(Solution Manual) Nuclear Systems Volume I Thermal Hydraulic Fundamentals

Todreas,Kazimi, 3

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Solution Manual for Nuclear Systems Volume I Thermal Hydraulic Fundamentals, The downloadable file consists of 396 pages, and the total size of the file is 4.84 MB. Purchase now only for highly discounted price.

Author(s): Todreas, Kazimi

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Chapter 1 Principal Characteristics of Power Reactors Contents Problem 1.1 Worldwide Utilization of Power Reactor Technology .................................... 6 5 rev 112420Solutions Manual for Nuclear Systems Volume I Thermal Hydraulic Fundamentals, 3e by Neil Todreas, Mujid Kazimi (All Chapters)Chapter 1 - Principal Characteristics of Power Reactors PROBLEM 1.1 QUESTION World Utilization o f Power Reactor Technology 1. For each position of Table 1.13 identify : (a) The principal technical reason for which this moderator-coolant combination cannot be exploited, and (b) The name of one (or more) terrestrial power reactor plants that have been built using this combination. TABLE 1. 13 Worldwide Utilization of Power Reactor Technology - Thermal Reactor Types Coolant Light water Heavy water Organic HB-40 Santowax -OM Gas hydrogen, nitrogen, CO₂, helium Liquid metal Pressurized Boiling Pressurized Boiling NaK Na Light water Heavy water Graphite Beryllium Organic References for thermal reactor types: 1. List of operational nuclear power plants. Nucl. News , August 1992. 2. Dietrich, J. R. and Zinn, W. H. Solid Fuel Reactors , Reading, MA: Addison-Wesley Pub. Co., 1958. 3. Directory of Nuclear Reactors , Vienna: International Atomic Energy Commission, published annually. 4. Kuljian, H. A. Nuclear Power Plant Design , Cranbury, NJ: A.S. Barnes & Co., 1968. 5. Meserve, R., Chairman, Safety Issues at the Defense Production Reactors: A Report to the U.S. Department of Energy , National Academy Press, Washington, DC, 1987. 6. Zinn, W. H., Pittman, F. K. and Hogerton, J. F. Nuclear Power , USA, McGraw -Hill, New York, NY, 1964. 6 rev 112420Chapter 1 - Principal Characteristics of Power Reactors 2. Which of these combinations would be best for submarine propulsion? Explain your choice. 3. State the single most important power cycle parameter affecting the cycle thermal efficiency. 4. Compare the primary side pressures of a PWR, a BWR and a MSR. Describe the reason for primary side pressure of the PWR being about twice the BWR and the advantage of MSR primary side pressure being close to atmospheric. 5. Comparing the core of a BWR with that of a PWR, explain the reason a BWR is considered a closed and a PWR an open core design. 6. Explain the reason that alloys of zirconium are used for fuel cladding. 7. Compare the number of coolant loops and fluids used in a BWR with those of a LMFR - describe the reason for the differences, if any. 8. What is the key difference in the VVER design from other LWR designs? 9. How do you classify a LFR with respect to neutron spectrum and thermal efficiency as compared with the SFR? 10. Describe the key difference with respect to core coolant circulation between NuScale and a typical LWR. PROBLEM 1.1 SOLUTION World Utilization of Power Reactor Technology Question 1 Note: Descriptions ar e made with (coolant type) – (moderator). The following notes are keyed to Table 1.13 A Pressurized Water Reactor (PWR). Used by: – USA – France – Germany – Former USSR B Boiling Water Reactor (BWR), Used by: – USA – Germany 7 rev 112420Chapter 1 - Principal Characteristics of Power Reactors – Sweden C Press. Heavy Water Heavy Water Vessel – AGESTA (Sweden) – R - 3/ ADAM (Sweden) – ATUCHA (Argentina) – MZFR (Germany) – BRUCE 1 -8 (Canada) – CP - 5 (USA) – NPD -2 (USA) D Boiling Heavy Water - Heavy Water Vessel – MARVIKEN (Sweden) – HALDEN BWR (Norway) – ATUCHA 2 (Argentina) E Press. Light Water Heavy Water Tube – NRX (Canada) – GENKILLY 2 (Canada) – SGHRW – CIRUS/TROMBAY (India) F Boiling Light Water Heavy Water Tube – CIRENE (Italy) – BLW 250 (Canada) – FUGEN (Japan) – VENUS (Belgium) G Press. Heavy Water - Heavy Water Tube – CANDU (Canada) – PRTR (Canada) – CVTR (Canada) – NPD (Canada) 8 rev 112420Chapter 1 - Principal Characteristics of Power Reactors – KARLSRUHE (Germany) – PICKERING (Canada) H Boiling Heavy Water Heavy Water Tube – NPD CONVERSION (Canada) I Organic - Heavy Water Tube – WR-1 – ESSOR – ORGEL – DON – HWOCR J Gas - Heavy Water Tube – BOHUNIZE (Czechoslovakia) – KKN (Germany) – EL-2, EL -4 (France) – LUCENS (Switzerland) K Press. Light Water Graphite – APS OBNINSK (Former USSR) – CHERNOBYL (Ukraine) – HANFORD (USA) – RBMK (Former USSR) L Boiling Light Water - Graphite – SOSNOVY BORINSK (Former USSR) – BELOYARSK (Former USSR) – FIRST NUCLEAR REACTOR OF USSR – N REACTOR (USA) M Gas - Graphite – MAGNOX (Great Britain) 9 rev 112420Chapter 1 - Principal Characteristics of Power Reactors – DRAGON (Great Britain) – HTGR (EGCR, PEACHBOTTOM, FT. ST. VRAIN, HINKLEY POINT B), (USA) – UTREX (USA) – AVR (Germany) – SIZEWELL NPS (Great Britain) – HECTOR – KIWI – NERVA – USA SPACE PROGRAM N Liquid Metal - Graphite – HALLAM (USA) – MOLTEN SALT REACTOR EXPERIMENT (ORNL, USA) O Gas - Beryllium – EBOR (USA) – DANIEL’S PILE (ORNL, USA) P Organic - Organic – PIQUA (USA) Q Press. Light Water -Beryllium – MIR (Russia) – GE TEST REACTOR (USA) – MR-2 – MIR (Russia) R Boiling Light Water Beryllium – BR 2 (Belgium) S Boiling Light Water Organic – AGN -211 (Switzerland) T Gas - Light Water – ESADA VESR (USA) – MOBILE LOW POWER PLANT (USA) – ML-1 (Idaho Falls, USA) – HEAT TRANSFER REACTOR EXPERIMENT (USAF) U Organic Heavy Water Vessel 10 rev 112420Chapter 1 - Principal Characteristics of Power Reactors – ECO (Italy) V Gas - Heavy Water Vessel – HWGCR (Former USSR) W Liquid Metal - Beryllium – SUBMARINE INTERMEDIATE REACTOR (USA) X Press, Light Water Organic – L-77 (Univ, of Nevada, USA) Answer to Question 1 continued. Comments on moderator-coolants which have been utilized worldwide. 1. There is no advantage gained here by using a light water moderator with a different coolant, because light water is already being used to moderate, it is simpler and more economical to use it as a coolant as well. 2. Liquid sodium will react with w ater. Using sodium as a coolant would require extensive (and expensive) efforts to separate the two materials (i.e., a tertiary loop). 3. Separation of light and heavy water is not feasible in a vessel-type reactor. If they were both placed in the vessel, the advantages of heavy water would be lost by dilution with light water. 4. In a vessel type reactor you cannot separate the coolant from the moderator. 5. It is unnecessarily expensive to use heavy water as a coolant if it is not being used for its excellent moderating ability. Since graphite is the moderator here it is more economical to use light water as the coolant (even though it has a higher absorption cross section than heavy water.) 6. This is an uneconomical combination. If the organic is being used as the coolant it should be used as the moderator as well. 7. Beryllium is expensive, toxic, brittle, and hard to work with. For a power reactor there are plenty of better choices. 8. Because the organic is the moderator, it is more economical to sue it as the coolant as well. The decomposition of organics under radiation will pose a problem if the organic is not being used as the moderator AND the coolant. If it is not circulating as a coolant its residence time in the core will be too long, leading to decomposition. 11 rev 112420Chapter 1 - Principal Characteristics of Power Reactors TABLE SM-1.1: Summary Coolant (right), Moderator (down) Pressurized light water Boiling light water Pressurized heavy water Boiling heavy water Organic (HB-40, Santowax - OM) Gas (Hydrogen, Nitrogen, Carbon dioxide, Helium) Liquid Metal (NaK, Na) Light Water A B 1 1 1 1 - (USADA VERS, USA) 2 Heavy water (vessel) 3 3 C D 4 4- (HWGCR, USSR) 2,4 Heavy water (tube) E F G H I J 2 Graphite K (APS) L 5 5 6 M M (MSRE) Beryllium Q (MIR, Russia - GE test reactor) R (BR -2, Belgium) 7 7 7 0 7 Organic 8 S (AGN - 211, Basel, Switzerland) 8 8 P 8 8 Question 2. Which of these combinations would be best for submarine propulsion? Explain your choice. Submarine propulsion reactors are generally designed for military ships. We may identify some characteristics which might be desirable for nuclear military submarines: – High reliability – Low weight: high power density – Low volume – Very low toxic release, especially in cruise space – Silent operation – Tolerance to acceleration (in case of attack) and battle damage – Power variation capability – Ease of operation 12 rev 112420Chapter 1 - Principal Characteristics of Power Reactors – Long fuel residence time – Need for both propulsion and electricity generation – Fuel cost is not a big concern In order to ensure reliability under acceleration, boiling water reactors should be avoided, because of the large density difference between the liquid and the vapor phase. Gas-cooled reactors should be avoided because of the large required volume, the risk of leaks and the high gas velocity, which may result into noisy operation. Since there is no limitation to fuel enrichment, there is no particular need for special moderating materials such as heavy water, graphite or organic. Therefore, pr essurized light water -cooled and moderated reactors are good options. Light water has several advantages, among which are: – Small slowing down length, which allows for a compact core. – Can be used in both the primary and secondary systems, simplifying the design. – Neutron activation products have a short half- life. – Can be easily distilled from seawater to provide makeup and safety function inventory. Alternatively, liquid metal-cooled reactors may be used if corrosion is under control. If liquid metal coolants are used, electromagnetic pumps may be installed which allow more silent operation. However, sodium coolants have the drawback of reacting violently with water, producing explosive/flammable hydrogen. Question 3. State the single most important power cycle parameter affecting the cycle thermal efficiency. Coolant Outlet Temperature—i t dictates reactor mission capability e.g. process heat and cycle thermal efficiency which a ffects capital cost. Also importantly it a ffects coolant corrosion performance. Question 4. Compare the primary side pressures of a PWR, a BWR and a MSR. Describe the reason for primary side pressure of the PWR being about twice the BWR and the advantage of MSR primary side pressure being close to atmospheric. PWR - 15.5MPa, BWR - 7.17 MPa, MSR - nominal 1 MPa PWR pressure is high to keep the primary coolant in the subcooled regime while still achieving high operating temperatures MSR nominal atmospheric primary operating pressure keeps stored energy in the salt coolant low which is a beneficial safety factor as well as allowing the primary coolant system pressure containment e.g. piping to be thin walled. Question 5. Comparing the core of a BWR with that of a PWR, explain the reason a BWR is considered a closed and a PWR an open core design. The BWR operates with a boiling mixture from about one third the distance from the core flow inlet. Particularly in the boiling region different axial pressure levels in adjacent coolant channels 13 rev 112420Chapter 1 - Principal Characteristics of Power Reactors develop due to the core radial neutron flux shape which in turn cause radial pressure gradients to develop between coolant channels. These radial pressure gradients in turn cause radial flows to develop between coolant channels which are countered in BWRs by use of smaller radial sized bundles than in PWRs (a closed core design) . Also since the PWR does not operate in the boiling region, radial pressure gradients of BWR magnitude do not develop so that PWR fuel assemblies do not have bounding walls (an open core design). Question 6. Explain the reason that alloys of zirconium are used for fuel cladding. Zirconium has low neutron capture cross section and sufficient strength and corrosion resistance at water cooled reactor operating temperatures Question 7. Compare the number of coolant loops and fluids used in a BWR with those of a LMFR - describe the reason for the differences, if any. The BWR uses a direct cycle, hence a single water coolant loop while the LMFR uses a three-coolant loop system. Both use the steam cycle. The LMFR loops are the primary sodium loop, the intermediate loop between the intermediate heat exchange and the steam generator typically to date also using sodium coolant and finally a water coolant in which steam is produced which flows t o the turbine in this third loop. Question 8. What is the key difference in the VVER design from other LWR designs? The VVER uses hexagonal shaped fuel assemblies and horizontal steam generators. Question 9. How do you classify a LFR with respect to neutr on spectrum and thermal efficiency as compared with the SFR? Both are fast neutron spectrum reactors. The primary coolant outlet temperatures are respectively 550 degrees Centigrade with thermal efficiencies of 43 -44 %. Question 10. Describe the key difference with respect to core coolant circulation between NuScale and a typical LWR. The Nuscale design is natural circulation using a helical coil steam generator operating in a once- thru producing superheated steam. A module produces 60 MWe with a plant composed of 12 modules. The typical LWR ( let’s take the PWR) is a forced circulation primary loop operating with a U tube pot type steam generator producing saturated steam (Westinghouse) or a steam generator of the once-thru design producing superheated st eam (the Areva design) . 14 rev 112420Chapter 2 Thermal Design Principles and Application Contents Problem 2.1 Relations among fuel element thermal properties in various power reactors ...... 16 Problem 2.2 Relationships between assemblies of different pin arrays ................................... 21 Problem 2.3 Minimum CHF ratio in a PWR for a flow coastdown transient ......................... 22 Problem 2.4 Minimum CPR in a BWR .................................................................................... 25 Problem 2.5 Primary cooling system pumping power for a PWR reactor .............................. 26 Problem 2.6 Relations among thermal design conditions in a PWR ....................................... 28 15 rev 112420Chapter 2 - Thermal Design Principles and Application PROBLEM 2.1 QUESTION Relations Among Fuel Elemen t Thermal Parameters in Various Power Reactors (Section 2.3) Compute the core average values of the volumetric energy- generation rate in the fuel ( q‴) and outside surface heat flux (q″) of cladding f or the reactor types BWR, PWR, PHWR, HTGR, AGR and SFBR . Use the core average linear power levels derivable from data in Table 2.1 and the geometric parameters in Table 1. 3. Answers (for BWR): – q‴ = 243.2 MW/m3 – 𝑞𝑞co″=500 .2 kW /m2 TABLE 2.1 Typical Core Thermal Performance Characteristics for Six Reference Power Reactor Types Characteristics BWR PWR(W) PHWRa HTGR AGR SFBRb Core Axis Vertical Vertical Horizontal Vertical Vertical Vertical No. of assemblies Axial 1 1 12 8 8 1 Radial 764 193 380 493 332 364 (C) 233 (BR) Assembly pitch (mm) 152 214 286 361 460 179 Active fuel height (m) 3.588 3.658 5.94 6.30 8.296 1.0 (C) 1.6 (C + BA) Equivalent diameter (m) 4.75 3.37 6.29 8.41 9.458 3.66 Total fuel weight (MT) 160 UO 2 101 UO 2 89.3 UO 2 1.56 U 34.0 Th 103.0 UO 2 29 UO 2 Reactor vessel Inside dimensions (m) 6.05 D × 21.6 H 4.83 D × 13.4 H 7.6D × 4L (5.94 L for EC 6) 11.3 D × 14.4 H 20.25 D × 21.87 H 21D × 19.5 H Wall thickness (mm) 152 224 28.6 4720 5800 25 Materialc SS-clad carbon steel SS-clad carbon steel Stainless steel Prestressed concrete Concrete helical prestressed Stainless steel Other features Pressure tubes Steel liners Steel lined Pool type Power density core average (kW/L) 52.3 104.5 12 8.4 2.66 280 Linear heat rate Core average (kW/m) 17.6 17.86 25.7 7.87 17.0 29 16 rev 112420Chapter 2 - Thermal Design Principles and Application TABLE 2.1 Typical Core Thermal Performance Characteristics for Six Reference Power Reactor Types Characteristics BWR PWR(W) PHWRa HTGR AGR SFBRb Core maximum (kW/m) 47.24 44.62 ~54d 23.0 29.8 45 Performance Equilibrium burnup (MWd/MT) 50,000 50,000 8300 105,000 20,000 110,000 Average assembly residence (full -power days) 2192 1644 470 1170 1320 640 (C) 320 (BR row 1) 640 (BR row 2) Refueling Sequence 1/4 per year 1/3 per year Continuous on - line 1/2 per year Continuous on - line Variable Outage time (days) 25 25 None for refueling Unavailable None for refueling 32 a Same for both CANDU 6 and Enhanced CANDU 6 (EC 6), except Reactor Vessel Inside Dimensions b SFBR: core (C), radial blanket (BR), axial blanket (BA) c SS = Stainless Steel d Assuming a 900 kW bundle, 37 elements and radial pin peaking factor of 1.12 Source : Principally Appendix K and (Knief, R. A. Nuclear Engineering: Theory and Technology of Commercial Nuclear Power , 2nd Ed. La Grange Park, IL: American Nuclear Society, 2008.) except for AGR data from M. A. H. G. Alderson (pers. comm., October 6, 1983) and A. A. Debenham (pers. comm., August 5, 1988) and PHWR data from CANDU 6 Technical Summary, CANDU 6 Program Team, Reactor Development Business Unit, May 2005. Table 1.3 Typical Characteristics of the Fuel for Six Reference Power Reactor Types Characteristics BWR PWR PHWR HTGR AGR SFBR Reference Design Manufacturer General Electric Westinghouse Atomic Energy of Canada, Ltd. General Atomic National Nuclear Corp. Novatome System (reactor station) BWR/5 (NMP2) (Seabrook) CANDU 6 Enhanced CANDU -6 (EC 6) (Fulton) (Heysham 2) (Superphenix 1) Moderator H2O H2O D2O / D 2O Graphite Graphite None Neutron energy Thermal Thermal Thermal / Thermal Thermal Thermal Fast Fuel production Converter Converter Converter / Converter Converter Converter Breeder 17 rev 112420Chapter 2 - Thermal Design Principles and Application Table 1.3 Typical Characteristics of the Fuel for Six Reference Power Reactor Types Characteristics BWR PWR PHWR HTGR AGR SFBR Fuelb Geometry Cylindrical pellet Cylindrical pellet Cylindrical pellet / Cylindrical pellet Microspheresc Cylindrical pellet Cylindrical pellet Dimensions (mm) 9.60 D × 10.0 L 8.192D × 9.8 L 12.2 D × 16.4L / 12.2 D × 16.4L 400–800 μm D 14.51D ×14.51 L 7.14 D Chemical form UO 2 UO 2 UO 2 / UO 2 UC/ ThO 2 UO 2 PuO 2 / UO 2 Fissile (first core avg. wt% unless designated as equilibrium core) 235U (3.5 eq. core) 235U (3.57 avg. eq. core) 235U (0.711) / 235U (0.711) 235U (93) 235U (2 zones at 2.1 and 2.7) 239Pu (2 zones at 16 and 19.7) Fertile 238U 238U 238U / 238U Th 238U Depleted U Fuel Rodsb Geometry Pellet stack in clad tube Pellet stack in clad tube Pellet stack Pellet stack in clad tube in clad tube Cylindrical fuel compacts Pellet stack in clad tube Pellet stack in clad tube Dimensions 11.20 mm D × 3.588 m H (× 4.09 m L ) 9.5 mm D ×3.658 m H (×3.876 m L ) 13.1 mm D × 13.1 mm D × 0.493 m L 0.493 m L 15.7 mm D × ~0.742 m H (× 0.793 m L ) 15.3 mm D × 0.987 m H (×1.04 m L) 8.5 mm D × 2.7 m H(C)d 15.8 mm D × 1.94 m H(RB)d Clad materiale Zircaloy -2 Zirlo™ Zircaloy -4 / Zircaloy -4 No clad Stainless steel Stainless steel Clad thickness (mm) 0.71 0.572 0.42 / 0.40 Not applicable 0.37 0.56 Fuel Assembly Geometry 9 × 9 square rod arraya 17 × 17 square rod array Concentric Concentric circles of rods circles of rods Cylindrical fuel compacts within a hex. graphite blockf Concentric circles of rods surrounded by graphite sleeve Hexagonal rod array Rod pitch (mm) 14.37 12.60 14.6 / 14.6 23 25.7 9.8 (C)/17.0 (RB) No. rod locations 81 289 37 / 37 132 (SA)/76 (CA)g 37 271 (C)/91 (RB) No. fuel rods 74 (8 part length) 264 37 / 37 132 (SA)/76 (CA)g 36 271 (C)/91 (RB) Outer dimensions (mm) 139 × 139 214 × 214 102D × 495L / 102 D × 495L 359F × 793L 190.4 (sleeve inner D) 173F (inner) × 5.4 m L Channel Yes No No / No No Yes Yes Total weight (kg) 273 ~640 ~25 / 23.7 — 395 — 18 rev 112420Chapter 2 - Thermal Design Principles and Application Table 1.3 Typical Characteristics of the Fuel for Six Reference Power Reactor Types Characteristics BWR PWR PHWR HTGR AGR SFBR a The most recent General Electric BWR fuel bundle designs, which have 9 × 9 and 10 × 10 rod arrays, have been introduced but d ata for many parameters are held proprietary. The table presents available data for a typical 9 × 9 GE11 design. Unique 11 × 11 and 12 × 12 rod arrays were employed at the early Big Rock Point BWR plant. b Fuel and fuel rod dimensions: diameter ( D), heated length ( H), total length ( L), (across the) flats (F ). c Blends of fuel microsph eres are molded to form fuel cylinders each having a diameter of 15.7 mm and a length of 5 –6 cm d SFBR -core (C), SFBR -radial blanket (RB). e Zircaloy and Zirlo are both trademarked alloys of zirconium. Zirlo stands for Zirconium low oxidation. f Early HTGRs (the German AVR and THTR) employed pebble fuel rather than prismatic graphite blocks. g HTGR -standard assembly (SA), HTGR -control assembly (CA). Source : BWR and PWR - Data from Appendix K. The BWR/5 is the Nine Mile Point 2 unit (NMP2) and the PWR is the four -loop Seabrook unit. PHWR - Adopted from Knief, R. A. Nuclear Engineering: Theory and Technology of Commercial Nuclear Power , pp. 707 –717. American Nuclear Society, La Grange Park, IL, 2008. CANDU 6 Technical Summary, CANDU 6 Program Team, R eactor Development Business Unit, May 2005. HTGR - Breher, W., Neyland, A. and Shenoy, A. Modular High -Temperature Gas -Cooled Reactor (MHTGR) Status . GA Technologies, GA- A18878, May 1987. AGR - Adopted from AEAT/R/PSEG/0405 Issue 3. Main Characteristics of Nuclear Power Plants in the European Union and Candidate Countries . Report for the European Commission, September 2001; Nuclear Engineering International . Supplement. August 1982. Alderson, M. A. H. G. (UKAEA, pers. comm., October 6, 1983 and December 6, 1983); Getting the most out of the AGRs, Nucl. Eng. Int ., 28(358):8 –11, August 1984. SFBR - Adopted from IAEA -TECDOC -1531. Fast Reactor Database 2006 Update . International Atomic Energy Agency. December 2006. 19 rev 112420Chapter 2 - Thermal Design Principles and Application PROBLEM 2.1 SOLUTION Relations Among Fuel Element Thermal Parameters in Various Power Reactors (Section 2.3) Compute the core average values of the volumetric energy- generation rate in the fuel ( q‴) and outside surface heat flux (q″) of cladding for the reactor types BWR, PWR, PHWR, HTGR, AGR and SFBR . Use the core average linear power levels derivable from data in Table 2. 1 and the geometric parameters in Table 1 .3. The volumetric energy generation rate is calculated using 𝑞𝑞‴=4𝑞𝑞′ 𝜋𝜋𝐷𝐷fo2 (1) and outside surface heat of the cladding is calculated with 𝑞𝑞co″=𝑞𝑞′ 𝜋𝜋𝐷𝐷co (2) where Dfo = Diameter of fuel pellet, D co = Diameter of fuel rod, q′ = Linear heat generation rate, 𝑞𝑞co″ = Heat flux at cladding surface and q‴ = Volumetric heat generation rate. Table SM -2.1 shows the data results for the different reactor types: TABLE SM -2.1: Data and Results for Various Power Reactors BWR PWR PHWR HTGR AGR SFBR Dco [mm] 11.20 9.5 13.1 15.7 15.3 8.5 Dfo [mm] 9.6 8.192 12.2 15.7* 14.51 7.14 𝑞𝑞co″�𝑘𝑘𝑘𝑘 𝑚𝑚2� 500.2 598.4 624.5 159.6 353.7 1086.0 𝑞𝑞‴�𝑀𝑀𝑘𝑘 𝑚𝑚2� 243.2 338.9 219.8 40.7 102.8 724.3 *Note: Fuel is coated with graphite and the coating thickness can be neglected. 20 rev 112420Chapter 2 - Thermal Design Principles and Application PROBLEM 2.2 QUESTION Relationships Between Assemblies of Different Pin Arrays (Section 2.3) A utility wishes to replace the fuel in its existing PWR from 15 ×15 fuel pin array assemblies to 17×17 fuel pin array assemblies. What is the ratio of the core average linear power , q′ in the new core to the old core, assuming reactor power, length, number of fuel assemblies and fuel mass are maintained constant. Repeat for the core average heat flux, q″. Answers: 𝑞𝑞17𝑥𝑥17′ 𝑞𝑞15𝑥𝑥15′ = 0. 779 𝑞𝑞17𝑥𝑥17″ 𝑞𝑞15𝑥𝑥15″ = 0. 882 PROBLEM 2.2 SOLUTION Relationships Between Assemblies of Different Pin Arrays (Section 2.3) A utility wishes to replace the fuel in its existing PWR from 15 ×15 fuel pin array assemblies to 17×17 fuel pin array assemblies. What is the ratio of the core average linear power and core average heat flux, q′ and q″, in the new core to the old core, assuming reactor power, length, number of fuel assemblies and fuel mass are maintained constant. The core average power is defined as 𝑄𝑄=𝑁𝑁𝑁𝑁𝑞𝑞′ (1) where Q is the core average power, N is the number of fuel pins, L is the length of the core and q′ is the core average linear power. If we keep the core power and length constant, we can determine the ratio of the linear power from new core to old core as 𝑞𝑞17𝑥𝑥17′ 𝑞𝑞15𝑥𝑥15′ =𝑄𝑄 𝑁𝑁17𝑥𝑥17𝑁𝑁 𝑄𝑄 𝑁𝑁15𝑥𝑥15𝑁𝑁=𝑁𝑁15𝑥𝑥15 𝑁𝑁17𝑥𝑥17=152 172= 0. 779 (2) The heat flux is defined as 𝑞𝑞″=𝑞𝑞′ 𝜋𝜋𝐷𝐷 (3) where q″ is the core average heat flux and D is the diameter of the fuel elements. We would like to not change the fuel mass in the core and therefore the diameter of the fuel elements must change when moving to the new core. To conserve mass or in this case just simply the area of fuel, we apply the following condition: 21 rev 112420Chapter 2 - Thermal Design Principles and Application 𝑁𝑁17𝑥𝑥17=𝜋𝜋 4𝐷𝐷17𝑥𝑥172=𝑁𝑁15𝑥𝑥15=𝜋𝜋 4𝐷𝐷15𝑥𝑥152 (4) We can then determine the ratio of the old fuel element diameter to new element diameter as 𝐷𝐷15𝑥𝑥15 𝐷𝐷17𝑥𝑥17=�𝑁𝑁17𝑥𝑥17 𝑁𝑁15𝑥𝑥15 (5) The ratio of the new to old core average heat flux can be determined by applying Equation (2) and the result from Equation (5): 𝑞𝑞17𝑥𝑥17″ 𝑞𝑞15𝑥𝑥15″ =𝑞𝑞17𝑥𝑥17′ 𝜋𝜋𝐷𝐷17𝑥𝑥17 𝑞𝑞15𝑥𝑥15′ 𝜋𝜋𝐷𝐷15𝑥𝑥15=𝑞𝑞17𝑥𝑥17′𝐷𝐷15𝑥𝑥15 𝑞𝑞15𝑥𝑥15′𝐷𝐷17𝑥𝑥17=𝑁𝑁15𝑥𝑥15 𝑁𝑁17𝑥𝑥17�𝑁𝑁17𝑥𝑥17 𝑁𝑁15𝑥𝑥15=152 172�172 152= 0. 882 (6) PROBLEM 2.3 QUESTION Minimum Critical Heat Flux Ratio in a PWR for a Flow Coastdown Transient (Section 2.4) Describe how you would determine the minimum critical heat flux ratio versus time for a flow coastdown transient by drawing the relevant channel operating curves and the CHF limit curves for several time values. Draw your sketches in relative proportion and be sure to state all assumptions. PROBLEM 2.3 SOLUTION Minimum Critical Heat Flux Ratio (MCHFR) in a PWR for a Flow Coastdown Transient (Section 2.4) Describe how you would determine the minimum critical heat flux ratio versus time for a flow coastdown transient by drawing the relevant channel operating curves and the CHF limit curves for several time values. Draw your sketches in relative proportion and be sure to state all assumptions. For this transient, we can have two different situations depending on whether the reactor will be shut down or not. In a flow coastdown transient, the mass flow through the reactor follows the trends of Figure SM-2.1. 22 rev 112420Chapter 2 - Thermal Design Principles and Application FIGURE SM-2.1: Mass flow rate behavior through reactor As mentioned, there are two poss ibilities for the reactor power as stated below and shown in Figure SM-2.2: 1. Curve 1 - the reactor is shut down. 2. Curve 2 - t he reactor is not shut down and the behavior of the power will be a function of reactivity feedback effects. FIGURE SM-2.2: Reactor power behavior There are two situations for change in operating heat flux with time. In both cases the critical heat flux decreases as the flow decreases. It is assumed that the inlet quality is constant. Condition 1 – operating heat flux decreases with time 23 rev 112420Chapter 2 - Thermal Design Principles and Application Figure SM -2.3: Condition 1, Heat flux as a function of quality At each instant we have a curve for the critical heat flux and for the maximum operating heat flux in the core. The CHF curve decreases with both decreasing heat flux and mass flux. The operating curve as shown in Figure SM -2.3 collapses primarily due to decrease in heat flux. The decrease in mass flux would cause the quality at any given position to increase if the heat flux were constant. Therefore, we can calculate the MCHFR at any time, t. The occurrenc e of CHF will depend on how the power and mass flux vary. Condition 2 – operating heat flux maximum is constant with time Figure SM -2.4: Condition 2, Heat flux as a function of quality In this case, we can assume that the maximum operating heat flux does not vary with time. 24 rev 112420Chapter 2 - Thermal Design Principles and Application PROBLEM 2.4 QUESTION Minimum Critical Power Ratio in a RWR (Section 2.4) Calculate the minimum critical power ratio for a typical 1062 MWe BWR operating at 100% power using the data in Appendix K. Assume that: 1. The axial linear power shape can be expressed as 𝑞𝑞′(𝑧𝑧)=𝑞𝑞ref′exp�−𝛼𝛼𝑧𝑧 𝑁𝑁�sin�𝜋𝜋𝑧𝑧 𝑁𝑁� where α = 1.96. Determine 𝑞𝑞ref′ such that 𝑞𝑞max′=47.24 kW m⁄. 2. The critical bundle power is 9319 kW. Answers: 𝑞𝑞ref′=104 .75 kW m⁄ MCPR = 1.28 PROBLEM 2.4 SOLUTION Minimum Critical Power Ratio in a BWR (Section 2.4 ) Calculate the minimum critical power ratio for a typical 1062 MWe BWR operating at 100% power using the data in Appendix K. The given parameters in the problem are listed below: – α = 1.96 – L = 3.588 m – 𝑞𝑞 max′=47.24kW m – Nrods = 74 – qcr = 9319 kW The linear heat rate profile is given as 𝑞𝑞′(𝑧𝑧)=𝑞𝑞ref′exp�−𝛼𝛼𝑧𝑧 𝑁𝑁�sin�𝜋𝜋𝑧𝑧 𝑁𝑁� (1) To begin, we must obtain the value of 𝑞𝑞ref′ such that the maximum linear heat rate is 𝑞𝑞max′. Note that for a standard cosine profile, 𝑞𝑞max′=𝑞𝑞ref′. To determine the axial location that yields the maximum linear heat rate, we can take the derivative of Equation (1) and solve for the roots, 𝑑𝑑𝑞𝑞′(𝑧𝑧) 𝑑𝑑𝑧𝑧= 0⟼ 𝑧𝑧max = 1. 157 m (2) 25 rev 112420Chapter 2 - Thermal Design Principles and Application Substituting 𝑧𝑧max into Equation 1, the parameter 𝑞𝑞ref′ can be solved for, noting that 𝑞𝑞′(𝑧𝑧max)= 𝑞𝑞max′. This results in 𝑞𝑞ref′=104 .75kw m⁄ (3) The linear heat rate (displayed in W/m) can be plotted as a function of axi al location as shown in Figure SM-2.5, to verify the maxi mum linear heat rate condition of about 4.7× 104 W/m . FIGURE SM-2.5: Axial Profile of the Linear Heat Rate The power of the bundle, assuming all fuel rods have the same linear profile (no local peaking), 𝑞𝑞=𝑁𝑁rods�𝑞𝑞′(z)𝑑𝑑𝑧𝑧=7270 .4 kw𝐿𝐿 0 (4) Then the Minimum Critical Power Ratio (MCPR) is MCPR =𝑞𝑞cr 𝑞𝑞= 1. 28 (5) PROBLEM 2.5 QUESTION Pumping Power for a PWR Reactor Cooling System (Section 2.5) Calculate the pumping power under steady -state operating conditions for a typical PWR reactor coolant system using only the following operating conditions: – Core thermal power, Q = 3411 MWth – Core temperature drop, ∆ Tcore = 33.7 °C – Inlet temperature, T in = 293 °C 26 rev 112420Chapter 2 - Thermal Design Principles and Application – Core system pressure, P = 15.5 MPa – Reactor coolant system pressure drop, ∆ P = 778 kPa – Pump efficiency, ηp = 0.85 Answer : Pumping power = 21.8 MW e PROBLEM 2.5 SOLUTION Pumping Power for a PWR Reactor Cooling System (Section 2.5) Calculate the pumping power under steady -state operating conditions for a typical PWR reactor coolant system using only the following operating conditions: – Core thermal power, 𝑄𝑄̇ = 3411 MWth – Core temperature drop, ∆ Tcore = 33.7 °C – Inlet temperature, T in = 293 °C – Core system pressure, P = 15.5 MPa – Reactor coolant system pressure drop, ∆ P = 778 kPa – Pump efficiency, η p = 0.85 The pumping power can be calculated using Eq uation 2.9a in the text which applies under ideal conditions 𝑘𝑘̇𝑝𝑝=∆𝑃𝑃𝑚𝑚̇ 𝜌𝜌=𝛥𝛥𝑃𝑃𝛥𝛥𝑓𝑓𝜐𝜐 (1) where Wp is the pumping power, A f is the flow area and υ is the velocity. We can perform a heat balance across the core to relate the core thermal power to the temperature drop with 𝑄𝑄̇=𝑚𝑚̇ 𝑐𝑐pΔ𝑇𝑇core =𝜌𝜌𝛥𝛥f𝜐𝜐𝑐𝑐pΔ𝑇𝑇core (2) To calculate the mass flow rate correctly and to relate the flow area and velocity correctly, we must determine the density of the fluid at inlet conditions, P = 15.5 MPa and T = 293 °C. This results in a density of ρ = 740.49.kg/m3. On the other hand the specific heat at constant pressure varies with temperature as the fluid is heated across the core, We may just take an average of this parameter, evaluating it at the core system pressure and the average coolant temperature, 𝑇𝑇core =309 .85 °𝐶𝐶. This result in an average specific heat of 𝑐𝑐p=5736 .03 J kg−K⁄ . Combing Equations 1 and 2, solving for the ideal pumping power, we obtain the following expression and solution: 27 rev 112420Chapter 2 - Thermal Design Principles and Application 𝑘𝑘̇𝑝𝑝=∆𝑃𝑃𝑄𝑄̇ 𝜌𝜌𝑐𝑐p∆𝑇𝑇core=18.5 MW (3) Taking into account the efficiency of the pump, we may divide by 0.85 to obtain the actual pumping power of 21.8 MW e. PROBLEM 2.6 QUESTION Relations Among Thermal Design Conditions in a PWR (Section 2.5) Compute the margin to failure defined as the ratio of linear power rate at failure to the maximum linear power ratio 𝑞𝑞fail′/𝑞𝑞max′ for a typical PWR having a core average linear power rate of 17.86 kW/m. Assume that the failure limit is established by centerline melting of the fuel at 70 kW/m. Use the following multiplication fa ctors: – Radial flux factor = 1.55 – Axial and local flux factor = 1.70 – Engineering uncertainty factor = 1.05 – Overpower factor = 1.15 Answer: Margin = 1.23 PROBLEM 2.6 SOLUTION Relations Among Thermal Design Conditions in a PWR (Section 2.5) Compute the margin for a typical PWR having the following parameters and multiplication factors: – Radial flux factor, F r = 1.55 – Axial and local flux factor, F z = 1.70 – Engineering uncertainty factor, F e = 1.05 – Overpower factor, F op = 1.15 The average linear power rate and the failure limit, established by centerline melting of the fuel are: – Average linear power rate, 𝑞𝑞 ave′=17.86 kW m⁄ – Failure limit, 𝑞𝑞fail′=70 kW m⁄ The maximum linear power rate is 𝑞𝑞max′=𝑞𝑞ave′𝐹𝐹r𝐹𝐹z𝐹𝐹e𝐹𝐹op=56.83kW m⁄ (1) 28 rev 112420Chapter 2 - Thermal Design Principles and Application The margin to failure is calculated as Margin =𝑞𝑞fail′ 𝑞𝑞max′= 1. 23 (2) 29 rev 112420