MOC numerical: discharge and head at P (30 cm pipe)

Use the finite difference form of the characteristic equations on the rectangular grid (A and B at time $t$, P at $t+\Delta t$):

$$C^+:\ H_P=H_A-B_p(Q_P-Q_A)-R_pQ_A|Q_A|,\qquad C^-:\ H_P=H_B+B_p(Q_P-Q_B)+R_pQ_B|Q_B|$$ $$B_p=\frac{a}{gA_p},\qquad R_p=\frac{f\,\Delta x}{2gDA_p^2}$$

$H$ is the piezometric head (it already contains the elevation), so the 1 m elevation difference between A and P is not needed to find $H_P$ and $Q_P$.

Step 1: Constants

$$A_p=\frac{\pi}{4}(0.3)^2=0.07069\ \text{m}^2$$ $$B_p=\frac{1200}{9.81\times0.07069}=1730.53\ \text{s/m}^2,\qquad R_p=\frac{0.02\times500}{2\times9.81\times0.3\times0.07069^2}=340.03\ \text{s}^2/\text{m}^5$$

(Here $a\Delta t=1200\times0.4=480$ m is close to $\Delta x=500$ m, so the characteristics are taken through A and B as in the figure.)

Step 2: Known quantities

$$K_1=H_A+B_pQ_A-R_pQ_A|Q_A|=26.5+1730.53(0.4)-340.03(0.16)=664.31$$ $$K_2=H_B-B_pQ_B+R_pQ_B|Q_B|=27.5-1730.53(0.45)+340.03(0.2025)=-682.38$$

Step 3: Solve for P

$$Q_P=\frac{K_1-K_2}{2B_p}=\frac{664.31-(-682.38)}{2\times1730.53}=0.3891\ \text{m}^3/\text{s}$$ $$H_P=\frac{K_1+K_2}{2}=-9.038\ \text{m}$$

Answer: $Q_P\approx0.3891$ m$^3$/s and $H_P\approx-9.038$ m. The negative head means the data give a pressure below the vapour pressure at P (column separation would occur), because the large impedance $B_p$ makes the $Q$ difference between A and B produce a big head change.