How is energy at the toe above the floor (RL Z) expressed?

E1 = 0.9 (50 - Z), because 10% is lost on the spillway

What condition fixes the floor level Z?

Tail water depth over the floor (25 - Z) equals y2

Why is a Type II basin chosen?

Fr1 = 7.85 > 4.5 and V1 = 24.7 m/s > 18 m/s

Which dimensions are correct for the Type II basin?

L about 4.3 y2 = 47 m, dentated end sill 0.2 y2 = 2.15 m

2078 Bhadra · 10 marks

Stilling basin invert level and USBR type

Short answer

Floor invert RL 14.27 m. y1=1.011 m, y2=10.73 m, Fr1=7.85. USBR Type II basin length 47 m, chute blocks 1.0 m, dentated end sill 2.15 m.

The discharge of water over a spillway 12 m wide is 300 m³/s into a stilling basin of the same width. The lake level behind the spillway has an elevation of 50 m and the river water surface elevation downstream of the stilling basin is 25 m. Assume a 10% energy loss in flow down the spillway, find the invert level elevation of the floor of the stilling basin so that a hydraulic jump forms in the basin. Select an appropriate USBR stilling basin and list all the dimensions.

Given: $Q=300$ m³/s, width $B=12$ m, so $q=\dfrac{300}{12}=25$ m³/s/m; lake level RL 50 m; tail water level RL 25 m; 10% of the energy is lost on the spillway.

Let $Z$ be the RL of the basin floor. Then the energy at the toe above the floor is

E_1=0.9\,(50-Z),\qquad y_1+\frac{q^2}{2g\,y_1^2}=E_1

and the jump forms at the toe if the tail water depth equals the sequent depth:

25-Z=y_2=\frac{y_1}{2}\left(\sqrt{1+8Fr_1^2}-1\right),\qquad Fr_1=\frac{q}{y_1\sqrt{gy_1}}

Solution by trial

Floor RL $Z$ (m)	$E_1$ (m)	$y_1$ (m)	$Fr_1$	$y_2$ (m)	$Z+y_2$ (m)
12.0	34.20	0.979	8.24	10.93	22.93
16.0	30.60	1.038	7.55	10.57	26.57
14.27	32.16	1.011	7.85	10.73	25.00

The condition $Z+y_2=25$ is met for

\boxed{Z=14.27\ \text{m}}\ \text{(invert level of the basin floor)},\quad y_1=1.011\ \text{m},\ V_1=24.7\ \text{m/s},\ Fr_1=7.85,\ y_2=10.73\ \text{m}

Selection of the USBR basin

$Fr_1=7.85>4.5$ , $V_1=24.7$ m/s $>18$ m/s and $q=25>18.5$ m³/s/m, so a Type II basin is suitable (Type III is limited to $V_1<18$ m/s; Type IV is for $2.5<Fr_1<4.5$ ).

Dimension	Value
Basin floor (invert) level	RL 14.27 m
Basin length $L\approx4.3\,y_2$	46.1 m, say 47 m
Chute blocks: height = width = spacing $\approx y_1$	1.011 m, say 1.0 m
Dentated end sill: height $0.2y_2$	2.15 m
End sill tooth width and spacing $0.15y_2$	1.61 m
Tail water depth over floor	10.73 m (= $y_2$ )
Width of basin	12 m (same as spillway)
Training walls	$\approx y_2+$ freeboard, i.e. about 12 m above floor

 lake 50 ___
           \ \
            \ \__[chute blocks]______[dentated sill]
  floor 14.27  \_|#|#|#|_______________|^|^|^|^|====> TWL 25
            |<-------- L = 47 m -------->|

Answer: floor invert RL 14.27 m; USBR Type II basin, length about 47 m, chute blocks 1.0 m, dentated end sill 2.15 m, with $y_1=1.011$ m and $y_2=10.73$ m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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