CHAPTER 7

ENERGY PRINCIPLE

Dr . Ercan Kahya

Engineering Fluid Mechanics 8/E by Crowe, Elger, and RobersonCopyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

General Energy Consideration

: Velocity coefficient and can be set to unity for regular & symmetrical cross-section like pipe

hp: head supplied by pumpht :head given up to a turbinehf : head loss• local losses (bends, expansions, valves)• frictional losses (function of pipe type, length)

tfP22

22

211

21

1T hhhzγ

P

2g

Vαz

γ

P

2g

VαE

Z: the positionP/ the pressure headV2/2g: the velocity head

Bernoulli vs. Energy

Z is the positionP/ is the pressure headV2/2g is the velocity head

22

22

11

21 z

γ

P

2g

Vz

γ

P

2g

V

Relates velocity and piezometric pressure along a streamline, steady, incompressible, inviscid flow.

tfP22

22

211

21

1T hhhzγ

P

2g

Vαz

γ

P

2g

VαE

Relates energy at two points for viscous, incompressible flow in a pipe, with accounting for additional energy addition / extraction

Energy Principle

• So far, mechanical forces on a fluid

– Pressure– Gravity– Shear Stress

• Considering Energy, we can solve:

– Power required to move fluids– Effects of pipe friction– Flow rates of fluids moving through pipes & orifices– Effects of obstacles, bends, and valves on flow

First Law of Thermodynamics

• E = energy of a system• Q = heat transferred to a system in a given time t• W = work done by the system on its surroundings during the same time

• Energy forms: Kinetic and Potential energy of a system as a whole and energy associated with motion of the molecules (atomic structure, chemical energy, electrical energy)

E = Ek + Ep + Eu

WQE

WQdt

dE + heat transferred to the system+ work done by the system- heat transferred from the system‐ work done on the systemInvolves sign convention:

First Law of Thermodynamics

Derivation of Energy Equation

cscv

dAVedVedt

d

dt

dE.

Reynolds Transport Theorem applied to First Law of Thermodynamics

E: extensive property of the systeme: intensive (energy per unit mass)

cscv

dAVedVedt

dWQ .

ueee pk

energy internalu

e

2

p

2

gz

Vek

Flow Work

Work is classified as: (work) = (flow work) +(shaft work)

Flow Work: Work done by pressure forces as the system moves through space

Force (F) = p AWork = F l = (pA) (Vt)

ApVWcs

f .

At section 2, work rate done on surrounding fluid is → V2 p2 A2

At section 1, work rate done by surrounding fluid is → - V1 p1 A1

Shaft Work (any work not associated with a pressure force!)

• Work done on flow by a pump– increases the energy of the system, thus the work is negative

• Work done by flow on a turbine– decreases the energy of the system, thus the work is positive

)2

()2

(22

iii

iooo

o hgzV

mhgzV

mWQ

.Q = rate of heat transfer TO the system (input) .W = rate of work transfer FROM the system (output) .m = rate of mass flow

h = specific enthalpy (h = u + p/ρ)

Steady-Flow Energy Equation

If the flow crossing the control surface occurs through a number of inlet andoutlet ports, and the velocity v is uniformly distributed (constant) across each port; then

Reynolds Transport Theorem: Simplified form

Example 7.2:If the pipe is 20cm and the rate of flow 0.06m3/s, what is the pressure in the pipe at L=2000m? Assume hl=0.02(L/D)V2/2g

lp hh t22

22

211

21

1 hzγ

P

2g

Vαz

γ

P

2g

Vα

This energy equation assumes steady flow & constant density

Power Equation

ppp ghmQhW

Both pump & turbine lose energy due to friction which is accounted for by the “efficiency” defined as the ratio of power output to power input.

Let’s relate “head” to “power & efficiency”

Pump power:

Power delivered to turbine: ttt ghmQhW

input

output

P

P

tts WW If mechanical efficiency of the turbine is ηt ,

the output power supplied by the turbine:

Example 7.4: Power produced by a turbine

Discharge Q = 14.1 m3/s ; Elevation drop = 61 m Total head loss = 1.5 m ; Efficiency = 87% Power = ?

lp hh t22

22

211

21

1 hzγ

P

2g

Vαz

γ

P

2g

Vα

Evaluations: V1 = V2 = 0 p1 = p2 = 0 z1-z2 = 61m

ht = (z1-z2) - hL = 61 – 1.5 = 59.5 m

Power equation:

tturbinetoinput QhP

= (9810 N/m3) (14.1 m3/s) (59.5m)= 8.23 MW

Efficiency equation:

turbinetoinputgeneratorfromoutput PP

= 0.87(8.23 MW) = 7.16 MW

Application of the Energy, Momentum and Continuity Principles in Combination

f22

22

211

21

1 hzγ

P

2g

Vαz

γ

P

2g

Vα

12 VmVmFs

Neglecting the force due to shear stress

)(VραγAApAp 1222221 VQLSin

g2

VVh

221

f

Sudden expansion head loss

Example 7.5: Force on a contraction in a pipe

Find horizontal force which is required to hold the contraction in place if P1=250kPa ; Q=0.707m3/s & head loss through the contraction

g2

V1.0h

22

f

Assume α1= α2 = 0 (kinetic energy correction factor)

SOLUTIONS:

11222211 VmVmFApAp x

f22

22

211

21

1 hzγ

P

2g

Vαz

γ

P

2g

Vα

2211 VAVAQ

To obtain unknown p2:

Q , p1, V1 and V2 : known

Fx and p2 : unknown

HYDRAULIC & ENERGY GRADE LINES

GRADE LINE INTERPRETATION - PUMP

GRADE LINE INTERPRETATION TURBINE

GRADE LINES - NOZZLE

GRADE LINES - PIPE DIAMETER CHANGE

GRADE LINES - SUB-ATMOSPHERIC PRESSURE

CLASS EXERCISE: Q7.32

Find the head loss btw the reservoirsurface and point C.

Assume that the head loss btw the reservoir surface and point B is three quater of the total head loss.

CLASS EXERCISE: Q7.36

CLASS EXERCISE: Q7.60

CLASS EXERCISE: Q7.71