Energy Conservation (Bernoulli’s Equation). Recall Euler’s equation:. Also recall that viscous forces were neglected, i.e. flow is invisicd. If one integrates Euler’s eqn. along a streamline, between two points , &. We get :. Which gives us the Bernoulli’s Equation.
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Recall Euler’s equation:
Also recall that viscous forces were neglected, i.e. flow is invisicd
If one integrates Euler’s eqn. along a streamline, between two points , &
We get :
Which gives us the Bernoulli’s Equation
Flow work + kinetic energy + potential energy = constant
ABernoulli’s Equation (Continued)
Flow Work (p/):
It is the work required to move fluid across the control volume boundaries.
Consider a fluid element of cross-sectional area A with pressure p acting on the control surface as shown.
Due to the fluid pressure, the fluid element moves a distance Dx within time Dt. Hence, the work done per unit time DW/Dt (flow power) is:
Flow work or Power
Flow work per unit mass
1/mass flow rate
Flow work is often also referred to as flow energy
Very Important: Bernoulli’s equation is only valid for :
incompressible fluids,steady flow along a streamline, no energy loss due to friction, no heat transfer.
Application of Bernoulli’s equation - Example 1:
Determine the velocity and mass flow rate of efflux from the circular hole (0.1 m dia.) at the bottom of the water tank (at this instant). The tank is open to the atmosphere and H=4 m
p1 = p2, V1=0
water height (m)
time (sec.)Bernoulli’s Eqn/Energy Conservation (cont.)
Example 2: If the tank has a cross-sectional area of 1 m2, estimate the time required to drain the tank to level 2.
First, choose the control volume as enclosed
by the dotted line. Specify h=h(t) as the water
level as a function of time.
Energy added, hA
(ex. pump, compressor)
Energy lost, hL
(ex. friction, valve, expansion)
Energy extracted, hE
(ex. turbine, windmill)
hL, friction loss
If energy is added, removed or lost via pumps turbines, friction, etc.then we useEnergy conservation(cont.)
Extended Bernoulli’s Equation
Example: Determine the efficiency of the pump if the power input of the motor
is measured to be 1.5 hp. It is known that the pump delivers 300 gal/min of water.
No turbine work and frictional losses, hence: hE=hL=0. Also z1=z2
Given: Q=300 gal/min=0.667 ft3/s=AV
V1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/s
6-in dia. pipe
Looking at the pressure term:
Mercury (m=844.9 lb/ft3)
water (w=62.4 lb/ft3)
1 hp=550 lb-ft/s
R: radius, D: diameter
L: pipe length
w: wall shear stress
Consider a laminar, fully developed circular pipe flow
Major Losses: due to friction, significant head loss is associated with the straight portions of pipe flows. This loss can be calculated using the Moody chart or
Minor Losses: Additional components (valves, bends, tees, contractions, etc) in
pipe flows also contribute to the total head loss of the system. Their contributions
are generally termed minor losses.
The head losses and pressure drops can be characterized by using the loss coefficient,
KL, which is defined as
One of the example of minor losses is the entrance flow loss. A typical flow pattern
for flow entering a sharp-edged entrance is shown in the following page. A vena
contracta region is formed at the inlet because the fluid can not turn a sharp corner.
Flow separation and associated viscous effects will tend to decrease the flow energy;
the phenomenon is fairly complicated. To simplify the analysis, a head loss and the
associated loss coefficient are used in the extended Bernoulli’s equation to take into
consideration this effect as described in the next page.
Let us now also account for energy transfer via Heat Transfer, e.g. in a heat exchanger
The most general form of conservation of energy for a system can be written as: dE = dQ-dW where (Ch. 3, YAC)Energy Conservation (cont.)
U = mu, u(internal energy per unit mass),
KE = (1/2)mV2 and PE = mgz
Flow work Wflow= m (p/)
It is common practice to combine the total energy with flow work.
The difference between energy in and out is due to heat transfer (into or out)
and work done (by or on) the system.
Heat in, =dQ/dt
Work out dW/dt
Example: Superheated water vapor enters a steam turbine at a mass flow rate
1 kg/s and exhausting as saturated steam as shown. Heat loss from the turbine is
10 kW under the following operating condition. Determine the turbine power output.Conservation of Energy – Application
From superheated vapor tables:
100% saturated steam
From saturated steam tables: hout=2748.7 kJ/kg
– rate of total heat transfer (J/s, W)
q – heat transfer per unit mass (J/kg)
– Heat Flux, heat transfer per unit area (J/m2)
Q, q … ?!%