In FTCS, T_i^{n+1} is found from:

Forward time difference and central space difference at the old time level

Backward time difference at new level

The stability limit for FTCS diffusion is:

r <= 0.5, where r = alpha dt/dx^2

Practice · 8 marks

Explicit scheme for transient conduction

Short answer

FTCS scheme: T_i^{n+1}=T_i^n + r(T_{i+1}^n-2T_i^n+T_{i-1}^n) with r=αΔt/Δx^2. Stability requires r≤0.5. For α=9.7e-5 m²/s, Δx=0.02 m, max Δt=2.06 s. With Δt=1 s (r=0.2425), after two steps interior temperatures: 49.39, 24.70, 18.82, 12.65 °C.

A metal bar has thermal diffusivity $\alpha = 9.7\times10^{-5}$ m²/s. It is modelled with nodes 0 to 5 at spacing $\Delta x = 0.02$ m. Initially the temperatures of nodes 0 to 5 are 100, 20, 20, 20, 20, 0 °C, and the end nodes are held at 100 °C and 0 °C. (a) Write the explicit FTCS scheme for $\partial T/\partial t=\alpha\,\partial^2T/\partial x^2$ and its stability limit. (b) Find the largest stable time step. (c) With $\Delta t = 1$ s, compute the interior temperatures after two time steps.

(a) FTCS scheme

Use a forward difference in time and a central difference in space:

\frac{T_i^{n+1}-T_i^n}{\Delta t}=\alpha\frac{T_{i+1}^n-2T_i^n+T_{i-1}^n}{\Delta x^2}

T_i^{n+1}=T_i^n+r\,(T_{i+1}^n-2T_i^n+T_{i-1}^n),\qquad r=\frac{\alpha\Delta t}{\Delta x^2}

The scheme is first order in time and second order in space. By von Neumann analysis the amplification factor is $G=1-4r\sin^2(\theta/2)$ ; $|G|\le1$ requires

r\le\frac12

(equivalently the coefficient of $T_i^n$ , $1-2r$ , must not be negative.)

(b) Maximum time step

\Delta t_{max}=\frac{0.5\,\Delta x^2}{\alpha}=\frac{0.5\times(0.02)^2}{9.7\times10^{-5}}=2.06\ \text{s}

(c) Two steps with $\Delta t=1$ s

r=\frac{9.7\times10^{-5}\times1}{(0.02)^2}=0.2425\quad(<0.5,\ \text{stable})

Step 1 (from 100, 20, 20, 20, 20, 0):

$T_1=20+0.2425(20-40+100)=20+19.40=39.40$
$T_2=20+0.2425(20-40+20)=20.00$
$T_3=20+0.2425(20-40+20)=20.00$
$T_4=20+0.2425(0-40+20)=20-4.85=15.15$

Step 2 (from 100, 39.40, 20, 20, 15.15, 0):

$T_1=39.40+0.2425(20-78.8+100)=39.40+9.99=49.39$
$T_2=20+0.2425(20-40+39.40)=20+4.70=24.70$
$T_3=20+0.2425(15.15-40+20)=20-1.18=18.82$
$T_4=15.15+0.2425(0-30.3+20)=15.15-2.50=12.65$

Node	0	1	2	3	4
t = 0	100	20	20	20	20
t = 1 s	100	39.40	20.00	20.00	15.15
t = 2 s	100	49.39	24.70	18.82	12.65

Heat diffuses in from the hot end, and node 4 cools towards the 0 °C end.

Answer: $r\le0.5$ , $\Delta t_{max}=2.06$ s; after 2 s the interior nodes are 49.39, 24.70, 18.82, 12.65 °C.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Explicit scheme for transient conduction

(a) FTCS scheme

(b) Maximum time step

(c) Two steps with Δt=1\Delta t=1Δt=1 s

Related questions

(c) Two steps with $\Delta t=1$ s