Question

What is the equivalent of (ρC ΦC)^t+Δt/2 using the second-order upwind Euler scheme for finite volume approach?

(a) \(\frac{3}{2}\) (ρC ΦC)^t+(ρC ΦC)^t-Δt

(b) (ρC ΦC)^t+\(\frac{1}{2}\) (ρC ΦC)^t-Δt

(c) \(\frac{3}{2}\) (ρC ΦC)^t+\(\frac{1}{2}\) (ρC ΦC)^t-Δ t

(d) \(\frac{1}{2}\)(ρC ΦC)^t+\(\frac{1}{2}\) (ρC ΦC)^t-Δ t

I have been asked this question by my college director while I was bunking the class.

My question is taken from Transient Flows topic in portion Transient Flows of Computational Fluid Dynamics

Answer 1

The correct answer is:

(b) (ρC ΦC)^t + 1/2 (ρC ΦC)^t-Δt

Explanation:

In the second-order upwind Euler scheme for the finite volume approach, the term (ρCΦC)t+Δt/2(ρC ΦC)^{t+Δt/2}(ρCΦC)t+Δt/2 represents the average value of the quantity at time levels $t$ and $t-\Delta t$. The method is designed to achieve second-order accuracy by using both the current and previous time steps.

The scheme typically takes the average of (ρCΦC)t(ρC ΦC)^t(ρCΦC)t and (ρCΦC)t−Δt(ρC ΦC)^{t-Δt}(ρCΦC)t−Δt, with the factor 1/2 to properly weight the values from the two time steps.

Thus, the expression becomes: (ρCΦC)t+Δt/2=(ρCΦC)t+12(ρCΦC)t−Δt(ρC ΦC)^{t + Δt/2} = (ρC ΦC)^t + \frac{1}{2} (ρC ΦC)^{t - Δt}(ρCΦC)t+Δt/2=(ρCΦC)t+21​(ρCΦC)t−Δt

Correct Answer: (b) (ρC ΦC)^t + 1/2 (ρC ΦC)^t-Δt