CALCULOS PARA PROBLEMA 10.

1 CLAUDIO MATAIX

A de cuadrado
A1=b(h)
60 m^2
63 m^2
= 21.7808517 66 m^2
b= 15 m 69 m^2
n= 0.025 72 m^2
s 0.0001
Q 100 (m^3)/s Perimetro mojado
h = variable Dando valoreS de 4 a 4.8 a h. Pm = b+2I
h= 4 m 36.56 m
h= 4.2 m 37.638 m
h= 4.4 m 38.716 m
h= 4.6 m 39.794 m
h= 4.8 m 40.872 m
h = variable Dando valoreS de 4 a 4.8 a h.
C =(23+1/n+0.00155/s)/(1+(23+0.00155/𝑆) 𝑛/√(𝑅_ℎ ))
𝑘=𝑄/√𝑠 =100/
√0.0001

interpolar valores de la rugosidad obteindos e

y
50.3267814
C
50.6514972

Con Rh obtenido interpolar pata ontener Pm

y
38.716
Pm
39.794
 CALCULO DEL AREA DEL CANAL
I=hipotenusa triangulo B=Base del triangulo A de 2 triangulos AC=Area del caudal
I=h/sen( ) B=I cos( ) At=B*h AC=At+A1
10.78 m 10.0104146 m 40.041658292 m^2 100.04165829 m^2
11.319 m 10.5109353 m 44.145928267 m^2 107.14592827 m^2
11.858 m 11.011456 m 48.450406534 m^2 114.45040653 m^2
12.397 m 11.5119768 m 52.955093091 m^2 121.95509309 m^2
12.936 m 12.0124975 m 57.659987941 m^2 129.65998794 m^2

Radio hidraulico Coeficiente C Rugosidad K

Rh=AC/Pm 𝑘=𝐶∗𝐴√(𝑅_ℎ )
2.73636921 m 49.6253397 (m^1/2)/s 8212.4307839 µm
2.84674872 m 49.9853139 (m^1/2)/s 9036.3420646 µm
2.95615266 m 50.3267814 (m^1/2)/s 9903.2997703 µm
3.06466033 m 50.6514972 (m^1/2)/s 10813.926221 µm
3.17234263 m 50.9609807 (m^1/2)/s 11768.843755 µm

=100/ ¨= 10,000 m^3/s

√0.0001

valores de la rugosidad obteindos en la tabla Con C obtenido interpolar pata ontener Rh

x y x
9903.29977 2.9561526644 50.326781388
10000 C= 50.3612632 Rh 50.361263246 Rh= 2.96767519
10813.9262 3.0646603286 50.651497176

tenido interpolar pata ontener Pm Con Pm obtenido interpolar pata ontener h

x y x
2.95615266 4.4 38.7160
2.96767519 Pm= 38.8304738 h 38.8305 h= 4.42
3.06466033 4.6 39.7940
𝑄=𝑘√𝑠

m
CALCULOS PARA PROBLEMA 10.7 CLAUDIO MATAIX

CALCULO DEL AREA DEL CANA

I=hipotenusa triangul
I=h/sen( )
0.70710678 m
0.77781746 m
= 45 0.84852814 m
1/n= 76.9 0.91923882 m
s 0.09445 0.98994949 m
v 1 m/s
Perimetro mojado
h = variable Dando valoreS de 4 a 4.8 a h. Pm = 2I
h= 0.5 m 1.41421356 m
h= 0.55 m 1.55563492 m
h= 0.6 m 1.69705627 m
h= 0.65 m 1.83847763 m
h= 0.7 m 1.97989899 m
h = variable Dando valoreS de 4 a 4.8 a h.
CALCULO DEL AREA DEL CANAL triangular
B=Base del triangulo AC=Area del caudal
b=I cos( ) AC=2b*h A partir de la v ontenidas interpolamos paraobtener h
0.5 m 0.5 m^2
0.55 m 0.605 m^2 0.55 0.78120417
0.6 m 0.72 m^2 h 1 h= 0.5704
0.65 m 0.845 m^2 0.6 1.31672753
0.7 m 0.98 m^2

Radio hidraulico Velocidad=1m/s

Rh=ab/2(a+b+c) 𝒗=𝟏/𝒏(𝑹_𝒉^(𝟐∕𝟑) (𝒔^(𝟏⁄𝟐) ))
0.60355339 m 0.44096939 (m^1/2)/s
0.80332956 m 0.78120417 (m^1/2)/s
1.04294026 m 1.31672753 (m^1/2)/s
1.3260068 m 2.128475 (m^1/2)/s
1.6561505 m 3.32029487 (m^1/2)/s
olamos paraobtener h