Maximum length of pipe portion AB

Data. $H = 15$ m, $D_1 = D_{AB} = 0.2$ m, $D_2 = D_{BC} = 0.15$ m, $L_1 + L_2 = 3000$ m, $f = 0.012$. Summit B is 2 m below the upper water level; pressure head at B $\ge -2$ m (gauge). Secondary (minor) losses are neglected.

Let $h_1 = \dfrac{V_1^2}{2g}$. By continuity $V_2 = V_1\left(\dfrac{D_1}{D_2}\right)^2 = 1.7778V_1$, so $\dfrac{V_2^2}{2g} = \left(\dfrac{D_1}{D_2}\right)^4h_1 = 3.1605\,h_1$.

Condition at B (limiting pressure)

Energy equation from the upper reservoir surface (datum, $p = 0$) to B, which is 2 m below the surface ($z_B = -2$), with $\dfrac{p_B}{\gamma} = -2$ m:

$$0 = -2 + (-2) + h_1\left(1 + f\frac{L_1}{D_1}\right) \;\Rightarrow\; h_1\left(1 + 0.060L_1\right) = 4 \quad\cdots(1)$$

(here $f/D_1 = 0.012/0.2 = 0.060$ per metre).

Condition for the total head

Level difference = friction in AB + friction in BC (no minor losses):

$$15 = h_1\,(0.060)L_1 + 3.1605\,h_1\left(\frac{0.012}{0.15}\right)(3000 - L_1)$$ $$15 = h_1\left[0.060L_1 + 0.080(3.1605)(3000 - L_1)\right] \quad\cdots(2)$$

Solve

Substitute $h_1 = \dfrac{4}{1 + 0.060L_1}$ from (1) into (2):

$$15\left(1 + 0.060L_1\right) = 4\left[0.060L_1 + 0.080(3.1605)(3000 - L_1)\right]$$ $$L_1 = 1806\ \text{m}$$

Check: $h_1 = \dfrac{4}{1 + 0.06(1806)} = 0.03657$ m, so $V_1 = 0.847$ m/s, $V_2 = 1.506$ m/s, $Q = 0.0266$ m³/s; friction in AB $= 3.963$ m and in BC $= 11.037$ m, which add to 15 m.

Answer: The maximum allowable length of portion AB $\approx 1806$ m (so BC is about 1194 m).