BERNOULLI’S EQUATION
• Bernoulli’s equation states that the sum of
all forms of energy in a fluid flowing along
an enclosed path is the same at any two
points in that path.
• Assumptions:
– Flow is steady
– Density is constant (incompressible)
– Friction losses are negligible
BERNOULLI’S EQUATION
• By conservation of energy,
(Energy)1 = (Energy)2
Σ(Energy) = 0
{D (Press. forces) + D (Kinetic Energy) +
D (Potential Energy)} = 0
•
•
•
•
Pressure forces = F S
Kinetic Energy = ½ m v2
Potential energy = m g h
Pressure forces @ 1 are given by,
F1 S1  p1 A1v1Dt

 A1v1  p1 Dt



m
p1 Dt

m

p1 Dt
 Dt
 F1 S1 
m p1

• Sub. all the values in the energy balance
equation:
1
1
2
2
F1 S1  mv1  mgh1  F2 S 2  mv 2  mgh 2
2
2
p1
p2 1 2
1 2
 m  m v1  m gh1  m
 m v2  m gh2
 2
 2
p1
p2 1 2
1 2

 v1  gh1 
 v2  gh2
 2
 2
p
2
v
OR

 gh  constant
 2
Applications of B.Eqn:
• B.E often combined with continuity
equation to find velocities & pressures in
the flow connected by a streamline
• Orifice meter; venturi meter
• Flow in pumps etc.,
Fluid friction……
• Fluid friction is defined as any conversion of
mechanical energy into heat in a flowing stream
• Denoted by the letter hf (J/kg)
• hf represents all the friction generated per unit
mass of fluid bet (1) & (2)
• B.E becomes……..
p1
p2 1 2
1 2
 v1  gh1 
 v2  gh2  h f
 2
 2
Pump work…….
• If a pump is used during flow, then the term “work
done by the pump” should be added to B.E
p1
p2 1 2
1 2
 v1  gh1  W p 
 v2  gh2
 2
 2
Prob 1
• A fluid of density 960 kg/m3 is flowing
steadily thro a tube as shown in the fig:
The sections diameters are d1=100mm &
d2=80mm. The press p1 =200kN/m2;
u1=5m/s. The tube is horizontal. What is
the pressure at section(2)?
• By continuity equation:

m  1v1 A1   2 v 2 A2
• v2 = 7.8125 m/s
• From B.E….
p1
p2 1 2
1 2
 v1  gh1 
 v2  gh2
 2
 2
• p2 = 182.703 x 103 N/m2
Prob 2
• Gasoline(680 kg/m3) flows from a 0.3m dia
pipe in which the pressure is 300kPa into
a 0.15m dia pipe in which the press is
120kPa. If the pipes are horizontal &
viscous effects are negligible, determine
the flow rate:
• By continuity equation:

m  1v1 A1   2 v 2 A2
• v2 = 4 v 1
• From B.E….
• v1= 5.94 m/s
• Flow rate, Q = A1v1 = 0.4199 m3/s
Prob 3
• Water flows steadily thro’ the pipe shown
in fig. such that the press @ sections 1 & 2
are 300kPa & 100kPa respectively.
Determine the dia of pipe @ section 2, if
the velocity at section 1 is 20m/s and
viscous effects are negligible:
• From B.E….
• v2= 42.2 m / s
• By continuity equation:

m  1v1 A1   2 v 2 A2
• D2 = 0.0688m
Prob 4
Water with a density of 998 kg/m3 is flowing at a steady
mass flow rate through a uniform-diameter pipe. The
entrance pressure of the fluid is 68.9 kN/m2 in the pipe,
which connects to a pump which actually supplies 155.4
J/kg of fluid flowing in the pipe. The exit pipe from the
pump is the same diameter as the inlet pipe. The exit
section of the pipe is 3.05 m higher than the entrance, and
the exit pressure is 137.8 kN/m2. The Reynolds number in
the pipe is above 4000 in the system. Calculate the
frictional loss hf in the pipe system.
• From B.Eqn:
p1
p2 1 2
1 2
 v1  gh1  W p 
 v2  gh2  h f
 2
 2
• Since dia of pipe is same…….v1 = v2
• hf = 56.44 J/kg
Prob 5
• A pump draws 69.1 gal/mm of a liquid solution
having a density of 114.8 lbm/ft3 from an open
storage feed tank of large cross-sectional area
through a 3.068”ID suction line. The pump
discharges its flow through a 2.067”ID line to an
open overhead tank. The end of the discharge
line is 50 ft above the level of the liquid in the
feed tank. The friction losses in the piping
system are 10.0 ft-lbf /lbm. What is the
horsepower of the pump if its efficiency is 65%?
What pressure must the pump develop?
• = 114.8 lbm/ft3
• =114.8 (0.454) / (0.3048)3
• Therefore,  = 1840.6 kg/m3
• hf = 10 ft-lbf / lbm
• =10(0.3048m)(4.4482N) / (0.454)
• Therefore, hf = 29.864 J/kg
• Apply B.Eqn bet (1) & (2)
p1
p2 1 2
1 2
 v1  gh1  W p 
 v2  gh2  h f
 2
 2
•
•
•
•
Since p1 = p2 (open to atmosphere)
h1 – h2 = 50’ = 15.24m
v2 = 2.0315 m/s (for 2.067”)
Since tank dia is very high…..v1<<v2 { v1/v2 =
A1/A2}
• v1 = 0
• Sub all values in B.Eqn….Wp = 2239.057 J/s
• Or Wp = 3.005 HP {1HP = 745 W}
• The press. developed in the pump (ie bet 3 & 4)
p3
p4 1 2
1 2
 v3  gh3  W p 
 v4  gh4  h f
 2
 2
• Since hf is only for piping system…..here for
pump…..hf = 0
• h3 = h4 and v4 = v2 = 2.035 m/s
• v3 = 0.91395 m/s (for 3.068”)
• Sub all the values in the B.Eqn….
• (p3 – p4) = 330.833 kPa…..pressure developed
by the pump
Time for emptying a tank
• Consider a steady flow of water from a
tank of height H
• Hi and Hf are initial & final height of water
in the tank
• Applying B.Eqn….
p1
p2 1 2
1 2
 v1  gh1 
 v2  gh2
 2
 2
 v2  2gH
• We know, Q1 = Q2
 dH 
  At
  A2 v2
dt 

Hf
 At

Hi
At
tf 
A2
dH
2 gH
1
2g

 A2 t f
2 Hi  H f

Prob 6
• Draining Cotton Seed Oil from a Tank
A cylindrical tank 1.52 m in dia and 7.62 m high
contains cotton seed oil having a density of 917
kg/m3. The tank is open to the atmosphere. A
discharge nozzle of inside diameter 15.8mm and
cross-sectional area A2 is located near the bottom of
the tank. The surface of the liquid is located at
H = 6.1 m above the center line of the nozzle. The
discharge nozzle is opened, draining the liquid level
from H=6.1 m to H=4.57 m. Calculate the time in
seconds to do this.
• We know,
At
tf 
A2
1
2g
• At = CSA of tank = (p/4)D2
• Therefore, t = 1387 sec

2 Hi  H f
