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Four-point implicit derivatives of Q

Short answer

Using the four-point implicit formulas with θ = 0.55: ∂Q/∂t = [(3580−3500)+(3470−3350)]/(2×3600) = 2.78×10⁻² m³/s², and ∂Q/∂x = 0.55×(−110/1000) + 0.45×(−150/1000) = −0.128 m²/s.

The value of flow rate QQ at four points in the space-time grid are shown in the figure below. Δt=1\Delta t = 1 h, Δx=1000\Delta x = 1000 m and θ=0.55\theta = 0.55. Calculate the values of ∂Q/∂t\partial Q/\partial t and ∂Q/∂x\partial Q/\partial x by the four-point implicit method. (θ\theta = weighting factor.)

[Figure: space-time grid with time levels j and j+1 and distance points i and i+1 (Δx=1000\Delta x = 1000 m, Δt=1\Delta t = 1 h); QQ at (i, j+1) =3580= 3580 cms, at (i+1, j+1) =3470= 3470 cms, at (i, j) =3500= 3500 cms, at (i+1, j) =3350= 3350 cms]

Four-point implicit method

∂Q∂t≈(Qij+1−Qij)+(Qi+1j+1−Qi+1j)2Δt\frac{\partial Q}{\partial t}\approx\frac{(Q_i^{j+1}-Q_i^j)+(Q_{i+1}^{j+1}-Q_{i+1}^j)}{2\Delta t} ∂Q∂x≈θ Qi+1j+1−Qij+1Δx+(1−θ) Qi+1j−QijΔx\frac{\partial Q}{\partial x}\approx\theta\,\frac{Q_{i+1}^{j+1}-Q_i^{j+1}}{\Delta x}+(1-\theta)\,\frac{Q_{i+1}^j-Q_i^j}{\Delta x}

Data: Qij+1=3580Q_i^{j+1}=3580, Qi+1j+1=3470Q_{i+1}^{j+1}=3470, Qij=3500Q_i^j=3500, Qi+1j=3350Q_{i+1}^j=3350; Δt=3600\Delta t=3600 s, Δx=1000\Delta x=1000 m, θ=0.55\theta=0.55.

Time derivative

∂Q∂t=(3580−3500)+(3470−3350)2×3600=2007200=2.7778×10−2 m3/s2\frac{\partial Q}{\partial t}=\frac{(3580-3500)+(3470-3350)}{2\times3600}=\frac{200}{7200}=2.7778\times10^{-2}\ \text{m}^3/\text{s}^2

Space derivative

∂Q∂x=0.55 3470−35801000+0.45 3350−35001000=−0.06050+(−0.06750)=−0.1280 m3/s per m\frac{\partial Q}{\partial x}=0.55\,\frac{3470-3580}{1000}+0.45\,\frac{3350-3500}{1000}=-0.06050+(-0.06750)=-0.1280\ \text{m}^3/\text{s per m}

Answer: ∂Q/∂t=2.7778×10−2\partial Q/\partial t=2.7778\times10^{-2} m3^3/s2^2 and ∂Q/∂x=−0.1280\partial Q/\partial x=-0.1280 m2^2/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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