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9 Fluid Dynamics
9.29 In prob. (9.28) if the tubes are connected in series then what quantity will flow
through the composite tube?
9.3 Solutions
9.3.1 Bernoulli’s Equation
9.1 From continuity equation
A 1 v 1 = A 2 v 2
∴ v 2 =
A 1 v 1
A 2
=
πr 2
1 v 1
πr 2
2
=
10 2 × 4
5 2
= 16 m/s
9.2 v x = 3x
2
− x y + 2z
2
v y = 2x
2
− 6x y + y
2
v z = −2x y − yz + 2y
2
∴
∂v x
∂ x
= 6x − y;
∂v y
∂ y
= −6x + 2y;
∂v z
∂z
= −y
∇ · v =
∂v x
∂ x
+
∂v y
∂ y
+
∂v z
∂z
= (6x − y) + (−6x + 2y) − y = 0
Thus the continuity equation for steady incompressible flow is satisfied.
9.3 Pressure difference across the wing
p =
1
2
ρ(v
2
1 − v
2
2 )
=
1
2
× 1.293 × (70
2
− 55
2
) = 1212 Pa
(a) Lift = (pressure difference) (area)
= 1212 × 4 = 4848 N
(b) Net force = Lift − Weight of plane
= 4848 − (300 × 9.8)
= 1908 N in the upward direction
9.4 P =
1
2
ρv
2
A 2
a 2 − 1
A
a
=
π R 2
πr 2 =
10
5
2
= 4