Operator Generic Fundamentals
Thermodynamics – Heat
Transfer
© Copyright 2014
Operator Generic Fundamentals
2
Heat Transfer Introduction
• Heat transfer is an important mechanism to understand and monitor
in all power plants.
An example of heat transfer:
• Traveling down a steep hill
in a vehicle towing a trailer
• Using the brakes to slow
the vehicle down will cause
the brakes to heat up due
to friction
• Could lead to the need
to pull over and allow
the brakes to cool
• Worst case would need
to use the emergency
run-away lane built into
many mountain roads.
© Copyright 2014
Introduction
Operator Generic Fundamentals
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Heat Transfer Introduction
• Heat transfer in a nuclear power plant.
– Reactor from fission and RCP heat
– Reactor coolant transports to Steam Generator (SG)
– Heat is transferred to secondary, while primary coolant back to
Reactor
– Steam produced in secondary used to turn Main Turbine /
Generator
– Extraction steam preheats feedwater
– Turbine exhausts to Main Condenser, where it is condensed
– Condensate flows through Condensate and Feed Pumps back to
SG
– Unused heat rejected to Cooling Towers or bay/lake
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Introduction
Operator Generic Fundamentals
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Heat Transfer Introduction
A loss of or reduction of heat transfer is a fundamental element of all
accidents that have occurred in the nuclear industry. One of the
primary design features is to ensure the heat produced in the nuclear
fuel is effectively removed.
Two of the Industry’s most significant events had severe plant
consequences due to the inability to remove heat from the reactor or
spent fuel pools.
How were Fukushima and Three Mile island affected by the loss of
heat transfer?
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Introduction
Operator Generic Fundamentals
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Fukushima Daiichi Units 1-6
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Introduction
Operator Generic Fundamentals
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Introduction
Operator Generic Fundamentals
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Introduction
Operator Generic Fundamentals
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Unit 1
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Introduction
Operator Generic Fundamentals
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Unit 1 Reactor Building
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Introduction
Operator Generic Fundamentals
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Control Room
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Introduction
Operator Generic Fundamentals
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Control Room
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Introduction
Operator Generic Fundamentals
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Unit 1 Supervisor
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Introduction
Operator Generic Fundamentals
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Unit 3 Reactor Building
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Introduction
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Unit 3 Reactor Building
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Introduction
Operator Generic Fundamentals
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Unit 4 Reactor Building
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Introduction
Operator Generic Fundamentals
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Unit 4 Damage Below Refuel Floor
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Introduction
Operator Generic Fundamentals
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Unit 4 Spent Fuel Pool – Fill Water
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Introduction
Operator Generic Fundamentals
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Introduction
Operator Generic Fundamentals
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Introduction
Operator Generic Fundamentals
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Heat Transfer Introduction
• These pictures are sobering reminder of why work in the Nuclear
Industry is considered ”Special and Unique”
• The production of hydrogen in each of these events is related to not
removing the heat produced in the nuclear fuel.
Heat produced in the nuclear fuel, must be removed during operation
and when shutdown, once fuel is used in a fuel cycle.
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Introduction
Operator Generic Fundamentals
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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. Describe heat transfer mechanisms and associated terminology.
2. Describe the heat transfer process in heat exchangers and factors
that reduce heat transfer.
3. Explain core thermal power and calculate its value using a
simplified heat balance.
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Introduction
Operator Generic Fundamentals
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Heat Transfer Mechanisms and
Terminology
TLO 1 – Describe heat transfer mechanisms and associated
terminology.
There are three modes or
mechanisms of heat transfer
that can occur between
materials. This section will
review these modes as well as
common terminology and
definitions associated with heat
transfer.
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TLO 1
Operator Generic Fundamentals
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Enabling Learning Objectives for TLO 1
1. Describe the relationship between heat, temperature, work, and the
Second Law of Thermodynamics.
2. Describe the three modes of heat transfer.
3. Define the following terms as they relate to heat transfer:
a. Heat flux
b. Thermal conductivity
c.
Log mean temperature difference
d. Convective heat transfer coefficient
e. Overall heat transfer coefficient
f.
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Bulk temperature
TLO 1
Operator Generic Fundamentals
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Heat, Temperature and Work
ELO 1.1 – Describe the relationship between heat, temperature, work,
and the second law of thermodynamics.
• Temperature and heat are distinctly different
Temperature
• Temperature measures amount of energy in the molecules of a
substance
• Predicts direction of heat transfer with known temperature conditions
• Symbol T, common temperature scales are Fahrenheit, Rankine,
Celsius, and Kelvin
Heat
• Heat is energy in transit, occurring at a molecular level.
• Heat energy transmits in three ways: conduction, convection, and
radiation
• Symbol Q, common units are the British Thermal Unit (BTU) and
calorie
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Heat, Temperature and Work
Heat and Temperature
• Heat is energy in transit, occurring at the molecular level because of
a temperature difference
• Heat energy transmits through
• Solids and fluids by conduction
• Fluids by convection
• Empty space by radiation
Figure: Heat Flow Direction
• The symbol for heat is Q, common units are
• British Thermal Unit (BTU) in the English system of units
• Calorie in the International System of Units (SI system).
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Heat, Temperature and Work
Heat and Work
• Both heat and work represent energy in transition, but there is a
distinction
• Work – the transfer of energy resulting from force acting through a
distance
• Heat – energy transferred as the result of a temperature difference
• Neither heat nor work is a thermodynamic property of a system.
• Transfer heat into or out of a system
• System can perform or receive work
• A system cannot contain or store either heat or work.
• Sign convention is that heat into a system and work out of a system
are positive quantities.
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Heat, Temperature and Work
Second Law of Thermodynamics
• Possible to convert work completely into heat, the converse not true
for a cyclic process.
• Certain natural processes always proceed in a certain direction (e.g.,
heat transfer occurs from a hot to a cold body).
The Second Law of Thermodynamics is explanation of these natural
phenomena. When a temperature difference exists across a boundary,
the second law indicates natural flow of energy is from hotter body to the
colder body.
The Second Law of Thermodynamics says it is impossible to convert all
the heat supplied to a system operating in a cycle into work.
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Heat, Temperature and Work
Second Law of Thermodynamics
The Second Law of Thermodynamics, described by Max Planck in 1903,
states it is impossible to construct an engine that will work in a complete
cycle and produce no other effect except the raising of a weight and the
cooling of a reservoir.
The second law says that if you draw heat from a reservoir to raise a
weight, lowering the weight will not generate enough heat to return the
reservoir to its original temperature, and eventually the cycle will stop.
If two blocks of metal at different temperatures, thermally insulated from
their surroundings, contact each other, heat will flow from the hotter to
the colder.
Eventually two blocks will reach same temperature, and heat transfer
ceases. Energy not lost, but some energy has transferred from one
block to another.
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Heat, Temperature and Work
Knowledge Check
____________ is a measure of the amount of energy possessed by the
molecules of a substance.
A. Heat
B. Work
C. Temperature
D. Convection
Correct answer is C.
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Heat Transfer Modes
ELO 1.2 – Describe the three modes of heat transfer.
The transfer of heat can occur by any one or a combination of three
modes; conduction, convection, and radiation. This chapter will
review each mode of heat transfer.
Figure: Modes of Heat Transfer
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Heat Transfer Modes
Conduction Heat Transfer
Conduction involves transfer of heat by interactions between adjacent
molecules of a material.
• Depends upon the driving "force" of temperature difference and the
resistance to heat transfer.
• The resistance to heat transfer depends upon the medium. All heat
transfer problems involve the temperature difference, the geometry,
and the physical properties of the object.
– Convection problems involve a fluid medium.
– Problems involving heat transfer by radiation study either solid or
fluid surfaces, separated by a gas, vapor, or vacuum.
There are several ways to correlate the geometry, physical properties,
and temperature difference of an object with the rate of heat transfer.
In conduction heat transfer, the most common correlation is Fourier’s
Law.
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Heat Transfer Modes
Conduction Heat Transfer – Fourier’s Law of Conduction
The law, in equation form, is used most often in its rectangular or cylindrical form
(pipes and cylinders), both are presented below.
Rectangular: 𝑄 = 𝑘𝐴
Cylindrical: 𝑄 = 𝑘𝐴
∆𝑇
∆𝑥
∆𝑇
∆𝑟
Where:
𝑄 = rate of heat transfer
𝐵𝑇𝑈
ℎ𝑟
A = cross-sectional area of heat transfer (ft2)
Δx = thickness of slab (ft)
Δr = thickness of cylindrical wall (ft)
ΔT = temperature difference (°F)
k = thermal conductivity of slab
𝐵𝑇𝑈
𝑓𝑡−ℎ𝑟−℉
The following examples demonstrate the use of the above equations in
determining the amount of heat transferred by conduction.
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Heat Transfer Modes
Conduction Heat Transfer
The following examples demonstrate
use of the equations in determining the
amount of heat transferred by
conduction.
Conduction-Rectangular Coordinates
Example:
1,000 BTU/hr is conducted through a
section of insulating material shown in
the figure below that measures 1 ft2 in
cross-sectional area. The thickness is 1
in. and the thermal conductivity is 0.12
BTU/hr-ft-°F.
Compute the temperature difference
across the material.
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ELO 1.2
Figure: Conduction through a Slab
Operator Generic Fundamentals
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Heat Transfer Modes
A concrete floor with a conductivity of 0.8 BTU/hr-ft-°F measures 30 ft
by 40 ft with a thickness of 4 inches. The floor has a surface
temperature of 70°F and the temperature beneath it is 60°F. What are
the heat flux and the heat transfer rate through the floor?
Heat Transfer Rate – Use
rectangular equation:
∆𝑇
𝑄 = 𝑘𝐴
= 𝑄"𝐴
∆𝑥
𝐵𝑇𝑈
= 24
1,200 𝑓𝑡 2
2
ℎ𝑟– 𝑓𝑡
𝐵𝑇𝑈
= 28,800
ℎ𝑟
Solution:
Heat Flux – Use Equation
𝑄
∆𝑇
𝑄" = = 𝑘
𝐴
∆𝑥
𝐵𝑇𝑈
10℉
= 0.8
ℎ𝑟– 𝑓𝑡– ℉ 0.333 𝑓𝑡
𝐵𝑇𝑈
= 24
ℎ𝑟– 𝑓𝑡 2
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Heat Transfer Modes
Convection Heat Transfer
Convection involves the transfer of heat by the motion and mixing of
"macroscopic" portions of a fluid (that is, the flow of a fluid past a solid
boundary).
• Natural convection – density variations resulting from temperature
differences within the fluid cause motion and mixing
• For example: transfer of heat from a hot water radiator to a room
• Forced convection – outside force, such as a pump, causes motion and
mixing
• For example: transfer of heat from the surface of a heat exchanger to
the bulk of a fluid pumped through the heat exchanger
•
More difficult to analyze heat transfer by convection than by conduction
•
•
No single property of the medium can describe the mechanism
Heat transfer by convection varies by situation (fluid flow conditions) and
mode of fluid flow
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Heat Transfer Modes
Convection Heat Transfer
Convection heat transfer is empirical because of the factors that affect the
stagnant film thickness:
• Fluid velocity
• Fluid viscosity
• Heat flux
• Surface roughness
• Type of flow (single-phase/two-phase)
Convection involves transfer of heat between a given surface temperature
(Ts) and fluid at a bulk temperature (Tb). The definition of Tb varies:
• For flow adjacent to a hot or cold surface, Tb = temperature of the fluid
"far" from the surface
• For boiling or condensation, Tb = saturation temperature of fluid
• For flow in a pipe, Tb = average temperature measured at a particular
cross-section
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Heat Transfer Modes
Convection Heat Transfer
The basic relationship for heat transfer by convection has same form as
conduction:
𝑄 = ℎ𝐴𝛥𝑇
Where:
𝑄 = rate of heat transfer (BTU/hr)
h = convective heat transfer coefficient (BTU/hr-ft2-°F)
A = surface area for heat transfer (ft2)
ΔT = temperature difference (°F)
Convective heat transfer coefficient (h) depends on physical properties of the
fluid situation.
• Typically, h for laminar flow < h for turbulent flow
• Turbulent flow has a thinner stagnant fluid film layer on the heat transfer
surface
• Values of h have been measured and tabulated for the commonly
encountered fluids and flow situations
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Heat Transfer Modes
Example:
A 22- foot long un-insulated steam line crosses a room. The outer diameter
of the steam line is 18 in. and the outer surface temperature is 280°F. The
convective heat transfer coefficient for the air is 18 BTU/hr-ft2-°F.
Calculate the heat transfer rate from the pipe into the room if the room
temperature is 72°F.
𝑄 = ℎ𝐴𝛥𝑇
= ℎ 2𝜋𝑟𝐿 ∆𝑇
𝐵𝑇𝑈
= 18
ℎ𝑟– 𝑓𝑡 2 – ℉
= 3.88 ×
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105
2 3.14 0.75 𝑓𝑡 22 𝑓𝑡
280℉ − 72℉
𝐵𝑇𝑈
ℎ𝑟
ELO 1.2
Operator Generic Fundamentals
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Heat Transfer Modes
Convective heat transfer takes place within pipes, tubes, or similar
cylindrical devices
• Equation 𝑄 = ℎ𝐴∆𝑇 varies as heat passes through the cylinder
• ∆𝑇 between inside and outside of the pipe, and the ∆𝑇 along the
pipe, necessitates the use of some average temperature value
• Called the log mean temperature difference (LMTD)
• The LMTD is ∆𝑇 at one end of heat exchanger minus ∆𝑇 at the other
end of the heat exchanger, divided by the natural logarithm of the
ratio of these two temperature differences
The definition for LMTD involves two important assumptions:
1. Fluid specific heats do not vary significantly with temperature
2. Coefficients of convection heat transfer are relatively constant
throughout the heat exchanger
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Operator Generic Fundamentals
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Heat Transfer Modes
Overall Heat Transfer Coefficient
Many heat transfer processes involve a combination of conduction and
convection
• For example: heat transfer in a steam generator involves
• Convection from the bulk of the fluid used to the steam generator
inner tube surface
• Conduction through the tube wall
• Convection from the outer tube surface to secondary side fluid
Combined heat transfer is a heat exchanger, involves
• h for the fluid film inside the tubes
• h for the fluid film outside the tubes
• Other key properties: thermal conductivity (k), tube wall thickness (Δx)
The overall heat transfer coefficient, used instead to relate the total rate
of heat transfer 𝑄 to the cross-sectional area for heat transfer (Ao)
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Heat Transfer Modes
Overall Heat Transfer Coefficient
The equation below, shows the relationship of the overall heat transfer
coefficient to the individual conduction and convection terms.
Figure: Overall Heat Transfer Coefficients
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Heat Transfer Modes
The figure shows “Combined Heat
Transfer”, illustrates an example of this
concept applied to cylindrical
geometry, which shows a typical
combined heat transfer situation.
1
𝑈𝑜 =
𝐴𝑜
∆𝑟𝐴𝑜
𝐴
+
+ 𝑜
ℎ1 𝐴1 𝑘𝐴𝑙𝑚 ℎ2 𝐴2
Figure: Combined Heat Transfer
© Copyright 2014
Heat transfer by
convection occurs
between
temperatures T1 and
T2; heat transfer by
conduction occurs
between
temperatures T2 and
T3; and heat transfer
occurs by convection
between
temperatures T3 and
T4. Thus, there are
three processes
involved.
ELO 1.2
Operator Generic Fundamentals
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Heat Transfer Modes
𝑈𝑜 =
1
𝐴𝑜
∆𝑟𝐴𝑜
𝐴
+
+ 𝑜
ℎ1 𝐴1 𝑘𝐴𝑙𝑚 ℎ2 𝐴2
The above equation for the overall heat transfer coefficient in cylindrical
geometry is relatively difficult to work with. We can simplify the equation
without losing much accuracy if we are analyzing a thin-walled tube; that
is, if the tube wall thickness is small compared to the tube diameter. For a
thin-walled tube, the inner surface area (A1), outer surface area (A2), and
log mean surface area (Alm), are all very close to being equal. Assuming
that A1, A2, and Alm are equal to each other and equal to Ao allows us to
cancel out all the area terms in the denominator.
This results in a much simpler expression that is similar to the one
developed for a flat plate heat exchanger in the figure, Overall Heat
Transfer Coefficients.
1
𝑈𝑜 =
© Copyright 2014
1 ∆𝑟 1
( +
+ )
ℎ1 𝑘 ℎ2
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Operator Generic Fundamentals
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Heat Transfer Modes
Convection Heat Transfer Calculations
Example: Consider a flat wall exposed to the environment. A layer of insulation
1 inch thick whose thermal conductivity is 0.8 BTU/hr-ft-°F covers the wall. The
temperature of the wall on the inside of the insulation is 600°F. The wall loses
heat to the environment by convection on the surface of the insulation. The
average value of the convection heat transfer coefficient on the insulation
surface is 950 BTU/hr-ft2-°F. Compute the bulk temperature of the environment
(Tb) if the outer surface of the insulation does not exceed 105°F.
Solution:
a. Find heat flux 𝑄" through
the insulation.
b. Find the bulk temperature of
the environment.
𝑄 = 𝑘𝐴(∆𝑇/∆𝑥)
𝑄 = ℎ𝐴 𝑇𝑖𝑛𝑠 − 𝑇𝑏
𝑄
𝐵𝑇𝑈
= 4,752
𝐴
ℎ𝑟– 𝑓𝑡 2
𝑄"
𝑇𝑏 = 𝑇𝑖𝑛𝑠 −
ℎ
𝑇𝑏 = 100℉
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Heat Transfer Modes
Radiant Heat Transfer and Thermal Radiation
Radiant heat transfer is thermal energy transferred by means of
electromagnetic waves or particles
• Involves transfer of heat by electromagnetic radiation
• Most energy in infrared region, though some is in visible region
The term thermal radiation distinguishes this form of electromagnetic
radiation from other forms(such as radio waves, x-rays, or gamma rays)
Radiant heat transfer does not need a medium to take place.
•
Any material with a temperature above absolute zero gives off some
radiant energy
•
For example: the transfer of heat from a fireplace across a room
•
For example: when a cloud covers the sun, both its heat and light
diminish
Due to high mass flow rates and relatively low temperatures in the
reactor, this mode of heat transfer is not significant in power production.
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Heat Transfer Modes
Radiant Heat Transfer and Thermal Radiation
Example:
During a loss-of-coolant accident, which one of the following heat
transfer mechanisms provides the most core cooling when fuel
elements are not in contact with the coolant?
A. Radiation
B. Emission
C. Convection
D. Conduction
The correct answer is A.
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Heat Transfer Modes
Knowledge Check
Refer to the drawing of a fuel rod and coolant flow channel at the
beginning of a fuel cycle (see figure below).
Which one of the following is the primary method of heat transfer
through the gap between the fuel pellets and the fuel cladding?
A. Conduction
B. Convection
C. Radiation
D. Natural Circulation
Correct answer is A.
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Operator Generic Fundamentals
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Heat Transfer Terms
ELO 1.3 – Define the following terms as they relate to heat transfer: heat flux,
thermal conductivity, log mean temperature difference, convective heat,
transfer coefficient, overall heat transfer coefficient, and bulk temperature.
Heat Flux
• The symbol 𝑄 represents the rate at which heat transfer occurs
• A common unit for heat transfer rate is BTU/hr
• Sometimes it is important to determine the heat transfer rate per unit
area, or heat flux, which has the symbol 𝑄"
• Units for heat flux are BTU/hr-ft2
• The heat flux can be determined by dividing the heat transfer rate by
the area through which the heat transfers
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Heat Transfer Terms
Heat Flux
• The symbol 𝑄 represents the rate at which heat transfer occurs
• A common unit for heat transfer rate is BTU/hr
• Sometimes it is important to determine the heat transfer rate per unit
area, or heat flux, which has the symbol 𝑄"
• Units for heat flux are BTU/hr-ft2
• The heat flux can be determined by dividing the heat transfer rate by
the area through which the heat transfers
𝑄
𝑄" =
𝐴
Note: One example of where heat flux is used is for core thermal limits (covered
in detail in a later module) where operating limits (average liner power density)
are imposed to ensure fuel cladding integrity is maintained, because actual heat
flux is always less than critical heat flux (CHF) and Departure from Nucleate
Boiling (DNB) does not occur.
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Heat Transfer Terms
Thermal Conductivity (k)
• Thermal conductivity is
• Describes heat transfer characteristics of a solid material
• A measure of a substance’s ability to transfer heat through a solid
by conduction
• Measured in BTU/hr-ft-°F
• Varies with temperature
• For vapors, it depends upon pressure
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Heat Transfer Terms
Log Mean Temperature Difference
In heat exchanger applications, the specified inlet and outlet
temperatures depend on fluid in tubes.
When the temperature change is not linear a precise temperature
change between two fluids across the heat exchanger is best
represented by log mean temperature difference (LMTD or ΔTlm)
(∆𝑇2 − ∆𝑇1 )
∆𝑇𝑙𝑚 =
∆𝑇2
ln
Where:
∆𝑇1
ΔT2 = Larger temperature difference between the two fluid streams at
either the entrance or the exit to the heat exchanger
ΔT1 = Smaller temperature difference between the two fluid streams at
either the entrance or the exit to the heat exchanger
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Heat Transfer Terms
Convective Heat Transfer Coefficient (h)
• Also referred to as a film coefficient
• Defines, in part, the heat transfer due to convection
• Represents thermal resistance of a relatively stagnant layer of fluid
between a heat transfer surface and the fluid medium
• Common units are BTU/hr-ft2-°F
Overall Heat Transfer Coefficient (Uo )
• Used in the case of combined heat transfer
• Relates the total rate of heat transfer, the overall cross-sectional area
for heat transfer (Ao), and the overall temperature difference (ΔTo)
• Combines the heat transfer coefficient of the two heat exchanger
fluids and the thermal conductivity of the heat exchanger tubes
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Heat Transfer Terms
Overall Heat Transfer Coefficient
𝑄 = 𝑈𝑜𝐴𝑜Δ𝑇𝑜
Where:
𝑄 = the rate heat of transfer (BTU/hr)
Uo = the overall heat transfer coefficient (BTU/hr-ft2-°F)
Ao = the overall cross-sectional area for heat transfer (ft2)
ΔTo = the overall temperature difference (°F)
Bulk Temperature (Tb)
• Varies according to the details of the situation
• For flow adjacent to a hot or cold surface, Tb = temperature of the
fluid that is "far" from the surface
• For boiling or condensation, Tb = saturation temperature
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Operator Generic Fundamentals
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Heat Transfer Terms
Knowledge Check
The heat transfer characteristics of a solid material are measured by a
property called ___________.
A. bulk temperature
B. heat flux
C. thermal conductivity
D. conduction
Correct answer is C.
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TLO 1 Summary
In this section you learned:
1.
Heat is energy transferred as a result of a temperature difference.
2.
Temperature is a measure of the amount of molecular energy
contained in a substance.
3.
Work is a transfer of energy resulting from force acting through a
distance.
4.
The Second Law of Thermodynamics implies that heat will not
transfer from a colder to a hotter body without some external source
of energy.
5.
Conduction involves the transfer of heat by the interactions of atoms
or molecules of a material through which the heat is being
transferred.
6.
Convection involves the transfer of heat by the mixing and motion of
macroscopic portions of a fluid.
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TLO 1 Summary
7.
Radiation, or radiant heat transfer, involves the transfer of heat by
electromagnetic radiation that arises due to temperature of a body.
8.
Heat flux is the rate of heat transfer per unit area.
9.
Thermal conductivity is a measure of a substance’s ability to transfer
heat through itself.
10. Log mean temperature difference – ΔT that most accurately represents
the ΔT for a heat exchanger
(∆𝑇 − ∆𝑇 )
∆𝑇𝑙𝑚 =
2
1
∆𝑇
ln ∆𝑇2
1
11. Local heat transfer coefficient – represents a measure of the ability to
transfer heat through a stagnant film layer
12. Overall heat transfer coefficient – measure of the ability of a heat
exchanger to transfer heat from one fluid to another
13. Bulk temperature - temperature of fluid that best represents majority of
the fluid which is not physically connected to the heat transfer site
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TLO 1
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TLO 1 Summary
Now that you have completed this lesson, you should be able to:
1. Describe the relationship between heat, temperature, work, and the
Second Law of Thermodynamics.
2. Describe the three modes of heat transfer.
3. Define the following terms as they relate to heat transfer:
a. Heat flux
b. Thermal conductivity
c.
Log mean temperature difference
d. Convective heat transfer coefficient
e. Overall heat transfer coefficient
f.
© Copyright 2014
Bulk temperature
TLO 1
Operator Generic Fundamentals
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Heat Transfer Characteristics in Heat
Exchangers
TLO 2 – Describe the heat transfer process in heat exchangers and
factors that reduce heat transfer.
• Heat exchangers are devices used to transfer thermal energy from one
fluid to another without mixing the two fluids
• The transfer of thermal energy between fluids is one of the most
important and frequently used processes in engineering.
– Usually occurs in a device known as a heat exchanger
– Common applications of heat exchangers are boilers, fan coolers,
cooling water heat exchangers, and condensers
• Basic design of a heat exchanger normally has two fluids of different
temperatures separated by conducting medium (for ex: one fluid flowing
inside metal tubes and the other fluid flowing around the tube exteriors)
– In the fluids flowing, heat transfers by convection
– Heat transfers through the tube wall by conduction
© Copyright 2014
TLO 2
Operator Generic Fundamentals
59
Heat Transfer Characteristics in Heat
Exchangers
• Heat exchangers fall into several categories. They transfer heat by
convection and by conduction through the wall
• Classified as either single or two-phase exchangers.
• Single-phase heat exchangers are usually of the tube-and-shell type;
that is, the exchanger consists of a set of tubes in a container called
a shell (see the figure below).
Figure: Typical Tube and Shell Heat Exchanger
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TLO 2
Operator Generic Fundamentals
60
Enabling Learning Objectives for TLO 2
1. Describe the difference in the temperature profiles for counter-flow
and parallel flow heat exchangers.
2. Describe the differences between regenerative and nonregenerative heat exchangers.
3. Given the temperature changes across a heat exchanger, calculate
the log mean temperature difference for the heat exchanger.
4. Describe how fluid films, steam, or gases can affect heat transfer
and fluid flow in heat exchangers.
5. Calculate heat transfer rates and temperatures in heat exchangers.
© Copyright 2014
TLO 2
Operator Generic Fundamentals
61
Counter and Parallel Flow Heat Exchangers
ELO 2.1 Describe the difference in the temperature profiles for counterflow and parallel flow heat exchangers.
Heat exchangers may be extremely different in design and construction,
however their modes of operation and effectiveness depend largely on
the direction of the fluid flow within the exchanger.
• Two most common arrangements for flow paths in heat exchangers
are:
• Counter-flow – direction of the flow of one of the working fluids is
opposite to the direction to the flow of the other fluid
• Parallel flow – both fluids in the heat exchanger flow in same
direction
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
62
Counter and Parallel Flow Heat Exchangers
• This figure represents the
directions of fluid flow in
parallel and counter-flow
exchangers.
• Under comparable
conditions, more heat transfer
occurs in a counter-flow than
in a parallel flow heat
exchanger.
Figure: Fluid Flow Direction
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ELO 2.1
Operator Generic Fundamentals
63
Counter and Parallel Flow Heat Exchangers
The temperature profiles of the heat exchangers indicate two major
disadvantages in the parallel-flow design.
1. Large temperature difference at the ends causes large thermal stresses.
The opposing expansion and contraction of the construction materials due
to high fluid temperatures differences can lead to eventual material failure.
2. Temperature of cold fluid exiting the heat exchanger never exceeds lowest
temperature of the hot fluid. This relationship is a distinct disadvantage if
design purpose is to raise temperature of old fluid.
Figure: Heat Exchanger Temperature Profiles
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ELO 2.1
Operator Generic Fundamentals
64
Counter and Parallel Flow Heat Exchangers
• Parallel flow is used when fluids should exit at nearly same
temperature
• Counter-flow has three significant advantages over parallel flow:
1. More uniform ∆T between fluids minimizes thermal stresses
2. Outlet temperature of cold fluid can approach highest
temperature of hot fluid (inlet temperature)
3. Uniform temperature difference produces a more uniform rate of
heat transfer
• In both designs, heat transfer consists of both conduction and
convection
• Process takes place over entire length of exchanger
• The temperature of fluids as they flow through the exchanger
varies over the entire exchanger length
• Rate of heat transfer varies along length of exchanger tubes
based on temperature difference between hot and cold fluid
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ELO 2.1
Operator Generic Fundamentals
65
Counter and Parallel Flow Heat Exchangers
Boundary Layers
• Difficult to depict heat transfer across many different layer conditions
• The heat transfer coefficient depends on
• Film thickness – the thinner the film, the higher the heat transfer
coefficient
• Fluid’s thermal conductivity – increase in thermal conductivity
corresponds to an increase in heat transfer
• The velocity of the fluid affects the thickness of the fluid film.
• The greater the velocity, the thinner the fluid film
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ELO 2.1
Operator Generic Fundamentals
66
Counter and Parallel Flow Heat Exchangers
Knowledge Check
All of the following are advantages of counter-flow heat exchangers
EXCEPT:
A. More uniform delta T, resulting in less thermal stresses
B. Uniform temperature difference produces a more uniform rate of
heat transfer
C. Large temperature difference at the ends causes less thermal
stress
D. Outlet temperature of the cold fluid can approach the highest
temperature of the hot fluid
Correct answer is C.
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
67
Regenerative and Non-regenerative Heat
Exchangers
ELO 2.2 – Describe the differences between regenerative and nonregenerative heat exchangers.
Another heat exchanger classification depends on their function in a
particular system and whether they are regenerative or non-regenerative.
The non-regenerative application is the most frequently used and uses
two separate fluids for the hot and cold fluid.
A regenerative heat exchanger typically uses one fluid within a system,
taking the fluid from different areas of the system to serve as both the hot
and the cold fluids.
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
68
Regenerative and Non-regenerative Heat
Exchangers
Non-regenerative Heat Exchanger
The non-regenerative application is the most frequently used and
involves two separate fluids. One fluid cools or heats the other fluid, with
no interconnection between the two fluids. Heat transferred from the
hotter fluid usually transfers again, either rejected to the environment or
some other heat sink.
Figure: Non-Regenerative Heat Exchanger
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ELO 2.2
Operator Generic Fundamentals
69
Regenerative and Non-regenerative Heat
Exchangers
Regenerative Heat Exchanger
In a regenerative heat exchanger there is less energy loss from the total
system, some systems such as the reactor water clean-up or purification
system may use a combination of both regenerative and non –
regenerative heat exchangers.
Figure: Regenerative Heat Exchanger
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ELO 2.2
Operator Generic Fundamentals
70
Regenerative and Non-regenerative Heat
Exchangers
Purification system as an example:
• Reactor coolant passes through a regenerative heat exchanger
• Non-regenerative heat exchanger
• Demineralizer
• Back through regenerative heat exchanger, and returns to the primary.
In the regenerative heat exchanger, the water entering purification system
pre-heats the water returning to the primary system.
• Minimizes thermal stress
• Reduces the temperature of water entering the purification system,
allowing use of a smaller heat exchanger to achieve the desired
temperature for purification.
The primary advantage of a regenerative heat exchanger application is
conservation of system energy (that is, less loss of system energy due to
the cooling of the fluid).
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
71
Regenerative and Non-regenerative Heat
Exchangers
Knowledge Check
A regenerative heat exchanger typically uses two separate system
fluids in the heat exchange process.
A. True
B. False
Correct answer is B.
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
72
Log Mean Temperature Application to
Heat Exchangers
ELO 2.3 – Given the temperature changes across a heat exchanger,
calculate the log mean temperature difference for the heat exchanger.
The temperature change that takes place across the heat exchanger
from the entrance to the exit is not linear. A precise temperature change
between two fluids across the heat exchanger is best represented by the
log mean temperature difference (LMTD or ΔTlm), defined below:
∆𝑇𝑙𝑚 =
(∆𝑇2 − ∆𝑇1 )
∆𝑇
ln ∆𝑇2
1
Where:
ΔT2 = Larger temperature difference between the two fluid streams at
either the entrance or the exit to the heat exchanger
ΔT1 = Smaller temperature difference between the two fluid streams at
either the entrance or the exit to the heat exchanger
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ELO 2.3
Operator Generic Fundamentals
73
Log Mean Temperature Application to
Heat Exchangers
To solve certain heat exchanger problems, use log mean temperature
difference (LMTD or ∆Tlm ) before the heat removal is determined.
Example:
A liquid-to-liquid counter-flow heat exchanger is part of an auxiliary
system at a facility. The heat exchanger functions to heat a cold fluid
from 120°F to 310°F. Assuming that the hot fluid enters at 500°F and
leaves at 400°F, calculate the LMTD for the exchanger.
© Copyright 2014
ELO 2.3
Operator Generic Fundamentals
74
Log Mean Temperature Application to
Heat Exchangers
Log Mean Temperature Application Step-by-Step Table
Step
Action
Use the Log Mean
equation.
∆𝑇𝑙𝑚 =
Solve for ∆T's
Solve for ∆Tlm
𝛥𝑇2 = 400°𝐹 − 120°𝐹 = 280°𝐹
𝛥𝑇1 = 500°𝐹 − 310°𝐹 = 190°𝐹
(280°𝐹 − 190°𝐹)
280°𝐹
ln
190°𝐹
= 232℉
∆𝑇𝑙𝑚 =
∆𝑇𝑙𝑚
© Copyright 2014
(∆𝑇2 − ∆𝑇1 )
∆𝑇
ln 2
∆𝑇1
ELO 2.3
Operator Generic Fundamentals
75
Log Mean Temperature Application to
Heat Exchangers
Student Demonstration # 1
A liquid-to-liquid counter-flow heat exchanger is part of a cooling water system at a
facility. The heat exchanger functions to cool the system discharge flow from 225°F
to 110°F. Assuming that the cooling water enters at 75°F and leaves at 120°F,
calculate the LMTD for the exchanger.
Step
Action
1. Use the Log Mean
equation.
∆𝑇𝑙𝑚 =
2. Solve for ∆T's
𝛥T2 = 225 − 110 = 115℉
(∆𝑇2 − ∆𝑇1 )
∆𝑇
ln ∆𝑇2
1
𝛥T = 120 − 75 = 45℉
1
3. Solve for ∆Tlm
© Copyright 2014
𝛥𝑇𝑙𝑚 =
115 − 45
= 75℉
115
ln
45
ELO 2.3
Operator Generic Fundamentals
76
Log Mean Temperature Application to
Heat Exchangers
Log Mean Temperature Application to Heat Exchangers
Demonstration
The solution to a heat exchanger problem may be simple so a straightforward
overall balance suffices, or may be complicated as to require integral calculus.
For example, it is possible to analyze a steam generator by an overall energy
balance from the feedwater inlet to the steam outlet in which the amount of heat
transferred is simply 𝑄 = 𝑚𝛥ℎ, where 𝑚 is the mass flow rate of the secondary
coolant and ∆h is the change in enthalpy of the fluid.
The same steam generator can also be analyzed by an energy balance on the
primary flow stream with the equation 𝑄 = 𝑚𝑐𝑝 𝛥𝑇, where 𝑚, cp, and ∆T are the
mass flow rate, specific heat capacity, and temperature change of the primary
coolant.
The heat transfer rate of the steam generator can also be determined by
comparing the temperatures on the primary and secondary sides with the heat
transfer characteristics of the steam generator using the equation 𝑄 = 𝑈𝑜 𝐴𝑜 𝛥𝑇𝑙𝑚
© Copyright 2014
ELO 2.3
Operator Generic Fundamentals
77
Log Mean Temperature Application to
Heat Exchangers
Condensers are also examples of components requiring the concept of
LMTD to address certain problems.
• When the steam enters the condenser, it gives up its latent heat of
vaporization to the circulating water and changes phase to a liquid.
• Because condensation is taking place, it is appropriate to term this
the latent heat of condensation.
• After steam condenses, the saturated liquid will continue to transfer
some heat to the circulating water system as it continues to fall to the
bottom of the condenser (hotwell). This continued cooling is called
subcooling and is necessary to prevent cavitation in the condensate
pumps.
Approach the solution to condenser problems in the same manner as
those for steam generators, as shown in the following example, which
first solves for the overall heat transfer coefficient Uo.
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ELO 2.3
Operator Generic Fundamentals
78
Log Mean Temperature Application to
Heat Exchangers
Overall Heat Transfer Coefficient
When dealing with heat transfer across heat exchanger tubes, first
calculate an overall heat transfer coefficient, Uo.
Earlier in this lesson, we looked at a method for calculating Uo for both
rectangular and cylindrical coordinates. Since the thickness of a
condenser tube wall is so small and the cross-sectional area for heat
transfer is relatively constant, we can use the equation below to calculate
U o.
1
𝑈𝑜 =
1 ∆𝑟 1
+
+
ℎ1 𝑘 ℎ2
© Copyright 2014
ELO 2.3
Operator Generic Fundamentals
79
Log Mean Temperature Application to
Heat Exchangers 𝑈 =
1
Calculate the heat transfer rate
per ft of tube from a condenser
under the following conditions:
• ∆Tlm = 232°F.
• Outer diameter of the copper
condenser tube is 0.75 in.
• Wall thickness of 0.1 in.
• The inner convective heat
transfer coefficient is 2,000
BTU/hr-ft2-°F
• The thermal conductivity of
copper is 200 BTU/hr-ft-°F
• The outer convective heat
transfer coefficient is 1,500
BTU/hr-ft2-°F
© Copyright 2014
𝑜
1 ∆𝑟 1
+
+
ℎ1 𝑘 ℎ2
1
=
1
0.1 𝑖𝑛 1 𝑓𝑡
1
+
×
+
2,000
200
12 𝑖𝑛 1,500
𝐵𝑇𝑈
= 827.6
ℎ𝑟– 𝑓𝑡 2 – ℉
𝑄 = Uo Ao ΔTlm
𝑄 Uo Ao ΔTlm
=
𝐿
L
= 𝑈𝑜 2𝜋𝑟ΔTlm
= 827.6
𝐵𝑇𝑈
ℎ𝑟– 𝑓𝑡 2 – ℉
= 37,700
ELO 2.3
2𝜋 0.375 𝑖𝑛
1 𝑓𝑡
12 𝑖𝑛
232℉
𝐵𝑇𝑈
ℎ𝑟– 𝑓𝑡
Operator Generic Fundamentals
80
Log Mean Temperature Application to
Heat Exchangers
Knowledge Check – NRC Question
Which one of the following pairs of fluids undergoing heat transfer in
similar cross-flow design heat exchangers will yield the greatest heat
exchanger overall heat transfer coefficient? (Assume comparable heat
exchanger sizes and fluid flow rates).
A. Oil to water in a lube oil cooler
B. Steam to water in a feedwater heater
C. Water to air in a ventilation heating unit
D. Water to water in a cooling water heat exchanger
Correct answer is B.
© Copyright 2014
ELO 2.3
Operator Generic Fundamentals
81
Log Mean Temperature Application to
Heat Exchangers
Knowledge Check
In a counter-flow, auxiliary cooling heat exchanger the cooling water
enters at 78˚F and exits at 99˚F, the system flow enters the heat
exchanger at 180˚F and exits at 110˚F. The log mean temperature
difference across this heat exchanger is:
A. 40.9˚F
B. 47.4˚F
C. 52.7˚F
D. 57.4˚F
Correct answer is C.
© Copyright 2014
ELO 2.3
Operator Generic Fundamentals
82
Factors that Impact Heat Transfer
ELO 2.4 – Describe how fluid films, steam, or gases can affect heat
transfer and fluid flow in heat exchangers.
• Heat exchangers depend on good heat transfer across the tube
bundles and film boundaries of the exchanger
• Any degradation to the heat transfer surfaces, film boundary, or fluid
flow will negatively affect
• Heat transfer rate
• Heat exchanger effectiveness
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ELO 2.4
Operator Generic Fundamentals
83
Factors that Impact Heat Transfer
Fluid Film Layers
One condition that will affect heat transfer is the formation of layers of
fluid films along the heat transfer surfaces. The thinner the fluid film, the
higher the convection heat transfer will be.
Several factors affect the heat transfer through a fluid film. The film
thickness and fluid’s thermal conductivity determine the heat transfer
coefficient.
The velocity of the fluid affects the thickness of the fluid film. The greater
the velocity the thinner the fluid film, and the higher the heat transfer
coefficient.
The fluid’s thermal conductivity will also affect the heat transfer
coefficient. An increase in thermal conductivity will correspond to an
increase in heat transfer coefficient.
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
84
Factors that Impact Heat Transfer
Fluid Gas and Air Impact
Venting of any air and non-condensable gasses is important for proper
operation of the heat exchanger. Pockets of air or gas can:
• Decrease the cooling surface area
• Disrupt flow through the heat exchanger
Boiling Impact
Nucleate boiling along the heat transfer surface will initially
• Break up the film layer and improve heat transfer in a heat exchanger
If boiling increases to the point where bulk boiling is occurring, the
turbulence will
• Restrict fluid flow across the tubes
• Reduce the heat transfer rate in the heat exchanger.
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
85
Factors that Impact Heat Transfer
HX Fouling Reduces Heat Transfer
Condition in a heat exchanger characterized by foreign material such as
algae, scale, or debris accumulating in a heat exchanger.
Fouling of tubes lowers the efficiency of heat exchanger by decreasing
the thermal conductivity of the tubes.
To transfer heat, the tubes must transfer through the fouling layer. There
are several methods to remove fouling from heat exchanger tubes.
• Hydro lancing
• Chemical cleaning
• Operating practices (the most effective)
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
86
Factors that Impact Heat Transfer
Heat Transfer Rate across a Heat Exchanger
One method of determining the heat added or removed across a heat
exchanger uses the mass flow rate and change in enthalpy.
The equation 𝑄 = 𝑚𝛥ℎ,
where:
𝑚 equals the mass flow rate of the coolant
∆h is change in enthalpy of the fluid
This type of problem will require use of either steam tables or the Mollier
diagram to determine enthalpies.
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
87
Factors that Impact Heat Transfer
Example Heat Transfer Problem:
A nuclear power plant is operating near 100 percent power. Extraction steam
from the main turbine supplies a feedwater heater. Extraction steam parameters
are as follows:
Steam pressure =
414 psia
Steam flow rate
=
7.5 x 105 lbm/hr
Steam enthalpy
=
1,150 BTU/lbm
The extraction steam condenses to saturated water at 414 psia, and then leaves
the feedwater heater via a drain line.
What is the heat transfer rate from the extraction steam to the feedwater in the
feedwater heater?
A.
3.8 x 107 BTU/hr
B.
8.6 x 107 BTU/hr
C. 4.0 x 108 BTU/hr
D. 7.2 x 108 BTU/hr
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
88
Factors that Impact Heat Transfer
To solve this problem, use equation 𝑄 = 𝑚𝛥ℎ, and solve for heat
transfer.
5
𝑄 = 7.5 × 10 𝑙𝑏𝑚/ℎ𝑟 (ℎ𝑠𝑡𝑚 − ℎ𝑤𝑎𝑡𝑒𝑟 ), use the steam tables to look up
the enthalpy of saturated water at 414 psia (extrapolated to 430
BTU/lbm)
5
= 7.5 × 10 𝑙𝑏𝑚/ℎ𝑟 (1,150 𝐵𝑇𝑈/𝑙𝑏𝑚 − 430 𝐵𝑇𝑈/𝑙𝑏𝑚)
5
= 7.5 × 10 𝑙𝑏𝑚/ℎ𝑟 (720 𝐵𝑇𝑈/𝑙𝑏𝑚)
8
= 5.4 × 10 𝐵𝑇𝑈/ℎ𝑟
Therefore, the correct answer is C.
© Copyright 2014
ELO 2.4
Operator Generic Fundamentals
89
Factors that Impact Heat Transfer
Knowledge Check – NRC Bank
A nuclear power plant is initially operating at a steady-state power level with the
following main condenser parameters:
Main condenser pressure
=
1.2 psia
Cooling water inlet temperature
=
60°F
Cooling water outlet temperature
=
84°F
Due to increased condenser air in-leakage, the overall heat transfer coefficient
of the main condenser decreases by 25 percent. Main condenser heat transfer
rate and cooling water temperatures are unchanged. Which one of the
following is the steady-state main condenser pressure resulting from the
reduced heat transfer coefficient?
A.
1.7 psia
B.
2.3 psia
C.
3.0 psia
D.
4.6 psia
© Copyright 2014
Correct answer is A.
ELO 2.4
Operator Generic Fundamentals
90
Solving Heat Exchanger Problems
ELO 2.5 – Calculate heat transfer rates and temperatures in heat
exchangers.
Heat transfer in a heat exchanger occurs by conduction and convection.
Calculate the rate of heat transfer, 𝑄, in a heat exchanger using the following
equation: 𝑄 = Uo Ao ΔTlm
The heat exchanger balance depends on a few key characteristics such as mass
flow, specific heat capacity of the fluids, and change in temperature.
𝑚1 𝐶𝑝1 ∆𝑡1 = 𝑚2 𝐶𝑝2 ∆𝑡2
Where:
𝑚 = mass flowrate
Cp = specific heat capacity
Δt = temp change across heat exchanger (∆Tlm is referred to as the log mean
temperature difference)
Using this heat balance equation it is possible to calculate the heat transfer rate,
change in mass flow rate or change in temperature of either fluid in a heat
exchanger.
© Copyright 2014
ELO 2.5
Operator Generic Fundamentals
91
Solving Heat Exchanger Problems
Example Heat Transfer Problem:
A plant auxiliary heat exchanger is providing cooling to a lube oil cooler
with the following parameters:
Toil in = 165°F
Toil out = 110°F
cp-oil = 1.1 BTU/lbm-°F
𝑚𝑜𝑖𝑙 = 3.0 x 104 lbm/hr
Twater in = 65°F
Twater out = 95°F
cp-water = 1.0 BTU/lbm-°F
What is the heat transfer rate of the oil across the heat exchanger?
© Copyright 2014
ELO 2.5
Operator Generic Fundamentals
92
Solving Heat Exchanger Problems
What is the heat transfer rate of the oil across the heat exchanger?
Step
Solve for 𝑄
of the oil
© Copyright 2014
Formula
𝑄𝑜𝑖𝑙 = 𝑚𝑜𝑖𝑙 𝑐 𝑝𝑜𝑖𝑙 ∆𝑡𝑜𝑖𝑙
Solution
𝑙𝑏𝑚
𝐵TU
𝑄 = (3.0 × 10
)(1.1
)(55°𝐹)
ℎ𝑟
𝑙𝑏𝑚– ℉
6 𝐵TU
= 1.815 × 10
ℎ𝑟
4
ELO 2.5
Operator Generic Fundamentals
93
Solving Heat Exchanger Problems
What is the heat transfer rate of the water across the heat exchanger?
Q Oil = Q Water = 1.815 x 106 BTU/hr
What is the mass flow
rate of the cooling water?
Step
Formula
Solve for 𝑄
𝑊𝑎𝑡𝑒𝑟
𝑄 of the = 𝑚
𝑐
∆𝑡𝑤𝑎𝑡𝑒𝑟
𝑤𝑎𝑡𝑒𝑟 𝑝𝑤𝑎𝑡𝑒𝑟
oil
Solution
𝐵TU
ℎ𝑟
6
𝐵TU
10 𝐵TU
1.0
30°𝐹 1.815 ×
𝑙𝑏𝑚– ℉
ℎ𝑟
𝐵TU
1.0
(30°𝐹)
𝑙𝑏𝑚– ℉
𝑄 = 1.815 × 10
= 𝑚𝑤𝑎𝑡𝑒𝑟
= 𝑚𝑤𝑎𝑡𝑒𝑟
6.05 × 10
© Copyright 2014
ELO 2.5
4
𝑙𝑏𝑚
ℎ𝑟
6
= (𝑚𝑤𝑎𝑡𝑒𝑟 )
Operator Generic Fundamentals
94
Solving Heat Exchanger Problems
Example # 2 Heat Exchanger Question (Typical of NRC Bank
Question)
During a nuclear power plant outage, personnel plugged 5 percent of all steam
generator (SG) tubes due to wall thinning. Full power reactor coolant system
flow rate and average reactor coolant temperature (Tavg) have not changed.
Here are the 100 percent power conditions before the outage:
Tavg = 578°F
TSG = 538°F
Which one of the following will be the approximate SG pressure after the outage
when the plant returns to 100 percent power? (Assume the overall heat transfer
coefficients for the SGs did not change.)
A.
960 psia
B.
930 psia
C. 900 psia
D. 870 psia
© Copyright 2014
ELO 2.5
Operator Generic Fundamentals
95
Solving Heat Exchanger Problems
Example # 2 Solution: Always write down initial conditions given for a problem.
SG has had 5% of tubes plugged during a plant outage. Reactor full system
flow and Tavg remain unchanged after the outage.
Given 100% power conditions before the outage: Tavg - 578˚F, and TSG - 538˚F
What is the SG pressure after the outage at 100% power?
Use 𝑄 = 𝑈𝐴𝛥𝑇. We know that the steam generator is at saturation, and we
were told that heat transfer coefficients did not change for this problem.
𝑈1 𝐴(𝑇𝑅𝑥 1– 𝑇𝑆𝐺 1) = 0.95 𝑈2 𝐴(𝑇𝑅𝑥 1– 𝑇𝑆𝐺 ),
The U and A terms cancel out, so
578 − 538 = 0.95(578– 𝑇𝑆𝐺 )
40 = 549.1 − 0.95𝑥
0.95𝑥 = 549.1 − 40
509.1
𝑥=
0.95
𝑇𝑆𝐺 = 535.9°𝐹, find 535.9 using the steam tables, which = 929.87 𝑝𝑠𝑖𝑎
The correct answer is B. 930 psia
© Copyright 2014
ELO 2.5
Operator Generic Fundamentals
96
Solving Heat Exchanger Problems
Knowledge Check – NRC Bank
Which one of the following will reduce the rate of heat transfer between
two liquids in a heat exchanger? (Assume single-phase conditions and
a constant specific heat for both liquids.)
A. The inlet temperatures of both liquids decrease by 20°F.
B. The inlet temperatures of both liquids increase by 20°F.
C. The inlet temperature of the hotter liquid increases by 20°F.
D. The inlet temperature of the colder liquid increases by 20°F.
Correct answer is D.
© Copyright 2014
ELO 2.5
Operator Generic Fundamentals
97
Solving Heat Exchanger Problems
Knowledge Check – NRC Bank
Refer to the drawing of an operating lube oil heat exchanger (see figure
below).
Increasing the oil flow rate through the heat exchanger will cause the oil
outlet temperature to __________ and the cooling water outlet
temperature to __________.
A. increase, increase
B. increase, decrease
C. decrease, increase
D. decrease, decrease
Correct answer is A.
© Copyright 2014
ELO 2.5
Operator Generic Fundamentals
98
TLO 2 Summary
In this section, you learned:
1.
The most common types of heat exchangers are counter-flow and parallel
flow. A counter-flow has the direction of the flow of one of the working fluids
opposite to the direction to the flow of the other fluid. In a parallel flow
exchanger, both fluids in the heat exchanger flow in the same direction.
2.
In a non-regenerative heat exchanger, one fluid cools or heats the other
fluid, with no interconnection between the two fluids. Heat that is removed
transferred from the hotter fluid is usually transfers again, either rejected to
the environment or some other heat sink.
3.
A regenerative heat exchanger typically uses the fluid from a different area
of the same system for both the hot and cold fluids
4.
The temperature change that takes place across the heat exchanger from
the entrance to the exit is not linear. The log mean temperature difference
(LMTD or ΔTlm) best represents the temperature change between two fluids
across the heat exchanger.
(∆𝑇2 − ∆𝑇1 )
∆𝑇𝑙𝑚 =
∆𝑇
ln ∆𝑇2
1
© Copyright 2014
TLO 2
Operator Generic Fundamentals
99
TLO 2 Summary
5. Heat exchangers depend on good heat transfer across the tube bundles and
film boundaries of the exchanger; any degradation to the heat transfer
surfaces, film boundary, or fluid flow will negatively affect the heat transfer
rate and heat exchanger effectiveness.
– The thinner the fluid film, the higher the convection heat transfer will be.
– Pockets of air or gas decrease the cooling surface area and the heat
exchanger may not function properly. These pockets of air or gases may
also disrupt flow through the heat exchanger.
– If boiling occurs in the heat exchanger and increases to the point where
bulk boiling is occurring, the turbulence will restrict fluid flow across the
tubes and reduce the heat transfer rate in the heat exchanger.
– One method of determining the heat added or removed across a heat
exchanger uses the mass flow rate and change in enthalpy across the
heat exchanger.
6. Overall heat transfer coefficient - The overall heat transfer coefficient
combines the heat transfer coefficient of the two heat exchanger fluids and
the thermal conductivity of the heat exchanger tubes. Uo is specific to the
heat exchanger and the fluids used in the heat exchanger.
© Copyright 2014
TLO 2
Operator Generic Fundamentals
100
TLO 2 Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Describe the difference in the temperature profiles for counter-flow
and parallel flow heat exchangers.
2. Describe the differences between regenerative and non-regenerative
heat exchangers.
3. If given the temperature changes across a heat exchanger, calculate
the log mean temperature difference for the heat exchanger.
4. Describe how fluid films, steam, or gases can affect heat transfer and
fluid flow in heat exchangers.
5. Calculate heat transfer rates and temperatures in heat exchangers.
© Copyright 2014
TLO 2
Operator Generic Fundamentals
101
Core Thermal Power
TLO 3 – Explain core thermal power and calculate its value using a
simplified heat balance.
Core Thermal Power (CTP) is a measure of heat input per unit time from
the reactor core to the reactor coolant.
Percent reactor power is the percent of rated core thermal power at
which the reactor is currently operating.
This section will cover calculations of core thermal power and percent
reactor power.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
102
Enabling Learning Objectives for TLO 3
1. Define the following terms:
a. Core thermal power
b. Percent reactor power
2. Explain the various methods
of calculating core thermal
power and calculate core
thermal power using a simple
heat balance.
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TLO 3
Operator Generic Fundamentals
103
Core Thermal Power and Percent Reactor
Power
ELO 3.1 Define the following terms: Core thermal power, Percent
reactor power.
Core thermal power (CTP) is an important parameter used to monitor
the reactor; the limits placed on reactor power distribution and power
level help ensure compliance with design and regulatory limits and
maintenance of health and safety of the public.
Core thermal power (CTP) is a measure of heat input per unit time from
the reactor core to the reactor coolant.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
104
Core Thermal Power and Percent Reactor
Power
Core Thermal Power
A heat balance sums all the energy inputs and outputs in a system
• Core thermal power uses the heat balance summation
• Plant process computer uses plant parameters to calculate
During steady state power operation, core thermal power is most
accurately determined by multiplying the total mass flow rate of the
feedwater by the change in enthalpy in the steam generators.
Nuclear instruments also derive and indicate core thermal power.
• Less accurate than the plant process computer, but serves as a rough
indication of reactor power.
The shape of the power distribution, the number, and placement of
nuclear instruments can affect the overall reactor power accuracy.
• Nuclear instruments will indicate the percent of reactor power, also
defined as a ratio actual reactor power to rated reactor power.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
105
Core Thermal Power and Percent Reactor
Power
Percent reactor power is the percent of rated core thermal power at which the
reactor is currently operating.
%𝑃𝑟𝑥 =
Example:
𝐶𝑇𝑃𝑎𝑐𝑡𝑢𝑎𝑙
× 100%
𝐶𝑇𝑃𝑟𝑎𝑡𝑒𝑑
A nuclear reactor is producing 500MW of core thermal power. The reactor
coolant pumps are adding 10MW of additional thermal power into the coolant
system based on heat balance calculations. The core carries a 1,500MW
thermal power rating. What is the core thermal power in percent?
𝐶𝑇𝑃𝑎𝑐𝑡𝑢𝑎𝑙
%𝑃𝑟𝑥 =
× 100%
𝐶𝑇𝑃𝑟𝑎𝑡𝑒𝑑
%𝑃𝑟𝑥 =
500 𝑀𝑊
× 100%
1,500 𝑀𝑊
%𝑃𝑟𝑥 = 33.3%
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
106
Core Thermal Power and Percent Reactor
Power
Knowledge Check – NRC Bank
During steady-state power operation, core thermal power can be most
accurately determined by multiplying the total mass flow rate of the...
A. reactor coolant by the change in temperature across the core.
B. reactor coolant by the change in enthalpy in the steam
generators.
C. feedwater by the change in enthalpy in the steam generators.
D. feedwater by the change in temperature across the core.
Correct answer is C.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
107
Core Thermal Power and Percent Reactor
Power
Knowledge Check – NRC Bank
A reactor is producing 200 MW of core thermal power. Reactor coolant
pumps are adding 10 MW of additional thermal power into the reactor
coolant system based on heat balance calculations. The core is rated
at 1,330 MW thermal power.
Which one of the following is the core thermal power in percent?
A. 14.0 percent
B. 14.3 percent
C. 15.0 percent
Correct answer is C.
D. 15.8 percent
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
108
Core Thermal Calculation
ELO 3.2 – Explain the various methods of calculating core thermal power
and calculate core thermal power using a simple heat balance.
Core thermal power is determined by calculating a heat balance across a
boundary
• The reactor vessel in the boundary used for the heat balance
• Total energy entering the system, plus the energy added to the
system should equal the energy leaving the system
𝑄in + 𝑄add = 𝑄out
• The energy entering the system is from the feedwater
• Energy added into the system comes from the heat from reactor
fission and the recirculation pump heat
• The energy leaving the system is from steam demand to the steam
system and ambient losses to atmosphere
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
109
Core Thermal Calculation
Using the original equation with these changes yields the following
equation:
𝑄fw + 𝐶𝑇𝑃 + 𝑄pmp = 𝑄stm + 𝑄amb
Now we can solve the equation for core thermal power:
𝐶𝑇𝑃 = 𝑄
stm
+𝑄
amb
−𝑄 −𝑄
fw
pmp
Normally, plant process computer performs
calculation, however, operators should
have understanding of how value is
calculated. The figure shows heat balance
terms relative to plant components. May
be necessary for plant personnel to perform
these calculations manually if computer is
unavailable, or as validation check of plant
computer value.
© Copyright 2014
ELO 3.2
Figure: Reactor Heat Balance
Operator Generic Fundamentals
110
Core Thermal Calculation
Example Heat Balance Question
Assume that the power range nuclear instruments have been adjusted to 100
percent based on a calculated heat balance. Which one of the following would
cause indicated reactor power to be greater than actual reactor power? Refer to
the figure above.
A.
Omitting the reactor coolant pump heat input term from the heat
balance calculation.
B.
The feedwater flow rate used in the heat balance calculation was lower
than actual feedwater flow rate.
C.
The steam pressure used in the heat balance calculation was 50 psi
higher than actual steam pressure.
D.
The enthalpy of the feed water was miscalculated to be 10 BTU/lbm
higher than actual feed water enthalpy.
NOTE: It is helpful to substitute Indicated Reactor Power for the CTP term as you
analyze the changes made to the heat balance.
𝐶𝑇𝑃 = 𝑄
stm
© Copyright 2014
+𝑄
amb
−𝑄 −𝑄
fw
pmp
ELO 3.2
Operator Generic Fundamentals
111
Core Thermal Calculation
Choice D analysis: if personnel miscalculated the enthalpy of the
feedwater higher than actual, indicated power would be higher than
actual.
Choice C analysis: if steam pressure value used is 50 psi higher than
actual steam pressure, the enthalpy will decrease for the higher steam
pressure, giving less Qstm (recall from the steam chapter and Mollier
diagram, that above ~ 500psi, enthalpy deceases as pressure is
increased), which will cause indicated power to indicate lower than
actual.
Choice B analysis: if feedwater flow rate used is lower than the actual
feed flow, then the Qfw term will indicate lower than actual value,
causing the indicated power to be less than actual CTP.
Choice A analysis: if the Qrcp term is left out of the calculation, then the
indicated power will be greater than the actual CTP.
Therefore, the correct answer is: A. The reactor coolant pump heat
input term was omitted from the heat balance calculation.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
112
Core Thermal Calculation
Example #2 – Core Thermal Power Problem
In a two-loop PWR nuclear power plant, feedwater flow rate to each
steam generator (SG) is 3.3 x 106 lbm/hr, at an enthalpy of 419
BTU/lbm. The steam exiting each SG is at 800 psia with 100 percent
steam quality.
Ignoring all other heat gain and loss mechanisms, what is the reactor
core thermal power?
A.
677 MW
B.
755 MW
C.
1,334 MW
D.
1,510 MW
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
113
Core Thermal Calculation
Solution: Core Thermal Calculation Step-by-Step Table
Ignoring all other heat gain and loss mechanisms:
1.
2.
Step
Action
Use the correct heat
transfer of CTP equation
CTP = 𝑄
−𝑄 −𝑄
fw
pmp
= 𝑚𝑓𝑤 × (ℎ𝑠𝑡𝑚 − ℎ𝑓𝑤 )
stm
+𝑄
amb
𝐵TU
6
Use steam tables if
= 3.3 × 10 (1,199.3 − 419
)
𝑙𝑏𝑚
necessary to get enthalpy,
𝐵TU
6 𝑙𝑏𝑚
solve for ∆h’s or ∆T
= 3.3 × 10
× (781.3
)
ℎ𝑟
𝑙𝑏𝑚
9 𝐵TU
= 2.578 × 10
(𝑝𝑒𝑟 𝑆𝐺)
ℎ𝑟9
9
= 2 × 2.578 × 10 (2 𝑆𝐺𝑠 ) = 5.16 × 10 (𝑡𝑜𝑡𝑎𝑙)
3.
Convert to MW if required 𝑄 = 5.16 × 109
1 𝑀𝑊
𝐵TU
3.413×106 ℎ𝑟
= 1,511 𝑀𝑊
The correct answer is D. 1,510 MW
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
114
Core Thermal Calculation
Student Practice
Example:
Reactor coolant enters a reactor core at 545°F and leaves at 595°F. The
reactor coolant flow rate is 6.6 x 107 lbm/hour and the specific heat
capacity of the coolant is 1.3 BTU/lbm-°F.
What is the reactor core thermal power?
A. 101 MW
B. 126 MW
C. 1,006 MW
D. 1,258 MW
𝑄 = 𝑚𝐶𝑝∆𝑡 = 6.6 x 107 lbm/hour × 1.3 BTU/lbm°F. x 50 ℉
= 4.29 x 109 BTU/hr
1 𝑀𝑊
3.413 × 106 𝐵𝑇𝑈/ℎ𝑟
= 1256.9 MW, Closest answer is D.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
115
Core Thermal Calculation
Knowledge Check – NRC Bank
A nuclear power plant is operating at power. Total feedwater flow rate
to all steam generators is 7.0 x 106 lbm/hr at a temperature of 440°F.
The steam exiting the steam generators is at 1,000 psia with 100
percent steam quality.
Ignoring all other heat gain and loss mechanisms, what is the reactor
core thermal power?
A. 1,335 MW
B. 1,359 MW
C. 1,589 MW
D. 1,612 MW
Correct answer is C.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
116
Core Thermal Calculation
Knowledge Check – NRC Bank
During a nuclear power plant outage, 6 percent of all steam generator
(SG) tubes were plugged. Full-power reactor coolant system flow rate
and average reactor coolant temperature (Tavg) have not changed.
Given the following 100 percent power conditions before the outage:
Tavg = 584°F
TS/G = 544°F
Which one of the following will be the approximate SG pressure after
the outage when the plant is returned to 100 percent power?
A. 974 psia
B. 954 psia
Correct answer is A.
C. 934 psia
D. 914 psia
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
117
TLO 3 Summary
1.
2.
3.
Core thermal power (CTP) is a measure of heat input per unit time from the
reactor core to the reactor coolant. Core thermal power is most accurately
determined by multiplying the total mass flow rate of the feedwater by the
change in enthalpy in the steam generators.
A heat balance sums all the energy inputs and outputs in a system;
determining core thermal power. The plant process computer uses plant
parameters to calculate core thermal power.
Percent reactor power is the percent of rated core thermal power at which the
reactor is currently operating.
𝐶𝑇𝑃𝑎𝑐𝑡𝑢𝑎𝑙
× 100%
𝐶𝑇𝑃𝑟𝑎𝑡𝑒𝑑
Core thermal power is determined by calculating a heat balance across a
boundary, the reactor vessel in the boundary used for the heat balance. The
total energy entering the system, plus the energy added to the system should
equal the energy leaving the system. 𝑄 + 𝑄 = 𝑄
%𝑃𝑟𝑥 =
4.
𝑖𝑛
𝑎𝑑𝑑
𝑜𝑢𝑡
𝑄𝑓𝑤 + 𝐶𝑇𝑃 + 𝑄𝑝𝑚𝑝 = 𝑄𝑠𝑡𝑚 + 𝑄𝑎𝑚𝑏
𝐶𝑇𝑃 = 𝑄 + 𝑄 − 𝑄 − 𝑄
𝑠𝑡𝑚
𝑎𝑚𝑏
𝑓𝑤
𝑝𝑚𝑝
© Copyright 2014
TLO 3
Operator Generic Fundamentals
118
TLO 3 Summary
5. Core thermal power problems are essentially heat transfer
problems.
a. One method of determining the heat added or removed across
a heat transfer boundary uses the mass flow rate and change in
enthalpy across the heat exchanger. The equation 𝑄 = 𝑚𝛥ℎ,
where 𝑚 equals the mass flow rate of the coolant and ∆h is the
change in enthalpy of the fluid expresses this method.
b. Depending on the parameters given for a specific problem, use
the equation 𝑄 = 𝑚𝑐𝑝 𝛥𝑇 to solve problems; remember that steam
tables and Mollier diagram may be necessary to obtain values to
solve equations.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
119
TLO 3 Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Define the following terms:
a. Core thermal power
b. Percent reactor power
2. Explain the various methods of calculating core thermal power and
calculate core thermal power using a simple heat balance.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
120
Heat Transfer Summary
Heat Transfer Highlights
1. Heat, temperature, work and the Second Law of Thermodynamics are inter-related.
Temperature is a measure of the amount of energy possessed by the molecules of a
substance. Heat is energy in transit, and will always flow from a hot body to a colder
body.
2. Work is the transfer of energy resulting from force acting through a distance. Heat is
energy transferred as the result of a temperature difference. Neither heat nor work is
a thermodynamic property of a system. It is possible to transfer heat into or out of a
system, and a system can perform or receive work, but a system cannot contain or
store either heat or work.
3. The Second Law of Thermodynamics states that it is impossible to construct an
engine that will work in a complete cycle and produce no other effect except the
raising of a weight and the cooling of a reservoir.
4. Heat transfer modes
a. Conduction involves the transfer of heat by the interactions of atoms or
molecules of a material through the material.
b. Convection involves the transfer of heat by the mixing and motion of
macroscopic portions of a fluid.
c. Radiation, or radiant heat transfer, involves the transfer of heat by
electromagnetic radiation that arises due to the temperature of a body.
© Copyright 2014
Summary
Operator Generic Fundamentals
121
Heat Transfer Summary
Heat Transfer Highlights
5. Log Mean Temperature Difference
In heat exchanger applications, the specified inlet and outlet temperatures depend
on the fluid in the tubes. The temperature change that takes place across the heat
exchanger from the entrance to the exit is not linear. A precise temperature change
between two fluids across the heat exchanger is best represented by the log mean
temperature difference (LMTD or ΔTlm).
6. Counter and Parallel Flow Heat Exchangers Details
The most common arrangements for flow paths within a heat exchanger are counterflow and parallel flow. A counter-flow heat exchanger is one in which the direction of
the flow of one of the working fluids is opposite to the direction to the flow of the
other fluid. In a parallel flow exchanger, both fluids in the heat exchanger flow in the
same direction.
7. Non-Regenerative Heat Exchanger
The non-regenerative application is the most frequently used and involves two
separate fluids. One fluid cools or heats the other fluid, with no interconnection
between the two fluids. Heat transferred from the hotter fluid usually transfers again,
either rejected to the environment or some other heat sink.
© Copyright 2014
Summary
Operator Generic Fundamentals
122
Heat Transfer Summary
Heat Transfer Highlights
8. Regenerative Heat Exchanger
A regenerative heat exchanger typically uses one fluid within a system, taking the fluid
from different areas of the system to serve as both the hot and the cold fluids.
9. Heat exchangers depend on good heat transfer across the tube bundles and film
boundaries of the exchanger; any degradation to the heat transfer surfaces, film
boundary, or fluid flow will negatively affect the heat transfer rate and heat exchanger
effectiveness.
a. The thinner the fluid film, the higher the convection heat transfer will be.
b. Pockets of air or gas decrease the cooling surface area and the heat exchanger
may not function properly, and can also disrupt flow through the heat exchanger.
c. If boiling occurs in the heat exchanger to the point where bulk boiling is occurring,
the turbulence restricts fluid flow across the tubes and reduces the heat transfer
rate.
d. One method of determining the heat added or removed across a heat exchanger
uses mass flow rate and change in enthalpy across the heat exchanger.
e. Overall heat transfer coefficient - The overall heat transfer coefficient combines
the heat transfer coefficient of the two heat exchanger fluids and the thermal
conductivity of the heat exchanger tubes. Uo is specific to the heat exchanger
and the fluids used in the heat exchanger: 𝑄 = 𝑈𝑜 𝐴𝑜 𝛥𝑇𝑜 .
© Copyright 2014
Summary
Operator Generic Fundamentals
123
Heat Transfer Summary
Heat Transfer Highlights
10. Heat transfer in a heat exchanger occurs by conduction and
convection. Calculate the rate of heat transfer, 𝑄, in a heat
exchanger using the following equation.
𝑄 = 𝑈𝑜 𝐴𝑜 𝛥𝑇lm
The heat exchanger balance is dependent on a few key characteristics
such as mass flow, specific heat capacity of the fluids, and change in
temperature.
11. A heat balance is used to determine core thermal power and
percent reactor power; power range monitors, adjusted based on
heat balance calculations, indicate reactor power.
© Copyright 2014
Summary
Operator Generic Fundamentals
124
Heat Transfer 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. Describe heat transfer mechanisms and associated terminology.
2. Describe the heat transfer process in heat exchangers and factors
that reduce heat transfer.
3. Explain core thermal power and calculate its value using a simplified
heat balance.
© Copyright 2014
Summary
Operator Generic Fundamentals