NUCE 402: Introduction to Nuclear System and Operation Lecture 4 PDF

NUCE 402: Introduction to Nuclear System and Operation Chapter 2-2 (Heat Removal from Nuclear Reactors – Conduction) Dr. Ahmed Alkaabi Conduction  Heat removal from a reactor  Conduction  Caused by the temperature difference...

NUCE 402: Introduction to Nuclear System and Operation Chapter 2-2 (Heat Removal from Nuclear Reactors – Conduction) Dr. Ahmed Alkaabi Conduction  Heat removal from a reactor  Conduction  Caused by the temperature difference  No macroscopic movement of matters  Convection  Caused by the temperature difference  Heat carried away by moving liquid or gas  Radiation  Heat removal by thermal radiation  Conduction  Fourier’s law q  kT Similarity with Fick’s law Heat flux Thermal conductivity J   D Btu/hr-ft2 Btu/hr-ft-oF q  n  qn Heat flow rate across a unit area with unit outward normal, n 2 Conduction  Energy conservation  In unit volume  Net rate of flow  Rate of heat   of heat out of V  production within V   0       q  ndA   div qdV A V   qdV     qdV V V   q  q  0 q  kT   q  k  T  k 2T q Then finally  2T  0 Steady state heat conduction equation, k or Poisson’s equation When there is no heat source  2T  0 Laplace’s equation 3 Solutions to Conduction Eq. - Plate  Plate-type fuel element  Fueled central strip of 2a  Surrounded by cladding of b  Uniform heat generation density, q"' Btu/hr-ft3  At steady state  Heat flow in x-direction only q d 2T q Poisson’s eq. T 2 0 2  0 k dx kf q 2 General solution T   x  C1 x  C2 2k f Boundary conditions T ( x  0)  Tm C 2  Tm dT 0 C1  0 dx x 0 4 Solutions to Conduction Eq. - Plate  Temperature distribution q 2 A T  x  Tm 2k f Temp. at the surface of the fuel q 2 Ts   a  Tm 2k f  Heat flow out of the surface of the fuel = heat produced within the fuel q  qAa Or, from the Fourier’s law q   k f dT q  k f ( 2 x)  qa dx 2k f x a then q  qA Tm  Ts rearrangement q Thermal resistance a / 2k f A 5 Solutions to Conduction Eq. - Plate  Temperature distribution in cladding  No heat source  T 0 2 d 2T 2 0 dx General solution T  C1 x  C2 Boundary conditions T ( x  a)  Ts C1a  C2  Ts xa T  Ts  (Ts  Tc ) b T ( x  a  b)  Tc C1 (a  b)  C2  Tc Heat balance dT Tc  Ts q  qA q   k c  kc dx b Ts  Tc q  kc A b 6 Solutions to Conduction Eq. - Plate  Temperature difference In fuel meat Tm  Ts a q Tm  Ts  q( ) a / 2k f A 2k f A In cladding Ts  Tc b q  kc A Ts  Tc  q ( ) b kc A therefore Tm  Tc a b q Tm  Tc  q(  ) a b 2k f A k c A  Total thermal 2k f A k c A resistance => Contribution of any materials can be simplified by linearly adding their thermal resistance 7 Solutions to Conduction Eq. - Cylinder  Cylindrical fuel rod  Fueled central rod of r=a  Surrounded by cladding of b  Uniform heat generation density, - q’” Btu/hr-ft3  At steady state  Heat flow in r-direction only q d 2T 1 dT q Poisson’s eq. T 2 0 2   0 k dr r dr k f q 2 General solution T  r  C1 ln r  C2 4k f Boundary conditions T ( x  0)  Tm C 2  Tm T is non  singular C1  0 8 Solutions to Conduction Eq. - Cylinder  Temperature distribution q 2 q 2 T  r  Tm Temp. at the surface of the fuel Ts   a  Tm 4k f 4k f  Heat flow out of the surface of the fuel = heat produced within the fuel Thermal resistance Tm  Ts q  qa 2 H q Rf  1 1 /( 4k f H ) 4k f H  Temperature distribution in cladding  No heat source General solution  T 0 2 d 2T 1 dT 2  0 T  C1 ln r  C2 dr r dr Boundary conditions T ( r  a )  Ts C1 ln a  C2  Ts T (r  a  b)  Tc C1 ln( a  b)  C2  Tc a Ts ln( a  b)  Tc ln a  (Ts  Tc ) ln r T b Then ln( 1  b / a) 9 Solutions to Conduction Eq. - Cylinder Heat balance dT q  qA q  qAsurface  kc 2 (a  b) H dr surface (Ts  Tc ) 1 q  kc 2 (a  b) H ( ) ln( 1  b / a) r r  a b (Ts  Tc ) Thermal resistance  kc 2H ln( 1  b / a ) ln( 1  b / a) Rc  2Hk c  Temperature difference In fuel rod Tm  Ts Tm  Ts  q( 1 q ) 1 /( 4k f H ) 4k f H Tm  Tc  q( R f  Rc ) In cladding (Ts  Tc ) ln( 1  b / a ) Tm  Tc q  kc 2H Ts  Tc  q q ln( 1  b / a) kc 2H R f  Rc 10 Space Dependent Heat Source  In finite cylindrical core z q(r )  qmax  (r ) cos( ) H  Negligible variation in r-direction within a given fuel rod  Non-uniform heat generation along z-direction => results in non-uniform temperature distribution along z-direction  Negligible heat flow along z-direction  Heat flow out of the surface of the fuel at z = heat produced within the fuel at z q( z)  2 (a  b) Hq( z)  q( z)a 2 H a2 Tm ( z)  Tc ( z )  q( z)( R f  Rc ) q( z )  q( z ) 2(a  b) q( z ) Surface area generalization q( z )  Tm  Tc  q( R f  Rc ) 2 (a  b) H Tm ( z )  Tc ( z )  2 (a  b) H ( R f  Rc )11 Exponential Heat Source  Heat generation in shield and structure  Flux attenuate exponentially  Radiative energy is deposited exponentially  Consider slab q  Se x q d 2T S  x T 2 0  e 0 k dx 2 k General solution S T  C1 x  C2  2 e  x k Boundary conditions T ( x  0)  T1 T ( x  a)  T2 x S x T  T1  (T2  T1 )  2 [1  e  (1  e  x )]  x a k a => T could be higher than both T1 and T2 12 Summary  Conduction Similarity with Fick’s law  Fourier’s law q  kT J   D  From energy balance q Steady state heat conduction equation,  2T  0 k or Poisson’s equation  Solutions d 2T q q 2  0 T  x  Tm  Plate-type dx 2 kf 2k f d 2T xa 0 T  Ts  (Ts  Tc ) dx 2 b Tm  Tc a b q Tm  Tc  q(  ) a  b 2k f A k c A  Cylindrical 2k f A k c A 1 Tm  Ts  q( ) 4k f H Tm  Tc Tm ( z )  Tc ( z ) q q( z )  ln( 1  b / a ) R f  Rc generalization 2 (a  b) H ( R f  Rc ) Ts  Tc  q kc 2H 13  Homework# 2  Due date: one week before class 14