/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 161 A \(3-\mathrm{m}^{2}\) black sur... [FREE SOLUTION] | 91影视

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A \(3-\mathrm{m}^{2}\) black surface at \(140^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(35^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(16 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat loss from the surface is (a) \(5105 \mathrm{~W}\) (b) \(2940 \mathrm{~W}\) (c) \(3779 \mathrm{~W}\) (d) \(8819 \mathrm{~W}\) (e) \(5040 \mathrm{~W}\)

Short Answer

Expert verified
Answer: (a) 5105 W

Step by step solution

01

Convert temperatures to Kelvin

To do the calculations, we need to convert the temperatures of the surface, air, and surrounding surfaces from Celsius to Kelvin. K = 掳C + 273.15 Surface temperature: 140掳C + 273.15 = 413.15 K Air temperature: 35掳C + 273.15 = 308.15 K Surrounding surfaces temperature: 15掳C + 273.15 = 288.15 K
02

Calculate the heat loss due to convection

Using the formula and given temperature values: Heat loss by convection = heat transfer coefficient 脳 Area 脳 螖T Heat loss by convection = 16 W/m虏路K 脳 3 m虏 脳 (413.15 K - 308.15 K) Heat loss by convection = 16 脳 3 脳 105 = 5040 W
03

Calculate the heat loss due to radiation

Apply the Stefan-Boltzmann law with given surface temperature and surrounding surfaces temperature: Heat loss by radiation = 蔚蟽 脳 Area 脳 (T1鈦 - T2鈦) Heat loss by radiation = (1)(5.67脳10鈦烩伕 W/m虏K鈦) 脳 3 m虏 脳 (413.15鈦 - 288.15鈦) Heat loss by radiation 鈮 65 W
04

Calculate the total heat loss

Add the heat loss due to convection and radiation to get the total heat loss: Total heat loss = heat loss by convection + heat loss by radiation Total heat loss = 5040 W + 65 W = 5105 W
05

Choose the correct option from the given choices

The total rate of heat loss from the surface is: (a) 5105 W Hence, the correct answer is (a) \(5105 \mathrm{~W}\).

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Key Concepts

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

Convection
Convection is the process of heat transfer through a fluid (which includes gases and liquids) caused by molecular motion. It occurs when a fluid is heated, causing it to expand and become less dense, thus rising in the surrounding cooler fluid. In the context of our exercise, the hot surface transfers heat to the cooler surrounding air through convection. This is calculated using the equation:
  • Heat loss by convection = heat transfer coefficient 脳 Area 脳 螖T
The heat transfer coefficient is a measure of how effectively heat is transferred between a solid surface and the fluid in contact with it. Convection can be categorized into two types:
  • Natural Convection: Caused by buoyancy forces that are due to density differences caused by temperature variations in the fluid.
  • Forced Convection: When a fluid is forced to flow over the surface by an external source like a fan or a pump.
In our problem, the surface loses heat to the air with a known heat transfer coefficient, helping us calculate the rate of heat loss efficiently.
Radiation
Radiation is heat transfer that occurs through electromagnetic waves, without the need for a medium. This means heat can be transferred through vacuum or transparent media (like air or glass), making it distinct from conduction and convection. In our exercise, the surface loses heat via radiation to the surrounding surfaces. Radiation heat transfer follows the Stefan-Boltzmann law, highlighting the emission power from a surface depending on its temperature. The formula used is:
  • Heat loss by radiation = 蔚蟽 脳 Area 脳 (T鈧佲伌 - T鈧傗伌)
Where:
  • 蔚 is the emissivity of the surface, a measure of how efficiently it radiates energy.
  • 蟽 is the Stefan-Boltzmann constant, which is approximately 5.67脳10鈦烩伕 W/m虏K鈦.
  • T鈧 and T鈧 are the absolute temperatures of the surface and the surroundings, respectively.
Unlike convection, radiation does not require a medium and operates at all temperatures. It becomes significant at high temperatures, such as those in our exercise, leading to distinct heat loss computations.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is foundational in understanding radiation heat transfer. It states that the total energy radiated per unit surface area is directly proportional to the fourth power of the absolute temperature of the surface. The formula we used in our calculation is a direct application of this law. The equations are:
  • Energy radiated = 蟽 脳 Area 脳 T鈦
  • For two bodies at different temperatures: Energy exchanged is determined by the temperature difference raised to the fourth power.
The law is instrumental in determining the radiative heat loss in various applications, whether it鈥檚 satellites in space or our black surface example. In our case, using this law allows us to calculate the net radiation heat transfer between the surface at a high temperature and its cooler surroundings. It helps highlight the efficiency of heat loss through radiation compared to convection.
Heat Transfer Coefficient
The heat transfer coefficient is a parameter used in calculations of convective heat transfer. It's derived from empirical data and signifies how effectively heat is transferred from a solid surface to a fluid (or vice versa). In this problem, it helps determine the rate at which heat moves from the hot surface into the cooler surrounding air. Defined as:
  • Heat transfer coefficient (h) = Heat transfer per unit area per degree of temperature difference
In formulas involving convection, the coefficient h, area (A), and temperature difference (螖T) together determine the rate of heat transfer:
  • Convective Heat Transfer = h 脳 A 脳 螖T
Values of the heat transfer coefficient can modify based on factors such as the nature of the fluid, flow conditions, and surface characteristics. A higher value indicates more efficient heat transfer, crucial in engineering applications where heat management is vital.

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Most popular questions from this chapter

A 300-ft-long section of a steam pipe whose outer diameter is 4 in passes through an open space at \(50^{\circ} \mathrm{F}\). The average temperature of the outer surface of the pipe is measured to be \(280^{\circ} \mathrm{F}\), and the average heat transfer coefficient on that surface is determined to be \(6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Determine \((a)\) the rate of heat loss from the steam pipe and (b) the annual cost of this energy loss if steam is generated in a natural gas furnace having an efficiency of 86 percent, and the price of natural gas is $$\$ 1.10 /$$ therm ( 1 therm \(=100,000\) Btu).

What is the driving force for \((a)\) heat transfer, \((b)\) electric current flow, and (c) fluid flow?

It is well-known that at the same outdoor air temperature a person is cooled at a faster rate under windy conditions than under calm conditions due to the higher convection heat transfer coefficients associated with windy air. The phrase wind chill is used to relate the rate of heat loss from people under windy conditions to an equivalent air temperature for calm conditions (considered to be a wind or walking speed of \(3 \mathrm{mph}\) or \(5 \mathrm{~km} / \mathrm{h})\). The hypothetical wind chill temperature (WCT), called the wind chill temperature index (WCTI), is an equivalent air temperature equal to the air temperature needed to produce the same cooling effect under calm conditions. A 2003 report on wind chill temperature by the U.S. National Weather Service gives the WCTI in metric units as WCTI \(\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 V^{0.16}+0.3965 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{C}\) and \(V\) the wind speed in \(\mathrm{km} / \mathrm{h}\) at \(10 \mathrm{~m}\) elevation. Show that this relation can be expressed in English units as WCTI \(\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{F}\) and \(V\) the wind speed in \(\mathrm{mph}\) at \(33 \mathrm{ft}\) elevation. Also, prepare a table for WCTI for air temperatures ranging from 10 to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from 10 to \(80 \mathrm{~km} / \mathrm{h}\). Comment on the magnitude of the cooling effect of the wind and the danger of frostbite.

The heat generated in the circuitry on the surface of a silicon chip \((k=130 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is conducted to the ceramic substrate to which it is attached. The chip is \(6 \mathrm{~mm} \times 6 \mathrm{~mm}\) in size and \(0.5 \mathrm{~mm}\) thick and dissipates \(5 \mathrm{~W}\) of power. Disregarding any heat transfer through the \(0.5-\mathrm{mm}\) high side surfaces, determine the temperature difference between the front and back surfaces of the chip in steady operation.

Can a medium involve \((a)\) conduction and convection, (b) conduction and radiation, or \((c)\) convection and radiation simultaneously? Give examples for the "yes" answers.

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