Operator Generic Fundamentals Thermodynamics – Heat Transfer

Operator Generic Fundamentals Thermodynamics – Heat Transfer
K1.01 Describe three mechanisms of heat transfer K1.02 Define thermal conductivity K1.03 Explain the manner in which fluid films affect heat transfer K1.04 Describe how the presence of gases or steam can affect heat transfer and fluid flow in heat exchangers K1.05 Define core thermal power K1.06 Explain methods of calculating core thermal power K1.07 Define percent reactor power K1.08 Calculate core thermal power using a simplified heat balance. Operator Generic Fundamentals Thermodynamics – Heat Transfer

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): Describe heat transfer mechanisms and associated terminology. Describe the heat transfer process in heat exchangers and factors that reduce heat transfer. Explain core thermal power and calculate its value using a simplified heat balance. Review TLOs with Class TLOs

Heat Transfer Mechanisms and Terminology
TLO 1 – Describe heat transfer mechanisms and associated terminology. 1.1 Describe the relationship between heat, temperature, work, and the Second Law of Thermodynamics. 1.2 Describe the three modes of heat transfer. 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 Bulk temperature TLO 1

Heat, Temperature and Work
ELO 1.1 – Describe the relationship between heat, temperature, work, and the second law of thermodynamics. Temperature Temperature - amount of energy in the molecules of a substance Predicts direction of heat transfer Hot to cold 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 ELO 1.1

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 ELO 1.1

Heat and Work 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 ELO 1.1

Second Law of Thermodynamics Possible to convert work completely into heat The converse not true for a cyclic process When a temperature difference exists across a boundary Second law indicates 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 Recall – increase in entropy ELO 1.1

Second Law of Thermodynamics Consider two blocks of metal at different temperatures Thermally insulated from their surroundings Come into contact with each other Heat will flow from the hotter to the colder Eventually two blocks will reach same temperature heat transfer ceases Energy not lost some energy has transferred from one block to another ELO 1.1

Knowledge Check ____________ is a measure of the amount of energy possessed by the molecules of a substance. Heat Work Temperature Convection Correct answer is C. Correct answer is C. This “temperature” energy is captured by the term “specific internal energy” (BTU/lbm) in Bernoulli’s Equation. ELO 1.1

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 - Transfer of heat the interactions between adjacent molecules Convection - Transfer of heat by a process of bulk motion and mixing of macroscopic portions of the fluid Radiation - The transfer of heat by electromagnetic radiation that arises due to the temperature of a body Most heat transfer processes are some combination of the above Several NRC bank questions will ask which of the following is the “primary mode of heat transfer” since most processes contain combinations of the above three modes. Figure: Modes of Heat Transfer ELO 1.2

Heat Transfer Modes Conductive Heat Transfer
Rate of heat transfer depends upon the driving "force" of the temperature difference the resistance to heat transfer the medium geometry Examples of conductive heat transfer processes Tubes in heat exchangers Fuel pellet, gap, cladding Gap pressurized with several hundred psia of helium Assists in conductive heat transfer (helium is not flowing) 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. ELO 1.2

The formula for conductive heat transfer depends on medium: Rectangular: 𝑄 =𝑘𝐴 ∆𝑇 ∆𝑥 Cylindrical: 𝑄 =𝑘𝐴 ∆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−℉ NOTE: main difference is either “thickness” of slab or “radius” of material A calculation using the above equation(s) are in the next section. ELO 1.2

Rectangular Coordinates Example: 1,000 BTU/hr is conducted through a section of insulating material 1 ft2 in cross-sectional area thickness is 1 in thermal conductivity is 0.12 BTU/hr-ft- °F. Compute the temperature difference across the material. Solution: Using rectangular equation: 𝑄 =𝑘𝐴 ∆𝑇 ∆𝑥 Solving for ΔT: ∆𝑇= 𝑄 ∆𝑥 𝑘𝐴 ∆𝑇= 1,000 𝐵𝑇𝑈 ℎ𝑟 𝑓𝑡 𝐵𝑇𝑈 ℎ𝑟–𝑓𝑡–℉ 1 𝑓 𝑡 2 ∆𝑇=694℉ Figure: Conduction through a Slab ∆𝑇= 1,000 𝐵𝑇𝑈 ℎ𝑟 𝑓𝑡 𝐵𝑇𝑈 ℎ𝑟–𝑓𝑡–℉ 1 𝑓 𝑡 2 𝑄 =𝑘𝐴 ∆𝑇 ∆𝑥 ∆𝑇=694℉ ELO 1.2

Heat Transfer Modes Convective Heat Transfer
Convection involves the transfer of heat by the motion and mixing of "macroscopic" portions of a fluid 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: natural circulation in RCS, cooling towers Forced convection – outside force, such as a pump, causes motion and mixing 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 ELO 1.2

Heat Transfer Modes 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 ELO 1.2

Heat Transfer Modes Convection Heat Transfer
The formula for convective heat transfer is: 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 ELO 1.2

Heat Transfer Modes Radiant Heat Transfer and Thermal Radiation
Radiant heat transfer - thermal energy transferred by means of electromagnetic waves or particles Involves transfer of heat by electromagnetic radiation Radiant heat transfer does not need a medium to take place Example, during core uncovery 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 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) Due to high mass flow rates and relatively low temperatures in the reactor, this mode of heat transfer is not significant in power production. ELO 1.2

Heat Transfer Modes Knowledge Check
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? Radiation Emission Convection Conduction Correct answer is A. The correct answer is: A. NRC Bank Question – P784 Analysis: Conductive - Transfer of heat by the interactions between adjacent molecules Convection - Transfer of heat by a process of bulk motion and mixing of macroscopic portions of the fluid Radiation - The transfer of heat by electromagnetic radiation that arises due to the temperature of a body During normal plant operations, convection heat transfer occurs between the cladding and bulk coolant. However, during a loss-of-coolant accident (LOCA), the cladding is no longer in contact with coolant, cladding temperature will rise rapidly; radiation heat transfer will occur between the cladding and surrounding voids. ELO 1.2

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? Conduction Convection Radiation Natural Circulation Correct answer is A. Correct answer is A. NRC Bank Question – P584 Analysis: Reactor fuel pellets are surrounded by helium gas, which transfers heat to the cladding via conduction heat transfer. Helium is chosen because it is a small atom (atomic mass of 4 amu), which makes it an effective medium for heat transfer. Keep in mind that the Helium gas is stationary; convective heat transfer is transfer of heat by a process of bulk motion and mixing of macroscopic portions of the fluid. Since the helium gas is NOT flowing, the primary mode of heat transfer in the gap is conductive heat transfer. ELO 1.2

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 The symbol 𝑄" represents the heat added per unit area Units for heat flux are BTU/hr-ft2 𝑄 "= 𝑄 𝐴 Related KA K1.02 Define thermal conductivity The heat flux can be determined by dividing the heat transfer rate by the area through which the heat transfers Concept used to determine HOT SPOT in core (kW/ft2 or kW/ft, which is linear power density) ELO 1.3

Heat Transfer Terms Conductive Heat Transfer Coefficient (k)
Defines, in part, the heat transfer due to conduction Units - BTU/hr-ft2-°F Thermal conductivity is: A measure of a substance’s ability to transfer heat through a solid by conduction Describes heat transfer characteristics of a solid material Varies with temperature For vapors, it depends upon pressure ELO 1.3

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 Units - BTU/hr-ft2-°F ELO 1.3

Heat Transfer Terms Log Mean Temperature Difference (LMTD)
Equation ( 𝑄 =ℎ𝐴∆𝑇) varies as heat passes through a cylinder Requires the use of some average temperature value LMTD is: ∆𝑇 at one end of heat exchanger minus ∆𝑇 at the other end of the heat exchanger divided by the natural log of the ratio of these two temperature differences Two important assumptions in LMTD: Fluid specific heats do not vary significantly with temperature Coefficients of convection heat transfer are relatively constant Log Mean Temperature Difference is NOT a concept tested by the NRC. You can assume the value for Delta-T provided in all equations is the LMTD. ELO 1.3

Heat Transfer Terms Log Mean Temperature Difference Formula:
∆ 𝑇 𝑙𝑚 = (∆ 𝑇 2 −∆ 𝑇 1 ) ln ∆ 𝑇 2 ∆ 𝑇 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 Note: "Since the cross-sectional area, A, is not a constant, the basic relationship for heat transfer by conduction in a slab cannot be used directly for a cylinder. ELO 1.3

Heat Transfer Terms Overall Heat Transfer Coefficient (Uo)
Many heat transfer processes involve a combination of conduction and convection 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 is some combination of heat transfer modes ELO 1.3

Heat Transfer Terms Overall Heat Transfer Coefficient 𝑄 =𝑈𝐴Δ𝑇 Where:
𝑄 = the rate heat of transfer (BTU/hr) U = the overall heat transfer coefficient (BTU/hr-ft2-°F) A = the overall cross-sectional area for heat transfer (ft2) ΔT = the overall temperature difference (°F) ELO 1.3

Heat Transfer Terms 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 ELO 1.3

Heat Transfer Terms The figure shows “Combined Heat Transfer”, illustrates an example of this concept applied to cylindrical geometry, which shows a typical combined heat transfer situation Convection occurs between temperatures T1 and T2 Conduction occurs between temperatures T2 and T3 Convection between temperatures T3 and T4 Figure: Combined Heat Transfer ELO 1.3

Heat Transfer Terms 𝑈 𝑜 = 1 𝐴 𝑜 ℎ 1 𝐴 1 + ∆𝑟 𝐴 𝑜 𝑘 𝐴 𝑙𝑚 + 𝐴 𝑜 ℎ 2 𝐴 2
𝑈 𝑜 = 1 𝐴 𝑜 ℎ 1 𝐴 1 + ∆𝑟 𝐴 𝑜 𝑘 𝐴 𝑙𝑚 + 𝐴 𝑜 ℎ 2 𝐴 2 Uo equation can be simplified, assuming: tube wall thickness is small compared to the tube diameter inner area (A1), outer area (A2), and log mean area (Alm) close to being equal Allows us to cancel out all the area terms in the denominator Results in this equation: 𝑈 𝑜 = 1 ( 1 ℎ 1 + ∆𝑟 𝑘 + 1 ℎ 2 ) The TOP 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. Keep in mind this is NOT an equation provided on the NRC Equation Sheet nor is it one required to be committed to memory. It is just an explanation that helps understand the factors that can affect the overall heat transfer coefficient (thickness of tube, thickness of laminar layer, etc.) ELO 1.3

Heat Transfer Terms 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 RCS flowing past fuel cladding, for example For boiling or condensation, Tb = saturation temperature Correct answer is C. ELO 1.3

Heat Transfer Terms Knowledge Check
The heat transfer characteristics of a solid material are measured by a property called ___________. bulk temperature heat flux thermal conductivity conduction Correct answer is C. Correct answer is C. ELO 1.3

Heat Transfer Characteristics in Heat Exchangers
TLO 2 – Describe the heat transfer process in heat exchangers and factors that reduce heat transfer. 2.1 Describe the difference in the temperature profiles for counter-flow and parallel flow heat exchangers. 2.2 Describe the differences between regenerative and non-regenerative heat exchangers. 2.3 Describe how fluid films, steam, or gases can affect heat transfer and fluid flow in heat exchangers. 2.4 Calculate heat transfer rates and temperatures in heat exchangers. The majority of the bank questions related to heat exchangers are in – Heat Exchangers. Depending on the order the material is presented, this section may be covered quickly. TLO 2

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 TLO 2 Intro

Counter and Parallel Flow Heat Exchangers
ELO 2.1 Describe the difference in the temperature profiles for counter- flow and parallel flow heat exchangers. 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 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. ELO 2.1

Flow Paths Parallel Flow Flow in same direction Hot In and Cold In on same side More thermal stress Counter Flow Flow in opposite direction Hot In and Cold Out on same side Less Delta-T across H/X Figure: Fluid Flow Direction ELO 2.1

Parallel Flow Heat Exchanger
Advantages Parallel flow is used when fluids should exit at nearly same temperature Disadvantages Large temperature difference at the ends large thermal stresses Cold Out can never exceed Hot Out Note that in – Heat exchangers, this is a concept tested by the NRC. Rate of heat transfer varies along length of exchanger tubes based on temperature difference between hot and cold fluid ELO 2.1

Counter Flow Heat Exchanger
Advantages More uniform ∆T minimizes thermal stresses Cold Out can be > Hot Out Uniform temperature difference produces a more uniform rate of heat transfer Disadvantages None noted Rate of heat transfer varies along length of exchanger tubes based on temperature difference between hot and cold fluid ELO 2.1

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 Recall, “radius” affects conductive heat transfer ELO 2.1

Knowledge Check All of the following are advantages of counter-flow heat exchangers EXCEPT: More uniform delta T, resulting in less thermal stresses Uniform temperature difference produces a more uniform rate of heat transfer Large temperature difference at the ends causes less thermal stress Outlet temperature of the cold fluid can approach the highest temperature of the hot fluid Correct answer is C. Correct answer is C. ELO 2.1

Regenerative and Non-Regenerative Heat Exchangers
ELO 2.2 – Describe the differences between regenerative and non- regenerative heat exchangers. Non-Regenerative H/X Uses two separate fluids for the hot and cold fluid Most common type of lube oil coolers Regenerative H/X Uses one fluid to serve as both the hot and the cold fluids Used in Letdown system, Feedwater heaters, and Moisture Separator reheaters 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. ELO 2.2

Non-Regenerative Heat Exchanger
One fluid cools (or heats) the other fluid No interconnection between the two fluids Temperature usually maintained by controlling Cold Out flow Figure: Non-Regenerative Heat Exchanger ELO 2.2

Regenerative Heat Exchanger
Less energy loss from the total system Hot In used to heat up Cold Out Reactor coolant purification system (letdown) Uses combination of both regenerative and non–regenerative H/X Purification system (letdown): Reactor coolant passes through a regenerative heat exchanger Non-regenerative heat exchanger Demineralizer Back through regenerative heat exchanger, and returns to the primary Figure: Regenerative Heat Exchanger ELO 2.2

Regenerative and Non-Regenerative Heat Exchangers
Reduces the temperature of water entering the purification system Minimizes thermal stress Conservation of system energy less loss of system energy due to the cooling of the fluid Allows use of a smaller heat exchanger to achieve the desired temperature for purification ELO 2.2

Factors that Impact Heat Transfer
ELO 2.3 – 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 ELO 2.3

Fluid Film Layers The thinner the fluid film, the higher the convection heat transfer Several factors affect the heat transfer through a fluid film Film thickness Fluid’s thermal conductivity The velocity of the fluid affects the thickness of the fluid film Greater the velocity the thinner the fluid film 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. ELO 2.3

Fluid Gas and Air Impact 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. Venting of any air and non-condensable gasses is important for proper operation of the heat exchanger. Keep in mind that there are two separate concepts related to boiling. Boiling occurring because water flashing to steam in a HX is bad because it restricts flow and Q-dot will decrease. Subcooled Nucleate Boiling (discussed in next chapter) results in a smaller Delta-T for a given heat flux, which will increase Q-dot. ELO 2.3

HX Fouling/Scaling Fouling of tubes lowers the efficiency Decreases the thermal conductivity of the tubes Tubes must transfer through the fouling layer Several methods to remove fouling from heat exchanger tubes Hydro lancing Chemical cleaning Operating practices the most effective Condition in a heat exchanger characterized by foreign material such as algae, scale, or debris accumulating in a heat exchanger. ELO 2.3

To INCREASE heat transfer rate across a heat exchanger: Increase mass flow rate of hot fluid Increase mass flow rate of cold fluid Increase flow rate of both fluids Increase temperature IN of hot fluid Decrease temperature IN of cold fluid This concept is actually tested in – Heat Exchangers. However, keep in mind that some questions in – Heat Exchangers test the INITIAL change in heat transfer rate, while some test the change from steady-state to steady-state. For example, when scaling occurs the heat transfer rate initially decreases resulting in less heat being transferred out of the hot into the cold. (Hotout increases, Coldout decreases). However, the HOTin will eventually increase bringing the heat transfer rate back up to where it was. The NET effect is the cold temperatures could return to normal, but the hot temperatures will be a little higher (but with the same Delta-T). ELO 2.3

Heat Transfer Rate Equations Once factors determined, heat transfer rate can be determined by: 𝑄 = 𝑚 𝛥ℎ (phase change on one side of HX) 𝑄 = 𝑚 𝑐𝑝𝛥𝑇 (no phase change on either side of HX) 𝑄 =𝑈𝐴𝛥𝑇 (across the HX with or without phase change) Where 𝛥𝑇 is Tave hot fluid – Tave cold fluid When determining “final” to “initial” conditions, consider the following: Final/Initial ratios will be utilized in the next ELO that calculates heat transfer rates from some initial condition to some final condition (like tube plugging). Although the dot is over the mass indicating mass flow rate, the rate term can be over the change in enthalpy or over the change in temperature. For the Fukishima type problems this is important to know since bank questions look for heat up rate values. 𝑄 𝐹𝑖𝑛𝑎𝑙= 𝑚 𝛥ℎ𝐹𝑖𝑛𝑎𝑙 𝑄 𝐼𝑛𝑖𝑡𝑖𝑎𝑙= 𝑚 𝛥ℎ𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑄 𝐹𝑖𝑛𝑎𝑙= 𝑚 𝑐𝑝𝛥𝑇𝐹𝑖𝑛𝑎𝑙 𝑄 𝐼𝑛𝑖𝑡𝑖𝑎𝑙= 𝑚 𝑐𝑝𝛥𝑇𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑄 𝐹𝑖𝑛𝑎𝑙=𝑈𝐴𝛥𝑇𝐹𝑖𝑛𝑎𝑙 𝑄 𝐼𝑛𝑖𝑡𝑖𝑎𝑙=𝑈𝐴𝛥𝑇𝐼𝑛𝑖𝑡𝑖𝑎𝑙 ELO 2.3

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). Oil to water in a lube oil cooler Steam to water in a feedwater heater Water to air in a ventilation heating unit Water to water in a cooling water heat exchanger Correct answer is B. Correct answer is B. NRC Question P3684 Analysis: There are many things that effect the overall heat transfer coefficient such as nature of the fluid, transport properties and temperature. Some approximate values of overall heat transfer coefficients between various mediums are as follows: Oil to water: (BTU/hr-ft2-F) Steam to water: (BTU/hr-ft2-F) Water to air: 3-50 (BTU/hr-ft2-F) Water to water: (BTU/hr-ft2-F) ELO 2.3

Solving Heat Exchanger Problems
ELO 2.4 – Calculate heat transfer rates and temperatures in heat exchangers. In addition to previous equations (provided on NRC Equation Sheet): The following can be used (across non phase change HX) 𝑚 1 𝐶 𝑝1 ∆ 𝑡 1 = 𝑚 2 𝐶 𝑝2 ∆ 𝑡 2 Heat transferred OUT of one side equals heat transferred IN the other side “Minus sign” can be seen if Delta-T’s are the same Out – In on both sides “Absolute” values shown if Hot – Cold used on both sides 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. It is not important for calculations to use the “negative” side on one side of the HX equals the “positive” side of the HX. Absolute values are all that is needed and it makes sense to use “HOT-COLD” on both sides since heat transfer is from hot to cold. The other concept can be used to determine that change in one temperature on one side based on a change on the other side and is discussed in more detail in – Heat Exchangers. ELO 2.4

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 𝑚 𝑜𝑖𝑙 = 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? ELO 2.4

What is the heat transfer rate of the oil across the heat exchanger? Step Formula Solution Solve for 𝑄 of the oil 𝑄 𝑜𝑖𝑙 = 𝑚 𝑜𝑖𝑙 𝑐 𝑝 𝑜𝑖𝑙 ∆ 𝑡 𝑜𝑖𝑙 𝑄 =(3.0× 𝑙𝑏𝑚 ℎ𝑟 )(1.1 𝐵TU 𝑙𝑏𝑚–℉ )(55°𝐹) = 1.815× 𝐵TU ℎ𝑟 ELO 2.4

What is the heat transfer rate of the water across the heat exchanger? Q Oil = Q Water = x 106 BTU/hr What is the mass flow rate of the cooling water? Step Formula Solution Solve for 𝑄 of the oil 𝑄 𝑊𝑎𝑡𝑒𝑟 = 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑝 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑡 𝑤𝑎𝑡𝑒𝑟 𝑄 =1.815× 𝐵TU ℎ𝑟 = 𝑚 𝑤𝑎𝑡𝑒𝑟 𝐵TU 𝑙𝑏𝑚–℉ (30°𝐹) 6.05× 𝑙𝑏𝑚 ℎ𝑟 =( 𝑚 𝑤𝑎𝑡𝑒𝑟 ) ELO 2.4

Knowledge Check 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.) 960 psia 930 psia 900 psia 870 psia Correct answer is B. ELO 2.4

Knowledge Check Solution: 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 Final to Initial ratio of 𝑄 =𝑈𝐴𝛥𝑇 equation: 𝑈 1 𝐴 𝑇 𝑅𝑥 – 𝑇 𝑆𝐺 1=0.95 𝑈 2 𝐴 𝑇 𝑅𝑥 – 𝑇 𝑆𝐺 2 The U and A terms cancel out, so 578−538=0.95(578– 𝑇 𝑆𝐺 ) 40=549.1−0.95𝑥 0.95𝑥=549.1−40 𝑥= 𝑇 𝑆𝐺 =535.9°𝐹, find using the steam tables, which = 𝑝𝑠𝑖𝑎 The correct answer is B (930 psia). Put the Final to Initial ratio equation discussed on a previous slide on the board to show the math for this example. Plugging steam generator tubes reduces the heat transfer coefficient by 5% (effective surface area is reduced): UAfinal = 0.95 UAinitial (UAfinal)( ) = (0.95UAinitial)(578-x) x = 535.9oF, which equates to a Psat of 930 psia NOTE: As a result of the SG tube blockage, the final Delta-T increased by 2 degrees (from 40oF to 42oF). ELO 2.4

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? 3.8 x 107 BTU/hr 8.6 x 107 BTU/hr 5.4 x 108 BTU/hr 7.2 x 108 BTU/hr ELO 2.4

To solve this problem, use equation 𝑄 = 𝑚 𝛥ℎ, and solve for heat transfer. 𝑄 =7.5× 𝑙𝑏𝑚/ℎ𝑟 ( ℎ 𝑠𝑡𝑚 − ℎ 𝑤𝑎𝑡𝑒𝑟 ), use the steam tables to look up the enthalpy of saturated water at 414 psia (interpolate between and 450 psia to obtain 430 BTU/lbm) =7.5× 𝑙𝑏𝑚/ℎ𝑟 (1,150 𝐵𝑇𝑈/𝑙𝑏𝑚−430 𝐵𝑇𝑈/𝑙𝑏𝑚) =7.5× 𝑙𝑏𝑚/ℎ𝑟 (720 𝐵𝑇𝑈/𝑙𝑏𝑚) =5.4× 𝐵𝑇𝑈/ℎ𝑟 Correct answer is C. Animated hidden rectangles to show answer. Remind students to use both Table 1 and Table 2, sometimes they can find data points in other tables. ELO 2.4

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? 1.7 psia 2.3 psia 3.0 psia 4.6 psia Correct answer is A. Work through solution on board, solution is located in the Student Guide. Correct answer is A. NRC Question P3384 NOTE: Since air is introduced in the system, vacuum will obviously decrease (absolute backpressure increases). In this case from 1.2 psia to 1,7 psia (or about a loss of 2 inches Hg vacuum) ELO 2.4

Core Thermal Power TLO 3 – Explain core thermal power and calculate its value using a simplified heat balance. 3.1 Define the following terms: Core thermal power Percent reactor power 3.2 Explain the various methods of calculating core thermal power and calculate core thermal power using a simple 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. TLO 3

Core Thermal Power and Percent Reactor Power
ELO 3.1 Define the following terms: core thermal power and percent reactor power. Core thermal power (CTP) - a measure of heat input per unit time from the reactor core to the reactor coolant Not all heat going to the SG comes from the core RCPs, Pzr htrs, etc. Not all heat coming from the core makes it to the SG Ambient losses Percent reactor power - the percent of rated core thermal power at which the reactor is currently operating Related KAs K1.05, Define core thermal power K1.07 Define 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. ELO 3.1

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 Core thermal power is most accurately determined by Mass flow rate of the feedwater times the change in enthalpy in the steam generators NI’s indicate the percent of reactor power Flux increases and shifts outward over core life Indicated reactor power (from NI’s) would tend to increase Heat balance calculation would determine actual reactor power NI’s adjusted accordingly ELO 3.1

Percent Reactor Power Ratio of actual core thermal power to rated core thermal power: % 𝑃 𝑟𝑥 = 𝐶𝑇 𝑃 𝑎𝑐𝑡𝑢𝑎𝑙 𝐶𝑇 𝑃 𝑟𝑎𝑡𝑒𝑑 ×100% Example: A nuclear reactor is producing 500MW of core thermal power. The core carries a 1,500MW thermal power rating. What is the core thermal power in percent? % 𝑃 𝑟𝑥 = 500 𝑀𝑊 1,500 𝑀𝑊 ×100% % 𝑃 𝑟𝑥 =33.3% ELO 3.1

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... reactor coolant by the change in temperature across the core. reactor coolant by the change in enthalpy in the steam generators. feedwater by the change in enthalpy in the steam generators. feedwater by the change in temperature across the core. Correct answer is C. Correct answer is C. NRC Question P585 Analysis: Core thermal power can be measured by several techniques: • Incore neutron detectors • Excore neutron detectors • Delta-T power (Reactor coolant system Thot – Tcold) • Secondary heat balance or calorimetric Of the various techniques, the secondary heat balance is the most accurate. This is because the parameters used to calculate heat balance power (including steam generator pressure, feedwater temperature, and feedwater flow) can be measured with a high degree of accuracy. ELO 3.1

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. 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 𝑄 in + 𝑄 add = 𝑄 out The energy entering the system is from the feedwater Energy added: heat from reactor fission and the reactor coolant pump heat The energy leaving steam demand to the steam system, ambient losses to atmosphere, blowdown in SG Related KAs – K1.06, Explain methods of calculating core thermal power ; K1.08 Calculate core thermal power using a simplified heat balance Note that there are other “additions” and/or “losses” as well, but currently none of them are tested by the NRC. For example, addition from PZR Htrs. Changes in mass flow rate from charging/letdown. Losses in SG system to blowdown. ELO 3.2

Using the original equation with these changes yields the following equation: 𝑄 fw +𝐶𝑇𝑃+ 𝑄 RCP = 𝑄 stm + 𝑄 amb + 𝑄 𝐵𝐷 Now we can solve the equation for core thermal power: 𝐶𝑇𝑃= 𝑄 STM + 𝑄 AMB + 𝑄 𝐵𝐷 − 𝑄 FW − 𝑄 RCP Recall equation: 𝑄 SG = 𝑚 stm(hstm-hfw) All we have done above is break 𝑄 SG into two terms: 𝑄 STM − 𝑄 FW , or; 𝑚 stm(hstm) - 𝑚 fw(hfw) Keeping in mind that: 𝑚 stm = 𝑚 fw - 𝑚 bd 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. Another variation on the above formula that can be used is: CTP = M-dotfeedwater (hsteam – hfeedwater) – m-dotBD (hsteam – hBD) – Q-dotRCP + Q-dotAmbient Figure: Reactor Heat Balance ELO 3.2

CTP (Test-Taking) Concepts If all values used are correct, CTPindicated calculation is correct CTPindicated = CTPactual If “indicated” power > “actual” power Something caused CTP calculation to increase Increase in “positive” term (SG, AMB, BD) Decrease in “negative” term (RCP, FW) Change in feedwater flow effects SG and FW terms ( 𝑚 stm = 𝑚 fw - 𝑚 bd) Since hstm > hfw, change to SG term > FW term Here are some concepts to think about when answering questions related to factors and how they effect CTP. The next couple of slides will test these concepts as they relate to an actual NRC bank question on CTP. ELO 3.2

Example Heat Balance Question Assume that the power range nuclear instruments have been adjusted to 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. Omitting the reactor coolant pump heat input term from the heat balance calculation. The feedwater flow rate used in the heat balance calculation was lower than actual feedwater flow rate. The steam pressure used in the heat balance calculation was 50 psi higher than actual steam pressure. The enthalpy of the feed water was miscalculated to be 10 BTU/lbm higher than actual feed water enthalpy. NOTE: Keep in mind the below equation and the “signs” of each term. 𝐶𝑇𝑃= 𝑄 STM + 𝑄 AMB + 𝑄 𝐵𝐷 − 𝑄 FW − 𝑄 RCP ELO 3.2

Since we are looking for “indicated” > “actual, then CTP would increase! Choice A: if Qrcp is left out of the calculation, (forget to subtract a negative term) then CTP would INCREASE. Choice B: if feedwater flow rate used < actual feed flow, then Qsg (positive term) would decrease > the Qfw (negative term), therefore CTP would DECREASE. Choice C: if steam pressure value used is 50 psi higher than actual steam pressure, the enthalpy will decrease for the higher steam pressure (recall Mollier diagram, > ~ 500psia, enthalpy deceases as pressure is increased); this would lower Qsg, then CTP would DECREASE. Choice D: if enthalpy of the feedwater was higher than actual, Qfw would increase, causing CTP to DECREASE. Therefore, the correct answer is A. The reactor coolant pump heat input term was omitted from the heat balance calculation. On Choice “C”, you MUST always assume that you are operating on the left hand side of the “hump” on the Mollier Diagram (PSG always > 500 psia). Therefore, an increase in SG pressure would mean a decrease in specific enthalpy. ELO 3.2

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 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? 677 MW 755 MW 1,334 MW 1,510 MW This is P84 in – Heat Transfer NRC bank. ELO 3.2

Solution: Core Thermal Calculation Step-by-Step Table Ignoring all other heat gain and loss mechanisms: The correct answer is D (1,510 MW). Step Action 1. Use the correct heat transfer of CTP equation CTP= 𝑄 stm + 𝑄 amb − 𝑄 fw − 𝑄 𝑅𝐶𝑃 = 𝑚 𝑓𝑤 ×( ℎ 𝑠𝑡𝑚 − ℎ 𝑓𝑤 ) 2. Use steam tables if necessary to get enthalpy, solve for ∆h’s or ∆T =3.3× 10 6 (1,199.3−419 𝐵TU 𝑙𝑏𝑚 ) =3.3× 𝑙𝑏𝑚 ℎ𝑟 ×(781.3 𝐵TU 𝑙𝑏𝑚 ) =2.578× 𝐵TU ℎ𝑟 (𝑝𝑒𝑟 𝑆𝐺) =2×2.578× 10 9 (2 𝑆𝐺𝑠 )=5.16× 10 9 (𝑡𝑜𝑡𝑎𝑙) 3. Convert to MW if required 𝑄 =5.16× 𝑀𝑊 3.413× 𝐵TU ℎ𝑟 =1,511 𝑀𝑊 NOTE: Good test-taking technique – when answers provided are “twice” or “half” of other choices, THERE IS A REASON! The STEM provides flow for one FW pump in a TWO-LOOP PLANT. Don’t forget to double the FW flow in the calculation. ELO 3.2

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? 14.0 percent 14.3 percent 15.0 percent 15.8 percent Correct answer is C. Correct answer is C. NRC Question P785 Analysis: NOTE: Since the STEM of the question said the reactor was producing 200 MW of “core thermal power”, the RCP heat has already been removed and should not be included in the calculation. ELO 3.2

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? 1,335 MW 1,359 MW 1,589 MW 1,612 MW Correct answer is C. Correct answer is C. NRC Question P2985 Analysis: Heat transfer across a single steam generator can be calculated using the equation: Q-dot = m-dot(Delta-h) This equation is used whenever a phase change occurs. Note that the stem already gives total feedwater flow to all steam generators so no additional calculation required. The mass flow rate is given in the stem Steam enthalpy can be looked up using the saturated vapor value at 1,000 psia: 1,193 Btu/lbm. Feedwater enthalpy can be found using the subcooled liquid approximation at 440ºF: 419 Btu/lbm. ELO 3.2

NRC KA to ELO Tie KA # KA Statement RO SRO ELO K1.01
Describe three mechanisms of heat transfer. 2.5 1.2 K1.02 Define thermal conductivity. 2.0 2.2 1.3 K1.03 Explain the manner in which fluid films affect heat transfer. 2.4 K1.04 Describe how the presence of gases or steam can affect heat transfer and fluid flow in heat exchangers. 2.8 3.0 K1.05 Define core thermal power. 2.7 2.9 3.1 K1.06 Explain methods of calculating core thermal power. 3.3 3.2 K1.07 Define percent reactor power. K1.08 Calculate core thermal power using a simplified heat balance. 3.4

Operator Generic Fundamentals Thermodynamics – Heat Transfer

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