# 物理代写|传热学代写Heat Transfer代考|MAE3240 BOUNDARY AND INITIAL CONDITIONS

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## 物理代写|传热学代写Heat Transfer代考|BOUNDARY AND INITIAL CONDITIONS

The physical conditions existing on the boundary should be known in order to solve the heat conduction equation and determine the temperature profile within the medium. Moreover, the initial condition $T(x, 0)=T_i$ should also be known if the heat transfer is time-dependent.
Boundary Conditions
There are three types of BCs commonly found in many heat transfer applications [1].

Specified surface temperature $T(0, t)=T_s$, as shown in Figure $1.8$

Specified surface heat flux, as shown in Figure $1.9 \mathrm{a}$ and $\mathrm{b}$
a. Finite heat flux
$$-k \frac{\partial T(0, T)}{\partial x}=q_s^{\prime \prime}$$
b. Adiabatic or insulated surface, which is a special case
$$-k \frac{\partial T(0, T)}{\partial x}=q_s^{\prime \prime}=0$$

## 物理代写|传热学代写Heat Transfer代考|SIMPLIFIED HEAT CONDUCTION EQUATIONS

Steady state, $\partial T / \partial t=0$, no heat generation, $\dot{q}=0$,
$$\begin{gathered} \text { 1-D } \frac{\partial^2 T}{\partial x^2}=0 \ \text { 2-D } \frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0 \ \text { 3-D } \frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}=0 \end{gathered}$$

Steady state, 1-D with a heat source (heater application)

$$\frac{\partial^2 T}{\partial x^2}+\frac{\dot{q}}{k}=0$$
Steady state, 1-D with heat sink-fin application
$$\frac{\partial^2 T}{\partial x^2}-\frac{\dot{q}}{k}=0$$

Transient, without heat generation $\dot{q}=0$
$$\begin{gathered} \text { 1-D } \frac{\partial^2 T}{\partial x^2}=\frac{1}{\alpha} \frac{\partial T}{\partial t} \ \text { 2-D } \frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=\frac{1}{\alpha} \frac{\partial T}{\partial t} \ \text { 3-D } \frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}=\frac{1}{\alpha} \frac{\partial T}{\partial t} \end{gathered}$$

## 物理代写|传热学代写热传导代考|边界和初始条件

a。有限热流
$$-k \frac{\partial T(0, T)}{\partial x}=q_s^{\prime \prime}$$
b。绝热或绝缘的表面，这是一种特殊情况
$$-k \frac{\partial T(0, T)}{\partial x}=q_s^{\prime \prime}=0$$

## 物理代写|传热学代写Heat Transfer代考|SIMPLIFIED Heat导热方程

.热传导方程

$$\begin{gathered} \text { 1-D } \frac{\partial^2 T}{\partial x^2}=0 \ \text { 2-D } \frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0 \ \text { 3-D } \frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}=0 \end{gathered}$$

1-D稳态，带热源(加热器应用)

$$\frac{\partial^2 T}{\partial x^2}+\frac{\dot{q}}{k}=0$$

$$\frac{\partial^2 T}{\partial x^2}-\frac{\dot{q}}{k}=0$$

$$\begin{gathered} \text { 1-D } \frac{\partial^2 T}{\partial x^2}=\frac{1}{\alpha} \frac{\partial T}{\partial t} \ \text { 2-D } \frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=\frac{1}{\alpha} \frac{\partial T}{\partial t} \ \text { 3-D } \frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}=\frac{1}{\alpha} \frac{\partial T}{\partial t} \end{gathered}$$

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