Dr . Ercan Kahya
Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved .

General Energy Consideration
E
T
 α
1
V
1
2
2g

P
1
γ
 z
1
 α
2
V
2
2
2g

P
2
γ
 z
2
 h
P
 h f
 h t
Z: the position
P/ g: the pressure head
V 2 /2g: the velocity head h p
: head supplied by pump h t
:head given up to a turbine h f
: head loss
• local losses (bends, expansions, valves)
• frictional losses (function of pipe type, length) a : V elocity coefficient and can be set to unity for regular & symmetrical cross-section like pipe

Bernoulli vs. Energy
V
1
2
2g

P
1
γ
 z
1

V
2
2
2g

P
2
γ
 z
2
Z is the position
P/ g is the pressure head
V 2 /2g is the velocity head
Relates velocity and piezometric pressure along a streamline, steady, incompressible, inviscid flow.
E
T
 α
1
V
1
2
2g

P
1
γ
 z
1
 α
2
V
2
2
2g

P
2
γ
 z
2
 h
P
 h f
 h t
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

Q

W
• 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 = E k
+ E p
+ E u dE

Q



W dt
Involves sign convention:
+ heat transferred to the system
+ work done by the system
- heat transferred from the system
‐ work done on the system

First Law of Thermodynamics

Derivation of Energy Equation
Reynolds Transport Theorem applied to First Law of
Thermodynamics e
 e k
 e p
 u
E : extensive property of the system e : intensive (energy per unit mass) dE dt
 d dt cv
 e
 dV
 cs
 e

V .
dA

Q


W
 d dt cv
 e
 dV
 cs
 e

V .
dA e e p k


V
2
2 gz u
 internal energy

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 A
Work = F
 l
= (pA) (V
 t )

W f
  cs pV .
A
At section 2, work rate done on surrounding fluid is
At section 1, work rate done by surrounding fluid is
V
→ - V
2
1 p
2 p
1
A
2
A
1

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

Reynolds Transport Theorem : Simplified form
If the flow crossing the control surface occurs through a number of inlet and outlet ports, and the velocity v is uniformly distributed (constant) across each port; then

Q

W

 
 m o
(
V o
2
2
 gz o
 h o
)
 
 m i
(
V i
2
2
 gz i
 h i
)
Steady-Flow Energy Equation
.
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)

Example 7.2:
If the pipe is 20cm and the rate of flow 0.06m
3 /s, what is the pressure in the pipe at L=2000m? Assume h l
=0.02(L/D)V 2 /2g
α
1
V
1
2
2g

P
1
γ
 z
1
 h p
 α
2
V
2
2
2g

P
2
γ
 z
2
 h t
 h l
This energy equation assumes steady flow & constant density

Power Equation
Let’s relate “head” to “power & efficiency”
Pump power:

W p
 g
Qh p

 m gh p
Power delivered to turbine:

W t
 g
Qh t

 m gh t
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.
 
P output
P input
If mechanical efficiency of the turbine is η t , the output power supplied by the turbine:

W s
 
 t
W t

Example 7.4: Power produced by a turbine
α
1
Discharge Q = 14.1 m 3 /s ; Elevation drop = 61 m
Total head loss = 1.5 m ; Efficiency = 87% Power = ?
V
1
2g
2

P
1
γ
 z
1
 h p
 α
2
V
2
2g
2

P
2
γ
 z
2
 h t
 h l
Evaluations:
V 1 = V 2 = 0 p 1 = p 2 = 0 z 1 -z 2 = 61m h t
= (z
1
-z
2
) - h
L
= 61 – 1.5 = 59.5 m
Power equation:
P input
 to
 turbine
 g
Qh t
= (9810 N/m3) (14.1 m3/s) (59.5m)
= 8.23 MW
Efficiency equation:
P output
 from
 generator
 
P input
 to
 turbine
= 0.87(8.23 MW) = 7.16 MW

Application of the Energy, Momentum and
Continuity Principles in Combination
α
1
V
1
2
2g

P
1
γ
 z
1
 α
2
V
2
2
2g

P
2
γ
 z
2
 h f

Fs

 m V
2

 m V
1
Neglecting the force due to shear stress p
1
A
2
 p
2
A
2
 γA
2
LSin α  ρ Q (V
2

V
1
)
Sudden expansion head loss h f


V
1

V
2

2
2 g

Example 7.5: Force on a contraction in a pipe
Find horizontal force which is required to hold the contraction in place if
P
1
=250kPa ; Q=0.707m
3 /s & head loss through the contraction h f

0 .
1
V
2
2
2 g
Assume α
1
= α
2 = 0 (kinetic energy correction factor)

SOLUTIONS:
Q , p
1
, V
1 and V
2
: known
Fx and p
2
: unknown
To obtain unknown p
2: p
1
A
1
 p
2
A
2

F x

 m
2
V
2

 m
1
V
1
α
1
V
1
2
2g

P
1
γ
 z
1
 α
2
V
2
2
2g

P
2
γ
 z
2
 h f
Q

A
1
V
1

A
2
V
2

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 reservoir surface 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