The three waterfilled tanks shown in Fig P8114 are connected

The three water-filled tanks shown in Fig. P8.114 are connected by pipes as indicated. If minor losses are neglected, determine the flow rate in each pipe (+ is out and – is into the reservoir)

Elevation=70 m Elevation = 30m 17 A Elevation 20 m D=0.10 m = 100m f = 0.015 D=0.08 m = 200m f =0.020 D=0.08 m e = 400m f =0.020

Solution

HEAD= F*L*V²/2*g*D

FOR A

H_{A =}F_A*L_A*V_A²/2*g*D_A

30=0.020*200*V_A²/2*9.81*0.08 =V_A² ₌  11.772 =V_A =3.43 m/sec

SIMILARLY

FOR B

H_B=F_BL_BV_B²/2*g*D_B

_70=0.05*100* V_B²/2*9081*0.10

V_B²=91.56 V_B =9.56m/sec

H_C=F_C*L_C*V_C²/2*G*D_C

20=0.020*400*VC2/2*9.81*0.08

V_C²=3.924

V_C=1.98m/sec

H_C=F_C*L_C*V_C²/2*G*D_C

20=0.020*400*VC2/2*9.81*0.08

V_C²=3.924

V_C=1.98m/

DISCHARGE IN A=Q_A=VELOCITY(V_A)*AREA (A_A)

Q_A = V_A*A_A

A_A=(/4)* D_A²=( /4)* 0.08²=0.0050265 M²   Q_A=0.0050265*3.43

Q_A=0.01724M³/SEC

Q_B = V_B*A_B

A_B=(/4)* DB²=( /4)* 0.010²=0.00785398M²

Q_B = 0.00785398*9.56=0.075084M³/SEC

Q_C = V_C*A_C

A_C=(/4)* D_C²=( /4)* 0.08²=0.0050265M²

Q_C=0.0050265*1.98=0.095256M³/SEC