0. Boundary Conditions for HT

Recall:

$$ \rho c_p \left[ \frac{\partial T}{\partial t}+ (\vec v \cdot \vec \nabla) T \right] = k\vec\nabla^2T + \dot{S_v} $$

1st Kind - Temperature defined @ Boundary

$$ T(0,t)=T_s $$

2nd Kind - Heat Flux defined @ Boudary

(a) Finite Heat Flux

$$ q_s = -k\vec \nabla T $$

For 1-dimension:

$$ q_s = -k\frac{\partial T}{\partial x}\vert_{boundary} $$

(b) Adiabatic or Insulated Surface

$$ \frac{\partial T}{\partial x}\vert_{boundary} = 0 $$

3rd Kind - Convection Surface Condition

$$ q = -k\frac{\partial T}{\partial x}\vert_{boundary}=h[T_\infty - T(0,t)] $$

1. 1-D Steady State Conduction

<aside> 💡 Assumptions:

• 1-Dimensional Conduction ONLY
• Steady State
• Uniform and Constant $k$
• NO Heat Generation $\dot S_v =0$ </aside>

Recall Differential Form of Heat Transport Equation:

$$ \rho c_p \left[ \cancel{\frac{\partial T}{\partial t}} + (\cancel{\vec{v}} \cdot \vec \nabla) T \right]

k\vec \nabla^2T + \cancel{\dot S_v} $$

First Term cancelled as Steady State

Second Term cancelled as NO Advection

Last Term cancelled as NO Heat Generation

$$ \therefore \frac{\partial^2T}{\partial x^2} = 0 \quad\to\quad T=C_1x+C_2 $$

Apply Boundary Conditions: $T = T_\text{s, 0}$ at $x=0$:

$$ C_2 = T_\text{s, 0} $$

$T=T_\text{s, L}$ at $x=L$:

$$ C_1 = \frac{T_\text{s, L} - T_\text{s, 0}}{L} $$

Therefore, the Temperature Profile is LINEAR:

$$ T = \frac{T_\text{s, L} - T_\text{s, 0}}{L}x + T_\text{s, 0} $$