LISTA DE EXERCÍCIOS LIVRO FUNDAMENTOS DE ENGENHARIA HIDRÁULICA
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LISTA DE EXERCÍCIOS LIVRO FUNDAMENTOS DE ENGENHARIA HIDRÁULICA


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Engenharia Civil Hidráulica Geral 
 
 
 
LISTA DE EXERCÍCIOS 
 
CAPITULO 02 
2.3. Um bocal convergente de 100 mm x 50 mm é colocado num sistema para 
assegurar uma velocidade de 5,0 m/s na extremidade menor do bocal. Calcular a 
velocidade a montante do bocal e a vazão escoada. 
\ud835\udc491 = 5,0\ud835\udc5a/\ud835\udc60 
\ud835\udc341 =
\ud835\udf0b \u2217 \ud835\udc512
4
 
\ud835\udc341 =
\ud835\udf0b \u2217 (0,05\ud835\udc5a)2
4
 
\ud835\udc341 = 1,9635 \u2217 10
\u22125\ud835\udc5a2 
 
\ud835\udc44 = \ud835\udc341 \u2217 \ud835\udc491 
\ud835\udc44 = 1,9635 \u2217 10\u22125\ud835\udc5a2 \u2217
5,0\ud835\udc5a
\ud835\udc60
 
\ud835\udc44 = 9,81 \u2217
10\u22125\ud835\udc5a3
\ud835\udc60
 
\ud835\udc44 = 9,81\ud835\udc59/\ud835\udc60 
 
Para encontra velocidade montante: v2 
Calcula-se primeiro área: A2 
\ud835\udc342 =
\ud835\udf0b \u2217 \ud835\udc512
4
 
\ud835\udc342 =
\ud835\udf0b \u2217 (0,1\ud835\udc5a)2
4
= 
\ud835\udc342 = 7,854 \u2217 10
\u22123\ud835\udc5a2 
\ud835\udc44 = \ud835\udc342 \u2217 \ud835\udc492 
\ud835\udc492 =
\ud835\udc44
\ud835\udc342
 
\ud835\udc492 =
9,81 \u2217
10\u22125\ud835\udc5a3
\ud835\udc60
7,854 \u2217 10\u22123\ud835\udc5a2
 
\ud835\udc492 = 1,25\ud835\udc5a/\ud835\udc60 
 
 
 
 
2.5. Um canal retangular com 5,0m de largura transporta uma vazão de 10m3/s ao 
longo de 1km de extensão. O canal tem início na conta 903,0 onde a lâmina d\u2019água 
e de 1,0m. Supondo que na seção final do canal a cota seja 890,0m e a velocidade 
média 3m/s, pede-se calcular a perda de carga total entre o início e o término do 
canal. 
Dados: 
Base=b=5m 
Z1=903m+h1 
h1=1m 
 
Z2=809m+h2 
V2=3m/s 
 
Engenharia Civil Hidráulica Geral 
 
 
 
Calculando área: A1 
\ud835\udc341 = \ud835\udc4f \u2217 \u210e1 
\ud835\udc341 = 5\ud835\udc5a \u2217 1\ud835\udc5a 
\ud835\udc341 = 5\ud835\udc5a
2
 
Calculando Velocidade: V1 
\ud835\udc44 = \ud835\udc341 \u2217 \ud835\udc491 
\ud835\udc491 =
\ud835\udc44
\ud835\udc341
 
\ud835\udc491 =
10\ud835\udc5a3/\ud835\udc60
5\ud835\udc5a2
 
\ud835\udc491 = 2\ud835\udc5a/\ud835\udc60 
Para encontra altura: h2 
\ud835\udc44 = \ud835\udc342 \u2217 \ud835\udc492 
\ud835\udc342 =
\ud835\udc44
\ud835\udc492
 
\ud835\udc4f \u2217 \u210e2 =
10 \ud835\udc5a3 \ud835\udc60\u2044
3 \ud835\udc5a \ud835\udc60\u2044
 
\u210e2 =
10 \ud835\udc5a3 \ud835\udc60\u2044
\ud835\udc4f \u2217 3 \ud835\udc5a \ud835\udc60\u2044
 
\u210e2 =
10 \ud835\udc5a3 \ud835\udc60\u2044
3 \ud835\udc5a \ud835\udc60\u2044 \u2217 5\ud835\udc5a
 
\u210e2 = 0,67\ud835\udc5a 
Usando a equação: 
\ud835\udc671 +
\ud835\udc431
\ud835\udefe
+
\ud835\udc491
2
2\ud835\udc54
= \ud835\udc672 +
\ud835\udc432
\ud835\udefe
+
\ud835\udc492
2
2\ud835\udc54
+ \u2206\ud835\udc3b 
(903m + 1m) + 0 +
(2 \ud835\udc5a \ud835\udc60\u2044 )2
2 \u2217 9,81
= (890\ud835\udc5a + 0,67\ud835\udc5a) + 0 +
(3 \ud835\udc5a \ud835\udc60\u2044 )2
2 \u2217 9,81
+ \u2206\ud835\udc3b 
904,203\ud835\udc5a = 891,129 + \u2206\ud835\udc3b 
\u2206\ud835\udc3b = 13,08\ud835\udc5a 
 
2.7. Uma tubulação de 500mm de diâmetro, assentada com uma inclinação de 1% 
ao longo de 1km do seu comprimento, transporta 250l/s. Sabendo-se que a pressão 
ao longo da tubulação é constante, determinar a perda de carga neste trecho. 
Dados: 
\ud835\udc37 = 500\ud835\udc5a\ud835\udc5a = 0,5\ud835\udc5a 
\ud835\udc43 = \ud835\udc50\ud835\udc61\ud835\udc52 
 
\ud835\udc3f = 1000\ud835\udc5a 
\ud835\udc44 = 250\ud835\udc59/\ud835\udc60 
 
Inclinação=1%, então h: 
\ud835\udc56% =
\ud835\udc3f
\u210e
 
\u210e = \ud835\udc56% \u2217 \ud835\udc3f 
\u210e = 1000\ud835\udc5a \u2217 0,1 
\u210e = 10\ud835\udc5a 
Usando a equação de Bernoulli 
\ud835\udc67\ud835\udc34 +
\ud835\udc43\ud835\udc34
\ud835\udefe
+
\ud835\udc49\ud835\udc34
2
2\ud835\udc54
= \ud835\udc67\ud835\udc35 +
\ud835\udc43\ud835\udc35
\ud835\udefe
+
\ud835\udc49\ud835\udc35
2
2\ud835\udc54
+ \u2206\ud835\udc3b\ud835\udc34\u2212\ud835\udc35 
0 + 0 + 0 = 10\ud835\udc5a + 0 + 0 + \u2206\ud835\udc3b\ud835\udc34\u2212\ud835\udc35 
 \u2206\ud835\udc3b\ud835\udc34\u2212\ud835\udc35 = 10\ud835\udc5a 
Engenharia Civil Hidráulica Geral 
 
 
 
 
2.8. Um tanque contém 0,50m de água e 1,20m de óleo cujo densidade relativa é 
de 0,80. Calcular a pressão no fundo do tanque e num ponto do líquido situado na 
interface entre dois líquidos. Expressar os resultados nos sistemas técnico e 
internacional. 
 
Dados: 
\u210eá\ud835\udc54\ud835\udc62\ud835\udc4e = 0,5\ud835\udc5a 
\ud835\udc37á\ud835\udc54\ud835\udc62\ud835\udc4e = 1\ud835\udc58\ud835\udc54/\ud835\udc5a
3 
 
\u210eó\ud835\udc59\ud835\udc52\ud835\udc5c = 1,20\ud835\udc5a 
\ud835\udc37ó\ud835\udc59\ud835\udc52\ud835\udc5c = 0,8\ud835\udc58\ud835\udc54/\ud835\udc5a
3 
 
Pressão relativa no ponto P: interface entre os líquidos 
\ud835\udc43\ud835\udc43 = \ud835\udc37ó\ud835\udc59\ud835\udc52\ud835\udc5c \u2217 \ud835\udc54 \u2217 \u210eó\ud835\udc59\ud835\udc52\ud835\udc5c 
\ud835\udc43\ud835\udc43 = (0,8\ud835\udc58\ud835\udc54 \ud835\udc5a
3)\u2044 \u2217 9,81\ud835\udc5a/\ud835\udc602 \u2217 1,20\ud835\udc5a 
\ud835\udc43\ud835\udc43 = 9,9418\ud835\udc3e\ud835\udc43\ud835\udc34 
 
Pressão relativa no ponto F: fundo do tanque 
\ud835\udc43\ud835\udc39 = \ud835\udc43\ud835\udc43 + \ud835\udc37á\ud835\udc54\ud835\udc62\ud835\udc4e \u2217 \ud835\udc54 \u2217 \u210eá\ud835\udc54\ud835\udc62\ud835\udc4e 
\ud835\udc43\ud835\udc39 = 9,9418\ud835\udc3e\ud835\udc43\ud835\udc34 + (1\ud835\udc58\ud835\udc54 \ud835\udc5a
3)\u2044 \u2217 9,81\ud835\udc5a/\ud835\udc602 \u2217 0,5\ud835\udc5a 
\ud835\udc43\ud835\udc39 = 14,308\ud835\udc3e\ud835\udc43\ud835\udc34 
 
 
CAPITULO 03 
3.1. Uma tubulação de 400mm de diâmetro e 2000m de comprimento parte de um 
reservatório de água cujo N.A. está na cota 90. A velocidade média no tubo é de 
1,0m/s; a carga de pressão e a cota no final da tubulação são 30m e 50m, 
respectivamente. 
a) Calcular a perda de carga provocada pelo escoamento nessa tubulação; 
 
Dados: 
D=400m 
L=2000m 
 
\ud835\udc671 +
\ud835\udc431
\ud835\udefe
+
\ud835\udc491
2
2\ud835\udc54
= \ud835\udc672 +
\ud835\udc432
\ud835\udefe
+
\ud835\udc492
2
2\ud835\udc54
+ \u2206\ud835\udc3b 
90 +
(1\ud835\udc5a/\ud835\udc60)2
2 \u2217 9,81
+ 0 = 50 + 30 +
(1\ud835\udc5a/\ud835\udc60)2
2 \u2217 9,81
+ \u2206\ud835\udc3b 
\u2206\ud835\udc3b = 90 \u2212 50 \u2212 30 
\u2206\ud835\udc3b = 10\ud835\udc5a 
 
 
b) Determinar a altura da linha piezométrica a 800m da extremidade da tubulação. 
 
Engenharia Civil Hidráulica Geral 
 
 
 
 
\ud835\udc671 = (\ud835\udc67\ud835\udc43 +
\ud835\udc43\ud835\udc5d
\ud835\udefe
) +
\ud835\udc49\ud835\udc5d
2
2\ud835\udc54
+ \u2206\ud835\udc3b1\u2212\ud835\udc43 
90\ud835\udc5a = \ud835\udc43\ud835\udc3c\ud835\udc38\ud835\udc4d +
(1\ud835\udc5a/\ud835\udc60)2
2 \u2217 9,81
+ 6\ud835\udc5a 
\ud835\udc43\ud835\udc56\ud835\udc52\ud835\udc67
\ud835\udc5d
= 90\ud835\udc5a \u2212 0,05\ud835\udc5a \u2212 6\ud835\udc5a 
\ud835\udc43\ud835\udc43\ud835\udc3c\ud835\udc38\ud835\udc4d = 83,95\ud835\udc5a 
 
\u2206\ud835\udc3b = \ud835\udc3d \u2217 \ud835\udc3f 
\ud835\udc3d =
\u2206\ud835\udc3b
\ud835\udc3f
 
\ud835\udc3d =
10\ud835\udc5a
2000\ud835\udc5a
 
\ud835\udc3d = 0,005 
 
\u2206\ud835\udc3b1\u2212\ud835\udc43 = \ud835\udc3d \u2217 \ud835\udc3f1\u2212\ud835\udc43 
\u2206\ud835\udc3b1\u2212\ud835\udc43 = 0,005 \u2217 1200\ud835\udc5a 
\u2206\ud835\udc3b1\u2212\ud835\udc43 = 6\ud835\udc5a 
 
3.3. Uma tubulação horizontal com 200mm de diâmetro, 100m de extensão, está 
ligada de um lado ao reservatório R com 15,0m de lâmina d\u2019água, e do outro a um 
bocal de 50mm de diâmetro na extremidade, conforme mostrado na figura a seguir. 
Este bocal foi testado em laboratório e apresentou um coeficiente de perda de carga 
de 0,10, quando referenciado à seção de maior velocidade. Calcular as velocidades 
na tubulação e na saída do bocal. 
 
\ud835\udc671 +
\ud835\udc431
\ud835\udefe
+
\ud835\udc491
2
2\ud835\udc54
= \ud835\udc672 +
\ud835\udc432
\ud835\udefe
+
\ud835\udc492
2
2\ud835\udc54
+ \u2206\ud835\udc3b1\u22122 
 
15 + 0 + 0 = 0 + 0 +
\ud835\udc492
2
2\ud835\udc54
+ (\u2206\ud835\udc3b\u2032 + \u2206\ud835\udc3b\u2032\u2032) 
15 =
\ud835\udc492
2
2\ud835\udc54
+
\ud835\udc53
\ud835\udc37
\u2217
\ud835\udc49\ud835\udc61
2
2\ud835\udc54
\u2217 \ud835\udc3f + (\ud835\udc3e\ud835\udc38\ud835\udc41\ud835\udc47 \u2217
\ud835\udc49\ud835\udc61
2
2\ud835\udc54
+ \ud835\udc3e\ud835\udc45\ud835\udc3a \u2217
\ud835\udc49\ud835\udc61
2
2\ud835\udc54
+ \ud835\udc3e\ud835\udc3a\ud835\udc3f\ud835\udc42\ud835\udc35\ud835\udc42 \u2217
\ud835\udc49\ud835\udc61
2
2\ud835\udc54
+ \ud835\udc3e\ud835\udc35\ud835\udc42\ud835\udc36\ud835\udc34\ud835\udc3f \u2217
\ud835\udc492
2
2\ud835\udc54
) 
 
 
 
\ud835\udc44\ud835\udc61 = \ud835\udc442 
\ud835\udc49\ud835\udc61 \u2217 \ud835\udc34\ud835\udc61 = \ud835\udc492 \u2217 \ud835\udc342 
 
\ud835\udc49\ud835\udc61 =
\ud835\udc492 \u2217 \ud835\udc342
\ud835\udc34\ud835\udc61
 
\ud835\udc49\ud835\udc61 =
\ud835\udc492 \u2217 \ud835\udf0b \u2217
\ud835\udc372
2
4
\ud835\udf0b \u2217
\ud835\udc37\ud835\udc61
2
4
 
 
\ud835\udc49\ud835\udc61 =
\ud835\udc492 \u2217 \ud835\udc372
2
\ud835\udc37\ud835\udc61
2 
 
 BOCAL 
 
 
Engenharia Civil Hidráulica Geral 
 
 
 
\ud835\udc49\ud835\udc61 =
\ud835\udc492 \u2217 0,05
2
0,22
 
 
\ud835\udc49\ud835\udc61 = 0,0625 \u2217 \ud835\udc492 
 
 
 
15 =
\ud835\udc492
2
2\ud835\udc54
+
\ud835\udc53
\ud835\udc37\ud835\udc61
\u2217
\ud835\udc49\ud835\udc61
2
2\ud835\udc54
\u2217 \ud835\udc3f +
\ud835\udc49\ud835\udc61
2
2\ud835\udc54
(\ud835\udc3e\ud835\udc38\ud835\udc41\ud835\udc47 + \ud835\udc3e\ud835\udc45\ud835\udc3a + \ud835\udc3e\ud835\udc3a\ud835\udc3f\ud835\udc42\ud835\udc35\ud835\udc42) +
\ud835\udc492
2
2\ud835\udc54
\u2217 \ud835\udc3e\ud835\udc35\ud835\udc42\ud835\udc36\ud835\udc34\ud835\udc3f 
 
15 =
\ud835\udc492
2
2\ud835\udc54
(1 + \ud835\udc3e\ud835\udc35\ud835\udc42\ud835\udc36\ud835\udc34\ud835\udc3f) +
\ud835\udc49\ud835\udc61
2
2\ud835\udc54
\u2217 (
\ud835\udc53 \u2217 \ud835\udc3f
\ud835\udc37\ud835\udc61
+ \ud835\udc3e\ud835\udc38\ud835\udc41\ud835\udc47 + \ud835\udc3e\ud835\udc45\ud835\udc3a + \ud835\udc3e\ud835\udc3a\ud835\udc3f\ud835\udc42\ud835\udc35\ud835\udc42) 
 
15 =
\ud835\udc492
2
2 \u2217 9,81
(1 + 0,1) +
(0,0625 \u2217 \ud835\udc492)
2
2 \u2217 9,81
\u2217 (
0,02 \u2217 100
0,2
+ 1,0 + 0,2 + 10,0) 
 
\ud835\udc492 = 15,75\ud835\udc5a/\ud835\udc60 
 
\ud835\udc49\ud835\udc61 = 0,0625 \u2217 \ud835\udc492 
 
\ud835\udc49\ud835\udc61 = 0,0625 \u2217
15,75\ud835\udc5a
\ud835\udc60
 
\ud835\udc49\ud835\udc61 = 0,98\ud835\udc5a/\ud835\udc60 
 
 
3.4. Determinar a altura \u201ch\u201d no reservatório, para que este abasteça 
simultaneamente aos três chuveiros mostrado na figura a segui utilizando tubos de 
PVC nas seguintes condições: 
\uf076 Vazão de cada chuveiro: 0,20l/s; 
\uf076 Diâmetro dos trechos 6-5 e 5-4: 21,6mm; 
\uf076 Diâmetro dos trechos 5-6, 4-2 e 4-1: 17mm; 
\uf076 Pressão dinâmica mínima no chuveiro: 0,2kgf/cm2. 
 
 
Engenharia Civil Hidráulica Geral 
 
 
 
h=? 
 
\ud835\udc44\ud835\udc50\u210e\ud835\udc62\ud835\udc63\ud835\udc52\ud835\udc56\ud835\udc5f\ud835\udc5c = 0,2\ud835\udc59/\ud835\udc60 
 
\ud835\udc432 = 0,2\ud835\udc58\ud835\udc54\ud835\udc53/\ud835\udc50\ud835\udc5a
2 
\ud835\udc432 = 19620\ud835\udc41/\ud835\udc5a
2 
\ud835\udefe\ud835\udc54\ud835\udc62\ud835\udc4e = 9810\ud835\udc41/\ud835\udc5a
2 
 
\ud835\udc491 =
\ud835\udc44\ud835\udc56 \u2217 4
\ud835\udf0b \u2217 \ud835\udc511
2 
 
\ud835\udc491 =
(0,2 \u2217 10\u22123) \u2217 4
\ud835\udf0b \u2217 0,0172
 
 
\ud835\udc491 = 0,881\ud835\udc5a/\ud835\udc60 
 
Calculando as velocidades: 
\ud835\udc496\u22125 =
\ud835\udc446\u22125 \u2217 4
\ud835\udf0b \u2217 \ud835\udc376\u22125
2 
 
\ud835\udc496\u22125 =
(0,0006) \u2217 4
\ud835\udf0b \u2217 0,02162
 
 
\ud835\udc496\u22125 = 1,64\ud835\udc5a/\ud835\udc60 
\ud835\udc495\u22124 =
\ud835\udc445\u22124 \u2217 4
\ud835\udf0b \u2217 \ud835\udc375\u22124
2 
 
\ud835\udc495\u22124 =
(0,0004) \u2217 4
\ud835\udf0b \u2217 0,2162
 
 
\ud835\udc495\u22124 = 1,09\ud835\udc5a/\ud835\udc60 
 
\ud835\udc494\u22121 =
\ud835\udc44\ud835\udc56 \u2217 4
\ud835\udf0b \u2217 \ud835\udc374\u22121
2 
 
\ud835\udc494\u22121 =
(0,0002) \u2217 4
\ud835\udf0b \u2217 0,0172
 
 
\ud835\udc494\u22121 = 0,881\ud835\udc5a/\ud835\udc60 
 
\ud835\udc494\u22122 =
\ud835\udc44\ud835\udc56 \u2217 4
\ud835\udf0b \u2217 \ud835\udc374\u22122
2 
 
\ud835\udc494\u22122 =
(0,0002) \u2217 4
\ud835\udf0b \u2217 0,0172
 
 
\ud835\udc494\u22122 = 0,881\ud835\udc5a/\ud835\udc60 
 
\ud835\udc495\u22123 =
\ud835\udc44\ud835\udc56 \u2217 4
\ud835\udf0b \u2217 \ud835\udc375\u22123
2 
 
\ud835\udc495\u22123 =
(0,0002 \u2217 10\u22123) \u2217 4
\ud835\udf0b \u2217 0,0172
 
 
\ud835\udc495\u22123 = 0,881\ud835\udc5a/\ud835\udc60 
 
Usando a formula de Fair-Whipple-Hsiao 
\ud835\udc3d = 0,000859
\ud835\udc441,75
\ud835\udc374,75
 
\u2206\u210e = \ud835\udc3d\ud835\udc3f 
\u2211\u2206\u210e\u2032 = \u2211 (0,000859 \u2217
\ud835\udc44\ud835\udc56
1,75
\ud835\udc37\ud835\udc56
1,75 \u2217 \ud835\udc3f\ud835\udc56) 
\u2206\u210e = 0,000859
\ud835\udc441,75
\ud835\udc374,75
\ud835\udc3f 
 
\u2206\u210e6\u22125
\u2032 = 0,000859
\ud835\udc446\u22125
1,75
\ud835\udc376\u22125
4,75 \ud835\udc3f6\u22125 
\u2206\u210e6\u22125
\u2032 = 0,000859
0,00061,75
0,2164,75
\u2217 10 
\u2206\u210e6\u22125
\u2032 = 1,61\ud835\udc5a 
 
\u2206\u210e5\u22124