Fluid Mechanics 2A (STR2A11) Date: 3/3/2017
FLUID MECHANICS 2A
STR2A11
General Energy Equation
CHAPTER 7
By: Thabang Mafokoane
tmafokoane@uj.ac.za
OBJECTIVES
1. Identify conditions under which energy losses
occur, and the means by which energy can be
added or removed from a fluid flow system
2. Expand Bernoulli’s eq. to form the general
energy equation
3. Apply the general energy equation to fluid flow
systems
4. Compute the power added to a fluid by pumps,
the power required to drive the pumps, and
efficiency of pumps
STR2A11 (2017)
By Thabang Mafokoane 1
Fluid Mechanics 2A (STR2A11) Date: 3/3/2017
OUTLINE
• Section 7.2 Energy losses and additions
• Section 7.3 Nomenclature of energy losses and
additions
• Section 7.4 General energy equation
• Section 7.5 Power required by pumps
STR2A11 (2017)
Energy Losses & Additions (7.2)
There are various types of devices and components
of fluid flow system that add energy, remove energy
or cause energy losses
1. Pumps
• Add energy
• Power device or electric motor drives
rotating shaft → delivers KE into the
fluid → causing fluid flow & increased
fluid pressure
STR2A11 (2017)
By Thabang Mafokoane 2
Fluid Mechanics 2A (STR2A11) Date: 3/3/2017
Energy Losses & Additions (7.2)
2. Fluid Motors
• Also turbines, rotary and linear actuators
• Take energy from a fluid → deliver it in the form of work →
causing rotation of shaft or linear movement of a piston
STR2A11 (2017)
Energy Losses & Additions (7.2)
3. Fluid Friction
• Fluid in motion offers friction resistance to flow
• Part of energy in system converted to thermal energy (heat)
• Dissipated through walls of pipe
• Magnitude of energy loss dependent on:
Properties of fluid
Flow velocity
Pipe size and length
Smoothness of the pipe wall
4. Valves and Fittings
• Control direction or flow rate of a fluid
• Cause local turbulence in fluid → energy dissipation
as heat
• These energy losses are called minor losses
STR2A11 (2017)
By Thabang Mafokoane 3
Fluid Mechanics 2A (STR2A11) Date: 3/3/2017
Nomenclature (7.3)
• Energy per unit weight → head → h
• hA → Energy added to the fluid → pumps
• hR → Energy removed from the fluid → fluid motor
• hL → Energy losses from the system → friction in
pipes or minor losses due to valves &
fittings
• Energy losses due to friction, valves and fittings is
directly proportional to velocity head
hL = K * (v2 / 2g)
K → resistance coefficient (Chp 8 & 10)
STR2A11 (2017)
General Energy Equation (7.4)
E1 + hA - hL - hR = E2
P1/γ + z1+v12/2g + hA - hR - hL = P2/γ + z2+v22/2g
Remember: Units meters (m)
Write equation in the direction of flow
STR2A11 (2017)
Sings are critical
By Thabang Mafokoane 4
Fluid Mechanics 2A (STR2A11) Date: 3/3/2017
Applications (6.8) 7.4
Procedure for applying Bernoulli’s equation:
1. Objective(s): what is to be found and what is given
2. Decide on 2 sections (1 know data – 1 unknown data)
General Energy
3. Write Bernoulli’s equation in the direction of flow
4. Note labelling of pressure, elevation and velocity head
(note where the reference points are on sketch)
5. Cancel terms and remove zero terms
6. Solve algebraically (without numbers)
7. Substitute known quantities and calculate the result
Remember: use consistent units
STR2A11 (2017)
Example 7.1 (SI)
Water flows from a large reservoir at the
rate of 0.034 m3/s through a pipe system as
shown in Fig. 7.7. Calculate the total amount
of energy lost from the system because of
the valve, the elbows, the pipe entrance,
and fluid friction.
hL = 4.707 m
STR2A11 (2017)
By Thabang Mafokoane 5
Fluid Mechanics 2A (STR2A11) Date: 3/3/2017
Example 7.2
The volume flow rate through the pump shown in Fig 7.8 is 0.014
m3/s. The fluid being pumped is oil with a specific gravity
of 0.86. Calculate the energy delivered by the
pump to the oil per unit weight of
oil flowing in the system. Energy
losses in the system are caused
by the check value and friction
losses as the fluid flows through
the piping. The magnitude of
such losses has been determined
to be 1.86 N.m/N.
hA = 42.9 m
STR2A11 (2017)
Power Required by Pumps (7.5)
• Power → the rate of doing work.
Fluid mechanics:
• Power → the rate at which energy is being
transferred
• Units → Watt (W) = 1 N.m/s or 1 J/s
• Power calculated by multiplying energy transferred
per newton of fluid by weight flow rate
PA = hAW = hAγQ
STR2A11 (2017)
By Thabang Mafokoane 6
Fluid Mechanics 2A (STR2A11) Date: 3/3/2017
Power Required by Pumps (7.5)
• Pump Efficiency → ratio of power delivered to the
fluid to power supplied to pump
Power delivered to fluid
• eM = =
Power put into the pump
• eM is always less then 1
STR2A11 (2017)
Example 7.3 (SI)
For the pump test arrangement shown in Figure 7.9, determine
the mechanical efficiency of the pump if the power input is
measured to be 2.87 kW when pumping 125 m 3/h of oil (γ = 8.8
kN/m3).
eM = 86 %
Section 7.6: Self Study
STR2A11 (2017)
By Thabang Mafokoane 7
