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Chapter 12: Problem 72

The 50 -mm peephole of a large furnace orerating a \(450^{\circ} \mathrm{C}\) is covered with a material having \(r=0.8\) and \(\rho=0\) for irradiation originating from the furnace. Tle material has an cmissivity of \(0.8\) and is opaque to imsdiation from a source at room temperature. The outer surface of the cover is exposed to surroundings and ambient air at \(27^{\circ} \mathrm{C}\) with a convection heat transet coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that convection effects on the inner surface of the cover are negligitle. calculate the heat loss by the furnace and the tempen: ture of the cover.

Short Answer

Expert verified
Use the steady-state energy balance equation to find the cover temperature and heat loss.

Step by step solution

01

Understand the Problem

We have a furnace operating at 450掳C, and the peephole of this furnace is covered by a material with certain properties. Radiant exchange and convection are occurring, and we need to determine both the heat loss and the surface temperature of the cover.
02

Note the Given Values

Given: - Furnace temperature, \( T_f = 450掳C + 273 = 723 \, K \)- Emissivity, \( \varepsilon = 0.8 \)- Reflectivity, \( r = 0.8 \), assume \( \rho = 0 \) for inner irradiation.- Ambient temperature, \( T_{\infty} = 27掳C + 273 = 300 \, K \)- Convection heat transfer coefficient, \( h = 50 \, W/m^2 \, K \)- Peephole diameter, 50 mm or 0.05 m (radius is 0.025 m).
03

Calculate the Heat Loss Due to Radiation

The heat loss due to radiation can be calculated using Stefan-Boltzmann law: \[Q_{rad} = A \varepsilon \sigma (T_f^4 - T_s^4)\]where \( \sigma = 5.67 \times 10^{-8} \, W/m^2 \, K^4 \) is the Stefan-Boltzmann constant and \( A = \pi (0.025)^2 \) is the area of the peephole.
04

Calculate the Heat Loss Due to Convection

The heat loss due to convection is given by: \[Q_{conv} = h A (T_s - T_{\infty})\]where \( T_s \) is the surface temperature of the cover which needs to be determined.
05

Establish the Energy Balance Equation

Set up an energy balance equation considering both radiation and convection. Since the system is at steady state, the heat loss by radiation should equal the heat gain by convection:\[Q_{rad} = Q_{conv}\]
06

Solve for the Cover Temperature

By combining the equations from Steps 3 and 4:\[A \varepsilon \sigma (T_f^4 - T_s^4) = h A (T_s - T_{\infty})\]Solve this equation for \( T_s \). This might require iterative calculation or numerical methods as the equation is nonlinear.
07

Compute Heat Loss

With the determined \( T_s \), compute the actual heat loss using the established equations, substituting back into either the radiation or convection equation.

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Key Concepts

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

Radiation Heat Transfer
Radiation heat transfer is the process by which thermal energy is emitted by a surface in the form of electromagnetic waves. Unlike conduction and convection, radiation does not require a medium to transmit heat. It can occur in a vacuum.

In our exercise, the furnace emits thermal radiation from its hot surface. The energy depends on the temperature of the surface and the material properties covering the peephole. The emissivity of the material, denoted by \(\varepsilon\), plays a crucial role. It indicates how effectively a material emits energy as thermal radiation as compared to a perfect black body, whose emissivity is 1.

To calculate the radiation heat transfer, we use the Stefan-Boltzmann law, which relates the radiation emitted by a black body to its temperature. This law is crucial in determining the heat loss because of radiation by the furnace in the given scenario.
Convection Heat Transfer
Convection heat transfer involves the movement of heat through a fluid, which could be a gas or a liquid. It relies on fluid moving around the surface to carry away heat. This process is more effective when there are temperature differences between the fluid and the surface.

In the problem, the material covering the furnace鈥檚 peephole loses heat to the surrounding air through convection. The rate of this heat transfer depends on the convection heat transfer coefficient, denoted by \(h\). This coefficient is specific to the conditions of the air and the properties of the surface.

The equation for convection heat transfer is straightforward: \(Q_{conv} = h A (T_s - T_\infty)\). Here, \(A\) is the surface area, \(T_s\) is the surface temperature, and \(T_\infty\) is the ambient temperature. Understanding this concept helps calculate how much heat is lost due to convection.
Stefan-Boltzmann Law
The Stefan-Boltzmann law is fundamental in the analysis of heat transfer by radiation. It quantifies the power emitted per unit area of a black body as proportional to the fourth power of the temperature of the body. Mathematically, it is expressed as:

\[Q_{rad} = A \varepsilon \sigma (T^4)\]

where \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \, W/m^2 \, K^4\), \(A\) is the surface area, \(\varepsilon\) is the emissivity, and \(T\) is the absolute temperature.

In this exercise, the Stefan-Boltzmann law allows us to calculate the heat radiated from the furnace to its surroundings through the peephole cover. The actual temperature difference between the furnace and the exterior comes into play, emphasizing the nonlinear nature of this heat transfer science. Utilizing this law, we can determine the balances needed for energy transfer and solve for the unknowns in this system.

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Most popular questions from this chapter

The neighborhood cat likes to sicep on the roof of our shed in the backyard. The roofing surface is weathered galvanized sheet metal \(\left(\varepsilon=0.65, \alpha_{S}=0.8\right)\). Consider a cool spring day when the ambient air temperature is \(10^{\circ} \mathrm{C}\) and the convection coefficient can be estimated from an empirical correlation of the form \(\bar{h}_{1}=1.0 \Delta T^{4 / 3}\), where \(\mathrm{T} T\) is the difference between the surface and ambient temperatures. Assume the sky temperature is \(-40^{\circ} \mathrm{C}\). (a) Assuming the backside of the roof is well insulated, calculate the roof temperarure when the solar irradiation is \(600 \mathrm{~W} / \mathrm{m}^{2}\). Will the cat enjoy sleeping under these conditions? (b) Consider the case when the hackside of the roof is not insulated, but is exposed to ambient air with the same convection cocfficient relation and experiences radiation exchange with the ground, also at the ambient air temperarure. Calculate the roof tempernture and comment on whether the roof will be a comfortable place for the cat to snooze.

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at \(2500 \mathrm{~K}\), a heated metal at \(1500 \mathrm{~K}\), human skin at \(305 \mathrm{~K}\). and a cryogenically cooled metal surface at \(60 \mathrm{~K}\). Estimate the fraction of the solar emission that is in the following spectral regions: the ultraviolet, the visible, and the infrared.

A radiator on a proposed satellite solar power station must dissipate beat being generated within the satellite by madiating it into space. The radiator surface has a solar absorptivity of \(0.5\) and an emissivity of \(0.95\). What is the equilibrium surface temperature when the solar irradiation is \(1000 \mathrm{~W} / \mathrm{m}^{2}\) and the required heat dissipation is 1500 W/m \({ }^{2}\) ?

A temperature sensor imbedded in the tip of a small tube having a diffuse, gray surface with an emissivity of \(0.8\) is centrally positioned within a large air-conditioned toom whose walls and air temperature are 30 and \(20^{\circ} \mathrm{C}\), respectively. (a) What temperature will the sensor indicate if the convection coefficient between the sensot tube and the air is \(5 \mathrm{~W} / \mathrm{m}^{2}\) - K? (b) What would be the effect of using a fan to induce airflow over the tube? Plot the sensor temperature as a function of the convection cocfficient for \(2 \leq h \leq 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and values of \(\mathrm{e}=0.2,0.5\). and \(0.8\).

A sphere \(\left(k=185 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=7.25 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{s}\right)\) of 30 -mm diameter whose surface is diffuse and gray with an emissivity of \(0.8\) is placed in a large oven whose walls are of uniform temperature at \(600 \mathrm{~K}\). The temperature of the air in the oven is \(400 \mathrm{~K}\), and the convection heat transfer coefficient between the sphere and the oven air is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the net heat transfer to the sphere when its temperature is \(300 \mathrm{~K}\). (b) What will be the steady-state temperature of the sphere? (c) How long will it take for the sphere, initially at \(300 \mathrm{~K}\), to come within \(20 \mathrm{~K}\) of the steady-state temperature? (d) For emissivities of \(0.2,0.4\), and \(0.8\), plot the elapsed time of part (c) as a function of the convection coefficient for \(10 \leq h \leq 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
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