1. Energy Generation

Differential Form of H.T. Equation Approach

To solve for Temperature Profile $T(x)$ if there is heat generation

Recall Differential Form of H.T. Equation:

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

k\vec \nabla^2T + \dot S_v $$

First Term cancelled as Steady State

Second Term cancelled as NO Advection

Therefore:

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

Solve to obtain:

$$ T = -\frac{\dot S_v}{2k}x^2+C_1x+C2 $$

Apply Boundary Conditions:

1. Symmetry Condition at $x=0$

$$ \frac{dT}{dx} = 0 \qquad @x=0 $$

Therefore,

$$ C_1=0 $$

2. Newton’s Law of Cooling at $x=L$

$$ -k\frac{dT}{dx} = h(T-T_\infty) \qquad @x=L $$

Substitute in $\frac{dT}{dx}$ and $T$:

$$ k(\frac{\dot S_v L}{k}) = h(-\frac{\dot S_vL^2}{2k} + C_2 - T_\infty)\\ C_2 = \frac{\dot S_vL}{h} + \frac{\dot S_v L^2}{2k} + T_\infty $$

Therefore:

$$ T = -\frac{\dot S_v}{2k}x^2+\frac{\dot S_vL}{h} + \frac{\dot S_v L^2}{2k} + T_\infty\\ T = \frac{\dot S_v L^2}{2k}(1-\frac{x^2}{L^2}) + \frac{\dot S_vL}{h} + T_\infty $$

At x = L:

$$ T_L = \frac{\dot S_v L}{h} + T_\infty\\ \therefore T = \frac{\dot S_v L^2}{2k}(1-\frac{x^2}{L^2})+T_L $$

At x = 0:

$$ T_0=\frac{\dot S_vL^2}{2k} + \frac{\dot S_v L}{h} + T_\infty = \frac{\dot S_vL^2}{2k} + T_L\\ \therefore T = -\frac{\dot S_v}{2k} x^2 +T_0 $$

Or,

$$ \frac{T(x)-T_0}{T_L-T_0} =(\frac{x}{L})^2 $$

Control Volume Analysis Approach

$$ \dot S_v (AL) = hA (T_L - T_\infty)\\ T_L - T_\infty = \frac{\dot S_v L}{h} $$

$$ \dot S_v (Ax) = -kA \frac{dT}{dx}\\ \frac{dT}{dx} = - \frac{\dot S_v x}{k}\\ T = -\frac{\dot S_v x^2}{2k}+C_3 $$

Apply Boundary Condition:

$$ T= T_0 \qquad @x=0\\ \therefore C_3 = T_0\\ \therefore T = -\frac{\dot S_v x^2}{2k}+T_0 $$

2. Mass Generation