Figure E3.2 (p. 101) Static and Dynamic Pressures

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Applications of Bernoulli’s Equation
p1 V12
p2 V2 2

 z1  
 z2
 2g
 2g
1
1
2
p1  V1  z1  p2  V2 2  z2
2
2
1
p2  p1  V0 2
2
What happens if the bicyclist is accelerating or decelerating?
Figure E3.2 (p. 101)
Static and Dynamic Pressures
• Along a streamline:
p
V 2
2g
  gz  constant
p= static pressure, thermodynamic pressure
V 2
=dynamic pressure, it represents the pressure
2g
rise when the fluid In motion is brought to a stop
sentropically.
ρg z= hydrostatic pressure, not actually a pressure,
but represent a change in pressure due to potential
energy variations of the fluid as a result of elevation
changes
The sum of these is called total pressure
Stagnation Pressure
2
pstag
V
 p
2g
Pitot Static Probe
Connected to a pressure transducer or a manometer
to measure the dynamic pressure
Cross-section of a directional finding
pitot-tube
For symmetric object, stagnation is clearly at the tip or front
For non-symmetric object, stagnation point is not very clear
Figure 3.5 (p. 108)
Stagnation points on bodies in flowing fluids.
Example 1. An airplane flies 100 mi/hr at an elevation of
10,000 ft in a standard atmosphere. Determine the
pressure at point (1) far ahead of the airplane, the pressure
at the stagnation point on the nose of the airplane (point 2)
Figure E3.6a (p. 110)
Figure E3.6b (p. 110)
Free Jets
Applying Bernoulli Equation between points (1) and (2),and
using p1=0=p2; z1=h; z2= 0, V1=0
h 
V 2
h

V 2
2
 2 gh at point 2
V= 2g(h  H ) at point 5
Figure 3.11 (p. 112)
Vertical flow from a tank.
An object falling a distance h
from rest, velocity is given as:
Page 113
Use the centerline velocity as the
average velocity as d<<h
Figure 3.12 (p. 113)
Horizontal flow from a tank.
Vena contracta is because of the
inability of the fluid to turn a sharp
corner
Figure 3.13 (p. 113)
Vena contracta effect for a sharp-edged orifice.
Figure 3.14 (p. 114)
Typical flow patterns and contraction coefficients for various
round exit configurations.
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