ENGINEERING OF NUCLEAR REACTORS

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Isabel Douglas
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1 .3 ENGINEEING OF NUCLEA EACOS OPEN BOOK QUIZ # SOLUIONS.5 HOUS Problem #: Startin from the Control Volume firt law r haft work equal to zero: U U Qn p Qc t () Gien the aumption:. Air, He, and water apor are perfect ae. Nelectin decay heat 3. he outide wall of the containment i perfectly inulated hi equation reduce to: or, expandin term: U U 0 () ( Ua Ua ) + ( U U ) + ( Uw Uw ) 0 (3) p of 7

2 reatin all pecie in equation 3 a perfect ae, and notin that at the end of the tranient, all ae will be at the ame temperature, thi equation may be re-expreed a: nc, ( ) nac, a( a ) nwac, wa( w, a ) (4) Notin that initially, the air and the water apor are in thermal-equilibrium: wa, a Now ole r : nc nac aa nwac waa, +, +, nc, + nac, a + nwac, wa Uin the Perfect Ga Law: ( n + na + nwa) P + Va where i the unieral a contant. (5) (6) Subtitutin equation 5 into equation 6 reult in: ( n + na + nwa) nc + nac aa + n c,, wa, wa a P + Va nc, + nac, a + nwac, wa (7) hi i the expreion r the final preure in the containment. he final preure limit i 4.5 atmophere. In MPa, thi i: atm 45. atm * 456 MPa (8). 033 * 05 Pa All alue are ien except r n, n wa, and V. Uin equation 7-9 from NS Vol I: p φp wa at ( ) a φ 8 a 3K Uin the team table: pat 07 MPa (9) reatin water a a perfect a: patva nwaa (0) p of 7

3 Solin r n wa, and ubtitutin ien and computed alue: n wa patva ( 08)( 34, 830) (. )( ) kmol () a Now the remainin unknown are n and V. hee quantitie are related throuh the Perfect Ga Law applied to the helium ytem at it initial tate. PV (. 888) n. 57 V () ( 0083)( 700) Subtitutin thi alue, and other computed or ien alue into equation 7 reult in: (. 57V )( ) (. 57 )(. 47)( 700) + ( 30)( 80)( 3) + ( 53. 6)( 5. 4)( 3) , 830 (. 57 )(. 47) + ( 30)( 80) + ( 53. 6)( 5. 4) (3) Simplified Further: V , * , , 965 (4) hi mut be oled iteratiely r V. he method of Biection wa ued in r the deelopment of thi olution. ith a preure tolerance of 00 MPa, the final anwer i: m3 (5) If the initial olume of the He i nelected in the final anwer, equation 4 become: V , * 06 34, , 965 (6) Uin the ame iteratie method, to a preure tolerance of 00 MPa, the anwer i: 56. m3 (7) p 3 of 7

4 Problem # A Initially Solid Pellet at q initial B Sintered Pellet at q initial C Sintered Pellet with maximum temperature of 500 C at q re > q D Increae linear power by 0% (i.e., q new,re.q re ) initial he aboe fiure, adapted from Fiure 8-6 in NS Vol I, decribe the loic of thi problem. An initially olid pellet (A) underoe interin (a decribed by the -zone interin model) (B). he initial linear power i increaed o that the pellet maximum temperature i 500 C (line C). hen, the linear power i further increaed 0% (D). Part a) he firt tep i to find the linear power r the initially olid pellet which upon interin, ha it linear power raied to q re o that max 500 C. Uin equation 8-99 and fiure 8- from NS Vol. I, obtain the relationhip between and q re. 800 k d q re π 700 () Uin equation 8-00 and fiure 8- from NS Vol., obtain the relationhip between q re,,, and : max k d q re 098. ρ 4π ρ 800 o ln () p 4 of 7

5 Now ole equation r the quantity q 4π re ρ ρ o q re 4π : + ln Subtitute thi into equation to eliminate q re : ln (3) (4) Perrmin ome alebraic manipulation, one arrie at:. 667 ln + From ma balance equation, 8-98 in NS Vol. : ρ ρo ρ 098. Subtitute the aboe expreion relatin and into equation 5 reultin in:. 667{ 0( + ln ( 9. 8) )} 776 Inertin thi reult into equation find the deired reult: q re 4π. 776 (5) 98. (6). 776 (7) q re (8) p 5 of 7

6 hi i the initial linear power. he new linear power will then be: q new. q re (9) Gien thi, find the new reultin max. Inert the new linear power into equation : k d π (0) he relationhip between and till hold, ubtitutin thee alue into equation : max? k098. d ( 6076) 0 ln( 9. 8) { ( + )} π 800 max? k 098. d () From NS Vol, fiure 8-: max 640 C () Part b) For annular fuel rod with no retructurin uin the ame linear power a arried at in part a, and maintainin the temperature limit of 500 C, we ue equation 8-69 in NS Vol. : max 500 q k d F , π 700 (3) Uin Fi 8-: max 500 k088. d (4) 700 p 6 of 7

7 and the linear power from part a: F 4π ( 4)( 4π), F (5) From Fi 8-8 r β (unirm eneration): Now: α, o 03 α. mm mm α 5 ( 03. ) 6. (6) p 7 of 7

1 inhibit5.mcd. Product Inhibition. Instructor: Nam Sun Wang

1 inhibit5.mcd. Product Inhibition. Instructor: Nam Sun Wang Product Inhibition. Intructor: Nam Sun Wang inhibit5.mcd Mechanim. nzyme combine with a ubtrate molecule for form a complex, which lead to product. The product can alo combine with an enzyme in a reerible