406
9 Fluid Dynamics
P +
1
2
ρ 1 v
2
1 = P + h 1 ρ 1 g + h 2 ρ 2 g
∴ v 1 =
2g
h 1 +
h 2 ρ 2
ρ 1
Substituting h 1 = 60 cm, h 2 = 40 cm, ρ 1 = 1, ρ 2 = 0.8 and g = 980, we
find v 1 = 425 cm/s or 4.25 m/s.
9.24 Volume efflux at A and B, Fig. 9.9
Fig. 9.9
Q A = v A S
Q B = v B S
(Mass efflux) A = ρv A S
(Mass efflux) B = ρv B S
Force F A = (rate of change of momentum) A
= ρv A Sv A = ρ Sv
2
A
= ρ S(2gh) = 2ρ Sgh
F B = 2ρ Sg(h + h)
F B − F A = 2 ρ S g h
(because the vector force is in the opposite direction)
= 2 × 1000 × 1.0 × 10
−4
× 9.8 × 0.51 = 1.0 N
9.3.3 Viscosity
9.25 Volume of liquid flowing per second
V =
πr 4 P
8ηl
9 Fluid Dynamics
P +
1
2
ρ 1 v
2
1 = P + h 1 ρ 1 g + h 2 ρ 2 g
∴ v 1 =
2g
h 1 +
h 2 ρ 2
ρ 1
Substituting h 1 = 60 cm, h 2 = 40 cm, ρ 1 = 1, ρ 2 = 0.8 and g = 980, we
find v 1 = 425 cm/s or 4.25 m/s.
9.24 Volume efflux at A and B, Fig. 9.9
Fig. 9.9
Q A = v A S
Q B = v B S
(Mass efflux) A = ρv A S
(Mass efflux) B = ρv B S
Force F A = (rate of change of momentum) A
= ρv A Sv A = ρ Sv
2
A
= ρ S(2gh) = 2ρ Sgh
F B = 2ρ Sg(h + h)
F B − F A = 2 ρ S g h
(because the vector force is in the opposite direction)
= 2 × 1000 × 1.0 × 10
−4
× 9.8 × 0.51 = 1.0 N
9.3.3 Viscosity
9.25 Volume of liquid flowing per second
V =
πr 4 P
8ηl