HGL and TEL for two-pipe system

Given. Reservoir level 8 m above the pipe axis (datum = pipe axis). Pipe 1: $D_1 = 0.15$ m, $L_1 = 25$ m. Pipe 2: $D_2 = 0.3$ m, $L_2 = 15$ m, $f = 0.01$. The pipe discharges freely, so the outlet velocity head stays as kinetic energy (no exit loss term, the HGL ends at the axis).

Velocities

Continuity: $V_2 = V_1\left(\dfrac{D_1}{D_2}\right)^2 = V_1/4$.

Energy equation from the reservoir surface to the outlet (all losses):

$$H = 0.5\frac{V_1^2}{2g} + f\frac{L_1}{D_1}\frac{V_1^2}{2g} + \frac{(V_1-V_2)^2}{2g} + f\frac{L_2}{D_2}\frac{V_2^2}{2g} + \frac{V_2^2}{2g}$$ $$8 = \frac{V_1^2}{2g}\left[0.5 + 0.01\left(\frac{25}{0.15}\right) + (1-0.25)^2 + 0.01\left(\frac{15}{0.3}\right)(0.25)^2 + (0.25)^2\right] = 2.8229\,\frac{V_1^2}{2g}$$ $$V_1 = \sqrt{\frac{2(9.81)(8)}{2.8229}} = 7.457\ \text{m/s},\qquad V_2 = 1.864\ \text{m/s}$$

Velocity heads: $V_1^2/2g = 2.834$ m, $V_2^2/2g = 0.177$ m. Discharge $= 131.8$ l/s.

Head losses

The total equals the 8 m available head.

Levels of TEL and HGL (m above the pipe axis)

 8.0 ___                        TEL
     |  |\ 0.5V²/2g
     |  | \__ TEL falls steeply in pipe 1 (small D)
     |  |    `-.__
     |  |  HGL     `-.  | drop (V1-V2)²/2g
     |  |  (well below) |`.__ TEL falls gently in pipe 2
     |  |               |    `-.__ HGL
 0 --|__|===[D1=0.15]===[D2=0.3]===> free outlet

Draw to scale: TEL: 8.0, 6.58 (entrance drop), 1.86 (end of pipe 1), 0.27 (after expansion drop), 0.18 (outlet). HGL is lower by 2.83 m in pipe 1 and by 0.18 m in pipe 2, and meets the pipe axis at the outlet. The HGL rises slightly at the expansion (from -0.97 to 0.09 m) because the velocity head falls.

Answer: $V_1 = 7.46$ m/s, $V_2 = 1.86$ m/s, TEL ends 0.177 m above the axis at the outlet and the HGL ends at the axis.