Un canal trapecial con ancho de plantilla de 4 m, coeficiente n = 0.

015 y taludes 1:1, conduc

las condiciones mostradas en la siguiente figura. Se desea conocer el perfil en
considerando un tirante al inicio del primer tramo de 2 m.

A  by  ky2 P  b  2 y 1  k 2 V2
Datos E  y
b (m) = 4 2g
n = 0.015
   by  ky   0
3
 Qn  2
2
2 5
k =1  1/ 2  b  2 y 1  k
s 
Q (m3/s) = 20
SoI = 0.00007 Q 2 A3 Permite hallar el tirante crítico

g B
SoII = 0.05
B  b  2ky

1) Hallamos el tirante normal:

ynI (m) = 3.306

46101.6749 A (m2) = 24 P (m) =
ynII (m) = 0.521
2.41495342 A (m2) = 2.35435523 P (m) =
yc (m) = 1.226 40.77 A (m2) = 6.40765779

Tramo I
A (m )
2
P (m) Rh2/3 V (m/s) V2/2g
y (m) 6.407 7.468 0.903 3.122 0.497
1.226 6.563 7.536 0.912 3.048 0.473
1.25 6.890 7.677 0.930 2.903 0.429
1.3 7.223 7.818 0.949 2.769 0.391
1.35 7.560 7.960 0.966 2.646 0.357
1.4 7.903 8.101 0.984 2.531 0.326
1.45 8.250 8.243 1.001 2.424 0.300
1.5 8.603 8.384 1.017 2.325 0.275
1.55 8.960 8.525 1.034 2.232 0.254
1.6 9.690 8.808 1.066 2.064 0.217
1.7 10.440 9.091 1.097 1.916 0.187
1.8 11.210 9.374 1.127 1.784 0.162
1.9 12.000 9.657 1.156 1.667 0.142
2
Tramo II 0.521
y (m) A (m2) P (m) Rh2/3 V (m/s) V2/2g
1.126 5.772 7.185 0.864 3.465 0.612
1.100 5.610 7.111 0.854 3.565 0.648
1.000 5.000 6.828 0.812 4.000 0.815
0.900 4.410 6.546 0.769 4.535 1.048
0.800 3.840 6.263 0.722 5.208 1.383
0.700 3.290 5.980 0.671 6.079 1.884
0.600 2.760 5.697 0.617 7.246 2.676
0.530 2.401 5.499 0.576 8.330 3.537

y(m) 12.00
10.00 2.00 0.00 8
10.00
9.90 1.90 177.88 7.988
9.80 1.80 313.25 7.978 8.00
9.70 1.70 413.81 7.971
9.60 1.60 485.88 7.966 6.00
9.55 1.55 513.19 7.964
4.00
9.50 1.50 535.22 7.963
9.45 1.45 552.50 7.961 2.00
9.40 1.40 565.51 7.960
9.35 1.35 574.68 7.960 0.00
0.00 100.00 200.00 30
9.30 1.30 580.41 7.959
-2.00
9.25 1.25 583.07 7.959
9.23 1.23 583.37 7.959 -4.00
9.05 1.10 583.58 7.949
8.86 1.00 585.29 7.863
8.52 0.90 590.06 7.625
7.89 0.80 600.65 7.095
6.69 0.70 622.73 5.991
4.12 0.60 672.24 3.516
-2.05 0.53 794.24 -2.584
= 0.015 y taludes 1:1, conduce un gasto de 20 m 3/s, bajo
esea conocer el perfil en los dos primeros tramos,

V2 2
1
E  y  S   Vn  V  Rh2 / 3 s1 / 2
2g f  R2/3  n
 h 

 ky 2 
5
0

r el tirante crítico

Para calcular los perfiles

13.350 0.00

5.473 0.00
B (m) = 6.45218034 0.00

E Sf (Sf1+Sf2)/2 Dx (m)
1.723 0.00269 0 583.37
1.723 0.00251 0.00260 -0.30 0.30 583.07
1.729 0.00219 0.00235 -2.66 2.96 580.41
1.741 0.00192 0.00205 -5.73 8.69 574.68
1.757 0.00169 0.00180 -9.17 17.86 565.51
1.776 0.00149 0.00159 -13.01 30.87 552.50
1.800 0.00132 0.00141 -17.28 48.15 535.22
1.825 0.00118 0.00125 -22.03 70.18 513.19
1.854 0.00105 0.00111 -27.30 97.49 485.88
1.917 0.00084 0.00095 -72.07 169.56 413.81
1.987 0.00069 0.00077 -100.56 270.12 313.25
2.062 0.00056 0.00063 -135.36 405.48 177.88
2.142 0.00047 0.00052 -177.88 583.37 0.00

E Sf (Sf1+Sf2)/2 Dx (m)
1.738 0.00362 0 583.37
1.748 0.00392 0.00377 0.21 0.21 583.58
 1.815 0.00545 0.00469 1.49 1.71 585.29
1.948 0.00784 0.00664 3.06 4.77 590.06
2.183 0.01172 0.00978 5.83 10.59 600.65
2.584 0.01844 0.01508 11.48 22.08 622.73
3.276 0.03105 0.02475 27.44 49.51 672.24
4.067 0.04714 0.03910 72.49 122.00 794.24
COMPORTAMIENTO DEL PERFIL

100.00 200.00 300.00 400.00 500.00 600.00 700.00 800.00

TIRANTE CRÍTICO AGUAS ARRIBA

TIRANTE CRÍTICO AGUAS ABAJO
ara calcular los perfiles