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Characteristic form FD solution (velocity, pressure)

Short answer

Converted to ODEs along dx/dt = ±a: dV/dt ± (1/ρa) dp/dt + g sinθ + (f/2D)V|V| = 0. Finite-difference integration over Δx = aΔt gives C⁺: p_P = p_A − ρa(V_P−V_A) − ρg(z_P−z_A) − (ρfΔx/2D)V_A|V_A|, and C⁻: p_P = p_B + ρa(V_P−V_B) + ρg(z_B−z_P) + (ρfΔx/2D)V_B|V_B|; solving yields V_P and p_P.

Develop a characteristic form of a finite-difference solution of the unsteady flow equations to get the solution in terms of velocity and pressure.

Governing equations (velocity VV, pressure pp)

For a pipe of diameter DD inclined at angle θ\theta to the horizontal (water density ρ\rho, wave speed aa, friction factor ff), neglecting the small convective terms:

Momentum:

L1=∂V∂t+1ρ∂p∂x+gsin⁡θ+f2DV∣V∣=0L_1=\frac{\partial V}{\partial t}+\frac1\rho\frac{\partial p}{\partial x}+g\sin\theta+\frac{f}{2D}V|V|=0

Continuity:

L2=∂p∂t+ρa2∂V∂x=0L_2=\frac{\partial p}{\partial t}+\rho a^2\frac{\partial V}{\partial x}=0

Characteristic equations

Form L1+λL2=0L_1+\lambda L_2=0:

[∂V∂t+λρa2∂V∂x]+λ[∂p∂t+1λρ∂p∂x]+gsin⁡θ+f2DV∣V∣=0\left[\frac{\partial V}{\partial t}+\lambda\rho a^2\frac{\partial V}{\partial x}\right]+\lambda\left[\frac{\partial p}{\partial t}+\frac{1}{\lambda\rho}\frac{\partial p}{\partial x}\right]+g\sin\theta+\frac{f}{2D}V|V|=0

For both brackets to be total derivatives, dxdt=λρa2=1λρ\dfrac{dx}{dt}=\lambda\rho a^2=\dfrac{1}{\lambda\rho}, so λ=±1ρa\lambda=\pm\dfrac{1}{\rho a} and dxdt=±a\dfrac{dx}{dt}=\pm a:

C+ (dxdt=+a):dVdt+1ρadpdt+gsin⁡θ+f2DV∣V∣=0C^+\ \left(\frac{dx}{dt}=+a\right):\quad \frac{dV}{dt}+\frac{1}{\rho a}\frac{dp}{dt}+g\sin\theta+\frac{f}{2D}V|V|=0 C− (dxdt=−a):dVdt−1ρadpdt+gsin⁡θ+f2DV∣V∣=0C^-\ \left(\frac{dx}{dt}=-a\right):\quad \frac{dV}{dt}-\frac{1}{\rho a}\frac{dp}{dt}+g\sin\theta+\frac{f}{2D}V|V|=0

Finite difference form

Use the rectangular grid with A, C, B at time tt and P at t+Δtt+\Delta t, with Δx=aΔt\Delta x=a\Delta t. Multiply by ρa\rho a, integrate along each characteristic (friction at the known point), and put Δxsin⁡θ=Δz\Delta x\sin\theta=\Delta z (rise of the pipe over Δx\Delta x):

 t+dt :         P
               / \
         C+  /     \  C-
 t    :    A----C----B
C+:pP=pA−ρa(VP−VA)−ρg(zP−zA)−ρfΔx2DVA∣VA∣C^+:\quad p_P=p_A-\rho a\left(V_P-V_A\right)-\rho g\left(z_P-z_A\right)-\frac{\rho f\Delta x}{2D}V_A|V_A| C−:pP=pB+ρa(VP−VB)+ρg(zB−zP)+ρfΔx2DVB∣VB∣C^-:\quad p_P=p_B+\rho a\left(V_P-V_B\right)+\rho g\left(z_B-z_P\right)+\frac{\rho f\Delta x}{2D}V_B|V_B|

Solution

Equating the two expressions for pPp_P:

VP=(pA−pB)+ρa(VA+VB)−ρg(zB−zA)−ρfΔx2D[VA∣VA∣+VB∣VB∣]2ρaV_P=\frac{(p_A-p_B)+\rho a(V_A+V_B)-\rho g(z_B-z_A)-\dfrac{\rho f\Delta x}{2D}\left[V_A|V_A|+V_B|V_B|\right]}{2\rho a}

and pPp_P is then obtained from either equation. At each time step, the interior points are found this way; boundary points use one characteristic plus the boundary condition (e.g. p=p= reservoir pressure, or V=0V=0 at a closed valve).

✓ Answer checked by Claude Opus 5.5 (high effort)

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