---
title: "Characteristic form FD solution (velocity, pressure)"
subject: "Computational Techniques in Civil Engineering"
chapter: 6
key: "semester_8/computational_techniques_in_civil_engineering:characteristic-fd-velocity-pressure"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/characteristic-fd-velocity-pressure
site: BCT Engineering
---

# Characteristic form FD solution (velocity, pressure)

Computational Techniques in Civil Engineering · Chapter 6: Method of Characteristics

- Asked in: 2079 Jestha, 6 marks; 2073 Magh, 6 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/characteristic-fd-velocity-pressure
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-6

## Question

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

## Answer

#### Governing equations (velocity $V$, pressure $p$)
For a pipe of diameter $D$ inclined at angle $\theta$ to the horizontal (water density $\rho$, wave speed $a$, friction factor $f$), neglecting the small convective terms:

**Momentum:**
$$
L_1=\frac{\partial V}{\partial t}+\frac1\rho\frac{\partial p}{\partial x}+g\sin\theta+\frac{f}{2D}V|V|=0
$$
**Continuity:**
$$
L_2=\frac{\partial p}{\partial t}+\rho a^2\frac{\partial V}{\partial x}=0
$$

#### Characteristic equations
Form $L_1+\lambda L_2=0$:
$$
\left[\frac{\partial V}{\partial t}+\lambda\rho a^2\frac{\partial V}{\partial x}\right]+\lambda\left[\frac{\partial p}{\partial t}+\frac{1}{\lambda\rho}\frac{\partial p}{\partial x}\right]+g\sin\theta+\frac{f}{2D}V|V|=0
$$
For both brackets to be total derivatives, $\dfrac{dx}{dt}=\lambda\rho a^2=\dfrac{1}{\lambda\rho}$, so $\lambda=\pm\dfrac{1}{\rho a}$ and $\dfrac{dx}{dt}=\pm a$:

$$
C^+\ \left(\frac{dx}{dt}=+a\right):\quad \frac{dV}{dt}+\frac{1}{\rho a}\frac{dp}{dt}+g\sin\theta+\frac{f}{2D}V|V|=0
$$
$$
C^-\ \left(\frac{dx}{dt}=-a\right):\quad \frac{dV}{dt}-\frac{1}{\rho a}\frac{dp}{dt}+g\sin\theta+\frac{f}{2D}V|V|=0
$$

#### Finite difference form
Use the rectangular grid with A, C, B at time $t$ and P at $t+\Delta t$, with $\Delta x=a\Delta t$. Multiply by $\rho a$, integrate along each characteristic (friction at the known point), and put $\Delta x\sin\theta=\Delta z$ (rise of the pipe over $\Delta x$):

```text
 t+dt :         P
               / \
         C+  /     \  C-
 t    :    A----C----B
```
$$
C^+:\quad p_P=p_A-\rho a\left(V_P-V_A\right)-\rho g\left(z_P-z_A\right)-\frac{\rho f\Delta x}{2D}V_A|V_A|
$$
$$
C^-:\quad p_P=p_B+\rho a\left(V_P-V_B\right)+\rho g\left(z_B-z_P\right)+\frac{\rho f\Delta x}{2D}V_B|V_B|
$$

#### Solution
Equating the two expressions for $p_P$:
$$
V_P=\frac{(p_A-p_B)+\rho a(V_A+V_B)-\rho g(z_B-z_A)-\dfrac{\rho f\Delta x}{2D}\left[V_A|V_A|+V_B|V_B|\right]}{2\rho a}
$$
and $p_P$ 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. $p=$ reservoir pressure, or $V=0$ at a closed valve).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. This formulation uses which dependent variables?
   A. Q and H
   B. V and p
   C. y and Q
   D. A and V
   Answer: B. Momentum and continuity are written in velocity and pressure. (Key point: Variables)

2. The characteristic lines have slope:
   A. dx/dt = +/- a
   B. dx/dt = +/- V
   C. dx/dt = +/- rho a
   D. dx/dt = 0
   Answer: A. lambda = +/- 1/(rho a). (Key point: Characteristics)

3. The finite-difference form uses a rectangular grid with:
   A. dx = a dt
   B. dx = V dt
   C. dt = a dx
   D. dx = dt
   Answer: A. C+ and C- lines then pass through grid points A and B. (Key point: Grid)

## Related in Computational Techniques in Civil Engineering

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