Heat Transfer: Physical Origins

<p>Heat transfer requires a temperature difference.</p> Signup and view all the answers

<p>Thermal energy transfer over a specified time interval.</p> Signup and view all the answers

<p>Temperature variations in a solid medium</p> Signup and view all the answers

<p>The relationship between heat flux and thermal conductivity</p> Signup and view all the answers

<p>Thermal conductivity</p> Signup and view all the answers

<p>It does not require a material medium.</p> Signup and view all the answers

<p>The difference in temperature across the wall per unit thickness</p> Signup and view all the answers

<p>The temperature difference between a surface and its surroundings to the convection heat transfer rate</p> Signup and view all the answers

<p>Surface emissivity</p> Signup and view all the answers

<p>$q^{''} = h (Ts - T∞) + hr (Ts - Tsur)$</p> Signup and view all the answers

<p>Stefan-Boltzmann constant</p> Signup and view all the answers

<p>It is considered a gray surface</p> Signup and view all the answers

<p>Conduction requires a material medium and occurs through direct molecular contact, convection involves bulk fluid motion and is influenced by temperature variations, while radiation transfers energy without a medium through electromagnetic waves.</p> Signup and view all the answers

<p>Fourier's Law quantifies the heat flux based on thermal conductivity and temperature gradient, and in one-dimensional conduction, it describes heat transfer through a material wall with defined thickness and temperature differences.</p> Signup and view all the answers

<p>You can calculate the rate of heat loss using the equation $q_x = q_x^{''} \cdot A$, where $q_x^{''}$ is determined from Fourier's Law and $A$ is the area of the wall.</p> Signup and view all the answers

<p>Thermal conductivity indicates how easily heat can pass through a material; higher thermal conductivity means more efficient heat transfer.</p> Signup and view all the answers

<p>Radiation can occur in a vacuum and is unaffected by material limitations, making it the most efficient form of heat transfer over long distances, while conduction and convection rely on material presence.</p> Signup and view all the answers

<p>The surface emissive power can be calculated using the formula $E = ext{ε} imes ext{σ} imes T^4$, where $T$ is the absolute temperature in Kelvin and $σ$ is the Stefan-Boltzmann constant.</p> Signup and view all the answers

<p>The rate of heat loss per unit length can be determined using the formula $Q = h imes A imes (T_s - T_a)$, where $h$ is the convection coefficient, $A$ is the surface area per unit length, and $T_s$ and $T_a$ are the surface and ambient temperatures respectively.</p> Signup and view all the answers

<p>Emissivity reflects a material's ability to emit thermal radiation; a value of 1 indicates a perfect black body, while values less than 1 indicate less effective emission.</p> Signup and view all the answers

<p>Irradiation refers to the total thermal radiation energy received by a surface per unit area, typically measured in W/m².</p> Signup and view all the answers

<p>The coefficient of free convection quantifies the heat transfer by natural convection, impacting the overall heat loss from surfaces to the surrounding fluid.</p> Signup and view all the answers

<p>It states that the heat transfer rate, $q''$, is proportional to the difference between the surface temperature, $T_s$, and the fluid temperature, $T_ ext{∞}$, which is expressed as $q'' = h(T_s - T_ ext{∞})$.</p> Signup and view all the answers

<p>Emissivity quantifies a surface's ability to emit thermal radiation, ranging from 0 to 1, where 1 represents a perfect blackbody emitter.</p> Signup and view all the answers

<p>The Stefan-Boltzmann constant, $ ext{σ} = 5.67 imes 10^{-8} ext{ W/m}^2 ext{K}^4$, is used to relate the total energy radiated per unit surface area of a black body to the fourth power of its absolute temperature.</p> Signup and view all the answers

<p>The absorbed incident radiation is given by $G_{abs} = ext{α} G$, where $G$ is the irradiation incident on the surface.</p> Signup and view all the answers

<p>The net radiation heat flux is given by $q'' = ext{ε} ext{E}<em>b(T_s) - ext{α} G = ext{ε} ext{σ} (T_s^4 - T</em>{sur}^4)$, capturing radiation exchanges with surroundings.</p> Signup and view all the answers