Transient Conduction FDM Solver
Solve 1D transient heat conduction numerically. Compare Explicit, Implicit Backward, and Crank-Nicolson formulations in real-time.
1D Grid Discretization
The solver discretizes the 1D space into $N$ control volumes (nodes). Nodal space steps are computed as $\Delta x = L / (N - 1)$.
- Radial Domains: For Cylindrical and Spherical systems, the left boundary ($i=1$) is automatically insulated representing symmetry at $r=0$.
- Variable Properties: Conductivities and specific heats are updated local to each node temperature at each step.
Solver Configuration
Results & Real-Time Animation
Dynamic Temperature Profile: T(x, t)
Nodal Temperature History: T vs Time
Simulation Metrics
Nodal temperature distribution reports will appear here after clicking "Generate Engineering Report".
Numerical Solver Methodology
Numerical Discretization
The 1D transient heat equation is discretized in space using central finite differences on $N$ node points. For interior nodes: $$\rho C_p \frac{T_i^{p+1} - T_i^p}{\Delta t} = \theta \cdot \text{RHS}_i^{p+1} + (1 - \theta) \cdot \text{RHS}_i^p$$ where $\theta$ determines the algorithm: - **Explicit Euler ($\theta = 0$):** Simple, marching step-by-step but requires stability limit $Fo \le 0.5$. - **Implicit Backward Euler ($\theta = 1$):** Unconditionally stable, 1st order accurate in time. - **Crank-Nicolson ($\theta = 0.5$):** Unconditionally stable, 2nd order accurate in time.
Thomas Algorithm (TDMA):
Implicit systems yield tridiagonal algebraic matrices solved efficiently in $O(N)$ operations via TDMA: $$A_i T_{i-1}^{p+1} + B_i T_i^{p+1} + C_i T_{i+1}^{p+1} = D_i$$
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer. Chapter 5.10.
- Patankar, S. V. (1980). Numerical Heat Transfer and Fluid Flow. McGraw-Hill.
- Ozisik, M. N. (1993). Heat Conduction (2nd Edition). Wiley. Chapter 13.
Nodal Energy Conservation Balance
A carbon steel plate ($k = 43$ W/m·K, density $\rho = 7850$ kg/m³, specific heat $C_p = 475$ J/kg·K) of thickness $L = 50$ mm is initially at $180$°C. The left face is insulated and the right face is cooled by convection ($h = 150$ W/m²·K, $T_\infty = 20$°C). Perform an energy balance validation at $t = 600$ s.
Step-by-step Solution:
1. Calculate the initial thermal energy stored per unit area:
$$E_{st,0} = \rho C_p L T_i = 7850 \times 475 \times 0.05 \times 180 = 33.56 \text{ MJ/m²}$$ 2. As time marches, heat escapes through the convection face. The cumulative energy leaving the boundary is:
$$E_{out} = \int_{0}^{t} h (T_{surface} - T_\infty) dt$$ 3. Compute energy remaining in the nodes at $t = 600$ s by numerical integration:
$$E_{st}(t) = \sum_{i=1}^N \rho C_{p,i} T_i(t) \Delta V_i$$ 4. Verify conservation of energy compliance:
$$E_{st}(t) - E_{st,0} = E_{in} - E_{out} = 0 - E_{out}$$ The numerical balance error is tracked to guarantee code conservation validity: $$\text{Error} = \frac{|(E_{st}(t) - E_{st,0}) + E_{out}|}{E_{st,0}} \times 100\% < 0.01\%$$