Planning of Water and Hydropower Intake Structures (GTZ, 1989, 122 p.) Appendix Annex 1: Hydrometric and hydrometeorological measuring instruments Annex 2: Flood probability calculation according to the recommendations of the DVWK (1979) Annex 3: Calculation example for a free overfall over a wooden beam weir Annex 4: Calculation example for a submerged overfall over a wooden beam weir Annex 5: Calculation example for a discharge over a weir with race door or stilling basin Annex 6: Calculation example for a straight side weir Annex 7: Calculation example for the outflow below a dam wall Annex 8: Calculation example for a bottom intake (Tyrolean weir) Annex 9: Numerical example of the proof of safety from hydraulic shear failure Annex 10: Numerical example of the proof of stability against sliding of a fixed weir Annex 11: Calculation example for the design of a sand trap

Planning of Water and Hydropower Intake Structures (GTZ, 1989, 122 p.)

Appendix

Annex 1: Hydrometric and hydrometeorological measuring instruments

The measuring instruments listed in this Annex are selected instruments for the measurement of precipitation and flow velocities. The selection of the makes is limited to German manufacturers:

(1) Wilhelm Lambrecht GmbH
Friedlander Weg 65/67
D-3400 Gottingen
Tel. 0551/49 580

(2) Adolf Thies GmbH + Co. KG
Postfach 3536
D-3400 Gottingen
Tel. 0551/79 20 52

(3) A. Ott GmbH
Postfach 2120
D-8960 Kempten
Tel. 0831/20 59-0
Telex 54 723

(4) SEBA Hydrometrie GmbH
Postfach
D-8950 Kaufbeuren 2
Tel. 08341/62 026
Telex SEBA 54 624

The prices indicated are as per October 1985 and include value-added tax and in some cases (depending upon the manufacturer) packing and transport insurance. All companies deliver their instruments free place of destination in the Federal Republic of Germany.

1. Precipitation measuring instruments

1.1 Rain gauge according to Hellmann

Figure

1.2 Rain recorder according to Hellmann {float principle)

Figure

1.3 Rain recorder according to Hellmann (float principle)

Figure

1.4 Rain recorder with tipping bucket

Figure

1.5 Prices

According to the manufacturer, the prices for rain recorders (float principle) are:

- without heater between about DM 1800 and DM 2850,
- with heater between about DM 2500 and DM 3450.

The price for a rain recorder with tipping bucket is about DM 3900.

2. Water-level gauges and accessories

2.1 Horizontal gauge/recording gauge

(drum-type recorder for the continuous recording of the water levels; protective shell necessary; low construction and maintenance requirements)

Figure

Figure

time of rotation of drum: 1 to 32 days
drum drive: mechanical or electrical
reversing indication (to cover extraordinary flood peaks): available
measuring ranges: unlimited
net weight (basic unit): 7 to 12 kg
Accessories:

- clockwork,
- toothed wheels,
- float,
- reversing indication,
- counterweights,
- float rope.

Whether the accessories belong to the scope of supply depends upon the manufacturer, or the gauge is delivered according to customer's data.

The price for the complete gauge varies between about DM 2 400 and DM 3 200.

2.2 Band recording gauge/horizontal gauge

(as under 2.1 but installation in the open air possible); assembly on float tube with a diameter of 4 to 6".

Technical data as under 2.1 but

- design: band recorder or drum recorder
- casing: plastics, cast material or cast aluminium
- weight: plastics: about 12 to 16 kg, cast material: about 28 kg

According to the manufacturer, the price for a basic unit (complete with accessories but without float tube) is about DM 3500 to DM 5700.

2.3 Staff gauge

- material: cast aluminium or sheet steel
- staff of cast aluminium, very resistant
- fixing to solid, vertical objects or walls possible
- dimensions: 200 to 1000 mm
- weight: sheet steel 2.0 kg/m, cast aluminium 6.2 kg/m.

According to the construction (sheet steel or cast aluminium) and the manufacturer, the price for a staff gauge is about DM 200 to DM 550.

Figure

2.4 Construction of limit value gauges

Easy construction of a limit value gauge which always indicates the highest water level after the passage of the flood waves; construction of the gauge by fixing small tins, etc., at definite intervals to a staff.

Figure

3. Current meter equipment for the measurement of the flow velocities

3.1 Compact current meter equipment

The compact current meter equipment can be used for measuring purposes at water depths of up to about 1.50 m. Adjustment of the measuring height (= current meter height) is made by displacing the current meter of the staff.

The scope of supply comprises:

- current meter,
- blade of plastics,
- counter with stop watch,
- connecting cable, 4 m long,
- bar, f 20 mm, 2 m long, two-part,
- extension bar, 1 m,
- pocket for the complete meter. The price is about DM 2800.

Universal current meter equipment

The universal current meter can be assembled individually. The complete scope of supply can consist, for example, of the following items (prices ace. to manufacturer):

- universal current meter with blade of plastics, between DM 2300 and DM 2650 - bar, 3 m, between DM 550 and DM 650
- connecting cable, 4 m, between DM 100 and DM 160
- canvas bag for the bars, between DM 120 and DM 160
- adjusting device for bar with intermediate piece (necessary for the rapid displacement of the blade on the bar), between DM 700 and DM 900
- counter (electromechanical), between DM 350 and DM 550
- stop watch, between DM 140 end DM 170
- instrument box, between DM 450 and DM 650.

The sum of these items is between about DM 4700 and DM 5900.

3.2 Float equipment

(without current meter; cf. prices under 3.1) The costs of a set of float equipment depend upon the amount of the loading weights (5 to 100 kg). As an example, a loading weight of 25 kg for a measuring range of v=0.025 to 3 m/s was used.

- Intermediate piece to the current meter, float helm and middle piece 25 kg in transport box, between about DM 4200 and DM 5150.
- Complete equipment incl. current meter, float, simple winch (cf: 3.3), carrying rope, counter, cable, etc., between about DM 13000 and DM 16000.

3.3 Winches and accessories

- Simple winches
(vertical displacement of the current meter from bridge)
Weight: 25 kg or 50 kg, depending upon manufacturer (depth counter, 25 m measuring cable, safety crank according to DIN, manual drive)


Figure

- Double winches
(vertical and horizontal displacement of the current meter for cable crane installation)
Double winches are necessary only in connection with a complete cable crane installation. The manufacturers supply these installations complete (winches, carrying cables, supports, current meter with float) according to the stated specific conditions of the respective river cross-section.
The price for

- simple winches is about DM 4000 to DM 4200,

- accessories (jib with return pulley) about DM 600 to DM 750.

Price example for a complete cable crane installation with a span of 50 m and float weights of 25 kg; scope of supply: current meter, carrying ropes and measuring cables, carrying rope attachments, return pulleys, crab, displacement ropes, double winch, tubular steel supports, counter according to manufacturer and equipment (without. assembly) between about DM 30000 and DM 37000.

Annex 2: Flood probability calculation according to the recommendations of the DVWK (1979)

Inclination coefficient CSX / CSY	Recurrence interval T in years
	1,01	2	2,5	3	5	10	20	25	40	50	100	200	500	1000
0	- 2,326	0,000	0,253	0,440	0,842	1,282	1,645	1,751	1,960	2,054	2,326	2,576	2,878	3,090
0,1	- 2,252	- 0,017	0,238	0,417	0,836	1,292	1,673	1,785	2,007	2,107	2,400	2,670	3,004	3,233
0.2	- 2,178	- 0.033	0,222	0,403	0,830	1,301	1,700	1,818	2,053	2,159	2,473	2,763	3,118	3,377
0,3	- 2,104	- 0,050	0,205	0,388	0,824	1,309	1,726	1,849	2,098	2,211	2,544	2,856	3,244	3,521
0,4	- 2,029	- 0,066	0,189	0,373	0,816	1,317	1,750	1,830	2,142	2,261	2,615	2,949	3,366	3,666
0,5	- 1,955	- 0,083	0,173	0,358	0,808	1,323	1,774	1,910	2,185	2,311	2,686	3,041	3,488	3,811
0,6	- 1,880	- 0,099	0,156	0,342	0,800	1,328	1,797	1,939	2,227	2,359	2,755	3,132	3,609	3,956
0,7	- 1,806	- 0,116	0,139	0,327	0,790	1,333	1,819	1,967	2,268	2,407	2,824	3,223	3,730	4,100
0,8	- 1,733	- 0,132	0,122	0,310	0,780	1,336	1,839	1,993	2,308	2,453	2,891	3,312	3,850	4,244
0,9	- 1,660	- 0,148	0,105	0,294	0,769	1,339	1,859	2,018	2,346	2,498	2,957	3,401	3,969	4,388
1,0	- 1,588	- 0,164	0,088	0,277	0,758	1,340	1,877	2,043	2,384	2,542	3,022	3,489	4,088	4,531
1,1	- 1,518	- 0,180	0,070	0,270	0,745	1,341	1,894	2,066	2,420	2,585	3.087	3,575	4,206	4,673
1,2	- 1,449	- 0,195	0,053	0,242	0,732	1,340	1,910	2,087	2,455	2,626	3,149	3,661	4,323	4,815
1,3	- 1,383	- 0,210	0,036	0,225	0,719	1,339	1,925	2,108	2,489	2,666	3,122	3,745	4,438	4,955
1,4	- 1,318	- 0,225	0,018	0,207	0,705	1,337	1,938	2,128	2,521	2,706	3,271	3,828	4,553	5,095
1,5	- 1,256	- 0,240	0,001	0,189	0,690	1,333	1,951	2,146	2,552	2,743	3,330	3,910	4,667	5,234
1,6	- 1,197	- 0,254	- 0,016	0,171	0,675	1,329	1,962	2,163	2,582	2,780	3,388	3,990	4,779	5,371
1,7	- 1,140	- 0,268	- 0,033	0,153	0,660	1,324	1,972	2,179	2,611	2,815	3,444	4,069	4,890	5,507
1,8	- 1,087	- 0,282	- 0,050	0,135	0,643	1,318	1,981	2,193	2,638	2,848	3,499	4,147	5,000	5,642
1,9	- 1,037	- 0,294	- 0,067	0,117	0,627	1,310	1,989	2,207	2,664	2,881	3,553	4,223	5,108	5,775
2,0	- 1,990	- 0,307	- 0,084	0,099	0,609	1,302	1,996	2,219	2,689	2,912	3,605	4,298	5,215	5,908
2,1	- 0,946	- 0,319	- 0,100	0,081	0,592	1,293	2,001	2,230	2,172	2,942	3,656	4,372	5,320	6,039
2,2	- 0,905	- 0,330	- 0,116	0,063	0,574	1,284	2,006	2,240	2,735	2,970	3,705	4,444	5,424	6,168
2,3	- 0,867	- 0,341	- 0,131	0,045	0,555	1,273	2,009	2,248	2,755	2,997	3,753	4,515	5,527	6,296
2,4	- 0,832	- 0,35l	- 0,147	0,027	0,537	1,262	2,011	2,256	2,775	3,023	3,800	4,584	5,628	6,423
2,5	- 0,799	- 0,360	- 0,161	0,010	0,518	1,250	2,012	2,262	2,793	3,048	3,845	4,652	5,728	6,548
2,6	- 0,769	- 0,369	- 0,176	- 0,007	0,499	1,238	2,013	2,267	2,811	3,071	3,889	4,718	5,827	6,672
2,7	- 0,740	- 0,377	- 0,189	- 0,024	0,480	1,224	2,012	2,272	2,827	3,093	3,832	4,783	5,923	6,794
2,8	- 0,714	- 0,384	- 0,203	- 0,041	0,460	1,210	2,010	2,275	2,841	3,114	3,973	4,847	6,019	6,915
2,9	- 0,690	- 0,390	- 0,215	- 0,057	0,440	1,195	2,007	2,277	2,855	3,134	4,013	4,909	6,113	7,034
3,0	- 0,667	- 0,396	- 0,227	- 0,073	0,420	1,180	2,003	2,278	2,867	3,152	4,051	4,970	6,205	7,152

Table 1: k values for the Pearson-3, log-Pearson-3 and Gumbel distribution

Figure

Course of calculation

x1,x2,...x_i,...x_N

observation values

arithmetic mean (1)

standard deviation (2)

variation coefficient (3)

inclination coefficient (4)

The distribution functions used are unilaterally limited. The limit of the feature range is

(5)

If the calculation is to be carried out by hand with the aid of tables, it is recommended to use the proportion quantities xi = x_i / x and h_i = y_i / y in the place of the values x_i and y_i.

(6)

(7)

The following calculation steps are to be carried out:

(1) Calculate y_i by taking the logarithm of the observation values x_i.
(2) Calculate y, C_vy and C_sy from the y_i values.
(3) If C_SY is greater than or equal to zero, calculate the sought T-yearly flood discharge x_T
for the recurrence interval T:

y_T = y + S_y (k (C_sy, T)) or (8)
y_T = y [(1 + c_vy · k (c_sy, T)] (9)

The k values according to Pearson are to be taken from Table 1 (linear interpolation).

x_T= 10^yT if decadic logarithms are used

and

x_T = e^yT if natural logarithms are used.

Here the course of calculation is terminated.

Year	HQ [m3/s]	Rank No. m	HQ [m3/s]	N + 1/m	y	n = y / y	n - 1		(n - 1)2	(n - 1)3
							-	+		-	+
1951	80.60	1	516.00	35.00	2.71	1.2387		0.2400	0.0576		0.01181
1952	164.00	2	296.00	17.50	2.47	1.1284		0.1300	0.0169		0.00110
1953	136.00	3	279.00	11.67	2.45	1.1167		0.1200	0.0144		0.00102
1954	82.50	4	274.00	8.75	2.44	1.1131		0.1100	0.0121		0.00100
1955	176.00	5	261.00	7.00	2.42	1.1035		0.1000	0.0100		0.00099
1956	203.00	6	242.00	5.83	2.38	1.0885		0.0900	0.0081		0.00073
1957	179.00	7	234.00	5.00	2.37	1.0818		0.0800	0.0064		0.00051
1958	136.00	8	210.00	4.38	2.32	1.0604		0.0600	0.0036		0.00022
1959	110.00	9	203.00	3.89	2.31	1.0537		0.0500	0.0025		0.00013
1960	160.00	10	190.00	3.50	2.28	1.0405		0.0400	0.0016		0.00006
1961	242.00	11	186.00	3.18	2.27	1,0363		0.0400	0.0016		0.00006
1962	234.00	12	179.00	2.19	2.25	1.0286		0.0300	0.0009		0.00003
1963	179.00	13	179.00	2.69	2.25	1.0287		0.0300	0.0009		0.00003
1964	152.00	14	176.00	2.50	2.25	1.0253		0.0300	0.0009		0.00003
1965	190.00	15	175.00	2.33	2.24	1.0242		0.0200	0.0004		0.00001
1966	279.00	16	164.00	2.19	2.21	1.0113		0.0100	0.0000		0.00000
1967	129.00	17	161.00	2.06	2.21	1.0077		0.0100	0.0010		0.00000
1968	261.00	18	160.00	1.94	2.20	1.0064		0.0100	0.0000		0.00000
1969	122.00	19	152.00	1.84	2.18	0.9963	(0.0037)		0.0000	(0.00000)
1970	274.00	20	149.00	1.75	2.17	0.9923	(0.0077)		0.0001	(0.00000)
1971	79.30	21	145.00	1.67	2.16	0.9869	(0.0131)		0.0002	(0.00000)
1972	68.70	22	136.00	1.59	2.13	0.9742	(0.0258)		0.0007	(0.00002)
1973	66.70	23	136.00	1.52	2.13	0.9742	(0.0258)		0.0007	(0.00002)
1974	161.00	24	129.00	1.46	2.11	0.9637	(0.0363)		0.0013	(0.00005)
1975	175.00	25	123.00	1.14	2.09	0.9543	(0.0457)		0.0021	(0.00010)
1976	94.30	26	122.00	1.35	2.09	0.9527	(0.0473)		0.0022	(0.00011)
1977	123.00	27	110.00	1.30	2.04	0.9321	(0.0679)		0.0046	(0.00031)
1978	87.30	28	94.30	1.25	1.97	0.9016	(0.0984)		0.0097	(0.00095)
1979	186.00	29	87.30	1.21	1.94	0.8863	(0.1137)		0.0129	(0.00147)
1980	210.00	30	82.50	1.17	1.92	0.8751	(0.1249)		0.0156	(0.00195)
1981	296.00	31	80.60	1.13	1.91	0.8705	(0.1295)		0.0168	(0.00217)
1982	145.00	32	79.30	1.09	1.90	0.8672	(0.1328)		0.0176	(0.00234)
1983	149.00	33	68.70	1.06	1.84	0.8388	(0.1612)		0.0260	(0.00419)
1984	516.00	34	66.70	1.03	1.82	0.8329	(0.1671)		0.0279	(0.00467)
							(1.2009)	1.2000	0.2763	(0.01834)	0.01772
											(0,0006)

Table 2: Calculation of the 10, 50, 100, 200 and 1000 - yearly flood discharge for an observation series of 34 years acc. to the DVWK recommendation in tabulated form (example gauge Betzdorf / Federal Republic of Germany). y decadic logarithms of observation values, N extent of data set, HQ max. annual discharges in m³/s, period of observation: 34 years.

(4) If C_sy is smaller than zero, calculate x, C_vx, C_sx and d from the x_i values.
(5) If C_sx or if d is smaller than zero, put c_sx = + 2 C_vx and continue with step 7.
(6) If C_sx and d are greater than or equal to zero, continue with step 7.

Steps 1 and 2 cf. Table 2

Parameters:	arithmetic mean y = 2.19
	variation coefficient Cvy = 0.0915
	inclination coefficient Csy = - 0.0252

YEAR	HQ[m3/s]	Rank No. m	HQ[m3/s]	N + 1/m	x = x / x	x - 1		(x - 1)2	(x - 1)3
1951	80.60	1	516.00	35.00	3.00		2.00	4.0000		8.0000
1952	164.00	2	296.00	17.50	1.72		0.72	0.5184		0.3732
1953	136.00	3	279.00	11.67	1.62		0.62	0.3844		0.2383
1954	82.50	4	270.00	8.75	1.59		0.59	0.3481		0.2054
1955	176.00	5	261.00	7.00	1.52		0.52	0.2704		0.1406
1956	203.00	6	242.00	5.83	1.41		0.41	0.1681		0.0689
1957	179.00	7	234.00	5.00	1.36		0.36	0.1296		0.0467
1958	136.00	8	210.00	4.38	1.22		0.22	0.0484		0.0106
1959	110.00	9	203.00	3.89	1.18		0.18	0.0324		0.0058
1960	160.00	10	190.00	3.50	1.10		0.10	0.0100		0.0010
1961	242.00	11	186.00	3.18	1.08		0.08	0.0064		0.0005
1962	234.00	12	179.00	2.92	1.04		0.04	0.0016		0.0001
1963	179.00	13	179.00	2.69	1.04		0.04	0.0016		0.0001
1964	152.00	14	176.00	2.50	1.02		0.02	0.0004		0.0000
1965	190.00	15	175.00	2.33	1.02		0.02	0.0004		0.0000
1966	279.00	16	164.00	2.19	0.95	0.05		0.0025	0.0001
1967	129.00	17	161.00	2.06	0.94	0.06		0.0036	0.0002
1968	261.00	18	160.00	1.94	0.93	0.07		0.0049	0.0003
1969	122.00	19	152.00	1.84	0.88	0.12		0.0144	0.0017
1970	274.00	20	149.00	1.75	0.87	0.13		0.0169	0.0022
1971	79.30	21	145.00	1.67	0.84	0.16		0.0256	0.0041
1972	68.70	22	136.00	1.59	0.79	0.21		0.0441	0.0093
1973	66.70	23	136.00	1.52	0.79	0.21		0.0441	0.0093
1974	161.00	24	129.00	1.46	0.75	0.25		0.0625	0.0156
1975	175.00	25	123.00	1.40	0.72	0.28		0.0784	0.0220
1976	94.30	26	122.00	1.35	0.71	0.29		0.0841	0.0244
1977	123.00	27	110.00	1.30	0.64	0.36		0.1296	0.0467
1978	87.30	28	94.30	1.25	0.55	0.45		0.2025	0.0911
1979	186.00	29	87.30	1.21	0.51	0.49		0.2401	0.1176
1980	210.00	30	82.50	1.17	0.48	0.52		0.2704	0.1406
1981	296.00	31	80.60	1.13	0.47	0.53		0.2709	0.1489
1982	145.00	32	79.30	1.09	0.46	0.54		0.2916	0.1575
1983	149.00	33	68.70	1.06	0.40	0.60		0.3600	0.2160
1984	516.00	34	66.70	1.03	0.39	0.61		0.3721	0.2270
						5.93	5.92	8.4485	1.2346	9.0913
										7.8567

Table 3

As C_sy < 0, step 3 follows: cf. Table 3

(7) Calculate the sought T-yearly flood discharge x_T for the recurrence interval T:

x_T = x + s_x · k 3 (C_sx, T) or (10)

x_T = x [1 + C_vx k (C_sx,T)] (11)

The k values according to Pearson are to be taken from Table 1 (linear interpolation). Here the course of calculation is terminated.

Fig.1: Flow chart

The flow chart (Fig. 1) shows once again the course of calculation.

Parameters: arithmetic mean x = 171.95 m³/s variation coefficient Cvx = 0.5059 standard deviation sx = 87.0 inclination coefficient Csx = 0.306

With k (Cx, T) from the k value table (for the Pearson-3, the log-Pearson-3 and the Gumbel distribution), the desired HQ value determination is carried out. For the present example, the HQ values are the following:

	HQ 10	HQ 50	HQ 100	HQ 200	HQ 1000
	(m³/s)	(m³/s)	(m³/s) .	(m³/s)	(m³/s)
Pearson III	284.1	423 2	484.4	546.5	694.1

For the graphical representation, cf. Fig. 2.

Fig. 2: Flood probability

When flood discharges with short recurrence periods (T < 15 years) are determined on the basis of annual series, in the case of the determination of the T-yearly flood according to the formula x_T = . . . or y_T = . . ., a correction of the recurrence interval must be made, for which the k values are determined. This is to be attributed to the use of annual series, whereas we are interested in the recurrence intervals between two floods of a definite extent irrespective of whether maximum annual discharges are concerned. The correction is made according to the formula

T = (e^1/T*) / (e^1/T* - 1) (12)

where T* is the sought actual recurrence period (Draschoff, 1972).
Owing to the data series which statistically are normally very short, the extrapolation to long recurrence periods should not be exaggerated. But practical requirements make it often necessary to calculate discharges of a specified (generally long) recurrence period from a small amount of data. In this case, one should be aware of the great uncertainty of the results. Confidence intervals around the distribution functions offer the possibility of making plain this uncertainty as a function of the length of the observation series, the recurrence interval and an error probability.

Partial series which are taken from an observation period of M years contain N values, N being generally greater than M. The course of calculation of the recommendation can be followed. It is, however, to be borne in mind that the equations x_T = . . . or y_T = . . . and, thus, the tables of the k values are to be entered into computationally with a recurrence interval T other than the sought recurrence period T*:

T = (N / M) x T* (13)

The sought 20-yearly flood, for example, is calculated from a partial series with 52 values from 21 years of observation according to the Pearson-3 distribution:

T = (52 / 21) x 20 = 50 years with x [1 + C_vx k · (C_sc; T = 50)]

Annex 3: Calculation example for a free overfall over a wooden beam weir

Fig. 1: Free overfall over a wooden beam weir (numerical example)

Fig. 2: Weir coefficient m for various weir forms (according to Press/Schroder [31)

Over a wooden beam weir an amount of water for power generation of at least 0.65 m³/s is to be evacuated. The water level in the power canal and in the river can be seen in Fig. 1. The weir has a width of 2 m. The necessary weir head h_ü is sought. After transformation of the weir formula

the weir head h results for the free overfall with c = 1:

(m)

with m = 0.64 (sharp-crested weir, cf. Fig. 2), B = 2.0 m, g = 9.81 m/s² (acceleration due to gravity), Q = 0.65 m³/s, c = correction factor. The following results:

This means that the discharge capacity of the wooden beam weir is about 0.65 m³/s in the example of a weir head of 0.31 m. If this minimum amount of water for power generation is to be evacuated in the power canal, the crest height of the weir must be 0.31 m below the lowest water level of the river!

Annex 4: Calculation example for a submerged overfall over a wooden beam weir

Owing to a flood event, the water level of the river has risen, and simultaneously considerably greater amounts of water flow into the power canal so that its water level has risen as well. This maximum water level results from the hydraulics of the canal and can be determined for each amount of water to be evacuated. For this numerical example the corresponding values have been entered in Fig. 1. For this condition of the submerged overfall the amount of water is to be determined which is evacuated by the weir. The weir width is 2 m.
- Discharge capacity of the weir

(m³/s)

with h_ü = weir head (m), g = acceleration due to gravity = 9.81 m/s², B = weir width (m), m = weir coefficient ( - ) (cf. Annex 3), c = correction coefficient.

- Correction factor c

h' = 0.30 m,
h_ü = 0.80 m,
h' / h_ü = 0.30 / 0.80 = 0.375

From the diagram, it thus follows for c (wooden beam weir): c ~ 0.86.

- Discharge capacity of the weir

(m³/s) =

Correction factor c for submerged overfall

Fig. 1: Submerged overfall over a wooden beam weir (numerical example)

Annex 5: Calculation example for a discharge over a weir with race door or stilling basin

The design of such an arrangement requires the following steps:

(1) The headwater depth ho and the tail water depth hu are given quantities. The headwater
depth results from the water level of the river and the tail water depth is dependent upon the

hydraulic conditions downstream, i.e. upon the evacuation of the amounts of water flowing
over the weir into a river course or into a power canal with the respective water level in the
river and in the power canal.

(2) From the weir formula the value h and, thus, the height of the weir crest are determined.

(3) Subsequently, the dimensions and a hydraulically favourable form of the weir are selected.

(4) It is presupposed that the bottom of the race floor is at the level of the river bottom. Then the water level h₁ at the beginning of the race floor can be determined as follows:

with

with

where Q = discharge over the weir (m³/s), B = weir width (m)

,

l = loss coefficient, about 0.1

Fig. 1: Fixed weir (elevation of water level and energy line)

(5) The water depth h₂ corresponding to the water depth h₁ is determined with the following formula

(m)

g = acceleration due to gravity = 9.81 m/s²

If h₂ just agrees with the tail water depth h_u, the assumption of the elevation of the stilling basin bottom is confirmed. (The tail water depth h_u results from the hydraulic conditions of the river or canal and can be roughly calculated with the formulas given in the relevant literature unless the water levels for definite discharges are known.) If the h₂ calculated in step 5 is smaller than the tail water depth, the nappe is dammed and the hydraulic jump drowned. The hydraulic jump is then shorter than calculated in step 6. Countermeasure: flatter design of the downstream face of the weir (cf. Fig. 2).

Fig. 2: Dammed hydraulic jump;Free hydraulic jump

In the two cases mentioned in step 5, now the stilling basin length can be determined with
step 6.

If h² is greater than the tail water depth h_u, the hydraulic jump will be so shifted downstream
that a longer race floor becomes necessary. This will mostly be the case if F₁ > 4.0 (cf. step 6).

Countermeasure: Construction of a stilling basin and arrangement of a countersill as described
in step 7 (cf. Fig. 3).

Fig. 3: Stilling basin with countersill

Fig. 4: Diagram for determining the hydraulic jump length l₂ for horizontal rectangular channels (source [3]); Fig. 5: Design values at the end of the stilling basin

(6) Determination of the stilling basin length: The stilling basin length can be taken from the diagram in Fig. 4. It is equal to the length of the hydraulic jump unless the latter is reduced by a sill as described in step 7. For the determination of the stilling basin length, Froude's number must be determined:

with (m/s),

where Q = discharge over the weir (m³/s), B = weir width = stilling basin width (m), h₁ = water level at beginning of stilling basin (m). From the diagram the stilling basin length is obtained as a function of h₁. (7) For the arrangement of the stilling basin bottom at a lower level, the measure s (cf. Fig. 5) is estimated as follows:

s > h₂—h_u (in m)

- determination of the new h₁ (now h₁ old + s) and h₂ according to steps 4 and 5
- determination of the stilling basin length with

L ~ 5 h_u +s (in m)

Numerical example (cf. Figs. 6 and 7 below)

(1) Data given

- headwater level (maximum impounding head) in the river:

h₀ = max. 2.0 m

- tail water level in the river at Q = 10 m³/s (max. amount of flood water to be evacuated):

h_u = 0.70 m

- weir width: B = 9 m
- weir coefficient: m = 0.60 (broad-crested weir)

(2) Determination of the weir crest height w

- weir formula:

Transformation with respect to h:

^(m)

Introduction of the values:

The weir crest height is determined from the maximum elevation of the water level less the weir head:

w = h₀ - h_u = 2.0 - 0.73 = 1.27 m

(3) Graphical representation of the dimensions

Fig. 6: Design of the stilling basin without sill or race floor (numerical example) (schematic representation of the weir)

(4) Determination of the elevation of the water surface h1

(m)

h₁ = 2.0 - 0.069 / h₁²

h₁³ - 2 · h₁² + 0.069 = 0 (solve equation iteratively!)

h₁ = 0.20 m

(5) Determination of the conjugate water depth h₂

h₂ = 1.00 m > h_u = 0.70 m

Dh = s = 1 - 0.70 m = 0.30 m

(6) Determination of the stilling basin length (without sill = length of race floor)

from diagram in Fig. 36.

L₂ / h₁ = 30 ® L₂ = 30 · 0.2 = 6 m

This means that the stilling basin length without sill would have to be 6 m.

(7) Arrangement of the stilling basin bottom at a lower level (shortening of the stilling basin) (with sill = stilling basin; reduced length of the structure)

Selected:

s = 0.50 m, h_u new = h_u + s = 0.70 + 0.50 = 1.20 m

w_new = 1.27 + 0.50 = 1.77 m

Fig. 7: Design of the stilling basin with sill (numerical example) (schematic representation of the weir)

h₁ new = 0.73 + 1.77 - 10² / (h₁² 9² 2 9,81) 1.1

h₁ = 2.5 - 0.069 / h₁²

h₁³ - 2.5 · h₁² + 0.069 = 0 (solve iteratively!)

h₁ new ~ 0.17 m

h₂ new = 1.13 m < h_u new = 1.20 m

Stilling basin length

L ~ 5 · h_u + s = 5 · 0.7 + 0.5 = 4 m

Annex 6: Calculation example for a straight side weir

I. With the aid of a side weir, an amount of water for power generation Q_A = 0.4 m³/s is to
be diverted from a river. In periods of low water, the river carries an amount of water of

Q₀ = 8 m³/s.

The tail water level tu for the residual amount of water of

Q_u = Q₀ - Q_A = 8 - 0.4 = 7.6 m³/s

is t_u = 1.20 m (quantity given by the rating curve = water level/discharge relation or determination with calculation methods given by the relevant literature).

The river width B is 6 m.
We seek the length of the side weir which should have a round crest.

First the minimum weir head h_u necessary in periods of low water is selected:

h_u = 0.20 m.

The weir crest height thus is

w = t_u - h_u = 1.20 m - 0.20 m = 1.00 m.

With the known values Q₀, B₀, w, h_Eu and t_u, h₀ is iteratively determined from the following formula:

with:

h_Eu = t_u + v_u² / 2g (m) (energy head)

v_u = Q_u / (t_u · B_u) = 7.6 / (1.2 6.0) = 1.06 m/s

h_Eu = 1.20 + 1.03² / (2 9.81) = 1.25 m

When the values are introduced into the formula, and with oe = 1.1, the following results:

The iterative solution yields:

h₀ = 0.17 m. .

Now the mean weir head is obtained:

h_m = (h₀ + h_u) / 2 = (0.17+0.20) / 2 = 0.19 m

From the diagram n is read as a function of

h_m / (h_m + w) = 0.19 / (0.19 + 1.00) = 0.16

and Q_u > 0 value for n:

n ~ 1.05

Correction factor n for side weir calculation

With this value the following is obtained:

a = n a = 1.05 1.1 = 1.16.

Introduction of the value a into the above formula and seeking of h₀ by iterative solution:

Introduction of the values into the formula:

The iterative solution is

h₀ = 0.17m

(Here the course of calculation is terminated as the 2nd value does not differ from the 1st value!)

- Mean weir head.

h_m = (h₀ + h_u) / 2 = (0.17 + 0.20) / 2 = 0.18 m

- The side weir length required for the evacuation of the amount of water for power generation

Q_A = 0.4 m³/s is obtained from the weir formula:

(m³/s)

- Transformation with respect to L (side weir length)

(m)

with m^x = 0.95 · 0.7 (round-crested weir) = 0.67

Hence it follows

Control whether the discharge to the side weir is flowing:

approach velocity

Froude's number

The required discharge is thus flowing!

Weir coefficient u for various weir forms (according to Press/Schroder [3])

II. During a flood event, the discharge can rise to Q₀ = 45 m³/s (cf. Fig. 1). In this case, the
water level to is 2.50 m, the river width B₀ = B_u = about 10 m.

For the weir dimensions fixed under item I, the discharge is to be determined which is
evacuated during the flood discharge via the side weir!

Note:

All values in the tail water are now unknown, as the evacuated discharge Q_A and, as a result,
also Q_u = Q₀ - Q_A must first be determined.

Course of calculation

- Elevation of water surface above side weir crest upstream of the side weir:

t₀' = t₀ - w = 2.50 m - 1.00 m = 1.50 m.

1st step:

- Estimate of the mean weir head:

h_m = about 1.30 m

- Determination of the discharge with hm estimated:

(values cf. I)

- Hence the residual water amount follows:

Q_u = Q₀ - Q_A = 45 - 7.8 = 37.2 m³/s

- The corresponding water depth t_u is:

t_u ~ 2.15 m (for the determination, cf. also course of calculation in the relevant literature)

- Energy head h_Eu:

- Determination of h₀ by iterative solution of the equation:

The solution follows for the value

h₀ = 1.0 m

- New mean weir head:

h_m new = (1.0 +1.15) / 2 = 1.08 m

with: h_u = t_u - w = 2.15 - 1 = 1.15 m

- New discharge Q_A to be evacuated over the side weir:

Q_A new = 6 m³

Fig. 1: Overfall over a straight side weir - flood event (numerical example)

2nd step:
(check whether the new estimated mean weir head must be improved)

- Residual amount of water:

Q_u = Q₀ - Q_A new = 45 - 6 = 39 m³/s

- Tail water depth:

t_u ~ 2.20 m (for the course of calculation, cf. relevant literature)

- Weir head h_u:

h_u = t_u - w = 2.20 - 1.00 = 1.20 m

- Energy head h_Eu:

- Determination of h₀:

The iterative solution yields:

h₀ = 1.12 m

- Mean weir head:

- New discharge Q_A to be evacuated:

Q_Anew = 6.6 m³/s

Figure

After this 2nd step of improvement, the procedure can be interrupted, as the values for Q_A are no
longer improved considerably. In this example, the exact value for Q_A lies between 6 and 6.6 m³/s!

- Check whether the inflow into the headwater is flowing:

The required flowing discharge is given!

Annex 7: Calculation example for the outflow below a dam wall

Case 1: Free outflow

At times of low water the headwater level before the dam wall should be at least h = 1.25 m. Which size must the sluice opening be in order that for a small hydroelectric power plant an amount of water for power generation Q_A = 0.40 m3/s might be evacuated in a canal with the width B= 1.5 m?

Further characteristic values of the canal:

gradient: I = 1‰
roughness coefficient: k_S = 40

This allows h_u to be calculated:

Q= A · v = B · h_u · v in m³/s

with v = k_S · R^2/3 · I^1/2 in m³/s

with R = (B h_u)/(B +2h_u) for the rectangular cross-section

By iterative introduction of different h_u values (estimates), the following results:

1st estimate

with hu = 0.45:	0.40 = 1.9 · 0.45 · (0.43)
	0.40 ¹ 0.37

2nd estimate

with hu = 0.5:	0.40 = 1.9 · 0.5 · 0.45
	0.40 ¹ 0.43

3rd estimate

with hu = 0.47:	0.40 = 1.9 · 0.47 · 0.437
	0.40¹ 0.39

h_u = 0.47m

By transformation of the equation for the outflow below sluices with respect to a, the following equations are obtained:

in m with k = 1 for free outflow

The check whether the assumption is correct that the outflow is free is carried out as follows:

h/a = 1.25/0.09 = 13.9

According to Fig. 38d), for d = 0.7 a limiting value h_u / a= 5.8 results.

h_u /a existing amounts to 0.47 / 0.09 = 5.22 < 5.8 limiting value, i.e. the assumption that the
outflow is free was correct.

Case 2: Submerged outflow

During the times of low water, the headwater level behind a dam wall is h = 1.25 m. It is to be determined which sluice opening will be necessary if in a canal with the parameters

k_S = 40
I = 0.5‰

and a sluice width B = 1.0 m which corresponds to the canal width, Q_A = 0.4 m³/s is to be evacuated. First h_u is calculated:

Iterative solution of h_u

1st estimate:

with hu = 0.9:	0.4 = 1.0 · 0.00051/2 0.9 (1 x 0.9/2.8)2/3
	0.4 = 0.8 · 0.47
	0.4 ¹ 0.38

2nd estimate:

with hu = 0.93:	0.4 = 1.0 x 40 x 0.00051/2 x 0.93 x (0.93 / 2.86)2/3
	0.4 = 0.832 · 0.473
	0.4 = 0.39

i.e. the 2nd estimate with hu = 0.93 was correct.

Assumption: submerged outflow opening a in first approximation with k = 0.8

Check: h / a = 1.25 / 0.17 = 7.35

with d = 0.7 according to Fig. 38d), a limiting value h_u / a= 4.3 results.

exist. h_u / a is: 0.93 / 0.17 = 5.47 > 4.3

i.e. the assumption that the outflow is submerged was correct.

According to Fig. 38e), for h_u / a = 5.47 and h_u / a= 7.35, a k value of 0.5 results.

New calculation of a:

Hence it follows:
h / a = 1.25 / 0.27 = 4.63 ® limiting value a = 3 for d = 0.7
h_u / a = 0.93 / 0.27 = 3.44 > 3 submerged outflow

With Fig. 38e), for h_u / a and h / a = 0.46, the following k value is obtained:
k ~ 0.5

The k value is in agreement. Hence it follows that in the case of submerged outflow, Q_A = 0.4 m³/s can be evacuated below the sluice with an opening width of 0.27 m in the tail water canal.

Annex 8: Calculation example for a bottom intake (Tyrolean weir)

At right angles with a river (stream) a Tyrolean weir is to be built in order to divert an amount of water for power generation Q_A = 0.85 m³/s. In this point the river width is about 8 m; at times of low water, the minimum headwater depth (= initial water height) is h₀ = 0.5 m. The Tyrolean weir with the collection canal is to be dimensioned in such a way that the diversion of the amount of water for power generation Q_A = 0.85 m³/s is always ensured at times of low water.

Selected quantities (numerical example)

- contraction coefficient for trash rack	m ~ 0.85 (round bars)
- internal width between bars	a = 2 cm
- centre distance between bars	d = 4 cm
- inclination of trash rack	b = 8°

Hence the following values result for

- h = 2/3 k h₀ = 2/3 0.927 0.5 = 0.31 m
- c = 0.6 a/b cos^3/2 b = 0.6 2/4 cos^3/2 8°

- c = 0.3

With these values the discharge through the trash rack first results as a function of the width b and the length L of the trash rack:

With Q_A = 0.85 m³/s, the following is obtained:

0.85 = 0.419 b L

0.85 / 0.419 = b L

b L = 2.03 or L= 203 / b (m)

width of trash rack b (m)	2	4	6
length of bash reck L(m)	1.00	0.51	0.34

- Selected width of trash rack: b = 4 m
- To this width the length L of the trash rack corresponds:

L = 2.03/b = 2.03/4 = 0.51 m

- The selection of the width and the corresponding length- of the trash rack should be governed by the following criteria:

adaptation of the Tyrolean weir to the local conditions,
selection of a sufficiently great trash rack length which determines the width of the collection canal arranged below. If the length of the trash rack is selected too small this collection canal for the evacuation of the water for power generation must be constructed deeper and, as a result, less economically, as for constructional reasons, it should have approximately the same width as the projection of the length of the trash rack onto the horizontal ground area.

Figure

During-operation, parts of the trash rack can be obstructed by wedged stones, leaves or branches so that the evacuation of the minimum amount of water can no longer be ensured. This is why the determined trash rack length L should be increased by 20%:

L = 1.2 · 0.51 = 0.61 m.

Dimensioning of the collection canal

Along the trash rack width b (= weir width), the amount of water for power generation falling through the trash rack increases linearly and reaches its maximum in the end cross section of the collection canal. For reasons of simplicity, this end cross-section is used for the dimensioning of the canal:

- Selected quantities (numerical example):

canal width	B = 0.65 m,
roughness	kS = 50 (concrete),
slope	I = 30‰

(The slope should at least be 30‰ in order to remove again from the collection canal the solid matter entrained, by a high tractive force of the water. For this purpose, a higher water velocity is necessary which chiefly depends upon the slope of the collection canal.)

- Sought: water depth t

Discharge formula for channels (rectangular)

(m³/s)

Introduction of the values:

0.85 = 5.6 t (0.65 t/(0.65 + 2 · t))^2/3

Iterative solution of the equation by inserting different values for t:

solution:	t = 0.46 m
freeboard:	0.25 · t = 0.12 m
total canal depth:	0.46 m + 0.12 m = 0.58 m

Annex 9: Numerical example of the proof of safety from hydraulic shear failure

(1) Representation of the weir with headwater and tail water level for the design case in question (lowest discharge, as in this case the hydraulic gradient Dh is greatest; cf. Fig. 1).

Fig. 1: Potential and stream line net for determining the safety from hydraulic shear failure. I, II flow channels, 1-27 potential lines

(2) Construction of the potential and streamline net:

- start with potential line No. 11 in the middle of the weir toe running vertically towards the bottom
- determination of the number of flow channels and entry of the approximate run of the streamlines
- entry of the potential lines beginning with the bottom flow channel in the direction of the weir foundation.
Attention: On the bottom flow channel the potential lines must be equidistant.
- Numerical example (cf. Fig. 1):
number of flow channels 2
number of potential steps n_S = 27
- Critical hydraulic gradient

I_crit. = (1 —n) (g_f / g_w - 1)
n ~ 0.35 (sand/gravel)
g_F~ 20 kN/m³ (sand/gravel)
g_w~ 10 kN/m³ (water)

I_crit. = (1 - 0.35) (20 / 10 - 1) = 0.65 · 1.0 = 0.65

- Existing gradient in the critical point (= point f)

I_exist. = Dh/(nS x DnS)

Dh = h₀ - h_u = 1.50 - 0.40 = 1.10 m

n_S = 27 potential lines

DnS ~ 0.50 m / 3 = 0.17 m (= mean distance between potential lines 26 and 27 = mean distance between all potential lines)

I_exist. = 1.10 / (27 0.17) = 0.24

- Safety from hydraulic shear failure

v = I_crit./Iexist. = 0.65 / 0.24 = 2.7 ³ 2.0

This means that in the critical zone (outlet into the tail water) no hydraulic shear failure will occur.

Annex 10: Numerical example of the proof of stability against sliding of a fixed weir

Fig. 1: Numerical example for determining the stability against sliding of a fixed weir

Figure

For the weir subject to underflow represented in Fig. 1, the stability against sliding is to be determined. The given quantities can be seen in the figure. All the forces are related to a weir width of 1 m.

Vertical forces:

- Weight of the structure (sketch 3) (g concrete ~ 25 kN/m³):

G = 1/2 · 25 · 2.00 · 1.00 + 25 · 3.20 · 0.5 + 25 · 0.7· 1.0 = 25 + 40 + 17.5 = 82.5 kN
(1) (2) (3)

- Water surcharge (estimate) (sketch 4):

W_V ~ 1.20 · 0.30 · 10 · 2.00 · 0.30 · 10 = 9.6kN

- Force due to seepage water pressure (sketch 2):

S= 4.95 · 07 + 1 / 2 · (6.75 - 4.95) · 0.7+ 1 / 2 (3.95 + 1.35) · 2.50 = 3.47 + 0.63 + 6.64 =10.74kN
(1) (2) (3)

- Force due to tail water pressure (sketch 2):

U = 19.0 · 0.7 + 9.0 · 2.50 = 13.3 + 22.5 = 35.8 kN
(1)^x (2)^x

- Sum of vertical forces:

= G +W_v - S - U

= 82.5 + 9.6 - 10.74 - 35.8 = 45.56 kN

Horizontal forces (sketch 1):

- Headwater:

W_H1e = 1/2 · 13.0 · 1.30 + 1/2 (25.75 + 13.0) · 1.50
(1)^xx (2)^xx
= 8.45 + 29.10 = 37.55 kN

- Tail water:

W_Hr = 1/2 · 4.0 · 0.4 + 1/2 · (4 + 10.35) · 0.50 + 1/2 (10.35 + 23.95) · 1.0
(1)^xxx (2)^xxx (3)^xxx

= 0.80 + 3.60 + 17.15 = 21.55 kN

- Bed load pressure:

P_G ~ 1/2 · (19 - 10) · 1² = 45kN

- Sum of horizontal forces:

SH = W_H1e + P_G - W_Hr
= 37.55 + 4.5 - 21.55 = 20.5 kN

Stability against sliding

; with tan j = 0.6 (rubble/gravel/sand)

= (45.56 x 0.60)/20.5 =1.3<1.50

With these given dimensions of the fixed weir, the necessary stability against sliding would not be ensured. To increase the stability against sliding the measures described in section 4.2 must be taken (extension of the stilling basin, deepening of the toe wall below the weir crest, etc.).

Annex 11: Calculation example for the design of a sand trap

(1) Determination of the grain-size limit

According to the operational requirements, the particle diameter of the suspended matter is
determined which is just to be caused to settle.
In practice, a grain-size limit of 0.2 mm has proved to be suitable (low-head and high-head
power plants). In the case of a head > 100 m and pure quartz sand, d_limit. = 0.05 mm is
selected (exceptional case).

(2) Determination of the flow velocity

The flow velocity vd in the sand trap must not exceed an upper limiting value if the limited grain size is to be deposited. v_d = a · d^1/2 in cm/s

d = particle diameter	a = coefficient as a function of d
	a = 36 at d > 1 mm
	a = 44 at 0.1 mm < d < 1 mm
	a = 51 at d < 0.1 mm

for d = 0.2 mm:
® v_d = 44 = 19.7 cm/s

In practice, for a limited grain size of 0.2 mm, a flow velocity of

v_d = 0.2 m/s

has proved to be suitable.

(3) Determination of sand trap dimensions

- Length of sand trap:

in m

with	L = effective settling length in m
	h = settling depth in m
	vd = flow velocity in m/s
	vS = sinking velocity of the limited grain size in m/s according to Fig. 1,
	in dependence upon gS / gW
	gS = specific weight of the particles
	gW = specific weight of water e.g. for sand: s = 2.7

Fig. 1: Sinking velocity of spherical particles in stilled water at 10 °C. (At other water temperatures, the values in the range of Stokes' law are to be multiplied by v/(1.31 · 10^-2).) Source [12]

Note: If the denominator is negative, other conditions will have to be selected. Settling is then not possible.

- Width of sand trap:

in m

t_d = L / v_d in s

with	Q = discharge in m³/s
	td = time of passage in s

In order to achieve a uniform approach of the water over the whole chamber width, the transition section is to be designed according to Fig. 2.

The selected settling depth is h = 1 m with limited grain size = 0.2 mm

v_d = 0.2 m/s
s = 2.7 ® v_s = 2.8 cm/s
a = 12º
®

Fig. 2

Determination of B:

B = (Q t_d )/ (L · h) with t_d = L / v_d = 10 / 0.2 = 50 s

® B = (1 · 50s )/ (10 · 1) = 5 m

Check for the sand trap dimensions:

i.e. l is too long.

New choice of the sand trap width:

h = 1.5 m

® L = 0.2 · 1.5 = 15 m

B = 1 · 74s / (15 · 1.5) = 3.33 m t_d = 15 / 0.2 = 75s

Check:

with a= 14°

As a result, the sand trap dimensions are selected as follows:

L = 15 m
h = 1.5 m
B = 3.4 m
1 = 4.7 m
a = 14º