Memoria De Calculo Se Un Sifon

1.- Dimensiones del Conducto

H/B= 1.25 ∴ H=1.25B V_(〖Real〗^ )=Q/A_(Real=(11.08/((2.11)(2.63)-4(1/2) (.1)^2 )))

Proponemos V= 2 m/s

Ac= Q/v=11.08/2= 5.54 m2 A_█(Real=5.5293 m^2@) V_(〖Real〗^ )=2.01 m/s

5.54=B(1.25B)

B=√(5.54/1.25)=2.11 m

H=1.25(2.11)=2.63

B=2.11M

H=2.63

2.-Longitud de Transición

T= B + 2td = 3 + 2(1.5)(2.3)= 9.9 m

t´=b= 2.11

L_T=((T-t´)/2)Cot(22.5)=((9.9-2.11)/2)Cot(22.5)=9.41 m

L_T=9.41m

3.- Pérdidas en transición de salida 1 – 2

z_1+y_1+ .8 〖v_1〗^2/2g= z_2+y_2+ .8 ((〖Q/( By_(2^ ) ))〗^2)/2g

1242.58+2.30+0.8(〖1.037〗^2/(2*9.81))=1242+y_2+0.8(((〖11.08/(2.11*y_2 ))〗^2)/(2*9.81))

y_2=2.7773m

h_(〖1-2〗^= ).2 (((V_1)^2)/2g-((V_2)^2)/2g )

h_(〖1-2〗^= ) 0.2 (0.1831-0.0548)

h_(1-2)= 0,02565

4.-Perdidas en Salida 2-3

h_(〖2-3〗^= ) 0.2 (((V_3)^2)/2g-((V_2)^2)/2g)

h_(〖2-3〗^= ) 0.2(0.20544-0.1831)

h_(〖2-3〗^= ) 0,00447

y_2+ ((V_2)^2)/2g+ h_(〖2-3〗^= )= y_3+p_(3^ )/γ+ ((V_3)^2)/2g

2.7773+0.1831+0.00447=2.63+p_(3^ )/γ+ 0.20544

p_(3^ )/γ=0.1282

5.- Perdidas en el conducto 3 – 4

h_(〖3-4〗^(= ) ) 〖Vn/R^(2/3) 〗^2 (L)+(Kv^2)/(2g√90)(√(∆_1 )+√(∆_2 )+√(∆_3 )+√(∆_4 )+√(∆_5 )+√(∆_6 )+√(∆_7 ))

R=((2.11)(2.63)-4(〖.1〗^2/2))/(2(2.11+2.63)-8(.1)+.4(.1)(√2))=.5980

h_(〖3-4〗^(= ) ) 0.28026

z_3+ p_(3^ )/γ+ h_(〖4-3〗^= )= Z_4+p_(4^ )/γ

.1242+ .1282+ 0.28026= 1242.4+p_(4^ )/γ

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