1. 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
Flow work + kinetic energy + potential energy = constant

2. Bernoulli’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. p A Dx
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

3. Bernoulli’s Equation (Cont)
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 1 H 2

4. 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. 1
First, choose the control volume as enclosed
by the dotted line. Specify h=h(t) as the water
level as a function of time.
h(t) 2 20 40 60 80
100 1 2 3 4
time (sec.)
water height (m)
2.5e-007 h ( ) t

5. Energy exchange (conservation) in a thermal system
Energy added, hA
(ex. pump, compressor)
Energy lost, hL
(ex. friction, valve, expansion)
Energy extracted, hE
(ex. turbine, windmill) hL
loss through
valves
heat exchanger hE hA
pump
turbine
hL, friction loss
through pipes hL
loss through
elbows
condenser

6. Energy conservation(cont.)
If energy is added, removed or lost via pumps turbines, friction, etc.then we use
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
6-in dia. pipe
4-in dia.pipe
Given: Q=300 gal/min=0.667 ft3/s=AV
V1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/s
pump
Z=15 in 1 2 zo
Looking at the pressure term:
Mercury (m=844.9 lb/ft3)
water (w=62.4 lb/ft3)
1 hp=550 lb-ft/s

7. Energy conservation (cont.)
Example (cont.)

8. Frictional losses in piping system
R: radius, D: diameter
L: pipe length
w: wall shear stress P2
Consider a laminar, fully developed circular pipe flow w p
P+dp
Darcy’s Equation:

11. Losses in Pipe Flows
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
Colebrook equation.
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.

12. Minor Loss through flow entranceV1 V2 V3
(1/2)V32
(1/2)V22
KL(1/2)V32
pp

13. Energy Conservation (cont.)
We will now also account for energy tranfer via Heat Transfer
The general most general form of conservation of energy for a system can be written as: dE = dQ-dW where (Ch. 3, YAC)
dE  Change in Total Energy, E
and E = U(internal energy)+Em(mechanical energy) (Ch. 1 YAC)
E = U + KE (kinetic energy) + PE(potential energy)
dW  Work done by the system where
W = Wext(external work) + Wflow(flow work)
dQ = Heat transfer into the system (via conduction, convection & radiation)
Convention: dQ > 0 net heat transfer into the system (Convention Q,q..)
dW > 0, positive work done by the system
Q: What is Internal Energy ?

14. 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 or flow work.
Thus:
The difference between energy in and out is due to heat transfer (into or out)
and work done (by or on) the system.

15. Energy Conservation (cont.)
Heat in, =dQ/dt
Work out dW/dt

16. Conservation of Energy – Application
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.
From superheated vapor tables:
hin= kJ/kg
P=1.4 Mpa
T=350 C
V=80 m/s
z=10 m
10 kw
P=0.5 Mpa
100% saturated steam
V=50 m/s
z=5 m
From saturated steam tables: hout= kJ/kg

17. Q, q … ?!%
Q – total heat transfer (J)
– 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)
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Internal Energy ?
Internal energy, U (total) or u (per unit mass) is the sum of all
microscopic forms of energy.
It can be viewed as the sum of the kinetic and potential energies of the molecules
Due to the vibrational, translational and rotational energies of the moelcules.
Proprtional to the temperature of the gas.
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