 Useful concepts associated with the Bernoulli equation
- Static, Stagnation, and Dynamic Pressures
Bernoulli eq. along a streamline
1 2
p 
v  z = constant
2
Static
Dynamic
(Unit of Pressure)
Hydrostatic
(Thermodynamic)
Piezometer
tube
 Physical meaning of each term
Velocity of
stream
i) 1st term (Static pressure, p)
: Only due to the fluid weight
 p1  p3  h31
 h43  h31  h
where p3  h43 (h4-3: Piezometer reading)
1 2
v )
2
: Pressure increase or decrease due to fluid motion
ii) 2nd term (Dynamic pressure,
iii) 3rd term (Gravitational potential, z )
: Pressure change due to elevation change
Between Point (1) and Point (2) on the same streamline,
1
1
p1  v12  z1  p2  v2 2  z 2  constant
2
2
because z1  z 2 (No elevation change)
v2  0 (Stagnated by a tube inserted)
 Difference in p between piezometer and tube inserted into a flow
(Stagnation pressure)
p2  H
1 2
v  p2  p1  H  h
2
(Elevation change, H  h : Due to dynamic pressure)
 Point (2): Stagnation point (V2  0 )
Line through point (2): Stagnation streamline
1
iv) Total pressure pT  p  V 2  z = constant
2
(Along a streamline)
 Special application of Static and Stagnation Pressure
- Determination of fluid speed (Pitot-static tube)
By measuring pressure at points (3) and (4)
and neglecting the elevation effect (e.g. gases)
1
At point (3), p3  p  V 2 (Stagnation Pressure)
2
where p and V: Static pressure and fluid velocity at (2)
At ponit (4) (Small holes), p4  p1  p (Static pressure ~ Piezometer)
1
Then, p3  p4  V 2
2
 V
2( p3  p4 )

: Pitot-static tube
Here
Here
Photo-detector
Wheel
How to detect the wheel speed
(e.g. Car and Computer mouse etc.)
 How to use the Bernoulli equation (Examples)
Choose 2 points (1) and (2) along a streamline
1
1
p1  V12  z1  p2  V2 2  z 2 : 6 variables ( p1, p2 ,V1,V2 , z1, z2 )
2
2
 5 more conditions from the problem description
E.g. 1 Free Jets (A jet of liquid from large reservoir)
Question: Find V of jet stream from a nozzle of container
Consider a situation shown
Step 1. Select one streamline
between (1) & (2)
Step 2. Apply the Bernoulli eq.
between (1) & (2)
1
1
p1  V12  z1  p2  V2 2  z 2
2
2
because p1= 0: Gauge pressure (The atmosphere)
V1 = 0 (Large container  Negligible change of fluid level)
2= 0 (Why?) [p2= p4 (Normal Bernoulli eq.)]
Then,
1
h  V 2
(If we choose z1 as h and z2 as 0)
2
or
2h
 2 gh : Velocity at the exit plane of nozzle
 V

 Velocity at any point [e.g. (5) in the figure] outside the nozzle,
 v  2 g (h  H ) (H: Distance from the nozzle)
: Conversion of P. E. to K.E. without viscosity (friction)
cf. Free body falling from rest without air resistance.
• In case of horizontal nozzle shown,
v1  v2  v3 (Elevation difference)
By assuming d << h,
v2 : Average velocity at the nozzle
E.g. 2 Confined Flows
- Fluid flowing within a container connected with nozzles and pipes
 What to know
- We can’t use the atmospheric p as a standard
- But, we have one more useful concept of conservation of mass
: No change of fluid mass in fixed volume (continuity equation)
 Mass flow rate, m (kg/s or slug/s) for a steady flow
Mass of fluid entering the container across inlet per unit time
= Mass of fluid leaving the container across outlet per unit time,
m 1 
Or
m 
m1 1V1tA1
 V tA m

(Inlet)  2 2 2  2  m 2 (Outlet)
t
t
t
t
m
(kg/s) = Q
t
(Q: Volume flowrate, m3/s)
For a time interval t ,
 1  1Q1  1 A1V1   2 A2V2   2Q2  m
2
m
For a incompressible fluid, 1   2
 A1V1  A2V2
or Q1 = Q2
 Change in area
→ Change in fluid velocity ( A1V1  A2V2
→ Change in pressure
1
1
( p1  V12  z1  p2  V2 2  z 2 )
2
2
 Increase in velocity → Decrease in pressure
: Blow off the roof by hurricane, Cavitation damage, etc.
E.g. 3 Flowrate Measurement
Case 1: Determine Q in pipes and conduits (Closed container)
Consider three typical devices (Depend on the type of restriction)
Step 1. At section (1)
: Low V, High p
At section (2)
: High V, Low p
Step 2. Apply the Bernoulli Eq.
For a horizontal flow (z1 = z2)
1
1
p1  V12  p2  V2 2
2
2

V2 2 
2( p1  p2 )
[1  (V1 / V2 ) 2 ]
Step 3. Use the continuity Eq.
Q  A1V1  A2V2
 A2
2( p1  p2 )
[1  ( A2 / A1 ) 2 ]
where A1 & A2: Known
p1  p2 : Measured by gages
Case 2: Determine Q in flumes and irrigation ditches (Open channer)
Typical devices: Sluice gate (Open bottom)
& Sharp-crested weir (Open top)
 Consider a sluice gate shown
Bernoulli Eq. & Continuity Eq. [(1)  (2)]
1
1
p1  V12  z1  p2  V2 2  z 2
2
2
2 g ( z1  z 2 )
 V2 
( p1 = p2 = 0)
1  (V1 / V2 ) 2
and
Q  A1V1  bz1V1  A2V2  bz2V2
Finally, the flowrate, Q  bz2
 In case of z1  z2 or
 Q  bz2 2gz1
2 g ( z1  z 2 )
1  ( z 2 / z1 ) 2
z1  z2  z1
or
or
z2 / z1  1
V2  2gz1
 The sharp-crested weir (Open top)
(If similar to horizontal free stream)
Then, Average velocity across the top of the weir 
Flow area for the weir  Hb
 Q = C1Hb 2 gH = C1b H3/2 2 g
2 gH
 Energy Line (EL) and the Hydraulic Grade Line (HGL)
- Geometrical Interpretation of the Bernoulli Eq.
Bernoulli eq.: Conserved total energy along streamline, i.e.
p V2

 z  constant on a streamline = H (Unit of Length)
 2g
 Heads:
p
: Pressure head,

H: Total head
v2
: Velocity head, z: Elevation head
2g
Energy line (EL): Representation of Energy versus Heads
(1) → (2): Elevation head↑
Pressure head↓
Velocity head↑
(2) → (3): Elevation head ↑
Pressure head ↓
Velocity head ↑
(1)
(2)
(3)
Total head: Constant
e.g. Flow from tank