Transcript Thermodynamic Properties Lesson Plan

Operator Generic Fundamentals
Thermodynamic Units and Properties
© Copyright 2016
Operator Generic Fundamentals
2
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%
score on the following topics (TLOs):
1. Describe thermodynamic properties and methods of measuring
intensive and extensive properties.
2. Explain the concepts of heat, work, and energy.
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TLOs
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Thermodynamic Properties
TLO 1 – Describe thermodynamic properties and methods of measuring
intensive and extensive properties.
1.1 Define the following properties: specific volume, density, mass,
weight, intensive, and extensive.
1.2 Define the thermodynamic properties of temperature and convert
between the Fahrenheit, Celsius, Kelvin, and Rankine scales.
1.3 Define the thermodynamic properties of pressure and convert
between pressure scales.
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Properties and Definitions
ELO 1.1 – Define the following properties: specific volume, density,
mass, weight, intensive, and extensive.
• Operators must be able to convert between units of measurement to
ensure plant operating within established limits
– RCS leak rates, pump surveillances, etc.
• Some conversions provided at bottom of NRC Equation Sheet
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Properties and Definitions
American Engineering System
Length
Mass
Time
Inch
Ounce
Second*
Foot*
Pound*
Minute
Yard
Ton
Hour
Mile
Day
NOTE: *Denotes standard unit of measure
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Properties and Definitions
International System (SI) – MKS Units
Length
Mass
Time
Millimeter
Milligram
Second*
Meter*
Gram
Minute
Kilometer
Kilogram*
Hour
Day
NOTE: *Denotes standard unit of measure
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Properties and Definitions
International System (SI) – CGS Units
Length
Mass
Time
Centimeter*
Milligram
Second*
Meter
Gram*
Minute
Kilometer
Kilogram
Hour
Day
NOTE: *Denotes standard unit of measure
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Properties and Definitions
Prefix
Symbol
Power of 10
Example
pico
p
10-12
1 picosecond (ps) = 10-12 seconds
nano
n
10-9
1 nanosecond (ns) = 10-9 seconds
micro
m
10-6
1 microsecond (ms) = 10-6 seconds
milli
m
10-3
1 millimeter (mm) = 10-3 meters
centi
c
10-2
1 centimeter (cm) = 10-2 meters
deci
d
10-1
1 decigram (dg) = 10-1 grams
hecto
h
102
1 hectometer (hm) = 102 meters
kilo
k
103
1 kilogram (kg) = 103 grams
mega
M
106
1 megawatt (MW) = 106 watts
giga
G
109
1 gigawatt (GW) = 109 watts
Figure: Metric System Prefixes
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Typical Conversion Table
Unit
English Units of
Measurement
1 yard (yd)
Meter-Kilogram-Second (MKS) Units of
Measurement
= 0.9144 meter (m)
12 inches (in.)
= 1 ft
5,280 feet (ft)
= 1 mi
1 (meter) m
= 3.281 ft
Time
1 in.
60 seconds (sec)
= 0.0254 m
= 1 minute (min)
Mass
3,600 sec
1 pound mass (lbm)
= 1 hour (hr)
0.4535 kg
2.205 lbm
= 1 kg
1 kilogram (kg)
1 square foot (ft2)
= 1,000 grams (g)
= 144 in.2
10.764 ft2
= 1 square meter (m2)
1 square yard (yd2)
= 9 ft2
1 square mile (mi2)
7.48 gallon (gal)
3.098 x 106 yd2
= 1 cubic foot (ft3)
1 gal
= 3.785 l (liter)
1 liter (l)
= 1,000 cubic centimeters (cm3)
Length
Area
Volume
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Properties and Definitions
Steps for Converting Units:
1. Identify units given and units required
2. Select the equivalence relationship
3. Arrange the ration in the appropriate manner such that (# desired
/current) = 1
4. Multiply quantity by ratio
5. Multiple conversion factors may be required
Practice conversions using “railroad track” technique to ensure proper
answers!
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Practice Unit Conversion
Convert 795 meters to feet:
• Step 1: Identify units given and units required (meters to feet)
• Step 2: Select equivalence relationship from conversion table:
1 𝑚𝑒𝑡𝑒𝑟(𝑔𝑖𝑣𝑒𝑛 𝑢𝑛𝑖𝑡𝑠) = 3.281 𝑓𝑡(𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠)
• Step 3: Arrange equivalence ratio in appropriate manner:
𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠
)
𝑔𝑖𝑣𝑒𝑛 𝑢𝑛𝑖𝑡𝑠
(
1=
3.281 𝑓𝑡
1𝑚
• Step 4: Multiply the quantity by the ratio:
3.281 𝑓𝑡
795 𝑚 3.281 𝑓𝑡
795 𝑚
=
= 795 × 3.281 𝑓𝑡
1𝑚
1
1𝑚
= 2608 𝑓𝑡
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Practice Unit Conversion
• Reactor core thermal power is 1,800 MWth. Convert to BTU/hr.
– Units given are megawatts and units desired are BTU/hr
– 1 Mw = 3.41 x 106 BTU/hr
1,800 MW
3.41 x 106 Btu
hr
1 MW
1,800 MW converts to 6.138 x 109 BTU/hr
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Properties and Definitions - Mass &
Weight
• Mass (m) - measure of amount
of material present in that body
• Weight (wt) - force exerted by
that body when its mass is
accelerated in a gravitational
field
• Mass and weight related
• gc has same numerical value as
acceleration of gravity
𝑚𝑔
𝑤𝑡 =
𝑔𝑐
• Where:
wt = weight (lbf)
m = mass (lbm)
g = acceleration due to gravity
= 32.17 ft/s
gc = gravitational constant
= 32.17 lbm-ft/lbf-s2
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Intensive vs. Extensive Properties
• Intensive
– independent of mass and does not depend on how much of
substance is present
– Temperature, pressure
• Extensive
– depends on mass (or how much of substance is present)
– Volume, weight
• For example
– Volume (V) is extensive
– Specific Volume (v) is intensive
o Volume/Mass – units are ft3/lbm
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Properties and Definitions – Specific Volume
• Volume - amount of space a
particular substance occupies
• Specific volume is total volume
(V) of that substance divided by
total mass (m) of that substance
• Specific volume values
provided in Steam Table
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𝑉
𝑣=
𝑚
• Where:
ELO 1.1
v = specific volume (ft3/lbm)
V = volume (ft3)
m = mass (lbm)
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Properties and Definitions - Density
𝑚 1
𝜌= =
𝑉 𝑣
• Density (ρ) is total mass (m) of
that substance divided by total
volume (V) occupied by that
substance
• Where:
– Describes how much stuff is
packed into specific volume
• Units of pound-mass per cubic
feet (lbm/ft3)
ρ = density (lbm/ft3)
m = mass (lbm)
V = volume (ft3)
v = specific volume (ft3/lbm)
• Density of a substance is
reciprocal of its specific volume
(ν)
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Properties and Definitions - Density
• Density can be changed by changing pressure or temperature
• Increasing pressure increases density of a material
• Increasing temperature decreases density
– Change in density greater at higher temperatures
• Pressure effect greater on steam
– Liquids essentially incompressible
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Properties and Definitions – Specific
Gravity
• Measure of relative density of a substance as compared to density of
water at a standard temperature
• Varies with temperature, specific gravities must be specified at
particular temperatures
𝜌𝑓𝑙𝑢𝑖𝑑
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐺𝑟𝑎𝑣𝑖𝑡𝑦 𝑆. 𝐺. =
𝜌𝑤𝑎𝑡𝑒𝑟
• Where:
rfluid = density of fluid being measured (lbm/ft3)
Rwater = density of water at standard temperature (68oF) and pressure
(14.7 psia) (62.4 lbm/ft3)
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Properties of Temperature
ELO 1.2 – Define the thermodynamic properties of temperature and
convert between the Fahrenheit, Celsius, Kelvin, and Rankine scales.
Figure: Comparison of Temperature Scales
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Temperature
• Temperature is a intensive measure of amount of energy stored in an
object of a standard mass
– Measure of average molecular kinetic energy of a substance
• The more molecular movement, the higher the temperature of the
substance will be
– Relative measure of how "hot" or "cold" a substance
• Used to predict direction of heat transfer
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Temperature Scales
• Two temperature scales
normally used for measurement
purposes
– Fahrenheit (F)
– Celsius (C)
• Based on number of increments
between freezing and boiling
– Celsius scale has 100 units
– Fahrenheit scale has 180
units
• Since both scales “relate”
temperature of a substance to a
recognized condition, they are
referred to as relative
temperature scales
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Figure: Boiling and Freezing Points of Water for
Celsius and Fahrenheit Temperature Scales
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Temperature Scales
• Zero points on the scales are arbitrary
– Celsius scale zero point is freezing point of water
– Fahrenheit scale zero point is coldest temperature achievable
with a mixture of ice and salt water
• Temperature at which water boils
– 100 on Celsius scale
– 212 on Fahrenheit scale
• Mathematical relationships
9
℉ = 32 +
°𝐶
5
℃ = (℉ − 32)
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Temperature Scales
• Relationship between scales represented by:
°𝑅 = ℉ + 460
°𝐾 = ℃ + 273
Figure: Comparison of Temperature Scales
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Scales of Pressure
ELO 1.3 – Define the thermodynamic properties of pressure and convert
between pressure scales.
Figure: Pressure Scale Relationships
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Scales of Pressure
• Force exerted per unit area on boundaries of substance (or system)
– Pascal’s Principle – “pressure felt undiminished throughout”
• Collisions of molecules of substance with boundaries of system
– Hit walls of their container or system pushing outward
– Forces resulting from these collisions cause pressure exerted by
a system on its surroundings
• Units
– lbf/in2 (psi)
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Scales of Pressure
• Scales use units of inches of H2O or Hg
• Height of column of liquid provides a certain pressure that can be
directly converted to force per unit area
– 0.491 psi = 1 inch of Hg
o NRC bank questions use 0.5 for simplicity
– 0.433 psi = 1 ft of water (based on lower temps – 60-80 degrees)
– 14.7 psia = 408 inches of water
– 14.7 psia = 29.9 inches of Hg
o NRC bank questions use 15 psi and 30.0 in Hg
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Scales of Pressure
• Absolute pressure (psia)
– Relative to a perfect vacuum
• Gauge pressure (psig)
– Relative to atmospheric pressure (14.7 psi)
– Pressure gauges register zero when open to atmosphere
• Pressure below atmospheric designated as vacuum
• Perfect vacuum corresponds to absolute zero pressure
– All values of absolute pressure are positive
• Gauge pressures:
– Positive if above atmospheric pressure
𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔𝑎𝑢𝑔𝑒
– Negative if below atmospheric pressure
𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 − 𝑃𝑣𝑎𝑐
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Scales of Pressure
• Relationships between absolute, gauge, vacuum, and atmospheric
pressures
15
30.0
30.0
15
Figure: Pressure Scale Relationships
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Inches Hg vs. PSIA- Relationships
• Sum of inches Hg and inches Absolute equal 30
− 30 in Hg vacuum = 0 in absolute
− 28 in Hg vacuum = 2 in absolute
• Sum of PSIA and PSIV equals 15
− 15 psia = 0 psiv
− 1 psia = 14 psiv
• 2 inches for every pound
− 15 psia = 30 inches absolute
− 14 psiv = 28 in Hg vacuum
• Based on above:
− 28 in Hg vacuum = 1 psia (this is a key relationship used
extensively!)
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Scales of Pressure
Knowledge Check
A pressure gauge on a condenser reads 27 inches of mercury (Hg)
vacuum. What is the absolute pressure corresponding to this vacuum?
(Assume an atmospheric pressure of 15 psia.)
A. 14.0 psia
B. 13.5 psia
C. 1.5 psia
D. 1.0 psia
Correct answer is C.
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Scales of Pressure
Pabs = Patm – Pvacuum
Pabs = 15 psia – 27”Hg (1psia/2”Hg)
Pabs = 15 psia – 13.5 psia = 1.5 psia
Absolute
Absolute
Vacuum
Figure: Gauge and Absolute Pressure Scale Relationship
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Scales of Pressure
• Pressure due to column of fluid is product of fluid’s density and height
𝜌𝑔𝑧
𝑃=
𝑔𝑐
• Where:
P = pressure (lbf/ft2, lbf/in2, psi)
r = density (lb/ft3)
g = acceleration of gravity (32.2 ft/s2)
z = height (ft, in)
𝑔𝑐 = gravitational conversion constant (32.2 ft lbm/lbf s2)
NOTE: the gc converts lbm to lbf
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PSI to Water Height Example
• A water storage tank is vented to atmosphere. The tank is located at
sea level and contains 100,000 gallons of water at 60°F. A pressure
gauge at the bottom of the tank reads 9.0 psig. What is the
approximate water level in the tank?
𝜌𝑔𝑧
Pg c
𝑃=
z
𝑔𝑐
rg
 9.0lb f
z  
2
in

 ft 3  s 2  32.2 ft  lbm  144in 2 





2
2


 62.4lbm  32.2 ft  lb f  s  ft 
z = 20.7 ft
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PSI to Water Height Example (cont.)
• Recall the unit conversion from psi to ft of water
– 0.433 psi = 1 ft H2O
o Therefore, 1 psi = 2.3 ft H2O (1/0.433 = 2.3)
• Where does this come from?
– Unit conversion from previous equation
• Note: Since density vs temperature curve is relatively straight at low
temps
– Density at standard temp/press of 62.4 can be used
• Only unit conversion required was
– 144 in2/ft2
 9.0lb f
z  
2
 in
 ft 3  s 2  32.2 ft  lbm  144in 2 





2
2


62
.
4
lb
32
.
2
ft
lb

s
ft



m 
f

• 9.0 x 2.3 = 20.7 ft; or, 9.0/0.433 = 20.7 ft
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Scales of Pressure
Knowledge Check – NRC Bank
Refer to the drawing of a tank with a differential pressure (D/P) level
detector (see figure below).
If the tank contains 30 feet of water at 60°F, what is the approximate
D/P sensed by the detector?
A. 7 psid
B. 13 psid
C. 20 psid
D. 28 psid
Correct answer is B.
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Heat, Work, and Energy
TLO 2 – Explain the concepts of heat, work, and energy.
2.1 State the First and Second Laws of Thermodynamics and how they
relate to the conservation of energy.
2.2 Define the following thermodynamic properties: potential energy,
kinetic energy, specific internal energy, specific P-V energy, specific
enthalpy, and specific entropy.
2.3 Explain the relationship between work, energy, and power.
2.4 Define the following terms: heat, sensible heat, latent heat, specific
heat, and super heat.
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Laws of Thermodynamics
ELO 2.1 – State the First and Second Laws of Thermodynamics and how
they relate to the conservation of energy
• The processes of our secondary cycle are either a transfer of heat or
a change in energy form
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Laws of Thermodynamics Introduction
• First Law of Thermodynamics states:
– Energy can be neither created nor destroyed, only altered in form
o Several energy conversions discussed in Thermodynamics
– Velocity energy to pressure energy (water hammer)
– Flow energy to internal energy (headloss)
– Heat energy to mechanical energy (turbine)
• Second Law of Thermodynamics states:
– No engine, actual or ideal, when operating in a cycle can convert
all the heat supplied it into mechanical work–heat must be
rejected
o Losses captured by term “entropy”
o Goal is to minimize the change in entropy
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Properties of Energy
ELO 2.2 – Define the following thermodynamic properties: potential
energy, kinetic energy, specific internal energy, specific P-V energy,
specific enthalpy, and specific entropy.
• Understanding energy terms will help understand the General Energy
equation
– Bernoulli’s Equation
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Properties of Energy
• Energy is capacity of a system to perform work
• Forms of stored energy important in analysis of systems:
– Potential energy
o Due to height
– Kinetic energy
o Due to velocity
– Internal energy
o Due to temperature
– P-V (flow) energy
o Due to pressure
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Energy - Potential
• Energy of position
• Potential energy will exist whenever an object which has mass has a
position within a force field
• Objects in the earth's gravitational field
• Using English system units
𝑚𝑔𝑧
𝑃𝐸 =
𝑔𝑐
• Where:
PE = potential energy (ft-lbf)
m = mass (lbm)
z = height above some reference level (ft)
g = acceleration due to gravity (ft/sec2)
gc = gravitational conversion constant 32.17 ft-lbm/lbf-sec2
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Energy - Potential
• Determine the potential energy of 50 lbm of water in a storage tank
100 ft above the ground.
𝑚𝑔𝑧
𝑃𝐸 =
𝑔𝑐
𝑓𝑡
2
100 𝑓𝑡
𝑠
𝑓𝑡 _ 𝑙𝑏𝑚
2
32.17
𝑙𝑏𝑓 _ 𝑠
50.0 𝑙𝑏𝑚
𝑃𝐸 =
32.17
3
𝑃𝐸 = 5.00 × 10 𝑓𝑡 _ 𝑙𝑏𝑓
• NOTE: Specific potential energy (pe) is the above value divided by
50
– 100 ft-lbf/lbm
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Energy - Kinetic
• Work needed to accelerate a
body from rest to its current
velocity
– Body maintains this kinetic
energy unless its speed
changes
• Kinetic energy (KE) is energy
that a body possesses as a
result of its motion
mv 2
KE 
2g c
• Where:
KE = kinetic energy (ft-lbf)
m = mass (lbm)
v = velocity (ft/sec)
gc = gravitational conversion
constant - 32.17 ft-lbm/lbf-sec2
• Using English system units
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Energy - Kinetic
• Determine the kinetic energy of 7 lbm of steam flowing through a pipe
at a velocity of 100 ft/sec.
2
𝑚v
𝐾𝐸 =
2𝑔𝑐
𝑓𝑡 2
7 𝑙𝑏𝑚 100.0
𝑠
𝐾𝐸 =
𝑓𝑡 _ 𝑙𝑏𝑚
2
2 32.17
𝑙𝑏𝑓 _ 𝑠
3
𝐾𝐸 = 1.088 × 10 𝑓𝑡 _ 𝑙𝑏𝑓
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Energy - Internal
• Potential and kinetic energy are macroscopic forms of energy
– Visualized in terms of position and velocity of objects
• Substances possess microscopic forms of energy including those
due to:
– Rotation
– Vibration
– Translation
– Interactions among molecules of a substance
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Energy - Internal
• Microscopic forms of energy collectively called internal energy
– Represented by symbol U
– British Thermal Unit (BTU) also unit of heat
o Function of temperature
• Specific internal energy (u) of a substance is its internal energy per
unit mass
– Total internal energy (U) divided by total mass (m)
𝑈
𝑢=
𝑀
– Where:
u = specific internal energy (Btu/lbm)
U = internal energy (Btu)
m = mass (lbm)
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Energy - Specific Internal
Example
• Determine the specific internal energy of 12 lbm of steam if the total
internal energy is 23,000 Btu.
𝑈
𝑢=
𝑀
2.300 × 104 𝐵𝑡𝑢
𝑢=
12 𝑙𝑏𝑚
103 𝐵𝑡𝑢
𝑢 = 1.917 ×
𝑙𝑏𝑚
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Energy – Flow Energy (PV)
• Energy arises from pressure (P) and volume (V) of a fluid
– Numerically equal to P x V
• A system where pressure and volume are permitted to expand
performs work on its surroundings (flow work)
– Energy defined as capacity of a system to perform work
– Fluid under pressure has capacity to perform work
• P-V energy (flow energy) foot-pounds force (ft-lbf)
• Specific P-V energy of a substance is P-V energy per unit mass
– Equals total P-V divided by total mass m, OR
– Product of pressure P and specific volume v written as Pv
o ft-lbf/lbm
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Energy - Pressure-Volume
Example
• Determine the specific P-V energy of 15 lbm of steam at 1000 psi in
an 18 ft3 tank.
𝑃𝑉
𝑃𝑣 =
𝑚
𝑃𝑣 =
3
𝑙𝑏𝑓
2
1,000
18 𝑓𝑡
𝑖𝑛
15.0 𝑙𝑏𝑚
2
144
𝑖𝑛
2
𝑓𝑡
𝑓𝑡 _ 𝑙𝑏𝑓
𝑃𝑣 = 1.73 × 10
𝑙𝑏𝑚
5
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Energy - Enthalpy
• Enthalpy (H) is measure of energy content of the fluid due to its
temperature, pressure, and volume
– Combination of internal and flow energies (H = U + PV)
• Specific enthalpy (h) defined as:
ℎ = 𝑢 + 𝑃𝜈
• Where:
u = specific internal energy (Btu/lbm) of system being studied
P = pressure of system (lbf/ft2)
ν = specific volume (ft3/lbm) of system
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Energy - Entropy
• Measure of inability to do work for a given heat transferred
– Quantifies energy of a substance that is no longer available
to perform useful work
– Represented by S
– Property of a substance like pressure, temperature, volume, and
enthalpy
• Steam tables include values of specific entropy (s = S/m) as part of
information tabulated
• Specific entropy (s) property is of no real value, but the Ds is
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Operator Generic Fundamentals
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Energy - Entropy
• Like enthalpy, entropy cannot
be measured directly
• Entropy of a substance is given
with respect to some reference
value – specific entropy of
water is zero at 32oF (492oR)
• Change in specific entropy (Δs),
not absolute value, important in
practical problems
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ELO 2.2
Figure: Entropy of Ice and Water
Operator Generic Fundamentals
53
Energy - Entropy
∆𝑄
∆𝑆 =
𝑇𝑎𝑏𝑠
∆𝑠 =
∆𝑞
𝑇𝑎𝑏𝑠
• Where:
ΔS = change in entropy of a system during some process (Btu/oR)
ΔQ = amount of heat transferred to or from system during process (Btu)
Tabs = absolute temperature at which heat was transferred (oR)
Δs = change in specific entropy of a system during some process
(Btu/lbm -oR)
Δq = amount of heat transferred to/from system during process (Btu/lbm)
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Operator Generic Fundamentals
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Properties of Energy
Knowledge Check
___________ is the measure of energy content of the fluid due to its
temperature, pressure, and volume.
A. Entropy
B. Kinetic energy
C. Enthalpy
D. Specific internal energy
Correct answer is C.
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ELO 2.2
Operator Generic Fundamentals
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Work, Energy, and Power
ELO 2.3 – Explain the relationship between work, energy, and power.
• Nuclear power plants transfer
thermal energy produced in nuclear
fuel into mechanical work of the
turbine-generator, and then
electrical energy
• Work is the applied force to move a
mass, multiplied by distance that
mass was moved; power is rate of
doing work (work done per unit time)
• Each term is related and must be
understood to solve thermodynamic
process equations
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ELO 2.3
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Work, Energy, and Power
• Work – measures completed task
• Energy – ability to do work
• Power – measures amount of work over time
• Power defined as time rate of doing work and is equivalent to
rate of energy transfer
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Work, Energy, and Power
• Work is:
– A form of energy in transit
– Not a property of a system
• Two types or work are mechanical and flow
• For mechanical systems, defined as action of a force on an object
through a distance
𝑊 = 𝐹𝑑
• Where:
W = work (ft-lbf)
F = force (lbf)
d = displacement (ft)
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Operator Generic Fundamentals
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Work, Energy, and Power
Example
• Determine the amount of work done if a force of 150 lbf is applied to
an object until it has moved a distance of 30 feet.
𝑊𝑜𝑟𝑘 = 𝐹𝑑
𝑊 = (150 𝑙𝑏𝑓)(30.0 𝑓𝑡)
𝑊 = 4.50 × 103 𝑓𝑡 _ 𝑙𝑏𝑓
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Work, Energy, and Power
• Flow is work required to maintain continuous steady movement of
fluid in a system
− Also force through a distance
• Flow work equivalent to force acting through a distance (such as
length) to maintain the flow
• 𝑊𝑓𝑙𝑜𝑤 = 𝐹𝐷
• Since 𝐹𝑜𝑟𝑐𝑒 = 𝑃𝐴 (𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 × 𝑎𝑟𝑒𝑎), 𝑊𝑓𝑙𝑜𝑤 = 𝑃𝐴𝐿
• Since 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝐴𝐿 (𝑎𝑟𝑒𝑎 × 𝑙𝑒𝑛𝑔𝑡ℎ), 𝑊𝑓𝑙𝑜𝑤 = 𝑃𝑉
Figure: Pipe Boundary Volume for Flow Energy and Related Formulas
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Operator Generic Fundamentals
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Work, Energy, and Power
• In dealing with work in relation to energy transfer systems, it is
important to distinguish between work done by the system on its
surroundings and work done on the system by its surroundings
– Work is done BY the system when used to turn a turbine and
thereby generate electricity in a turbine-generator (+ Work)
– Work is done ON the system when a pump is used to move
working fluid from one location to another (– Work)
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Operator Generic Fundamentals
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Work, Energy, and Power
• Units of various forms of energy are different but equivalent
• Potential, kinetic, internal, P-V, work, and heat may be measured in
numerous basic units
• Three types of units used to measure energy:
– Mechanical units such as foot-pound-force (ft-lbf)
– Thermal units such as British thermal unit (Btu)
– Electrical units such as watt-second (W-sec)
• In the mks and cgs systems (not testable):
– Mechanical units of energy are joule (j) and erg
– Thermal units are kilocalorie (kcal) and calorie (cal)
– Electrical units are watt-second (W-sec) and erg
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ELO 2.3
Operator Generic Fundamentals
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Work, Energy, and Power
• Power is the rate at which work is done, or work per unit time
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑃𝑜𝑤𝑒𝑟 =
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
• Power has units of energy per unit time
– mechanical units of power are ft-lbf/s or ft-lbf/hr and hp
– thermal units are BTU/hr
– electrical units are watts (W) or kilowatts (kW = 103 W)
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Operator Generic Fundamentals
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Work, Energy, and Power
• One horsepower is equivalent to 550 ft-lbf/s
𝐹𝑑
𝑃=
𝑡
• Where:
P = power (W or ft-lbf/s)
F = force (lbf)
d = distance (ft)
t = time (sec)
𝐹v
𝑃=
550
• Where:
P = power (hp)
F = force (lbf)
v = velocity (ft/s)
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Work, Energy, and Power
• Mechanical equivalent of heat often denoted by J, Joule’s constant
𝑓𝑡 _ 𝑙𝑏𝑓
𝐽 = 778
𝐵𝑡𝑢
• Other useful conversions, see formula sheet
1 𝐵𝑡𝑢 = 778 𝑓𝑡 _ 𝑙𝑏𝑓
– In other words, 1 BTU of heat can do 778 ft-lbf of work
𝐵𝑡𝑢
6
1 𝑀𝑊 = 3.41 × 10
ℎ𝑟
𝐵𝑡𝑢
1 ℎ𝑝 = 2.54 × 103
ℎ𝑟
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Operator Generic Fundamentals
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Properties of Heat
ELO 2.4 – Define the following terms: heat, sensible heat, latent heat,
specific heat, and super heat.
• Heat is energy in transition caused by a difference in temperature
• Sample heat transfer process
– Steam Generator
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Properties of Heat
• Heat is energy in transit
• Transfer of energy as heat occurs at molecular level as a result of a
temperature difference
• Symbol Q used to denote heat
• Unit of heat is the British thermal unit (Btu)
– Specifically called 60 degree Btu since it is measured by a one
degree temperature change from 59.5oF to 60.5oF
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Properties of Heat
• Amount of heat transferred depends upon path
• Important to distinguish between heat added to a system from its
surroundings and heat removed from a system to its surroundings
– Positive value for heat indicates heat is added to system by its
surroundings (+Q)
o Steam Generator
– Negative value for heat indicates heat is removed from system
by its surroundings (-Q)
o Condenser
– Contrast with work - positive when energy is transferred from the
system and negative when transferred to the system
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Operator Generic Fundamentals
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Properties of Heat
• q indicates heat added to or removed from a system per unit mass
– Equals total heat (Q) added or removed divided by mass (m)
– Quantity represented by q referred to as heat transferred per unit
mass
𝑄
𝑞=
𝑚
– Where:
q = heat transferred per unit mass (Btu/lbm)
Q = heat transferred (Btu)
m = mass (lbm)
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Properties of Heat
Example
• Determine heat transferred per unit mass if 1,500 Btu’s are
transferred to 40 lbm of water
Solution
𝑄
𝑚
1,500 𝐵𝑡𝑢
𝑞 =
40.0 𝑙𝑏𝑚
𝐵𝑡𝑢
𝑞 = 37.5
𝑙𝑏𝑚
𝑞=
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Operator Generic Fundamentals
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Properties of Heat - Sensible Heat
• Heat added to or removed from a substance to produce a change in
its temperature
– Units of heat often defined in terms of changes in temperature
they produce
• Basically the heat added to a subcooled liquid
– Feedwater entering the SG, for example
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ELO 2.4
Operator Generic Fundamentals
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Properties of Heat - Latent Heat
• Amount of heat added to or removed from a substance to produce a
change in phase
• When latent heat added/removed, no temperature change occurs
• Three types of latent heat
– Latent heat of vaporization - heat added or removed to change
phase between liquid and vapor
o Removal part normally called latent heat of condensation
– Latent heat of fusion - heat added or removed to change phase
between solid and liquid
o Not applicable to our thermodynamic processes
– Latent heat of sublimation - heat added or removed to change
phase between solid and vapor
o Not applicable to our thermodynamic processes
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Operator Generic Fundamentals
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Properties of Heat - Specific Heat
• Ratio of heat (Q) added to or removed from a substance to change in
temperature (ΔT) produced called heat capacity (Cp) of substance
• Heat capacity of a substance per unit mass called specific heat (cp) of
substance
– Cp and cp apply when heat is added or removed at constant
pressure
© Copyright 2016
ELO 2.4
Operator Generic Fundamentals
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Properties of Heat - Specific Heat
𝐶𝑝 =
𝑄
∆𝑇
𝑐𝑝 =
𝑄
𝑚∆𝑇
𝑐𝑝 =
𝑞
∆𝑇
• Where:
Cp = heat capacity at constant pressure (Btu/°F)
cp = specific heat capacity at constant pressure (Btu/lbm-°F)
Q = heat transferred (Btu)
q = heat transferred per unit mass (Btu/lbm)
m = mass (lbm)
∆𝑇 = temperature change (°F)
• One lbm of water raised 1°F and one Btu of heat added
– Implies specific heat (cp) of water is 1 Btu/lbm-°F
– cp of water equal to 1 Btu/lbm-°F at 39.1°F
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Properties of Heat
Specific Heat Example
Solution
• How much heat is required to
raise the temperature of 5 lbm
of water from 50oF to 150oF?
𝑄
𝐶𝑝 =
𝑚∆𝑇
𝑄 = 𝑚𝐶𝑝 ∆𝑇
Assume specific heat (cp) for
water is constant at 1.0 Btu/lbmoF
𝑄
1.0 𝐵𝑡𝑢
= (5 𝑙𝑏𝑚)
(150℉
𝑙𝑏𝑚_ ℉
− 50℉)
1.0 𝐵𝑡𝑢
𝑄 = (5 𝑙𝑏𝑚)
(100℉)
𝑙𝑏𝑚_ ℉
𝑄 = 500 𝐵𝑡𝑢
© Copyright 2016
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Operator Generic Fundamentals
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Properties of Heat - Super Heat
• Number of degrees above saturation temperature at a specific
pressure
• From previous discussions on heat and work, similarities evident:
– Heat and work both transient phenomena
– Systems never possess heat or work, but either or both may
occur when a system undergoes a change of energy state
– Both heat and work are boundary phenomena in that both are
observed at boundary of system
– Both represent energy crossing system boundary
© Copyright 2016
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Operator Generic Fundamentals
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Properties of Heat
Knowledge Check
Which of the following must be added to or removed from a substance
to produce a temperature change?
A. Latent heat
B. Specific heat
C. Sensible heat
D. Thermal heat
Correct answer is C.
© Copyright 2016
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Operator Generic Fundamentals
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Units and Properties Summary
• Relationships between inHg vac, inHg abs, and psia
– Majority of bank questions on this concept
• Relationship between psi and ft H20
– Recall 0.433 psi = 1 ft H2O (at low temps)
• DP Level Detector concepts discussed in 191002 – Sensors and
Detectors
• The first and second laws of thermodynamics
• Thermodynamic properties related to energy (PE, KE, internal
energy, enthalpy and entropy)
• The relationship of work, energy, and power
• Types of heat (sensible, latent, specific, and super)
© Copyright 2016
Summary
Operator Generic Fundamentals
78
NRC KA to ELO Tie
KA #
KA Statement
RO SRO
ELO
K1.01 Convert between absolute and gauge pressure and vacuum scales.
2.5
2.7
1.3
K1.02 Recognize the difference between absolute and relative (Kelvin) temperature scales.
1.9
2.0
1.2
K1.03 Describe how pressure and level sensing instruments work.
2.6
2.6
NOTE
K1.04 Explain relationships between work, power, and energy.
2.2
2.3
2.3
K1.05 Explain the law of conservation of energy.
2.1
2.1
2.1
NOTE K1.03 is covered (and tested) in 191002 - Sensors and Detectors
© Copyright 2016
Operator Generic Fundamentals