Penstock pressure rise and pressure-time sketch

Data. $V = 4$ m/s, $L = 4500$ m, $c = 1500$ m/s, sudden valve closure at the downstream end.

Maximum pressure rise (Joukowsky)

Sudden closure means $t_c \le 2L/c = 6.0$ s (rapid closure), so the full rise occurs:

$$\Delta H = \frac{cV}{g} = \frac{1500(4)}{9.81} = 611.6\ \text{m of water}$$ $$\Delta p = \rho cV = 1000(1500)(4) = 6.0\times10^{6}\ \text{Pa} = 6.0\ \text{MPa}$$

Pressure-time variation at 400 m upstream of the valve

Distance from the valve $d = 400$ m. Wave arrival times:

• $t_1 = \dfrac{d}{c} = \dfrac{400}{1500} = 0.267$ s: the pressure wave arrives; pressure rises by $+611.6$ m ($+6.0$ MPa)
• $t_2 = \dfrac{2L - d}{c} = 5.733$ s: the negative wave reflected from the reservoir arrives; pressure returns to normal
• $t_3 = \dfrac{2L + d}{c} = 6.267$ s: the negative wave reflected from the valve arrives; pressure drops by $611.6$ m
• $t_4 = \dfrac{4L - d}{c} = 11.733$ s: the positive wave reflected from the reservoir arrives; pressure returns to normal
• The cycle repeats every $4L/c = 12.0$ s.

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(top = 1.0, bottom = -1.0; '-' is zero line)

(+1 = $+\Delta H$, $-1$ = $-\Delta H$ relative to the normal pressure; time from 0 to $4L/c = 12$ s.)

Answer: Maximum pressure rise $= 612$ m of water ($6.0$ MPa). At 400 m from the valve the head is raised by 612 m from 0.267 s to 5.733 s and lowered by 612 m from 6.267 s to 11.733 s; this stepped wave repeats every 12 s.