need to calculate the area of the plate and the heat transfer coefficient. Area of the plate, A = L * W = 0.16 m * 0.2 m = 0.032 m² Assuming natural convection, we can use the heat transfer coefficient for a horizontal plate facing up, known as the average Nusselt number (Nu) formula: Nu = 0.54 * (Ra * Pr)^0.25 Note that Ra and Pr are the Rayleigh number and Prandtl number, respectively, and can be found using air properties (thermal diffusivity and kinematic viscosity) as follows: Ra = (g * β * ΔT * L³) / (α * ν) Pr = ν / α where g is the gravitational acceleration, β is the coefficient of thermal expansion, ΔT is the difference between plate and air temperatures, α is the thermal diffusivity, and ν is the kinematic viscosity. For a rough estimate of these parameters for air at 20°C, we can use the values: β = 1 / (273 + T_inf) K^-1= 1 / (273 + 20) K^-1 ≈ 0.0034 K^-1 ν = 15.89 * 10^-6 m²/s α = 22.1 * 10^-6 m²/s Pr = ν / α ≈ 0.72 Now we can compute the heat transfer coefficient (h) using the Nusselt number relation: h = Nu * k_air / L where k_air is the thermal conductivity for air (≈ 0.026 W/m·K) and L is the length of the plate.
03

Calculate the steady-state temperature

At steady-state, the heat gained from the heater will equal the heat lost through radiation and convection. We can write the energy balance equation as follows: Q_heater = Q_radiation + Q_convection Since radiation heat transfer can be described by the Stefan-Boltzmann Law, we have: Q_radiation = ε * A * σ * (T_plate^4 - T_surr^4) where σ is the Stefan-Boltzmann constant (5.67 x 10^−8 W/m²·K^4). For convection heat transfer, we have: Q_convection = h * A * (T_plate - T_inf) We can solve this set of equations for T_plate using a numerical method (e.g., the Newton-Raphson method) or by using a trial and error approach.
04

Evaluate the initial guess

Once the steady-state temperature T_plate is determined, we can check if our initial guess T_guess = 50°C is reasonable or not by comparing the two values. If the calculated T_plate is close to the T_guess, then the initial assumption is reasonable. However, if the difference is significant, we should refine our initial guess and re-calculate the steady-state temperature.

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Key Concepts

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

Natural Convection
Natural convection is a heat transfer mechanism where fluid motion is generated by buoyancy forces. These forces arise from density variations in the fluid due to temperature gradients. In our example, the air surrounding the heated plate becomes warmer and less dense than the cooler air away from the plate. Consequently, the warm air rises, and cooler air moves in to replace it, creating a natural circulation pattern.

For the thin horizontal plate in the problem, the natural convective heat transfer can be described by the average Nusselt number. The Nusselt number is a dimensionless quantity that characterizes the efficacy of convection and is calculated using specific properties of air such as the Rayleigh and Prandtl numbers. These numbers, in turn, depend on various parameters including gravity, thermal expansion coefficient, temperature difference, plate dimensions, thermal diffusivity, and kinematic viscosity.

When the plate heats up, we can determine the heat transfer by using the Nusselt number to find the heat transfer coefficient (h), which reveals how effectively the plate transfers heat to the surrounding air. This coefficient is vital for computing the convective heat loss from the plate to its environment.
Radiation Heat Transfer
Radiation heat transfer refers to the energy emitted by matter in the form of electromagnetic waves due to the material's temperature. This mode of heat transfer does not require a medium to occur, which means it can take place even in a vacuum. In the problem context, the heated plate emits thermal radiation due to its temperature being higher than the temperature of the surrounding surfaces.

For the evaluation of radiation heat transfer, we employ the Stefan-Boltzmann Law, which states that the power radiated per unit area of a black body is proportional to the fourth power of its absolute temperature. However, as most materials aren't perfect black bodies, we incorporate the emissivity (ε) of the plate, which in this case has a value of 0.90, indicating that the plate is a good emitter of radiation.

To embody the effect of this type of heat transfer in our calculation, we use the equation for radiative heat transfer that includes the plate's surface area (A), the Stefan-Boltzmann constant (σ), and the temperature difference between the plate and its surroundings, raised to the fourth power. As the plate operates, it maintains an energy balance, losing heat through radiation, as well as through natural convection.
Steady-State Temperature
Steady-state temperature is a term used to describe a condition where the temperature of an object doesn't change with time, meaning that the temperature has stabilized. It's achieved when the energy added to the system equals the energy lost. In our textbook exercise, the plate reaches a steady-state temperature when the power supplied by the electric resistance heating elements is equal to the sum of heat lost through radiation and convection.

At the steady state, the calculation involves the use of an energy balance equation. The power from the heater is the energy input, while the energy outputs are the heat losses through radiation and convection. To find the steady-state temperature, we need to solve the energy balance equation, which can often require iterative methods or numerical solutions, especially when multiple modes of heat transfer are involved.

The concept of steady-state temperature is crucial in many engineering applications, such as in heating and cooling systems, where maintaining a specific temperature is essential. In our scenario, determining the steady-state temperature allows us to evaluate the appropriateness of the initial guess and inform us whether our assumptions were valid or if further refinement is necessary.

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Most popular questions from this chapter

Consider a double-pane window whose air space width is \(20 \mathrm{~mm}\). Now a thin polyester film is used to divide the air space into two 10-mm-wide layers. How will the film affect \((a)\) convection and (b) radiation heat transfer through the window?

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(1.5 \mathrm{~m}\) wide and \(4.5 \mathrm{~m}\) long, and the average temperature of the exposed surface of the collector is \(42^{\circ} \mathrm{C}\). Determine the rate of heat loss from the collector by natural convection during a calm day when the ambient air temperature is \(8^{\circ} \mathrm{C}\). Also, determine the heat loss by radiation by taking the emissivity of the collector surface to be \(0.85\) and the effective sky temperature to be \(-15^{\circ} \mathrm{C}\).

A spherical tank \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with an inner diameter of \(3 \mathrm{~m}\) and a wall thickness of \(10 \mathrm{~mm}\) is used for storing hot liquid. The hot liquid inside the tank causes the inner surface temperature to be as high as \(100^{\circ} \mathrm{C}\). To prevent thermal burns on the skin of the people working near the vicinity of the tank, the tank is covered with a \(7-\mathrm{cm}\) thick layer of insulation \((k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the outer surface is painted to give an emissivity of \(0.35\). The tank is located in a surrounding with air at \(16^{\circ} \mathrm{C}\). Determine whether or not the insulation layer is sufficient to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\) to prevent thermal burn hazards. Discuss ways to further decrease the outer surface temperature. Evaluate the air properties at \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

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 \(0.2-\mathrm{m}\)-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 \(T_{s, c}\).

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