CALCULOS

1 POBLACION FUTURA

Método Geometrico

Pf = Pa (1 + r)

Pf = Poblacion futura Pf

Pa = Población Actual del 1993 15800 proyectado al 2004r = Tasa de crecimiento 1.1t = Periodo 20

Pf = 15800(1+1.1/100)

Pf = 19664.38 Habitantes

2 DISEÑO DE CAUDAL

Dot = 150 Lt/hab/diaK1 = 1.3K2 = 2.0

Caudal Promedio Qp = Pf + Dot86400

Qp = 4739 * 15086400

Qp = 34.139544 Lps

Caudal Maximo Diario Qmd = K1 x Qp

Qmd = 1.3 * 8.2276

Qmd = 44.3814071 Lps

Caudal Maximo Horario Qmh = K2 x Qp

Qmh = 2.0 * 8.2276

Qmh = 68.2790879 Lps

t

2020

3 VOLUMEN DE RESERVORIO

Volumen de Regulación Vreg = 18% * Qmd * 86.4

Vreg = 0.18 * 10.6959 * 86.4

Vreg = 690.219644 M3/dia

Volumen de Reserva Vres = 7% * Qmd * 86.4

Vres = 0.07 * 1'.6959*86.4

Vres = 268.41875 M3/dia

Volumen de Reservorio Vreser = 25% * Qmd * 86.4

Vreser = 0.25 * 10.6959 * 86.4

Vreser = 958.638394 M3/dia

4 LINEA DE CONDUCCION Y ADUCCION

Linea de Conducción

Distancia de captacional reservorio : 1.050 KmCota Captacion : 395.5 m.s.n.m.Cota Reservorio : 367.5 m.s.n.m.

S = 395.50 - 367.5 / 1050

S = 26.6666667

Hallando pendiente S = 26.6666667

Hallando diametro Qm = 4.26 * 10.-4 * c * d 2.63 * s 0.5444.381407 4.26 * 10.-4 * c * d 2.63 * 26.794 0.54

d = 3.66

Diametro 2" 4"

Velocidad V 2" = 1.974 * 10.6959/4V 2" = 21.9022244

V 4" = 1.974 * 10.6959/16V 4" = 5.47555611

Hallando la gradienteS 2" = Qm = 4.26 * 10.-4 * c * d 2.63 * s 0.54

10.6959 4.26 * 10.-4 * c * 2 2.63 * s 0.54S = 509.75

S 4" = Qm = 4.26 * 10.-4 * c * d 2.63 * s 0.5410.6959 4.26 * 10.-4 * c * 4 2.63 * s 0.54

S = 17.12

Hallando perdida de carga HFD 2" = Hf = 509.75*1.045

Hf = 535.2375 mt

D 4" = Hf = 17.12*1.045Hf = 17.976 mt

Hallando cota piezometricaCp1 = Cp2 + Hf1-2

D 2" = 395.5 Cp2 + 532.6888Cp2 -139.7375 No mt

D 4" = 395.5 Cp2 + 17.89Cp2 377.524 Si mt

Por lo que asumimos un diametro de 4"

Linea de Aduccion

Distancia del reservorioal distrito de Jauja 0.265 KmCota CP2 : 377.524 m.s.n.m.Cota Dist. Jauja 15 + : 350 m.s.n.m.

S = 377.61 - 350 / 0.265

S = 30 / 0.265

Hallando pendiente S = 103.864

Hallando diametro Qm = 4.26 * 10.-4 * c * d 2.63 * s 0.5416.4553 4.26 * 10.-4 * c * d 2.63 * 104.189 0.54

d = 3.264

Diametro 2" 4"

Velocidad V 2" = 1.974 * 16.4553/4V 2" = 33.6957299

V 4" = 1.974 * 16.4553/16V 4" = 8.42393247

Hallando la gradienteS 2" = Qm = 4.26 * 10.-4 * c * d 2.63 * s 0.54

16.4553 4.26 * 10.-4 * c * 2 2.63 * s 0.54S = 1131.91

S 4" = Qm = 4.26 * 10.-4 * c * d 2.63 * s 0.5416.4553 4.26 * 10.-4 * c * 4 2.63 * s 0.54

S = 38.70

Hallando perdida de carga HFD 2" = Hf = 1131.91*0.265

Hf = 299.95615 mt

D 4" = Hf = 38.70*0.265Hf = 10.2555 mt

Hallando cota piezometricaCp1 = Cp3 + Hf1-2

D 2" = 377.61 Cp2 + 1131.91Cp3 77.65385 No mt

D 4" = 377.61 Cp2 + 10.255Cp3 367.3545 Si mt

Lo cual utilizamos con el diametro de 4"

0.000426

140

5.88874998

6.29655183