Saturday, July 6, 2013

physical aging

https://docs.google.com/file/d/0By-jEDtazz8cT21peHo5Smk5QVE/edit?usp=sharing

Saturday, May 18, 2013

1D heat equation applied to a rod, Dirichlet + Robins boundary condition

The temperature profile of a rod, subjected to a heat load is derived. The heat is dissipated by convective heat transfer. The rod has a length \(L\) , and a cross-sectional area, \(A\). The thermal loading \(P\) is applied at \(x=L\), at \(x=0\), the Robins boundary condition is applied to account for the convective heat transfer.
The one-dimensional heat equation is given by:
$$\frac{d^2T}{dx^2}=0$$
The boundary conditions are as follows:
$$\frac{dT}{dx}(x=L)=\frac{P}{\lambda A}$$
$$\lambda A \cdot \frac{dT}{dx}(x=0)=h\cdot (T(x=0)-T_{water})$$
where, \(P\) is the aplied thermal loading in \([W]\), \(A\) is the cross-sectional area of the beam, and \(\lambda\) the thermal conductivity of the involved material in \([W/mK]\). \(h\) is the convective constant of the coolant fluid and \(T_{water}\) is the temperature of the coolant fluid, for example water.

The solution of equation 1 is found by integration.
$$\frac{dT}{dx} = C_1$$
$$T(x)=C_1 x + C_2$$

Solving for the boundary conditions, we find the integrations constants \(C_1\) and \(C_2\) to be:
$$C_1 = \frac{P}{\lambda A}$$
$$C_2 = \frac{P}{h} + T_{water}  $$

Hence, the temperature of the rod is:

$$T(x) =  \frac{P}{\lambda A} x +\frac{P}{h} + T_{water} $$

Saturday, May 11, 2013

1D heat equation applied to a rod. Dirichlet + Neumann boundary conditions

The temperature profile of a rod, subjected to a thermal loading is derived. The rod has length \(L\), and a cross-sectional area, \(A\).  The thermal loading \(P\) is applied at \(x=0\). At \(x=L\), the temperature is kept constant at \(T_C\).

The one-dimensional heat equation is given by:
$$\frac{d^2T}{dx^2}=0$$
The boundary conditions are as follows:
$$T(x=L)=T_C$$
$$\frac{dT}{dx}(x=0)=-\frac{P}{\lambda A}$$

where, \(P\) is the aplied thermal loading in \([W]\), \(A\) is the cross-sectional area of the beam, and \(\lambda\) the thermal conductivity of the involved material in \([W/mK]\).

The thermal gradients are obtained by integration of equation 1.
$$\frac{dT}{dx} = C_1$$
$$T(x)=C_1 x + C_2$$

Solving for the boundary conditions, we find the temperature profile of the beam to be:
$$T(x) =- \frac{P}{\lambda A} x+T_C+\frac{pL}{\lambda A}$$