Page 221 - Schaum's Outline of Theory and Problems of Applied Physics

206 FLUIDS IN MOTION [CHAP. 17

(b) From v 1 A 1 = v 2 A 2 we have
A 1 r 2 (0.16 m) 2
1
v 2 = v 1 = v 1 2 = (0.99 m/s) 2 = 3.96 m/s
A 2 r 2 (0.08 m)
(c) We now substitute the known quantities p 1 , v 1 , and v 2 with h 1 = 0 and h 2 = 2 m into Bernoulli’s equation
1
2
1
p 1 + dgh 1 + dv = p 2 + dgh 2 + dv 2
2 1 2 2
This yields
1
2

p 2 = p 1 + d v − v 2 − dgh 2
2 1 2
2
5
2
2
1
3
3
3
3
= 1.8 × 10 Pa + (10 kg/m )[(0.99 m/s) − (3.96 m/s) ] − (10 kg/m )(9.8 m/s )(2m)
2
5
= 1.53 × 10 Pa
PRESSURE AND VELOCITY
When the flow is horizontal, so h 1 = h 2 , Bernoulli’s equation becomes
2
1
1
p 1 + dv = p 2 + dv 2
2 1 2 2
The pressure is greatest where the fluid velocity is least, and the pressure is least where the fluid velocity is
greatest. The lift developed by an airplane wing is an example of this effect: The air moving across the curved
upper surface of the wing travels faster than that moving across the lower surface because it must cover a greater
distance. Hence the pressure is less on the upper surface, and the net result is an upward force on the wing.
SOLVED PROBLEM 17.12
2
An airplane whose mass is 40,000 kg and whose total wing area is 120 m is in level flight. What is the
difference in pressure between the upper and lower surfaces of its wings?
The lift developed by the wings in level flight is equal to the airplane’s weight. Hence
5
2
4
F = w = mg = (4 × 10 kg)(9.8 m/s ) = 3.92 × 10 N
The pressure difference is therefore
5
F 3.92 × 10 N
3
p = = = 3.27 × 10 Pa
A 120 m 2
2
which is equal to 0.032 atm or 0.47 lb/in. .
SOLVED PROBLEM 17.13
5
At what velocity does water emerge from an orifice in a tank (a) in which the gauge pressure is 3×10 Pa
2
3
3
3
and (b) in which the gauge pressure is 40 lb/in. (d water = 10 kg/m = 1.94 slugs/ft )?
(a) Let point 1 be at the orifice and point 2 be inside the tank at the same level. In this situation h 1 = h 2 and v 2 = 0
(very nearly). Substituting in Bernoulli’s equation yields

2(p 2 − p 1 )
1 2
p 1 + dv = p 2 v 1 =
1
2 d
5
The quantity p 2 − p 1 is the gauge pressure of 3 × 10 Pa since p 1 is atmospheric pressure. Hence

5
(2)(3 × 10 Pa)
v 1 = 3 = 24.5 m/s
3
10 kg/m