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Energy Conservation (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

Dx

p

A

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

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

4

3

water height (m)

h

(

t

)

2

1

2.5e-007

0

0

20

40

60

80

100

0

t

100

time (sec.)

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

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

2

1

pump

zo

Z=15 in

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

Given: Q=300 gal/min=0.667 ft3/s=AV

V1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/s

4-in dia.pipe

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

Example (cont.)

P1

R: radius, D: diameter

L: pipe length

w: wall shear stress

P2

Consider a laminar, fully developed circular pipe flow

w

P+dp

p

Darcy’s Equation:

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.

V1

V2

V3

(1/2)V32

(1/2)V22

KL(1/2)V32

pp

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)

- 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 (Symbols Q,q..)
- dW > 0, positive work done by the system
- Q: What is Internal Energy ?

mechanical energy

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.

Thus:

The difference between energy in and out is due to heat transfer (into or out)

and work done (by or on) the system.

- Hence, a system exchanges energy with the environment due to:
- Flow in/out 2) Heat Transfer, Q and 3) Work, W
- This energy exchange is governed by the First Law of Thermodynamics

Heat in, =dQ/dt

system

Work out dW/dt

“Enthalpy”

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=3149.5 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=2748.7 kJ/kg

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)

Q, q … ?!%

Back

- 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.
- Proportional to the temperature of the gas.

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