Series-parallel pipes with control valve

Method. Use $h_f = rQ^2$ for each pipe, with $r = \dfrac{8fL}{\pi^2gD^5}$ ($f = 0.024$). Only friction is considered. Heads: $H_A = 100$ m, $H_E = 40$ m, $H_F = 60$ m. For equal discharge let $Q_E = Q_F = q$, so the flow in $BC$ is $2q$.

Resistance coefficients (s²/m⁵)

Parallel pair A to B (same head loss in both): $Q = \sqrt{h/r_1} + \sqrt{h/r_2}$, so

$$\frac{1}{\sqrt{r_{AB}}} = \frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}} \;\Rightarrow\; r_{AB} = 788.4$$

Discharge

Energy balance from A to F (pair AB, then BC carrying $2q$, then CF carrying $q$):

$$H_A - H_F = r_{AB}(2q)^2 + r_{BC}(2q)^2 + r_{CF}q^2$$ $$100 - 60 = q^2\left[4(788.4 + 1566.9) + 5949.1\right] = 15370.2\,q^2 \;\Rightarrow\; q = 0.05101\ \text{m}^3/\text{s}$$ $$Q = 2q = 0.10203\ \text{m}^3/\text{s} = 102.0\ \text{l/s}$$

Head at C

$$H_C = H_F + r_{CF}q^2 = 60 + 5949.1(0.002602) = 75.48\ \text{m}$$

(Check from A: $H_B = H_C + r_{BC}(2q)^2 = 91.79$ m; the two parallel pipes then carry 0.03714 and 0.06489 m³/s, which add to 0.10203 m³/s.)

Valve opening

Branch C to E has pipe CD and the valve DE in series, carrying $q$ with head loss $H_C - H_E = 75.48 - 40 = 35.48$ m:

$$(r_{CD} + K_{DE})q^2 = 35.48 \;\Rightarrow\; K_{DE} = \frac{35.48}{0.002602} - 2974.6 = 10659.7$$ $$K_{DE} = \left(\frac{4000}{n}\right)^2 \;\Rightarrow\; n = \frac{4000}{\sqrt{10659.7}} = 38.7\ \%$$

Answer: Head at C $= 75.48$ m; total discharge $= 0.1020$ m³/s ($102.0$ l/s); valve opening $n \approx 38.7\ \%$.