9.3 Solutions
399
60,000 − 45,000 =
1
2
× 1000 × 15v
2
or v = 1.414 m / s (throat)
Rate of flow of water
Q = v A = (1.414)(π × 0.01
2
) = 0.0444 m
3
/s
9.5 v =
2ghρ
ρ
=
2 × 9.8 × 0.15 × 810
1.293
= 42.9 m/s = 154.5 km/h
9.6 Total area of the holes
A = 80 × 2.5 × 10
−6 m
2
= 2 × 10
−4 m
2
Q = Av
v =
Q
A
=
2 × 10 −3
2 × 10 −4 = 10 m/s
9.7 (a) v x = 3x y + y
2
v y = 5x y + 2x
∂v x
∂ x
= 3y;
∂v y
∂ y
= 5x
∂v x
∂ x
+
∂v y
∂ y
= 3y + 5x = 0
Therefore, steady incompressible flow is not possible.
(b) v x = 3x
2
+ y
2
v y = −6x y
∂v x
∂ x
= 6x;
∂v y
∂ y
= −6x
∂v x
∂ x
+
∂v y
∂ y
= 6x − 6x = 0
Thus, steady incompressible flow is possible.
9.8 P =
1
2
ρ(v
2
1 − v
2
2 )
v 1 =
2P
ρ
+ v 2
2 =
2 × 1000
1.293
+ 100 2 = 107.45 m/s