Presentation on theme: "BERNOULLIS EQUATION Bernoullis 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."— Presentation transcript:
BERNOULLIS EQUATION Bernoullis 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
Pressure forces = F S Kinetic Energy = ½ m v 2 Potential energy = m g h Pressure 1 are given by,
Sub. all the values in the energy balance equation: Bernoullis eqn: All are having an units of 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 h f (J/kg) h f represents all the friction generated per unit mass of fluid bet (1) & (2) B.E becomes……..
Pump work……. If a pump is used during flow, then the term work done by the pump should be added to B.E
Prob 1 A fluid of density 960 kg/m 3 is flowing steadily thro a tube as shown in the fig: The sections diameters are d 1 =100mm & d 2 =80mm. The press p 1 =200kN/m 2 ; u 1 =5m/s. The tube is horizontal. What is the pressure at section(2)?
By continuity equation: v 2 = m/s From B.E…. p 2 = x 10 3 N/m 2
Prob 2 Gasoline(680 kg/m 3 ) 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: v 2 = 4 v 1 From B.E…. v 1 = 5.94 m/s Flow rate, Q = A 1 v 1 = m 3 /s
Prob 3 Water flows steadily thro the pipe shown in fig. such that the sections 1 & 2 are 300kPa & 100kPa respectively. Determine the dia of section 2, if the velocity at section 1 is 20m/s and viscous effects are negligible:
From B.E…. v 2 = 42.2 m / s By continuity equation: D 2 = m
Prob 4 Water with a density of 998 kg/m 3 is flowing at a steady mass flow rate through a uniform-diameter pipe. The entrance pressure of the fluid is 68.9 kN/m 2 in the pipe, which connects to a pump which actually supplies 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 kN/m 2. The Reynolds number in the pipe is above 4000 in the system. Calculate the frictional loss h f in the pipe system.
From B.Eqn: Since dia of pipe is same…….v 1 = v 2 hf = J/kg
Prob 5 A pump draws 69.1 gal/mm of a liquid solution having a density of lb m /ft 3 from an open storage feed tank of large cross-sectional area through a 3.068ID suction line. The pump discharges its flow through a 2.067ID 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-lb f /lb m. What is the horsepower of the pump if its efficiency is 65%? What pressure must the pump develop?
= lbm/ft 3 = (0.454) / (0.3048) 3 Therefore, = kg/m 3 h f = 10 ft-lb f / lb m = 10(0.3048m)(4.4482N) / (0.454) Therefore, h f = J/kg
Apply B.Eqn bet (1) & (2) Since p1 = p2 (open to atmosphere) h 1 – h 2 = 50 = 15.24m v 2 = m/s (for 2.067) Since tank dia is very high…..v 1 <
The press. developed in the pump (ie bet 3 & 4) Since h f is only for piping system…..here for pump…..h f = 0 h 3 = h 4 and v 4 = v 2 = m/s v 3 = m/s (for 3.068) Sub all the values in the B.Eqn…. (p 3 – p 4 ) = kPa…..pressure developed by the pump
Time for emptying a tank Consider a steady flow of water from a tank of height H H i and H f are initial & final height of water in the tank
We know, Q1 = Q2
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/m 3. The tank is open to the atmosphere. A discharge nozzle of inside diameter 15.8mm and cross-sectional area A 2 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, A t = CSA of tank = ( /4)D 2 Therefore, t = 1387 sec