BERNOULLIS EQUATION Bernoullis equation states that the sum of all forms of energy in a fluid...

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

(Press. forces) + (Kinetic Energy) +

(Potential Energy)} = 0

• Pressure forces = F S• Kinetic Energy = ½ m v2

• Potential energy = m g h• Pressure forces @ 1 are given by,

tvApSF 11111

1

11

pmSF

tpvA 111

tpt

m

1

tpm

1

• Sub. all the values in the energy balance equation:

222221

2111 2

1

2

1mghmvSFmghmvSF

222

21

21

1

2

1

2

1mghmv

pmmghmv

pm

222

21

21

1

2

1

2

1ghv

pghv

p

tconsghvp

OR tan2

2

Bernoulli

’s

eqn: All

are

havin

g an

units o

f

J/kg

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……..

fhghvp

ghvp

222

21

21

1

2

1

2

1

Pump work…….

• If a pump is used during flow, then the term “work done by the pump” should be added to B.E

222

21

21

1

2

1

2

1ghv

pWghv

pp

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:

• v2 = 7.8125 m/s

• From B.E….

• p2 = 182.703 x 103 N/m2

222111 AvAvm

222

21

21

1

2

1

2

1ghv

pghv

p

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:

• v2 = 4 v1

• From B.E….

• v1= 5.94 m/s

• Flow rate, Q = A1v1 = 0.4199 m3/s

222111 AvAvm

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:

• D2 = 0.0688m

222111 AvAvm

Prob 4Water 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:

• Since dia of pipe is same…….v1 = v2

• hf = 56.44 J/kg

fp hghvp

Wghvp

222

21

21

1

2

1

2

1

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)

• 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}

fp hghvp

Wghvp

222

21

21

1

2

1

2

1

• The press. developed in the pump (ie bet 3 & 4)

• 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

fp hghvp

Wghvp

424

43

23

3

2

1

2

1

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….

222

21

21

1

2

1

2

1ghv

pghv

p

gHv 22

• We know, Q1 = Q2

22vAdt

dHAt

f

i

H

H

ft tAgH

dHA 2

2

fit

f HHgA

At 2

2

1

2

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 = CSA of tank = (/4)D2

• Therefore, t = 1387 sec

fit

f HHgA

At 2

2

1

2