ENGINEERING OF NUCLEAR REACTORS OPEN BOOK FINAL EXAM 3 HOURS

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1 .31 ENGINEERING OF NUCLEAR REACTORS Tuesay, December 16 th, 014, 9:00 a.m. 1:00 p.m. OPEN BOOK FINAL EXAM 3 HOURS Problem 1 (0%) Power Uprate in a PWR Core Consier the hot channel in a PWR core. You wish to increase the power in that channel by 30%, without changing the outlet temperature an pressure. You are to evaluate two alternative approaches: A. Reuce inlet temperature; same mass flow rate B. Increase mass flow rate; same inlet temperature Please answer the following questions: i) If the value of the MDNBR in the reference case is 1.6, what is the value of the MDNBR in approaches A an B? You may assume the heat flux in the channel is axially uniform. (15%) ii) Which approach woul you choose? In answering this question, please consier the results in Part i, but also any other esign an operation aspects that you eem appropriate. (5%) Problem (15%) Use of a Spring to Reuce the Claing Stresses As a means to reuce the stresses in the claing of a BWR fuel ro, a nuclear engineer proposes to use a spring that compresses the fuel pellets, as shown in Figure 1. The ro outer iameter an claing thickness are 11. an 0.5 mm, respectively. During operations the pressure insie the ro (ue to the filler gas an fission gases) is 3 MPa an the coolant pressure is 7 MPa. At these conitions the spring is compresse by 1 cm. Calculate the rigiity constant of the spring [N/m] that will result in a zero axial stress ( z) in the claing. Spring P in =3 MPa P out=7 MPa Claing Fuel pellet Figure 1. Schematic of the fuel ro en (not to scale) 1

2 Problem 3 (65%) Debris Transport following a LOCA in a PWR Containment The occurrence of a Large-Break Loss of Coolant Accient (LB-LOCA) in a Pressurize Water Reactor (PWR) woul cause the generation of large amounts of ebris within the containment, i.e. mostly fibers from the insulation aroun primary system pipes an components. Such ebris woul be washe to the containment sump along with any irt normally present within the containment. Water that accumulates in the sump is use by the Emergency Core Cooling System (ECCS), i.e. the sump pump injects that water into the reactor vessel. This situation is shown schematically in Figure. There is a concern that the ebris coul be transporte to the core by the sump pump an there clog the fuel channels, thus preventing core cooling in the long term. To minimize the probability of ebris transport, it is esirable to limit the coolant flow rate from the sump pump to the minimum flow rate require to remove the ecay heat. Containment steel shell Steam RPV Conensate Break Core 4 m 4 m Debris Sump 6 m Sump pump Figure. Post-LOCA recirculation within containment In answering the following questions, please state an justify all your assumptions. The properties neee for this problem are reporte in the table on the last page of this exam. i) Calculate the minimum flow rate require to remove the ecay heat by boiling, one hour after shutown, for a PWR of nominal power equal to Q 0 =4000 MWt. Assume the sump water is saturate at MPa. Assume the reactor ha operate for an infinite perio of time prior to shutown. (10%)

3 Now focus on the ebris in the water that accumulates in the sump. The total volume of water in the sump is 100 m 3 an its epth is 4 m. ii) Calculate the ebris terminal velocity for gravity settling, V. You may assume the ebris are spherical, have a ensity of 3500 kg/m 3, an a constant rag coefficient, C F /( A V / ), equal to 3, where F is the rag force, A D / 4 is the frontal f area of the ebris, D=100 m is the effective average iameter of the ebris, an f = 958 kg/m 3 is the ensity of water in the sump. (10%) iii) Using the results in Parts i an ii, compare the time scale for ebris settling in the sump to the time scale for ebris flow through the sump, an etermine if the ebris actually settle to the bottom of the sump. (5%) Steam exits the primary system through the break an is conense somewhere in the containment; the conensate rips back into the sump, from which it is pumpe to the core, effectively realizing a loop (see Figure ). iv) Calculate the power require by the sump pump to eliver a coolant mass flow rate of 3 kg/s to the core. (10%) In answering your question you may make use of the following assumptions: o Steay-state operation o The pump has an isentropic efficiency of 85% o Neglect all friction an form pressure losses in the loop except for the pressure loss in the core, which can be moele as a single form loss with coefficient K core = 150. Use the ensity of steam g an the flow area of the core (6 m ) to calculate the form pressure loss. o Neglect all acceleration pressure changes in the loop. o Consier gravity pressure changes throughout the loop. Relevant elevations are shown in Fig.. o Assume the steam quality varies linearly in the core from zero (inlet) to one (outlet); use HEM to calculate the average ensity in the core. Now focus on the containment shell, which is 3-cm thick an mae of steel with thermal conuctivity 35 W/m C. There is air at 40 C circulating on the outer surface with a heat transfer coefficient of 40 W/m C. Conensation of steam at 100 C occurs on the inner surface with a heat transfer coefficient of 800 W/m C. v) What is the minimum containment surface area require to conense 3 kg/s of steam? (10%) vi) What esign changes woul you consier if you wishe to reuce the minimum surface area requirement in Part v? A qualitative answer is acceptable. (5%) 3

4 Now focus on the containment as a whole. Its free volume is V c = 60,000 m 3. The total mass of air is M a = kg. At a certain time after the LB-LOCA the temperature in the containment atmosphere is 100 C an the total amount of water (steam + liqui) is M w = kg. vii) viii) What is the volume occupie by the liqui an what is the volume occupie by the steam? In answering this question you may assume the liqui water/steam mixture in the containment is saturate. (10%) What is the total pressure in the containment? Treat air as a perfect gas (R=86 J/kg-K, c v=719 J/kg-K). (5%) Properties of saturate water at MPa Parameter Value T sat 100 C f 958 kg/m 3 g 0.6 kg/m 3 h f h g c f c g f g k f k g 419 kj/kg 676 kj/kg 4. kj/(kg C).03 kj/(kg C) Pa s Pa s W/(m C) 0.05 W/(m C) N/m 4

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