Operator Generic Fundamentals
Thermodynamics – Fluid Statics and
Dynamics
© Copyright 2014
Operator Generic Fundamentals
2
Fluid Statics and Dynamics Introduction
• This module provides the student with an understanding of:
– Fluid system fundamental properties
– Conservation of mass
– Continuity equation
– Application of Bernoulli’s equation
• Discussions include:
– Two-phase flow, including steam generator recirculation and
water hammers.
– Causes and methods for water hammer prevention.
© Copyright 2014
Introduction
Operator Generic Fundamentals
3
Terminal Learning Objectives
At the completion of this training session, the trainee will demonstrate
mastery of this topic by passing a written exam with a grade of 80
percent or higher on the following Terminal Learning Objectives (TLOs):
1. Explain the fundamental properties of fluids and closed fluid
systems.
2. Describe laminar and turbulent flows.
3. Explain Bernoulli’s equation then use it to determine fluid properties
at different points in a system.
4. Explain two-phase flow and water hammers, including the
problems they can cause in plant operation, and the means of
controlling them.
© Copyright 2014
Introduction
Operator Generic Fundamentals
4
Fundamental Properties of Fluids
TLO 1 – Explain the fundamental properties of fluids and closed fluid
systems.
• The study of fluid behaviors and characteristics are important for the
analysis and understanding of fluid system responses.
• The next two sections address fundamental properties and analysis
of incompressible fluids in simple closed system applications.
© Copyright 2014
TLO 1
Operator Generic Fundamentals
5
Enabling Learning Objectives for TLO 1
1. Explain the following terms:
a. Buoyancy
b. Static pressure
c. Dynamic pressure
d. Total pressure
2. Explain the following:
a. Pascal’s Law
b. Mass flow rate
c. Volumetric flow rate
d. Density compensation
e. Conservation of mass
f. Continuity equation
© Copyright 2014
TLO 1
Operator Generic Fundamentals
6
Fundamental Properties of Fluids
ELO 1.1 – Explain the following terms: buoyancy, static pressure,
dynamic pressure, and total pressure.
• A fluid is any substance that flows freely and its molecules lack rigid
attachment within a crystalline structure.
– Liquids, gases, and materials normally considered solids, such as
glass.
• Incompressible or compressible describe fluid properties.
– Maintain a constant density with changes in pressure; for
example, liquids.
© Copyright 2014
ELO 1.1
Operator Generic Fundamentals
7
Fundamental Properties of Fluids
• Compressible fluids like gases change density easily.
– Gases expand and completely fill their container.
• Properties of fluids affect (or measures) the way in which the fluid
behaves.
– Temperature
– Pressure
– Mass
– Specific volume
– Density
© Copyright 2014
ELO 1.1
Operator Generic Fundamentals
8
Fundamental Properties of Fluids
Buoyancy:
• Describes how an object responds when placed in a fluid.
– Our bodies float in water, as do wood and ice.
– A rock underwater is lighter than when removed from the water.
• Ships rely on buoyant force to float.
• The Greek philosopher Archimedes calculated buoyant forces.
– An object placed in a fluid is buoyed by a force equal to the
weight of the fluid it displaces.
• An object weighing more than the liquid it displaces sinks, but loses
an amount of weight equal to the displaced liquid.
• A floating object displaces its own weight in the fluid in which it floats.
© Copyright 2014
ELO 1.1
Operator Generic Fundamentals
9
Fundamental Properties of Fluids
Static Pressure:
• A confined, non-moving fluid is a static fluid.
• Static pressure exerted due to presence of a substance and the
random movement of its molecules (PV Energy).
• Pressure exerted by a static fluid is the force exerted by the fluid on
the walls of the container per unit area.
Dynamic Pressure:
• The pressure due to flow is called the dynamic pressure (Kinetic
Energy).
• Fluid flow is a dynamic event.
Total Pressure:
• Total pressure is the combination of the static pressure and the
dynamic pressure.
© Copyright 2014
ELO 1.1
Operator Generic Fundamentals
10
Fundamental Properties of Fluids
Knowledge Check
How a body responds when placed in a fluid is ...
A. buoyancy
B. static pressure
C. dynamic pressure
D. total pressure
Correct answer is A.
© Copyright 2014
ELO 1.1
Operator Generic Fundamentals
11
Closed System Fluid Properties
ELO 1.2 – Explain the following: Pascal’s Law, mass flow rate, volumetric
flow rate, density compensation, conservation of mass, and continuity
equation.
• To understand fluid response in complex open boundary systems first
requires a basic understanding of:
– Fluid characteristics within a closed system
– Terms describing quantities and rates of fluid flow
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
12
Pascal’s Law
Pressure applied to a confined fluid transmits undiminished
throughout the confining vessel of the system.
• All pistons have same crosssectional area
• Piston A exerting 50 lbf
• Based on Pascal’s law, this
pressure equal throughout
container
• Piston B, at top has 50 lbf
applied from force of piston A
• Piston C has 50 lbf plus 10 lbf
force from weight of water above
• Pistons D and E have 50 lbf
force plus weight of water based
on location
© Copyright 2014
ELO 1.2
Figure: Pascal's Law
Operator Generic Fundamentals
13
Closed System Fluid Properties
Volumetric Flow Rate:
• The volumetric flow rate of a fluid system is a measure of the volume
of fluid passing a point in the system per unit of time.
• It is the product of the cross-sectional area (A) for flow and the
average flow velocity (v).
𝑉 = 𝐴𝑣
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
14
Closed System Fluid Properties
Example: Volumetric Flow Rate
• A pipe with an inner diameter of 4 inches contains water that flows at
an average velocity of 14 feet per second. Calculate the volumetric
flow rate of water in the pipe.
• Solution:
𝑉 = 𝜋𝑟 2 𝑣
𝑉 = 3.14
2
𝑓𝑡
12
2
14
𝑓𝑡
𝑠𝑒𝑐
𝑓𝑡 3
𝑉 = 1.22
𝑠𝑒𝑐
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
15
Closed System Fluid Properties
Volumetric Flow Rate and Pressure Relationship:
• Velocity is a function of the square root of the change in pressure
drop.
• Doubling velocity doubles the flow rate, which quadruples pressure
drop.
• This relationship holds well for water, but less accurately for gases.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
16
Closed System Fluid Properties
Example:
A 20 gpm leak to atmosphere has developed from a cooling water
system that is operating at 200 psig. If pressure decreases to 50 psig,
what is the new leak rate?
Solution:
The change in pressure is now a quarter of what it was. The square
root of .25 = .5. Therefore, flow (and velocity) will decrease by half or to
10 gpm.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
17
Closed System Fluid Properties
Mass Flow Rate:
• A measure of the mass of fluid passing a point in the system per unit
time.
• Mass flow rate relates to the volumetric flow rate:
𝑚 = 𝜌𝑉
Where:
ρ = density of the fluid.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
18
Closed System Fluid Properties
Mass Flow Rate:
• If the volumetric flow rate is in cubic feet per second (ft3/sec) and the
density is in pounds-mass per cubic foot (lbm/ft3)
– Mass flow rate measured in pounds-mass per second.
• A common term used for mass flow rate is pounds-mass per hour
(lbm/hr).
𝑚 = 𝜌𝐴𝑣
• Value frequently used for the density of water is 62.4
𝑙𝑏𝑚
;
𝑓𝑡 3
however,
density depends on pressure and temperature.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
19
Example: Mass Flow Rate
With a volumetric flow rate of 1.22 ft3/sec, calculate the mass flow rate.
Assume a density of 62.44 lbm/ft3.
Solution:
𝑚 = 𝜌𝑉
𝑙𝑏𝑚
𝑚 = 62.44 3
𝑓𝑡
𝑙𝑏𝑚
𝑚 = 76.2
𝑠𝑒𝑐
© Copyright 2014
𝑓𝑡 3
1.22
𝑠𝑒𝑐
ELO 1.2
Operator Generic Fundamentals
20
Density Compensation
• Mass flow rate calculations are density dependent.
– If the instrument measures volume, density compensation is
unnecessary since volume is not density related.
• When compensation for density is required, it is usually performed
electronically.
– An example is steam flow (lbm/hour) electronically corrected by a
signal input from steam pressure.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
21
Density Compensation
• As the temperature of the fluid increases, density decreases and
specific volume increases (inverse of density).
• Incompressible liquids will not change significantly in density with
pressure change.
– But do change in density with a change in temperature
• Density of a vapor, steam, for example, or gas DOES change
significantly with temperature and pressure.
– Steam flow compensation is an example of vapor density.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
22
Density Compensation
• The relationship of density to mass flow rate is directly proportional.
• If the density increases, assuming a constant indication of mass flow
rate, the actual mass flow rate increases.
– A density decrease results in lower actual flow.
• However, temperature and density are inversely related.
– An increase in temperature results in a decrease in density, and a
decrease in actual mass flow rate.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
23
Conservation of Mass
The basic principles of fluid flow:
1. Principle of momentum (leads to equations of fluid forces).
2. Conservation of energy (leads to the First Law of
Thermodynamics).
3. Conservation of mass (leads to the continuity equation).
Conservation of mass states for any system closed to all transfers of
matter and energy, the mass must remain constant over time, as
system mass cannot change quantities unless added or removed.
– Therefore, mass is conserved.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
24
Conservation of Mass
• Regardless of the type of the flow, all flow situations are subject to
the established basic laws of nature as expressed in equation form.
• Conservation of mass and conservation of energy are always
satisfied in fluid problems, along with Newton’s laws of motion.
• Each problem also has physical constraints, referred to
mathematically as boundary conditions, that must be satisfied before
a solution to the problem is consistent with the physical results.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
25
Continuity Equation
• Conservation of mass states that all mass flow rates into a control
volume equal all mass flow rates out of the control volume plus the
rate of change of mass within the control volume.
𝑚𝑖𝑛
∆𝑚
= 𝑚𝑜𝑢𝑡 +
∆𝑡
• Where:
∆𝑚 = increase or decrease of the mass within the control volume
∆𝑡 over a specified time period
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
26
Continuity Equation
• The continuity equation is a mathematical expression of the
conservation of mass principle.
• For a control volume that has a single inlet and a single outlet, the
mass flow rate into the volume must equal the mass flow rate out.
𝑚𝑖𝑛𝑙𝑒𝑡 = 𝑚𝑜𝑢𝑡𝑙𝑒𝑡
(𝜌𝐴𝑣)𝑖𝑛𝑙𝑒𝑡 = (𝜌𝐴𝑣)𝑜𝑢𝑡𝑙𝑒𝑡
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
27
Continuity Equation
• For a control volume with multiple inlets and outlets, the principle of
conservation of mass requires that the sum of the mass flow rates
into the control volume equals the sum of the mass flow rates out of
the control volume.
𝑚𝑖𝑛𝑙𝑒𝑡𝑠 =
© Copyright 2014
ELO 1.2
𝑚𝑜𝑢𝑡𝑙𝑒𝑡𝑠
Operator Generic Fundamentals
28
Continuity Equation – Piping Expansion
Example # 1:
• Steady-state flow exists in a pipe that undergoes a gradual expansion
from a diameter of 6 inches to a diameter of 8 inches.
• The density of the fluid in the pipe is constant at 60.8 lbm/ft3.
If the flow velocity is 22.4 ft/sec in the 6 inch section, what is the flow
velocity in the 8 inch section?
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
29
Solution: Continuity Equation – Piping
Expansion
• From the continuity equation, the mass flow rate in the 6 inch section
must equal the mass flow rate in the 8 inch section.
𝑚1 = 𝑚2
𝜌1 𝐴1 𝑣1 = 𝜌2 𝐴2 𝑣2
𝑣1 𝐴1 𝜌1
𝑣2 =
𝐴2 𝜌2
𝑣1 𝐴1 𝜌1
𝑣2 =
𝐴2 𝜌
𝜋𝑟1 2
𝑣2 = 𝑣1
𝜋𝑟2 2
𝑓𝑡 3 𝑖𝑛
𝑣2 = 22.4
𝑠𝑒𝑐 4 𝑖𝑛
𝑓𝑡
𝑣2 = 12.6
𝑠𝑒𝑐
© Copyright 2014
The increase in pipe diameter from 6
inches to 8 inches caused a decrease
in flow velocity from 22.4 to 12.6 ft/sec
2
2
ELO 1.2
Operator Generic Fundamentals
30
Continuity Equation – Centrifugal Pump
Example:
• The inlet diameter of the cooling pump shown in the figure below is
28 inches.
• The outlet flow through the pump is 9,200 lbm/second. The density
of the water is 49 lbm/ft3.
What is the velocity at the pump inlet?
Figure: Centrifugal Pump in Continuity Equation
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
31
Solution: Continuity Equation - Centrifugal
2
Pump
1 𝑓𝑡
2
𝐴𝑖𝑛𝑙𝑒𝑡 = 𝜋𝑟 = 3.14
14 𝑖𝑛
12 𝑖𝑛
= 4.28 𝑓𝑡 2
𝑙𝑏𝑚
𝑚𝑖𝑛𝑙𝑒𝑡 = 𝑚𝑜𝑢𝑡𝑙𝑒𝑡 = 9,200
𝑠𝑒𝑐
𝑙𝑏𝑚
(𝜌𝐴𝑣)𝑖𝑛𝑙𝑒𝑡 = 9,200
𝑠𝑒𝑐
𝑙𝑏𝑚
9,200
𝑠𝑒𝑐
𝑣𝑖𝑛𝑙𝑒𝑡 =
𝐴𝜌
𝑙𝑏𝑚
9,200
𝑠𝑒𝑐
=
𝑙𝑏𝑚
4.28 𝑓𝑡 2 49 3
𝑓𝑡
𝑓𝑡
𝑣𝑖𝑛𝑙𝑒𝑡 = 43.9
𝑠𝑒𝑐
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
32
Continuity Equation – Multiple Outlets
Example:
• A piping system has a Y configuration for separating the flow, shown
below in the figure. The diameter of the inlet leg is 12 in., and the
diameters of the outlet legs are 8 inches and 10 inches. The density
of water is 62.4 lbm/ft3.
• The velocity in the 10 inch leg is 10 ft/sec. The flow through the main
portion is 500 lbm/sec.
What is the velocity out of
the 8 inch pipe section?
Figure: Y Configuration Example
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
33
Solution: Continuity Equation – Multiple
Outlets
1 𝑓𝑡
𝐴8 = 𝜋 4 𝑖𝑛
12 𝑖𝑛
2
𝐴10
2
= 0.545 𝑓𝑡 2
𝑚𝑖𝑛𝑙𝑒𝑡𝑠 =
8
= 𝑚12 − 𝜌𝐴𝑣
𝑚12 − 𝜌𝐴𝑣
𝑣8 =
𝜌𝐴 8
= 0.349 𝑓𝑡 2
1 𝑓𝑡
= 𝜋 5 𝑖𝑛
12 𝑖𝑛
𝜌𝐴𝑣
𝑚𝑜𝑢𝑡𝑙𝑒𝑡𝑠
10
10
𝑙𝑏𝑚
𝑙𝑏𝑚
500 𝑠𝑒𝑐 − 62.4 3 0.545 𝑓𝑡 2
𝑓𝑡
=
𝑙𝑏𝑚
62.4 3 0.349 𝑓𝑡 2
𝑓𝑡
𝑓𝑡
10 𝑠𝑒𝑐
𝑓𝑡
𝑣8 = 7.3
𝑠𝑒𝑐
𝑚12 = 𝑚10 + 𝑚8
𝑚8 = 𝑚12 − 𝑚10
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
34
Fluid Properties of a Closed System
Knowledge Check – NRC Bank
A heat exchanger has the following initial cooling water inlet temperature and differential
pressure (ΔP) parameters:
– Inlet Temperature = 70 °F
– Heat Exchanger ΔP = 10 psi
Six hours later, the current heat exchanger cooling water parameters are:
– Inlet Temperature = 85 °F
– Heat Exchanger ΔP = 10 psi
In comparison to the initial cooling water mass flow rate, the current mass flow rate is...
A.
lower, because the density of the cooling water has decreased
B.
higher, because the velocity of the cooling water has increased
C.
the same, because the changes in cooling water velocity and density offset
D.
The same, because the heat exchanger cooling water ΔP is the same
Correct Answer is A.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
35
TLO 1 Summary
1.
Fundamental properties of fluids include buoyancy, compressibility,
static pressure, dynamic pressure, and total pressure.
– Buoyancy describes how an object responds when placed in a fluid.
o While underwater, a rock is lighter than when it removed from the
water.
o Boats rely on buoyant force to float.
o Archimedes calculated buoyant forces; an object placed in a fluid,
is buoyed up by a force equal to the weight of the fluid it displaces.
– Static Pressure - a confined, non-moving fluid is static fluid.
o Static pressure is the pressure exerted due to the presence of a
substance and the random movement of its molecules (PV
Energy).
o Pressure exerted by a static fluid is the force exerted by the fluid
on the wall of the container per unit area.
© Copyright 2014
TLO 1
Operator Generic Fundamentals
36
TLO 1 Summary
– Dynamic Pressure - pressure due to flow is called the dynamic
pressure (kinetic energy). Fluid flow is a dynamic event.
– Total Pressure - the combination of the static pressure and the
dynamic pressure.
2.
Fluid properties in a closed system help describe properties in an
open system.
– Pascal’s law states that pressure applied to a confined fluid is
transmitted undiminished throughout the confining vessel of a
system.
– Volumetric flow rate is the volume of fluid per unit of time passing a
specific point in a fluid system.
o Calculated by the product of the average fluid velocity and the
cross-sectional area for flow.
– Velocity is a function of the square root of the change in pressure
drop; doubling velocity doubles the flow rate, which quadruples the
pressure drop.
© Copyright 2014
TLO 1
Operator Generic Fundamentals
37
TLO 1 Summary
– Mass flow rate is the mass of fluid per unit time passing a point in
a fluid system.
o Calculated by the product of the volumetric flow rate and the
fluid density.
– Density compensation - mass flow rate calculations are density
dependent.
– Density compensation is required when measuring mass flow
rates if the measured fluid has large changes in pressure or
temperature affecting actual mass flow values.
– If the instrument measures volume, density compensation is
unnecessary since volume is not density related.
– Incompressible liquids do not change significantly in density with
pressure change (only temperature)
– Vapor density, steam, for example, changes significantly with both
temperature and pressure.
© Copyright 2014
TLO 1
Operator Generic Fundamentals
38
TLO 1 Summary
– The relationship of density to mass flow rate is directly
proportional. An increase in temperature results in a decrease in
density and a decrease in actual mass flow rate.
– The principle of conservation of mass states that all mass flow
rates into a control volume equal all mass flow rates out of the
control volume plus the rate of change of mass within the control
volume.
– For a control volume with a single inlet and outlet, the continuity
equation can be expressed as follows:
𝑚𝑖𝑛 = 𝑚𝑜𝑢𝑡 +
© Copyright 2014
∆𝑚
∆𝑡
TLO 1
Operator Generic Fundamentals
39
TLO 1 Summary
– For a control volume with multiple inlets and outlets, the continuity
equation is:
𝑚𝑖𝑛𝑙𝑒𝑡𝑠 =
© Copyright 2014
𝑚𝑜𝑢𝑡𝑙𝑒𝑡𝑠
TLO 1
Operator Generic Fundamentals
40
TLO 1 Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Explain the following terms:
a. Buoyancy
b. Static pressure
c. Dynamic pressure
d. Total pressure
2. Explain the following:
a. Pascal’s Law
b. Mass flow rate
c. Volumetric flow rate
d. Density compensation
e. Conservation of mass
f. Continuity equation
© Copyright 2014
TLO 1
Operator Generic Fundamentals
41
Laminar and Turbulent Flows
TLO 2 – Describe laminar and turbulent flows.
• Fluid flow is either laminar or turbulent
– Each has different flow behavior and heat transfer characteristics.
• Some fluids are thick and some are thin in terms of how they resist
flow.
– Described as viscosity of the fluid, it contributes to the head loss
in a fluid system.
© Copyright 2014
TLO 2
Operator Generic Fundamentals
42
Enabling Learning Objectives for TLO 2
1. Describe the characteristics and flow velocity profiles of both
laminar flow and turbulent flow.
2. Explain the following terms:
a. Viscosity
b. Ideal fluid
© Copyright 2014
TLO 2
Operator Generic Fundamentals
43
Laminar and Turbulent Flow
ELO 2.1 – Describe the characteristics and flow velocity profiles of
laminar flow and turbulent flow.
• Laminar and turbulent flow are different.
• To understand the desirability of each of these flow types, it is
necessary to understand their characteristics
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
44
Flow Regimes
• Two categories or regimes classify fluid flow types.
– Laminar flow
– Turbulent flow
• The amount of fluid friction determines the amount of energy required
to maintain the desired flow, which depends on the mode of flow.
• Of concern are:
– Heat transfer to the fluid
– Energy required to create the desired flow
– Mixing within the fluid
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
45
Laminar Flow
• Laminar flow, also known as streamline or viscous flow, is descriptive
of this flow type.
• Layers of water flow over one another at different speeds, with
virtually no mixing between layers
• Fluid particles move in definite and observable paths or streamlines.
• Laminar flow is characteristic of viscous (thick) fluid, or is one in
which the fluid’s viscosity plays a significant role.
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
46
Turbulent Flow
• Characterized by irregular movement of particles of the fluid
• No definite frequency as with wave motion.
• Fluid particles travel in irregular paths with no observable pattern and
no distinct layers.
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
47
Flow Velocity Profiles
• Fluid particles do not travel at the same velocity within a pipe.
• The shape of the velocity profile depends on whether the flow is
laminar or turbulent.
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
48
Flow Velocity Profiles
Figure: Laminar and Turbulent Flow Velocity Profiles
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
49
Average (Bulk) Velocity
• In many fluid flow problems, instead of determining exact velocities at
different locations in the same flow cross-section, a single average
velocity is used to represent the velocity of the entire fluid flow in the
pipe.
– Simple for turbulent flow since the velocity profile is flat over the
majority of the pipe cross-section.
– Reasonable to assume that the average velocity is the same as
the velocity at the center of the pipe.
• If the flow is laminar, it is still necessary to represent the average
velocity at any given cross-section since fluid flow equations use an
average value.
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
50
Average (Bulk) Velocity
• It is important that the temperature profile within the fluid mimics the
velocity profile.
• The regime of fluid flow, whether laminar or turbulent, may be
determined from the Reynolds number.
– Laminar flow exists where the Reynolds number is less than
2,000.
– Turbulent flow exists with Reynolds numbers greater than 3,500.
– Transition flow identifies with values between these two numbers.
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
51
Laminar and Turbulent Flow
Knowledge Check
What type of flow is characteristic of viscous (thick) fluid or is one in
which viscosity of the fluid plays a significant part?
A. Laminar flow
B. Turbulent flow
C. Viscous flow
D. Reverse flow
Correct Answer is A.
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
52
Characteristics of Fluids
ELO 2.2 – Explain the terms viscosity and ideal fluid.
Fluids include gases and liquids. Fluids have certain characteristics
that are different from solids.
Certain characteristics of fluids help solve flow problems and assist in
understanding the requirements for system design. One of these
important characteristics is fluid viscosity.
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
53
Viscosity
• Viscosity is a fluid property that measures the resistance of the fluid
to deforming due to a shear force.
• The internal friction of fluids resists flow past a solid surface or other
layers of the fluid.
• A measure of the resistance of a fluid to flowing.
– Thick oil has a high viscosity
– Water has a low viscosity.
• Unit of measurement for viscosity:
µ = 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 (𝑙𝑏𝑓–𝑠𝑒𝑐/𝑓𝑡2).
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
54
Viscosity
• Usually depends on the temperature of the fluid and is relatively
independent of the pressure.
• As the temperature of most fluids increases, their viscosity decreases
• Lubricating oil of engines is an example of viscosity.
– When oil is cold, the oil is viscous or thick.
– At operating temperatures, viscosity of oil decreases significantly.
• On the other hand, gases increase in viscosity with increasing
temperatures.
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
55
Viscosity
Viscosity
(centipoises)
Karo® Corn Syrup or
Honey
2,000-3,000
Water at 70 degrees °F
1-5
Blackstrap Molasses
5,000-10,000
Blood or Kerosene
10
10,000-25,000
Antifreeze or Ethylene
Glycol
15
Hershey's®
Chocolate Syrup
Material
Heinz® Tomato
50,000-75,000
Ketchup or French's®
Mustard*
Motor Oil Society of
50-100
Automotive Engineers
(SAE) 10 or Corn Syrup
Motor Oil SAE 30 or
Maple Syrup
150-200
Motor Oil SAE 40 or
Castrol® Motor Oil
250-500
Motor Oil SAE 60 or
Glycerin
1,000-2,000
Tomato Paste or
Peanut Butter
150,000200,000
Crisco® Vegetable
Shortening or Lard
1,000,0002,000,000
Caulking Compound
5,000,00010,000,000
Window Putty
100,000,000
*Material is thixotropic: fluids thick when static, flow or become thin and less
viscous over time when shaken or agitated.
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
56
Ideal Fluid
• An ideal fluid is one that is incompressible and has no viscosity.
• Ideal fluids do not exist, but are sometimes useful to simplify the
problem’s fluid flow calculations.
• Bernoulli’s (simplified) equation applies only to fluid flow treated as
ideal, with no fluid friction.
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
57
Characteristics of Fluids
Knowledge Check
Viscosity is a fluid property that measures ...
A. a fluid’s density.
B. the resistance of a fluid to flowing.
C. specific volume of a fluid.
D. temperature times density of a fluid.
Correct Answer is B.
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
58
TLO 2 Summary
1.
Fluids have two flow regimes, laminar or turbulent.
– Laminar Flow
o Layers of fluid flow over one another at different speeds with virtually no
mixing between layers.
o Flow velocity profile for laminar flow in circular pipes is parabolic in
shape.
o Maximum flow in the center of the pipe; minimum flow at pipe walls
o Average flow velocity is approximately half of the maximum velocity.
– Turbulent Flow
o Irregular movement of particles of the fluid.
o Flow velocity profile for turbulent flow is fairly flat across the center
section of a pipe and drops rapidly extremely close to the walls.
o Average flow velocity approximately equals the velocity at the center of
the pipe and improves internal mixing of fluid.
© Copyright 2014
TLO 2
Operator Generic Fundamentals
59
TLO 2 Summary
– An increasing Reynolds number indicates an increasing
turbulence of flow.
o Laminar flow exists where the Reynolds number is less than
2,000.
o Turbulent flow exists with Reynolds numbers greater than
3,500.
o Transition flow identified with values between these two
numbers.
2. Fluids have unique characteristics, including viscosity.
– Viscosity is the fluid property that measures the resistance of the
fluid to deforming due to a shear force.
– For most fluids, temperature and viscosity are inversely
proportional.
– An ideal fluid is one that is incompressible and has no viscosity.
© Copyright 2014
TLO 2
Operator Generic Fundamentals
60
TLO 2 Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Describe the characteristics and flow velocity profiles of laminar
flow and turbulent flow.
2. Explain the following terms:
a. Viscosity
b. Ideal fluid
© Copyright 2014
TLO 2
Operator Generic Fundamentals
61
Crossword Puzzle
• It’s crossword puzzle time!
© Copyright 2014
TLOs
Operator Generic Fundamentals
62
Bernoulli’s Equation and Energy
TLO 3 – Explain Bernoulli’s equation and use it to determine fluid
properties at different points in a system.
• Bernoulli’s equation is a widely used tool for solving fluid flow
problems.
• It expresses each of the energies possessed by a fluid, elevation,
velocity, and pressure, in equivalent heads.
• With the principle of conservation of mass, this equation makes it
possible to:
– Examine individual components of piping systems
– Determine what fluid properties vary
– Show how the energy balance is affected
© Copyright 2014
TLO 3
Operator Generic Fundamentals
63
Enabling Learning Objectives for TLO 3
1. Using the simplified form of Bernoulli’s equation, explain the
interrelationships of these energies or heads:
a. Elevation head
b. Velocity head
c. Pressure head
2. Using the extended form of Bernoulli’s equation, explain both
head loss and pump head.
3. Explain the following causes of head loss:
a. Friction
b. Viscosity
c. Minor losses
4. Describe methods used to control fluid flow rates and the resulting
pressure and flow rate changes.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
64
Simplified Bernoullis Equation
ELO 3.1 – Using the simplified form of Bernoulli’s equation, explain the
interrelationships of these types of energy: elevation head, velocity head,
and pressure head.
• Special case of the general energy equation
– Probably the most used tool for solving fluid flow problems.
• Easy way to relate the energies for elevation head, velocity head, and
pressure head of a fluid.
• Bernoulli’s equation can be modified in a way that accounts for both
head losses and pump work.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
65
General Energy Equation
Conservation of energy principle states that energy can be neither
created nor destroyed. Equivalent to the First Law of Thermodynamics.
Following equation is the general energy equation for an open system:
𝑄 + 𝑈 = 𝑃𝐸 + 𝐾𝐸 + 𝑃𝑉 𝑖𝑛
= 𝑊 + 𝑈 + 𝑃𝐸 + 𝐾𝐸 + 𝑃𝑉
𝑜𝑢𝑡
+ 𝑈 + 𝑃𝐸 + 𝐾𝐸 + 𝑃𝑉
𝑠𝑡𝑜𝑟𝑒𝑑
Where:
• Q = heat (BTU)
• U = internal energy (BTU)
• PE = potential energy (ft-lbf)
• KE = kinetic energy (ft-lbf)
• P = pressure (lbf/ft2)
• V = volume (ft3)
• W = work (ft-lbf)
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
66
Simplified Bernoulli Equation
𝑃𝐸 + 𝐾𝐸 + (𝑃𝑉)1 = 𝑃𝐸 + 𝐾𝐸 + (𝑃𝑉)2
Substituting appropriate expressions for potential energy and kinetic
energy, the above equation rewritten:
𝑚𝑔𝑧1 𝑚𝑣12
𝑚𝑔𝑧2 𝑚𝑣22
+
+ 𝑃1 𝑉1 =
+
+ 𝑃2 𝑉2
𝑔𝑐
2𝑔𝑐
𝑔𝑐
2𝑔𝑐
Where:
• m = mass (lbm)
• z = height above reference (ft)
• v = average velocity (ft/sec)
• g = acceleration due to gravity (32.17 ft/sec2)
• gc = gravitational constant, (32.17 ft-lbm/lbf-sec2)
• P = pressure (lbf/ft2)
• V = volume (ft3)
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
67
Simplified Bernoulli Equation
• The gravitation constant (gc ) is only required when using the English
System of Measurement and mass is measured in pounds mass (lbm).
– Essentially a unit conversion factor.
– Not required if mass is in slugs or in the metric system.
• Terms represent form of energy (potential, kinetic, and pressure related
energies).
• Bernoulli’s equation represents a balance of the KE, PE, PV energies
so that if one increases, one or more of the others decreases to offset
the change.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
68
Simplified Bernoulli Equation
Another form of Bernoulli’s equation:
𝑣12
𝑣22
𝑔𝑐
𝑧1 +
+ 𝑃1 𝑣1 = 𝑧2 +
+ 𝑃2 𝑣2
2𝑔
2𝑔
𝑔
Where:
• z =
• 𝑣 =
• P =
• ν =
• g =
• gc =
© Copyright 2014
height above reference level (ft)
average velocity of fluid (ft/sec)
pressure of fluid (lbf/ft2)
specific volume of fluid (ft3/lbm)
acceleration due to gravity (ft/sec2)
gravitational constant, (32.17 ft-lbm/lbf-sec2)
ELO 3.1
Operator Generic Fundamentals
69
Simplified Bernoulli Equation – Energy
• The different forms of energy in Bernoulli’s equation, referred to as
heads, are:
– Elevation head
– Velocity head
– Pressure head
• Head references height, typically in feet of a water column, equivalent
to a supporting pressure and expresses the energy possessed by a
fluid.
– Elevation head represents the potential energy of a fluid due to its
elevation above a reference level.
– Velocity head represents the kinetic energy of the fluid due to its
velocity; column height in feet a flowing fluid would rise if all of its
kinetic energy was converted to potential energy.
– Pressure head is the flow energy of a column of fluid whose weight
equals the pressure of the fluid.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
70
Simplified Bernoulli Equation – Energy
• The total head is the sum of the elevation head, velocity head, and
pressure head of a fluid.
• Therefore, Bernoulli’s equation states that the total head of the fluid is
constant.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
71
Energy Conversions in Fluid Systems
• Bernoulli’s equation allows examination of energy transfers among
elevation head, velocity head, and pressure head.
• This equation makes it possible to examine individual components of
piping systems and determine what fluid properties vary and the affect
on energy balance.
• A pipe containing an ideal fluid undergoes a gradual expansion in
diameter, the continuity equation tells us:
– As diameter and flow area increase, the flow velocity must decrease
to maintain the same mass flow rate.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
72
Energy Conversions in Fluid Systems
• With outlet velocity less than inlet velocity, the velocity head of the flow
decreases. If pipe is horizontal, no change in elevation head occurs.
– Therefore, an increase in pressure head compensates for the
decrease in velocity head.
• As an incompressible ideal fluid, specific volume does not change.
– Only way that the pressure head for an incompressible fluid can
increase is for the pressure to increase.
• Bernoulli’s equation illustrates this.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
73
Energy Conversions in Fluid Systems
• For a constant diameter pipe containing an ideal fluid undergoing a
decrease in elevation, the same net effect occurs. However:
– The flow velocity and the velocity head must be constant to satisfy
the mass continuity equation.
– Increase in pressure head compensates for decrease in elevation
head.
• Again, the fluid is incompressible so the increase in pressure head
must result in an increase in pressure.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
74
Energy Conversions in Fluid Systems
• As in the conservation of mass, the Bernoulli equation applies to
problems where more than one flow may enter or leave the system
simultaneously.
– Bernoulli’s equation solves series and parallel piping system
problems.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
75
Example: Bernoulli’s Equation
• Assume frictionless flow in a long, horizontal, conical pipe.
• The diameter is 2.0 ft at one end and 4.0 ft at the other.
• The pressure head at the smaller end is 16 ft of water.
If water flows through this cone at a rate of 125.6 ft3/sec, find the
velocities at the two ends and the pressure head at the larger end.
Figure: Conical Pipe Shows Flow of Pressure Energy
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
76
Example: Bernoulli’s Equation Solution
𝑉1 = 𝐴1 𝑣1
𝑣1 =
𝑉1
𝐴1
𝑉2
𝐴2
𝑓𝑡 3
𝑣2 =
𝑓𝑡 3
125.6
125.6
𝑠𝑒𝑐
𝑠𝑒𝑐
𝑣1 =
𝑣
=
2
𝜋 1 𝑓𝑡 2
𝜋 2 𝑓𝑡 2
𝑓𝑡
𝑓𝑡
𝑣1 = 40
𝑣2 = 10
𝑠𝑒𝑐
𝑠𝑒𝑐
𝑣12
𝑔𝑐
𝑣22
𝑔𝑐
𝑧1 +
+ 𝑃1 𝑣1 = 𝑧2 +
+ 𝑃2 𝑣2
2𝑔
𝑔
2𝑔
𝑔
𝑔𝑐
𝑔𝑐
𝑣12 − 𝑣22
𝑃2 𝑣2 = 𝑃1 𝑣1 + 𝑧1 − 𝑧2 +
𝑔
𝑔
2𝑔
= 16 𝑓𝑡 + 0 𝑓𝑡 +
𝑓𝑡
40
𝑠𝑒𝑐
2
𝑓𝑡
− 10
𝑠𝑒𝑐
𝑓𝑡
2 32.17
𝑠𝑒𝑐 2
2
= 39.3 𝑓𝑡
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
77
Simplified Bernoulli's Equation
Knowledge Check
The elevation head represents the ______________ of a fluid due to its
elevation above a reference level. The velocity head represents the
____________ of the fluid due to its velocity; column height in feet a
flowing fluid would rise if all of its kinetic energy were converted to
potential energy.
A. kinetic energy, total energy
B. potential energy, kinetic energy
C. total energy, kinetic energy
D. kinetic energy, potential energy
Correct Answer is B.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
78
Extended Bernoulli's Equation
ELO 3.2 – Using the extended form of Bernoulli’s equation, explain both
head loss and pump head.
• Practical applications of the simplified Bernoulli Equation to real
piping systems are not possible:
– Calculating fluid friction applies only to ideal fluids.
– Calculating work on the fluid because no work is allowed on or by
the fluid.
o Prevents two points in a fluid stream from analysis if a pump
exists between the two points.
• Modifying the simplified Bernoulli equation makes it possible to deal
with both head losses and pump work.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
79
Extended Bernoulli's Equation
• Takes into account gains and losses of head.
• Often used to solve most fluid flow problems. The below equation is
one form:
𝑣12
𝑔𝑐
𝑣22
𝑔𝑐
𝑧1 +
+ 𝑃1 𝑣1 + 𝐻𝑝 = 𝑧2 +
+ 𝑃2 𝑣2 + 𝐻𝑓
2𝑔
𝑔
2𝑔
𝑔
Where:
• z =
• 𝑣 =
• P =
• ν =
• Hp =
• Hf =
• g =
© Copyright 2014
height above reference level (ft)
average velocity of fluid (ft/sec)
pressure of fluid (lbf/ft2)
specific volume of fluid (ft3/lbm)
head added by pump (ft)
head loss due to fluid friction (ft)
acceleration due to gravity (ft/sec2)
ELO 3.2
Operator Generic Fundamentals
80
Extended Bernoulli's Equation
• Most techniques for evaluating head loss due to friction are empirical.
– based almost exclusively on experimental evidence, and
– are based on a proportionality constant called the friction factor
(f), discussed later in this module.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
81
Example: Extended Bernoulli's Equation
Water from a large reservoir is pumped to a point 65 feet higher than
the reservoir.
How many feet of head must the pump add if 8,000 lbm/hr flows
through a 6 inch pipe and the frictional head loss is 2 feet?
The density of the fluid is 62.4 lbm/ft3, and the cross-sectional area of a
6-inch pipe is 0.2006 ft2.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
82
Example: Extended Bernoulli's Equation
• To use the modified form of Bernoulli’s equation, reference points
chosen at the surface of the reservoir (point 1) and the outlet of the pipe
(point 2). The pressure at the surface of the reservoir is the same as
the pressure at the exit of the pipe, such as, atmospheric pressure.
The velocity at point 1 is essentially zero (0).
• Using the equation for the mass flow rate, determine the velocity at
point 2:
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
83
Example: Extended Bernoulli's Equation
𝑚2 = 𝜌𝐴2 𝑣2
𝑚2
𝑣2 =
𝜌𝐴2
8,000
𝑣2 =
𝑙𝑏𝑚
ℎ𝑟
𝑙𝑏𝑚
0.2006 𝑓𝑡 2
3
𝑓𝑡
𝑓𝑡
1 ℎ𝑟
𝑣2 = 639
ℎ𝑟 3,600 𝑠𝑒𝑐
𝑓𝑡
𝑣2 = 0.178
𝑠𝑒𝑐
62.4
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
84
Example: Extended Bernoulli's Equation
𝑣12
𝑔𝑐
𝑣22
𝑔𝑐
𝑧1 +
+ 𝑃1 𝑣1 + 𝐻𝑝 = 𝑧2 +
+ 𝑃2 𝑣2 + 𝐻𝑓
2𝑔
𝑔
2𝑔
𝑔
𝐻𝑝 = 𝑧2 − 𝑧1
= 65 𝑓𝑡 +
𝑣22 − 𝑣12
𝑔𝑐
+
+ 𝑃2 − 𝑃1 𝑣 + 𝐻𝑓
2𝑔
𝑔
𝑓𝑡
0.1778
𝑠𝑒𝑐
2
𝑓𝑡
− 0
𝑠𝑒𝑐
𝑓𝑡
2 32.17
𝑠𝑒𝑐 2
2
+ 0 𝑓𝑡 + 2 𝑓𝑡
𝐻𝑝 = 67 𝑓𝑡
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
85
Applying Bernoulli’s Equation to a Venturi
A Venturi is a flow measuring device consisting of a gradual contraction
followed by a gradual expansion.
Figure: Venturi Tube Flow Meter
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
86
Applying Bernoulli’s Equation to a Venturi
• Bernoulli’s equation states that the total head of the flow must be
constant.
• Elevation does not change between points 1 and 2; the elevation
head at the two points is the same and cancels each other.
• Bernoulli’s equation simplifies to the following for a horizontal Venturi:
𝑣12
𝑔𝑐
𝑣22
𝑔𝑐
+ 𝑃1 𝑣1
=
+ 𝑃2 𝑣2
2𝑔
𝑔
2𝑔
𝑔
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
87
Applying Bernoulli’s Equation to a Venturi
• Applying the continuity equation to expresses flow velocity at point 1
as a function of the flow velocity at point 2 and the ratio of the two
flow areas:
𝑣12
𝑔𝑐
𝑣22
𝑔𝑐
+ 𝑃1 𝑣1
=
+ 𝑃2 𝑣2
2𝑔
𝑔
2𝑔
𝑔
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
88
Applying Bernoulli’s Equation to a Venturi
• Applying the continuity equation to expresses flow velocity at point 1
as a function of the flow velocity at point 2 and the ratio of the two
flow areas.
𝜌1 𝐴1 𝑣1 = 𝜌2 𝐴2 𝑣2
𝜌2 𝐴2 𝑣2
𝑣1 =
𝜌1 𝐴1
𝐴2
𝑣1 = 𝑣2
𝐴1
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
89
Applying Bernoulli’s Equation to a Venturi
Solve for v2 using Bernoulli’s equation,
substituting previous results for v1
The result:
• Flow velocity at the throat of the
Venturi and the volumetric flow rate
are directly proportional to the square
root of the differential pressure.
• Measured pressure drop, P1-P2, across
the Venturi and proportional to flow
rate, is useful for measuring flow.
© Copyright 2014
ELO 3.2
𝑣22 − 𝑣12
𝑔𝑐
= 𝑃1 − 𝑃2 𝑣
2𝑔
𝑔
𝐴2
𝑣22 − 𝑣2
𝐴1
𝑣22
2
= 𝑃1 − 𝑃2 2𝑣𝑔𝑐
𝐴2
1 − 𝑣2
𝐴1
𝑣22 =
𝐴2
𝐴1
2
𝑃1 − 𝑃2 2𝑣𝑔𝑐
1 − 𝑣2
𝑣2 =
= 𝑃1 − 𝑃2 2𝑣𝑔𝑐
𝑃1 − 𝑃2 2𝑣𝑔𝑐
1 − 𝑣2
𝑣2 =
2
𝑃1 − 𝑃2
𝐴2
𝐴1
2
2𝑣𝑔𝑐
𝐴
1 − 𝑣2 2
𝐴1
2
Operator Generic Fundamentals
90
Applying Bernoulli’s Equation to a Venturi
• Pressures at the upstream section and throat are actual pressures,
and velocities from Bernoulli’s equation without a loss term are
theoretical velocities.
– When losses are considered, the velocities are actual velocities.
• Multiplying the theoretical velocity by a Venturi correction factor (Cv),
accounting for friction losses, (0.98 for most Venturi), the actual
velocity is obtained.
• The actual velocity times the actual area of the throat determines the
actual discharge volumetric flow rate.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
91
Venturi Sonic Flow
• If flow through a typical ∆P (change in pressure) Venturi becomes
excessively large, such as a downstream piping break, creating a
high ΔP (P1-P2), the velocity through the nozzle can reach sonic
speeds such as Mach 1.
– Referred to as a choked or sonic state of operation.
• In this state any further decrease in downstream pressure (P2), has
no effect to force flow to increase through the nozzle (throat); for
example, choking the flow.
• Some Venturi designed as Sonic Chokes or Critical Flow Nozzles
– Used in control systems produce a desired fixed flow rate
unaffected by downstream pressure.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
92
Example: Venturi
• A main steam line break downstream of the Venturi causes the main
steam mass flow rate through the Venturi to increase.
• Soon, the steam reaches sonic velocity in the throat of the Venturi.
How is main steam mass flow rate through the Venturi affected as the
steam pressure downstream of the Venturi continues to decrease?
Correct Answer:
It will not continue to increase because the steam velocity cannot
increase above sonic velocity in the throat of the Venturi.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
93
Bernoulli’s Equation Extended
Knowledge Check – NRC Bank
Refer to the drawing of a section of pipe that contains flowing subcooled water.
Given:
Pressure at P1 is 24 psig.
Pressure at P2 is 16 psig.
Pressure change due to change in velocity is 2 psig.
Pressure change due to change in elevation is 10 psig.
The pressure decrease due to friction head loss between P1 and P2 is __________;
and the direction of flow is from __________.
A.
2 psig; left to right
B.
2 psig; right to left
C.
4 psig; left to right
D.
4 psig; right to left
Correct Answer is D.
© Copyright 2014
Figure: Pipe Section of Flowing
Subcooled Water
ELO 3.2
Operator Generic Fundamentals
94
Causes of Head Loss
ELO 3.3 – Explain the following causes of head loss: friction loss,
viscosity loss, and minor losses.
• Head loss that occurs in pipes depends on:
– Flow velocity
– Pipe length and diameter
– Friction factor based on the roughness of the pipe and the
Reynolds number of the flow
• Head loss that occurs in the components of a flow path correlates to
a length of pipe that causes an equivalent head loss.
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
95
Causes of Head Loss
• Head loss is a measure of the reduction in the total head of the fluid
as it moves through a fluid system.
• Unavoidable in real fluid systems.
– Friction between the fluid and the walls of the pipe.
– Friction between adjacent fluid particles as they move relative to
one another.
– Turbulence caused by factors such as flow directional changes,
piping entrances and exits, pumps, valves, flow reducers, and
fittings disrupting the flow.
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
96
Head Loss
• Frictional loss is that part of the total head loss that occurs as the
fluid flows through straight pipes.
• Head loss for fluid flow is directly proportional to the length of pipe,
the square of the fluid velocity, and the friction factor, a term
accounting for fluid friction.
• Head loss is inversely proportional to the diameter of the pipe.
𝐿𝑣 2
𝐻𝑒𝑎𝑑𝑙𝑜𝑠𝑠 ∝ 𝑓
𝐷
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
97
Friction Factor
• The friction factor depends on the Reynolds number for the flow and
the degree of roughness of the pipe’s inner surface.
• The quantity used to measure the roughness of the pipe is the
relative roughness term, which equals the average height of surface
irregularities (ε) divided by the pipe diameter (D).
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑅𝑜𝑢𝑔ℎ𝑛𝑒𝑠𝑠 =
© Copyright 2014
ELO 3.3
𝜀
𝐷
Operator Generic Fundamentals
98
Moody Chart
Figure: Moody Chart
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
99
Moody Chart
Figure: Moody Chart
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
100
Viscosity Loss
• Viscosity describes a fluid's internal resistance to flow and is a
component of fluid friction.
• A fluid’s viscosity depends on pressure and temperature.
• Viscosity decreases as the temperature increases.
• Viscosity also affects head loss; the more viscous a fluid is, the more
resistant to flow it is.
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
101
Darcy’s Equation
• Combines pipe dimension, flow velocity, and the friction factor into a
equation for frictional head loss calculations.
𝐿𝑣 2
𝐻𝑓 = 𝑓
𝐷2𝑔
Where:
• f = friction factor (dimensionless) - Rougher piping walls or higher
fluid viscosity, the greater the friction
• L = length of pipe (ft) - Longer the pipe, the greater the friction
• D = diameter of pipe (ft) - Greater the pipe diameter, the lower the
friction
• V = fluid velocity (ft/sec) - Greater the fluid velocity, the higher the
friction. Exponentially, head loss varies with the square of fluid
velocity.
• G = gravitational acceleration (ft/sec2) - fixed
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
102
Minor Losses
• Losses that occur in pipelines due to bends, elbows, joints, valves,
etc.
– These losses often are more important than the losses due to
pipe friction.
• For all minor losses in turbulent flow, the head loss varies as the
square of the velocity.
• A convenient method of expressing the minor losses in flow is the
loss coefficient (k).
• Standard engineering and/or technical handbooks provide values of
the loss coefficient (k) for typical situations and fittings.
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
103
Minor Losses
• The following equation calculates minor losses of individual fluid
system components:
𝑣2
𝐻𝑓 = 𝑘
2𝑔
Where:
• k = loss coefficient
• v = fluid velocity (ft/sec)
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
104
Equivalent Piping Length
• Minor losses expressed in terms of the equivalent length (Leq) of pipe
having the same head loss for the same discharge flow rate.
𝐿𝑒𝑞 = 𝑘
𝐷
𝑓
Where:
• f = friction factor (dimensionless)
• D = diameter of pipe (ft)
• k = loss coefficient
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
105
Equivalent Piping Length
Item
Globe Valve
Description
90° Standard Elbow
400
160
10
35
150
900
10
60
30
45° Standard Elbow
16
Return Bend
50
Gate Valve
Standard Tee
© Copyright 2014
Conventional
Y-Pattern
Fully Open
75% Open
50% Open
25% Open
Flow Through Run
Flow Through Branch
Equivalent Length
ELO 3.3
𝑳𝒆𝒒
𝑫
Operator Generic Fundamentals
106
Causes of Energy Loss
Knowledge Check
The energy loss that occurs in pipes depends on all of the following
attributes EXCEPT:
A. Flow velocity
B. Pipe diameter
C. Saturation temperature
D. Roughness of pipe
Correct answer is C.
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
107
Flow Control Methods
ELO 3.4 – Describe methods used to control fluid flow rates as well as
the resulting pressure and flow rate changes.
• Multiple methods of controlling fluid flow exist.
• System application determines whether pump configuration changes
or if throttle valves provide fluid flow control.
© Copyright 2014
ELO 3.4
Operator Generic Fundamentals
108
Flow Control Methods
• Multiple methods of system flow control exist, depending on the
equipment installed.
• Pump configurations allow for changing the number of operating
pumps or their speed control.
• Pump design characteristics and the fluid system itself determine the
effect on system pressure and flow.
• Positive displacement pumps - system flow rate varies with pump
speed.
– Raising pump speed increases flow rate and decreasing pump
speed decreases flow rate.
– Pump head is relatively constant.
© Copyright 2014
ELO 3.4
Operator Generic Fundamentals
109
Flow Control Methods
• Starting an additional centrifugal pump in parallel with the running
centrifugal pump normally causes a larger increase in system flow
and a relatively small increase in pressure.
• Exact changes are variable with system design - system pressure
may exceed shutoff head of the centrifugal pump.
• System designs often utilize throttle valves to control flow rate.
– Opening the valve increases flow
– Closing the valve decreases the flow.
© Copyright 2014
ELO 3.4
Operator Generic Fundamentals
110
Flow Control Methods
• Changing throttle valve position effectively changes the frictional
head loss in the system.
– Closing the throttle valve increases Hf
– Opening the throttle valve decreases Hf
• As an example, if Hf decreases, flow should increase (all other factors
held constant). Actual system flow and pressure changes depend on
system design, pump capacity, etc.
© Copyright 2014
ELO 3.4
Operator Generic Fundamentals
111
Flow Control Methods
• Varying the speed of a pump or changing the number of operating
pumps within the system affects the pump head (Hp) term.
• If Hf remains constant, increasing pump head (Hp) increases system
pressure and flow.
– However, head loss varies with the square of velocity.
– Increasing the system flow causes an exponential change in head
loss, resulting in restrictions to flow gains.
© Copyright 2014
ELO 3.4
Operator Generic Fundamentals
112
Flow Control Methods
Knowledge Check – NRC Bank
A four-loop PWR nuclear power plant uses four identical reactor coolant
pumps (RCPs) to supply reactor coolant flow through the reactor
vessel. The plant is currently operating at 20% power with all RCPs in
operation.
Which one of the following describes the stable RCS flow rate through
the reactor vessel following the trip of one RCP? (Assume that no
operator actions are taken and the reactor does not trip.)
A. Less than 75 percent of the original flow rate.
B. Exactly 75 percent of the original flow rate.
C. Greater than 75 percent of the original flow rate.
D. Unpredictable without pump curves for the RCPs.
Correct Answer is C.
© Copyright 2014
ELO 3.4
Operator Generic Fundamentals
113
Flow Control Methods
Knowledge Check - NRC Bank
Two identical centrifugal pumps (CPs) and two identical positive displacement pumps
(PDPs) are able to take suction on a vented water storage tank and provide makeup
water flow to a cooling water system. Pumps can be cross-connected to provide multiple
configurations. In single pump alignment, each pump will supply 100 gpm at a system
pressure of 1,000 psig. Given the following information:
Centrifugal Pumps
– Shutoff head = 1,500 psig
– Maximum design pressure = 2,000 psig
– Flow rate with no backpressure = 180 gpm
Positive Displacement Pumps
– Maximum design pressure = 2,000 psig
Which one of the following pump configurations will supply the lowest makeup water flow
rate to the system if system pressure is 1,700 psig?
A.
Two CPs in series
B.
Two CPs in parallel
C.
One PDP and one CP in series (CP supplying PDP)
D.
One PDP and one CP in parallel
© Copyright 2014
ELO 3.4
Correct Answer is B.
Operator Generic Fundamentals
114
TLO 3 Summary
• Bernoulli’s equation is used tool for solving fluid-flow problems.
– Bernoulli’s equation results from the application of the general
energy equation and the First Law of Thermodynamics to a
steady flow system in which no work is done on or by the fluid, no
heat is transferred to or from the fluid, and no change occurs in
the internal energy (for example, no temperature change) of the
fluid.
𝑃𝐸 + 𝐾𝐸 + (𝑃𝑉)1 = 𝑃𝐸 + 𝐾𝐸 + (𝑃𝑉)2
– Units for the different forms of energy in Bernoulli’s equation,
referred to as heads, are:
o Elevation head
o Velocity head
o Pressure head
© Copyright 2014
TLO 3
Operator Generic Fundamentals
115
TLO 3 Summary
– The elevation head represents the potential energy of a fluid due
to its elevation above a reference level.
– The velocity head represents the kinetic energy of the fluid due to
its velocity
o Column height in feet a flowing fluid rises if all of its kinetic
energy was converted to potential energy.
– The pressure head is the flow energy of a column of fluid whose
weight equals the pressure of the fluid.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
116
TLO 3 Summary
– Bernoulli’s equation makes it easy to examine how energy transfers
take place among elevation head, velocity head, and pressure head
o Allows examination of individual components of piping systems
and determines what fluid properties vary and how the energy
balance is affected.
• Modifying and extending the Bernoulli equation makes it applicable to
real piping systems.
– Modifying the Bernoulli equation to take into account gains and
losses of head results in an equation, referred to as the extended
Bernoulli equation.
o Accounts for both friction losses and pump work
– A Venturi determines mass flow rates due to changes in pressure
and fluid velocity.
o Volumetric flow rate through a Venturi is directly proportional to the
square root of the differential pressure between the Venturi’s inlet
and its throat.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
117
TLO 3 Summary
• Head Loss
– Head loss is the reduction in the total head (sum of potential head,
velocity head, and pressure head) of a fluid caused by the friction
present in the fluid’s motion.
– Head loss is unavoidable:
o Friction between the fluid and the walls of the pipe.
o Friction between adjacent fluid particles as they move relative to
one another.
o Turbulence caused by flow directional changes, piping entrances
and exits, pumps, valves, flow reducers, and fittings, disrupting the
flow.
o Head loss for fluid flow is directly proportional to the length of pipe,
the square of the fluid velocity, and friction factor, a term
accounting for fluid friction.
o Head loss is inversely proportional to the diameter of the pipe.
o Minor losses is the head loss that occurs due to bends, elbows,
joints, valves, and other components in the system.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
118
TLO 3 Summary
– Friction factor for fluid flow determined using a Moody chart of the
relative roughness of the pipe and the Reynolds number.
– Darcy’s equation for head loss combines pipe dimension, flow
velocity, and the friction factor into a useful equation for frictional
head loss calculations.
– Equivalent length of pipe that would cause the same head loss as a
valve or fitting can be determined by multiplying the value of L/D for
the component found in handbooks or vendor manuals by the
diameter of the pipe.
• Flow Control Methods
– Pump configurations may allow changes in the number of operating
pumps or speed control.
o Pump design characteristics and the fluid system itself determine
the effect on system pressure and flow.
o The installed pump may be a positive displacement or centrifugal
pump; each has different operational characteristics.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
119
TLO 3 Summary
• System flow rate varies with pump speed in positive displacement
pumps. Pump head is relatively constant.
• Starting an additional centrifugal pump in parallel with the running
centrifugal pump often causes a larger increase in system flow and a
relatively small increase in pressure.
• Exact changes vary with system design - system pressure may
exceed shutoff head of the centrifugal pump.
• Changing the throttle valve position effectively changes the frictional
head loss in the system.
– Closing the throttle valve increases Hf, while opening the throttle
valve decreases Hf.
– For example, if Hf decreases, flow should increase (with all other
factors held constant).
– Actual system flow and pressure changes depend on system
design, pump capacity, etc.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
120
TLO 3 Summary
• Varying the speed of a pump or changing the number of operating
pumps within the system affects the pump head (Hp) term.
– If Hf remains constant, increasing pump head (Hp) increases
system pressure and flow.
– However, head loss varies with the square of velocity. Increasing
system flow causes an exponential change in head loss and
results in restrictions to flow gains.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
121
TLO 3 Summary
Now that you have completed this lesson, you should be able to do the
following:
1.
Using the simplified form of Bernoulli’s equation, explain the
interrelationships of these energies or heads:
a.
Elevation head
b.
Velocity head
c.
Pressure head
2.
Using the extended form of Bernoulli’s equation, explain both head
loss and pump head.
3.
Explain the following causes of head loss:
4.
a.
Friction
b.
Viscosity
c.
Minor losses
Describe methods used to control fluid flow rates and the resulting
pressure and flow rate changes.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
122
Two-Phase Flow and Water Hammers
TLO 4 – Explain two-phase flow and water hammers, including the
problems they may cause in plant operation and the ways of controlling it.
• Water at saturation conditions may exist as a fluid and a vapor.
• Two-phase flow is beneficial for enhanced heat transfer from the fuel
to the reactor coolant.
– However, two-phase flow has greater friction losses and requires
more powerful pumps to maintain flow.
• Water hammers are specific two-phase flow problems.
– Extreme pressure surges may occur.
– Equipment damage and personal harm.
• Steam generators utilize two-phase flow.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
123
Enabling Learning Objectives for TLO 4
1. Explain the following terms, including initiating events:
a. Two-phase flow
b. Pressure spike
2. Explain steam generator recirculation ratio and its purpose.
3. Describe water hammers, including severity and resulting pressure
spikes.
4. List actions that operators can take to minimize the occurrence of
water and steam hammers.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
124
Two-Phase Flow and Pressure Spikes
ELO 4.1 Explain the following terms, including initiating events: twophase flow, and pressure spike.
• Water at saturation conditions may exist as both a fluid and a vapor.
• Mixing steam and water can cause unusual flow characteristics and
even damage fluid piping systems.
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
125
Two-Phase Flow
• Flow in the reactor core and in the Steam Generator (SG) tube
bundle region are some examples of two-phase flow.
• A number of two-phase flow patterns exist:
– Channel flow in the reactor includes bubbly flow, slug flow, and
annular flow.
– Lessons in thermal hydraulics provide details of these two-phase
flow types.
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
126
Two-Phase Flow
• Two-phase fluid flow may cause operational problems.
• Vapor formation and a subsequent collapse in the piping may cause
sudden changes in system pressure.
– These pressure surges may produce severe water hammers.
– Water hammers are potentially damaging to equipment and may
also cause personal injury.
• Flow oscillations may also occur, including problems in plant stability
such as power and thermal transients.
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
127
Two-Phase Flow Friction Losses
• Several techniques are available for predicting the head loss due to
fluid friction for two-phase flow.
• Two-phase flow friction is greater than single-phase friction for the
same pipe dimensions, in some cases by a factor of 100.
• Difference is a function of the type of flow, greater turbulence, and
vaporization, which increases velocity.
• Two-phase friction losses are experimentally determined by
measuring pressure drops across different piping components.
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
128
Two-Phase Flow Friction Losses
• One technique for determining the two-phase friction loss involves
the two-phase friction multiplier (R).
𝐻𝑓 , 𝑡𝑤𝑜– 𝑝ℎ𝑎𝑠𝑒
𝑅=
𝐻𝑓 , 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑 𝑙𝑖𝑞𝑢𝑖𝑑
Where:
• R = two-phase friction multiplier (no units)
• Hf , two-phase = two-phase head loss due to friction (ft)
• Hf , saturated liquid = single-phase head loss due to friction (ft)
• Two-phase friction multiplier (R) is higher at lower pressures than at
higher pressures
• Two-phase head loss can be many times greater than the singlephase head loss.
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
129
Pressure Spikes
• A water hammer may cause a rapid rise in pressure above static
pressure.
• Peak pressure spike attained occurs at the instant of flow change and
magnitude depends on the actual velocity change.
• The following equation calculates the pressure peak:
𝜌𝑐∆𝑣
∆𝑃 =
𝑔𝑐
Where:
∆P = Pressure spike (lbf/ft2)
ρ = Density of the fluid (lbm/ft3)
c = Velocity of the pressure wave (Speed of sound in the fluid) (ft/sec)
∆v = Change in velocity of the fluid (ft/sec)
gc = Gravitational constant (lbm-ft/lbf-sec2)
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
130
Example: Pressure Spike Problem
• Water at a density of 62.4 lbm/ft3 and a pressure of 120 psi is flowing
through a pipe at 10 ft/sec.
• The speed of sound in the water is 4,780 ft/sec.
A check valve suddenly closes. What is the maximum pressure of the
fluid in psi?
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
131
Example: Pressure Spike Problem
𝑃𝑀𝑎𝑥 = 𝑃𝑠𝑡𝑎𝑡𝑖𝑐 + ∆𝑃𝑠𝑝𝑖𝑘𝑒
𝑙𝑏𝑓 𝜌𝑐∆𝑣
𝑃𝑀𝑎𝑥 = 120 2 +
𝑖𝑛
𝑔𝑐
𝑙𝑏𝑓
𝑓𝑡
𝑓𝑡
62.4
×
4,780
×
10
𝑙𝑏𝑓
𝑠𝑒𝑐
𝑠𝑒𝑐
𝑓𝑡 3
𝑃𝑀𝑎𝑥 = 120 2 +
𝑙𝑏𝑚– 𝑓𝑡
𝑖𝑛
32.17
𝑙𝑏𝑓– 𝑠𝑒𝑐 2
𝑙𝑏𝑓
𝑙𝑏𝑓 1 𝑓𝑡 2
𝑃𝑀𝑎𝑥 = 120 2 + 92,717 2
𝑖𝑛
𝑖𝑛 144 𝑖𝑛2
𝑙𝑏𝑓
𝑙𝑏𝑓
𝑃𝑀𝑎𝑥 = 120 2 + 6,447 2
𝑖𝑛
𝑖𝑛
𝑃𝑀𝑎𝑥 = 763.8 𝑝𝑠𝑖
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
132
Two-Phase Flow and Pressure Spikes
Knowledge Check – NRC Bank
Reactor coolant system (RCS) hot leg temperature is constant at 568
°F while RCS pressure is decreasing due to a small reactor coolant
leak. Which one of the following RCS pressure ranges includes the
pressure at which two-phase flow will first occur in the hot leg?
A. 1,250 to 1,201 psig
B. 1,200 to 1,151 psig
C. 1,150 to 1,101 psig
D. 1,100 to 1,051 psig
Correct Answer is B.
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
133
Steam Generator Recirculation Ratio
ELO 4.2 – Explain steam generator recirculation ratio and its purpose.
• Feedwater returns to the SG at a temperature below saturation.
– Requires heating of the feedwater inside the SG.
• Inside the SG feedwater flows through the downcomer region to the
bottom of the tube bundle.
• Feedwater heating occurs from the primary coolant flowing through
the U-tubes.
• Feedwater flows upward through tube bundle heats to saturation,
then to a steam-water vapor mixture.
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
134
Two-Phase Flow
• Steam leaving tube bundle region enters moisture separators and
removes moisture from the steam.
• Leaving the SG, the steam flows through the main steam lines to the
turbine as dry, saturated steam.
• Liquid from the moisture separators flows down around the separator
shroud, mixing with incoming feedwater.
• Feedwater and moisture from the separators mixed together
approach saturation temperature, unlike the feedwater directly from
the inlet header.
– Reheating effect to the feedwater
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
135
Steam Generator Recirculation Ratio
Recirculation is the
mixture of incoming
feedwater and moisture
separator drainage.
Figure: Typical U-tube Steam Generator from Westinghouse Electric Company LLC
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
136
Recirculation Ratio (RCR)
• The recirculation ratio is the amount of recirculated moisture to the
amount of steam produced.
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑢𝑏𝑒 𝑏𝑢𝑛𝑑𝑙𝑒
𝑅𝐶𝑅 =
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑠𝑡𝑒𝑎𝑚 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑢𝑏𝑒 𝑏𝑢𝑛𝑑𝑙𝑒
Or
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑟𝑒𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑡𝑜 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟
𝑅𝐶𝑅 =
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑠𝑡𝑒𝑎𝑚 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑚 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
137
Purpose of Steam Generator Recirculation
• Two important functions:
1. Preheats feedwater to near saturation prior to entering the bottom
of the tube bundle.
2. Minimizes thermal stress on the nozzles and steam generator
shell by reduction of their temperature differences.
• Recirculation ratio decreases dramatically as power level increases,
reaching a minimum value at 100% power.
• Typical numbers are
– RCR of about 3 at 100% power
– 25 to 30 at 10% power.
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
138
Steam Generator Recirculation Ratio
RCR decreases as power increases
because:
• Steam flow increases as power
increases
• Thermal driving head in the S/Gs
maintains liquid recirculation flow
constant at low power levels
versus high power levels.
• Liquid flow decreases
proportionally to the increase in
steam flow as power level
increases; causing an decrease in
RCR
• Result is more steam flow
compared to recirculation flow.
Figure: Recirculation Ratio Versus Power Level
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
139
Steam Generator Recirculation Ratio
Knowledge Check
Which of the following is NOT a purpose of steam generator
recirculation?
A. preheats feedwater to near saturation at entry to tube bundles
B. minimizes thermal stress on the steam generator nozzles
C. reduces solids buildup on the steam generator tube sheet
D. reduces temperature differences between the steam generator
nozzles and shell.
Correct Answer is C.
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
140
Steam Generator Recirculation Ratio
Knowledge Check
Which one of the following describes the relationship of the moisture
removed by the steam generator moisture separators to the feedwater inlet
flow rate? Assume steady state conditions at 100 percent power.
A. The mass flow rate of the feedwater inlet is equal to the steam
leaving the steam generator plus the moisture removed by the
moisture separators.
B. The mass flow rate of the moisture removed by the steam
generators plus the steam flow rate equals the mass flow rate of
the feedwater inlet.
C. The mass flow leaving the steam generator tube bundle area
equals the feedwater inlet mass flow rate.
D. The mass flow rate leaving the steam generator tube bundle area
minus the moisture removed by the moisture separators equals the
feedwater inlet mass flow rate.
Correct Answer is D.
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
141
Water Hammers
ELO 4.3 – Describe water hammers, including severity and resulting
pressure spikes.
• When slugs of liquid are driven, or drawn by vacuum, through a pipe,
these fluid slugs move at high speeds (speed of sound).
• The result is that liquid can slam into valves, pipe bends, tees, or
other piping components, break piping supports, straightening piping
bends, knocking pumps out of alignment, or breaching the system
piping.
• In extreme cases, pressure spikes may exceed piping ratings,
rupturing the piping.
© Copyright 2014
ELO 4.3
Operator Generic Fundamentals
142
Water Hammers
• All fluid hammers come from the momentum of a moving higher
density liquid.
– Results of a sudden change in the fluid’s momentum when the
fluid’s velocity changes quickly.
• Most water hammers result from factors such as the fast operation of
valves, the rapid warming of steam lines, and operator
misunderstanding, or impatience.
© Copyright 2014
ELO 4.3
Operator Generic Fundamentals
143
Water Hammers
• The severity of water hammer caused by valve operation depends
on:
– Initial system pressure
– Density of the fluid
– Speed of sound in the fluid
– Elasticity of the fluid and the pipe
– Change in the velocity of the fluid
– Diameter and thickness of the pipe
– Valve stroking time
© Copyright 2014
ELO 4.3
Operator Generic Fundamentals
144
Water Hammers
• Closing a valve converts the
kinetic energy of the moving
fluid to potential energy.
• Elasticity of fluid and pipe wall
produces a wave of positive
pressure toward fluid’s source.
• When this wave reaches the
source, the mass of fluid will be
at rest, but under tremendous
pressure.
• The compressed liquid and
stretched pipe walls start
releasing the liquid in the pipe
back to the source and return to
static pressure of the source.
© Copyright 2014
ELO 4.3
Figure: Typical Water Hammer
Operator Generic Fundamentals
145
Water Hammers
• This energy release forms another
pressure wave to valve.
• When shockwave reaches valve,
pipe wall begins contracting due to
momentum of fluid.
• The contraction transmits to the
source, placing pressure in piping less
than static pressure of the source.
• The pressure waves travel back and
forth several times until fluid friction
dampens alternating pressure waves.
• Duration of pressure peaks depends
on length of pipe and speed of sound
in fluid.
• The entire process usually occurs in
less than one second.
Figure: Typical Water Hammer
© Copyright 2014
ELO 4.3
Operator Generic Fundamentals
146
Water and Steam Hammers
Water and steam hammers may come from multiple causes. The most
common are:
– Quickly closing a valve – rapidly stopping a mass of fluid from
flowing causes pressure waves through the system, resulting in
mechanical shock.
– Quickly opening a valve – admits fluid under pressure into an
empty line – a particular cause of steam hammers, the rapid entry
of cool water into a steam-filled pipe.
o Steam rapidly condenses, creating shock waves as the fluid
fills the pipe.
© Copyright 2014
ELO 4.3
Operator Generic Fundamentals
147
Occurrence of Water & Steam Hammers
The most common causes of water hammer (cont):
– Cold condensate sitting in a steam line – causes steam to
collapse and create a vacuum void into which a mass of fluid
travels at a high speed toward piping components.
– Hot condensate sitting in steam line – could be blown into
collisions with piping components by the movement of steam.
o A confined mass of incompressible liquid moving at high
speeds creates pressure when it strikes the piping components
and is a common occurrence in the condensate system.
© Copyright 2014
ELO 4.3
Operator Generic Fundamentals
148
Water Hammer
Knowledge Check – NRC Bank
The primary reason for slowly opening the discharge valve of a large
motor-driven centrifugal cooling water pump after starting the pump is
to minimize the...
A. net positive suction head requirements.
B. potential for a water hammer.
C. motor running current requirements.
D. potential for pump cavitation.
Correct Answer is B.
© Copyright 2014
ELO 4.3
Operator Generic Fundamentals
149
Preventing Water and Steam Hammers
ELO 4.4 – List actions that operators can take to minimize the occurrence
of water and steam hammers.
• Water and steam hammers are common occurrences in industrial
plants, including nuclear.
• Water and steam hammers are preventable.
• Good operator practices include making slow flow changes in piping
systems.
© Copyright 2014
ELO 4.4
Operator Generic Fundamentals
150
Preventing Water and Steam Hammers
For water and steam hammer prevention, operators should follow these
precautions:
– Ensure piping systems are properly filled and vented.
– Operate manual system valves slowly.
– Ensure compliance with all heatup rate limitations.
– Drain and warm system steam lines prior to initiating flow.
– When possible, initiate pump starts against a closed discharge
valve.
– Open the discharge valve slowly to initiate system flow.
o Some systems designs include auto opening of discharge
valves.
© Copyright 2014
ELO 4.4
Operator Generic Fundamentals
151
Preventing Water and Steam Hammers
• Start smaller capacity pumps before larger capacity pumps, if
possible.
• Use warm-up valves around main steam stop valves.
• Close pump discharge valves before stopping pumps, if possible.
• Periodically verify proper function of moisture traps and steam traps
during operation.
• Use high point pump casing vents.
– Primes the pump and removes non-condensable gases.
© Copyright 2014
ELO 4.4
Operator Generic Fundamentals
152
Preventing Water and Steam Hammers
Knowledge Check – NRC Bank
Which one of the following will minimize the possibility of water
hammer?
A. Draining the discharge line of a centrifugal pump after
shutdown.
B. Draining condensate out of steam lines before and after
initiating flow.
C. Starting a centrifugal pump with its discharge valve fully open.
D. Starting a positive displacement pump with its discharge valve
partially closed.
Correct Answer is B.
© Copyright 2014
ELO 4.4
Operator Generic Fundamentals
153
TLO 4 Summary
• Water saturation conditions may exist as both a fluid as well as a vapor, and
results in two-phase flow.
– Two-phase flow, the simultaneous flow of liquid water and steam, occurs
at certain locations in fluid flow systems.
o Examples are flow in the nuclear reactor and in the tube bundle region
of the steam generators.
– A number of two-phase flow patterns exist
o Examples include bubbly flow, slug flow, and annular flow.
– Two-phase fluid flow can cause operational problems.
o Vapor formation and subsequent collapse in the piping cause sudden
changes in system pressure.
o Pressure surges result in severe water hammers.
o Water hammers can potentially damage equipment and may cause
personal injury.
o Flow oscillations may also occur. Process control, temperature, and
level swings can cause problems in plant stability, including power and
thermal transients.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
154
TLO 4 Summary
– Simple relationships used for analyzing single-phase flow are
invalid for analyzing two-phase flow.
– Two-phase flow friction is greater than single-phase friction for the
same conduit dimensions, in some cases by a factor of 100.
– Two-phase friction losses are experimentally determined by
measuring pressure drops across different piping components.
– To determine two-phase flow friction, the two-phase friction
multiplier (R), defined as the ratio of the two-phase head loss
divided by the head loss evaluated using saturated liquid (singlephase) properties.
– A water hammer may cause a rapid rise in pressure above static
pressure. The peak pressure spike attained occurs at the instant
of flow change and the magnitude depends on the actual velocity
change.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
155
TLO 4 Summary
• Steam generator recirculation ratio uses a mixture of incoming
feedwater and moisture separator drainage.
– Recirculation is mixing incoming feedwater and moisture
separator drainage. Recirculation ratio (RCR) is the ratio of
recirculated moisture to the amount of steam produced.
𝑅𝐶𝑅 =
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑢𝑏𝑒 𝑏𝑢𝑛𝑑𝑙𝑒
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑠𝑡𝑒𝑎𝑚 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑢𝑏𝑒 𝑏𝑢𝑛𝑑𝑙𝑒
Or
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑟𝑒𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑡𝑜 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟
𝑅𝐶𝑅 =
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑠𝑡𝑒𝑎𝑚 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑚 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟
© Copyright 2014
TLO 4
Operator Generic Fundamentals
156
TLO 4 Summary
– Recirculation ratio decreases as power level increases and
reaches minimum value at 100% power. Typically a RCR is
about 3 at 100% power and 25 to 30 at 10% power.
• All fluid hammers come from the momentum of a moving higher
density liquid
– Fluid hammers result from a sudden change in the fluid’s
momentum occurring when the fluid’s velocity changes quickly.
– Results from factors such as the quick operation of valves, rapid
warming of steam lines, and operator misunderstanding or
impatience.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
157
TLO 4 Summary
– Water and steam hammers commonly caused by:
o Quickly closing a valve: rapidly stopping a mass of fluid causes
pressure waves through the system, resulting in mechanical
shock.
o Quickly opening a valve: admits fluid under pressure into an
empty line and the rapid entry of cool water into a steam filled
pipe. The steam rapidly condenses creating shock waves as the
fluid fills the pipe.
o Cold condensate sitting in a steam line: collapsing steam create a
vacuum in which fluid travels at high speed toward piping
components.
o Hot condensate sitting in the steam line: could be blown by the
movement of steam into collisions with piping components. A
confined mass of incompressible liquid moving at high speeds
creates pressure when it strikes the piping components. This
often happens in the condensate system.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
158
TLO 4 Summary
• To prevent water and steam hammers, operators should do the following:
– Ensure piping systems are properly filled and vented.
– Operate manual system valves slowly.
– Ensure compliance with all heatup rate limitations.
– Drain and warm system steam lines before initiating flow.
– Initiate pump starts against a closed discharge valve when possible.
– Open the discharge valve slowly to initiate system flow. Some systems
designs include auto opening of discharge valves.
– Start smaller capacity pumps before larger capacity pumps, if possible.
– Use warm-up valves around main steam stop valves.
– Close pump discharge valves before stopping pumps when possible.
– Periodically verify proper function of moisture traps and steam traps
during operation.
– Use high point pump casing vents, which allows the pump to be primed,
and permits removal of non-condensable gases.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
159
TLO 4 Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Explain the following terms, including initiating events:
a. Two-phase flow
b. Pressure spike
2. Explain steam generator recirculation ratio and its purpose.
3. Describe water hammers, including severity and resulting pressure
spikes.
4. List actions that operators can take to minimize the occurrence of
water and steam hammers.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
160
Crossword Puzzle
• It’s crossword puzzle time!
© Copyright 2014
TLOs
Operator Generic Fundamentals
161
Fluid Statics and Dynamics Summary
In this module, we learned about Fluid Statics and Dynamics, the
thermodynamic study of fluid systems, including their design and
responses.
A fluid is a substance that flows freely because its molecules lack rigid
attachment to one another within a crystalline structure including liquids,
gases, and materials normally considered solids, such as glass. Fluids
are compressible or incompressible and may be buoyant.
• Pascal's law states that pressure applied to a confined fluid transmits
undiminished throughout the confining vessel of the system. Factors
such as static pressure, dynamic pressure, and total pressure influence
the volumetric flow rates of fluids.
• The three basic principles of fluid flow include the:
– Principle of momentum, which leads to equations of fluid forces;
– Conservation of energy, which leads to the First Law of
Thermodynamics, and
– Conservation of mass, which leads to the continuity equation.
© Copyright 2014
Summary
Operator Generic Fundamentals
162
Fluid Statics and Dynamics Summary
• Bernoulli's equation is a special case of the general energy equation
and is probably the most used tool for solving fluid-flow problems.
• Modifications can be made to Bernoulli's equation for both head
losses and pump work.
𝑄 + 𝑈 = 𝑃𝐸 + 𝐾𝐸 + 𝑃𝑉 𝑖𝑛
= 𝑊 + 𝑈 + 𝑃𝐸 + 𝐾𝐸 + 𝑃𝑉 𝑜𝑢𝑡 + 𝑈 + 𝑃𝐸 + 𝐾𝐸 + 𝑃𝑉 𝑠𝑡𝑜𝑟𝑒𝑑
• The energy loss that occurs in pipes depends on:
o flow velocity
o pipe length
o pipe diameter
o friction factor based on the roughness of the pipe, and
o the Reynolds number of the flow.
The friction factor depends on the Reynolds number for the flow and
the degree of roughness of the pipe's inner surface.
© Copyright 2014
Summary
Operator Generic Fundamentals
163
Fluid Statics and Dynamics Summary
• Darcy's equation for head loss combines pipe dimension, flow
velocity, and the friction factor into a useful equation for frictional
head loss calculations.
𝐿𝑣 2
𝐻𝑓 = 𝑓
𝐷2𝑔
• Minor losses describe head loss that occurs in pipelines due to
bends, elbows, joints, valves, etc.; however, this is a misnomer
because in many cases these losses are more important than the
losses due to pipe friction.
𝑣2
𝐻𝑓 = 𝑘
2𝑔
• It is common to express minor losses in terms of the equivalent
length (Leq) of pipe having the same head loss for the same
discharge flow rate:
𝐷
𝐿𝑒𝑞 = 𝑘
𝑓
© Copyright 2014
Summary
Operator Generic Fundamentals
164
Fluid Statics and Dynamics Summary
• Two-phase flow is the simultaneous flow of liquid water and steam,
which can cause operational problems including vapor formation and
subsequent collapse in the piping. This causes sudden changes in
system pressure and is capable of producing severe water hammers.
The peak pressure spike attained occurs at the instant of flow change
and the magnitude depends on the actual velocity change.
• Water hammer, also known as a steam hammer, is a physical shock to
the components of a fluid system, liquid, or vapor caused by impact of
high-velocity liquids.
• All fluid hammers come from momentum of a moving higher density
liquid, resulting in a sudden change in the fluid’s momentum that occurs
when the fluid’s velocity changes quickly. Most fluid hammers result
from factors such as the fast operation of valves, rapid warming of steam
lines, and operator misunderstanding or impatience.
• Water and steam hammers commonly occur in industrial plants,
including nuclear plants. Water and steam hammers are preventable.
© Copyright 2014
Summary
Operator Generic Fundamentals
165
Fluid Statics and Dynamics Summary
Now that you have completed this module, you should be able to
demonstrate mastery of this topic by passing a written exam with a
grade of 80 percent or higher on the following TLOs:
1. Explain the fundamental properties of fluids and closed fluid
systems.
2. Describe laminar and turbulent flow.
3. Explain Bernoulli’s equation and use it to determine fluid properties
at different points in a system.
4. Explain two-phase flow and water hammers, including the
problems they can cause in plant operation, and the means to
control it.
© Copyright 2014
Summary
Operator Generic Fundamentals