Characteristic curve; velocity and pressure at P

Characteristic curve

In the $x$-$t$ plane, a characteristic curve is a line $dx/dt=\pm a$ along which the partial differential equations of unsteady pipe flow reduce to ordinary differential equations. It is the path of a small pressure disturbance: the $C^+$ curve ($dx/dt=+a$) carries information downstream and the $C^-$ curve ($dx/dt=-a$) upstream. The solution at a point $P$ is found from the values at the two points A and B where the $C^+$ and $C^-$ curves through $P$ meet the previous time line.

 t+dt :         P
               / \
         C+  /     \  C-
 t    :    A----C----B

Numerical part

Equations used (velocity and pressure form, $\rho=1000$ kg/m$^3$, $\Delta x=500$ m so $a\Delta t=1000\times0.5=500$ m):

$$C^+:\ p_P=p_A-\rho a(V_P-V_A)-\rho g(z_P-z_A)-\frac{\rho f\Delta x}{2D}V_A|V_A|$$ $$C^-:\ p_P=p_B+\rho a(V_P-V_B)+\rho g(z_B-z_P)+\frac{\rho f\Delta x}{2D}V_B|V_B|$$

Assumption: the pipe has a uniform slope, so P is 2.50 m above A and 2.50 m below B ($z_B-z_A=5.0$ m, $z_P-z_A=2.5$ m, $z_B-z_P=2.5$ m).

Constants

• $\rho a=1000\times1000=1.0\times10^{6}$ kg/(m$^2$s)
• $\dfrac{\rho f\Delta x}{2D}=\dfrac{1000\times0.02\times500}{2\times0.35}=14285.71$ kg/m$^3$ (friction coefficient)
• $\rho g(z_B-z_A)=9810\times5.0=49050$ N/m$^2$

Velocity at P

$$V_P=\frac{(p_A-p_B)+\rho a(V_A+V_B)-\rho g(z_B-z_A)-\frac{\rho f\Delta x}{2D}\left[V_A|V_A|+V_B|V_B|\right]}{2\rho a}$$ $$V_P=\frac{(102000-124000)+10^6(12.25)-49050-14285.71(36+39.0625)}{2\times10^6} =\frac{11106629}{2\times10^6}=5.5533\ \text{m/s}$$

Pressure at P (from $C^+$)

$$p_P=102000-10^6(5.5533-6)-9810(2.5)-14285.71(36)=9875\ \text{N/m}^2=9.88\ \text{kN/m}^2$$

Check with $C^-$: $p_P=124000+10^6(5.5533-6.25)+9810(2.5)+14285.71(39.0625)=9875$ N/m$^2$ (same).

(The same result follows from the head form: $B_p=a/g=101.94$ s, $H_A=10.398$ m, $H_B=17.640$ m, giving $H_P=3.507$ m and $p_P=\gamma(H_P-z_P)$.)

Answer: $V_P\approx5.5533$ m/s and $p_P\approx9.88$ kN/m$^2$.