Lectures on Applied Reactor Technology and Nuclear Power Safety. Lecture No 7

Transcription

1 ectures on Nuclear Power Safety ecture No 7 itle: hermal-hydraulic nalysis of Single-Phase lows in Heated hannels Department of Energy echnology KH Spring 005 Slide No

2 Outline of the ecture lad-oolant Heat ransfer in hannels with Single Phase lows oolant Enthalpy Distribution in Heated hannels emperature Distribution in uel Elements Pressure Distribution in hannels with Single-Phase low Slide No

3 lad-oolant Heat ransfer in hannels with Single Phase lows () z sc Subcooled flow region z nb Single phase (non-boiling) flow region sup sub Saturation temp Wall temp w Bulk liquid temp f Slide No 3

4 lad-oolant Heat ransfer in hannels with Single Phase lows () In ight Water Reactors, coolant is subcooled at the inlet to the reactor core he subcooling is defined as the difference between the saturation temperature and the actual coolant temperature or example, if the inlet temperature and pressure of the water coolant are 549 K and 7 MPa, respectively, then the inlet subcooling is equal to 559 K 549 K 0 K, since the saturation temperature of water at 7 MPa pressure is equal to 559 K Slide No 4

5 lad-oolant Heat ransfer in hannels with Single Phase lows (3) In the single-phase region, when 0 < z < z nb, the surface temperature w of the heated wall and the liquid bulk temperature f are related to each other as follows, w f wf q h wf where h is the heat transfer coefficient and is the temperature difference between the surface of the heated wall and the bulk liquid Slide No 5

6 lad-oolant Heat ransfer in hannels with Single Phase lows (4) he heat transfer coefficient h is evaluated from correlations, which, in turn, are based on experimental data and are using the principles of the dimensionless analysis he following general relationship are employed Nu f (Re, Pr, ), where: Nu Nusselt number D Re µ c p µ c µ Pr, Prw λ λ Reynolds number hd λ p, Prandtl number w Slide No 6

7 lad-oolant Heat ransfer in hannels with Single Phase lows (5) Many different correlations have been proposed Example of a correlation for laminar flows (Re < 000) is as follows Nu 0.7 Re 0.33 Pr 0.43 Pr Pr w 0.5 D 3 ρ gβ µ 0. Slide No 7

8 lad-oolant Heat ransfer in hannels with Single Phase lows (6) or turbulent flows (Re > 000) the following correlation, proposed by Dittus-Boelter, is used: Nu 0.03 Re 0.8 Pr 0.33 Slide No 8

9 oolant Enthalpy Distribution in Heated hannels () ssume a heated channel as shown in the figure to the right. he channel is uniformly heated along its length with heat flux q [W/m ], it has a flow crosssection area and heated perimeter P h. he energy balance for a portion of channel in green is as follows: q H c (z)+dh c H c (z) z z+dz z c H [ H ( z dh ] W H ( z) + q ( z) P ( z) dz W ) + c c dh ( z) q ( z) P dz W ( z c H ) W, H ci Slide No 9

10 oolant Enthalpy Distribution in Heated hannels () hus, the enthalpy distribution of coolant is described by the following differential equation: z dhc ( z) q ( z) PH ( z) dz W Integration yields H c ( z) z H ci + q z PH z dz W ( ) ( ) 0 q H c (z)+dh c H c (z) z+dz z W, H ci Slide No 0

11 oolant Enthalpy Distribution in Heated hannels (3) or small temperature and pressure changes the enthalpy can be expressed as a linear function of temperature: dh c p *d Using W and assuming constant channel area and heat flux distribution, the coolant temperature can be found as, f ( z) fi + q PH z c p Slide No

12 oolant Enthalpy Distribution in Heated hannels (4) he temperature is thus linearly distributed between the inlet and the outlet of the assembly z fo he outlet temperature becomes q fo fi + q PH c p z0 fi W i, fi Slide No

13 oolant Enthalpy Distribution in Heated hannels (5) Usually the axial power distribution is not uniform. In a bare reactor the axial power distribution is given by the cosine function or with z 0 at the inlet sinus function: q z q z) πz q sin ( 0 he differential equation for the enthalpy (temperature) distribution is now dh c z) dz q 0 PH W ( z) πz sin, or d dz ( z) ( f 0 q PH ( z) πz sin W c p W i, fi z0 Slide No 3

14 oolant Enthalpy Distribution in Heated hannels (6) fter integration, the coolant enthalpy (temperature) distribution is as follows q z fo H f c ( z) ( z) q 0 PH πz + H W cos π q 0 PH πz + W c cos π p fi ci, or f (z) he outlet enthalpy (temperature) can be found as: q0 PH q0 PH H co H c( ) + H ci, or fo f ( ) + π W π W cp fi W i, fi z0 Slide No 4

15 emperature Distribution in uel Elements () he stationary heat conduction equation for an infinite cylindrical fuel pin is as follows r d dr d λ r q dr oolant integration yields d r λ r q dr lad ap uel r Slide No 5

16 emperature Distribution in uel Elements () or constant fuel conductivity the equation can be readily integrated as follows r 4λ () r q + However, the fuel conductivity is a function of a temperature, and the integration has to be performed as follows: λ d r q q dr λd r 0 0 rdr r q 4 Slide No 6

17 emperature Distribution in uel Elements (3) Introducing the average fuel conductivity given as: λ 0 0 λ d he total temperature drop in the the fuel can be found as: 0 q r 4 λ Or, using the so-called linear power density q π r q q 4π λ Slide No 7

18 emperature Distribution in uel Elements (4) he temperature drop across the gap can be found assuming that the conductivity is the major heat transfer mechanisms: r d dr λ d r 0 r ln r + dr dr λ he integration constant is found as: d λ dr r r r λ d π q q π r nd the temperature drop in the gap is found as: ( r ) ( r ) q π λ ln r r Slide No 8

19 emperature Distribution in uel Elements (5) Similar derivation for the clad region yields the following temperature drop: ( r ) ( r ) q π λ ln r r here r c and λ c are the outer clad radius and clad material conductivity, respectively Heat convection from the clad surface to the coolant is given as q h ( ) f Slide No 9

20 emperature Distribution in uel Elements (6) Since q q ( π ) r temperature drop in the coolant thermal boundary layer is as follows: f f q π r h he total temperature drop is thus f q r + ln + π λ λ r λ ln r r + r h Slide No 0

21 Slide No emperature Distribution in uel Elements (7) or non-uniform ( cosine ) heat flux distribution Substituting the above to yields z q z q π sin ) ( 0 fi p H f z c W P q z + π π cos ) ( 0 ( ) h q h q f f z h q z c W P q z fi p H π π π sin cos ) ( 0 0

22 emperature Distribution in uel Elements (8) igure to the right shows the clad temperature distribution assuming the cosine axial power distribution q z max It should be noted that the clad temperature gets its maximum value max at a certain location z max (z) W i, fi z0 Slide No

23 emperature Distribution in uel Elements (9) he location of the maximum clad temperature can be found as: d dz z max 0 : tan π q 0 W P c H p πz sin + π W c h P H p q 0 π πz cos 0 h Substituting z zmax in the equation for the clad temperature yields the maximum clad temperature q P πz 0 H max + W c sec max π p fi Slide No 3

24 Slide No 4 emperature Distribution in uel Elements (0) It can be shown that the maximum fuel temperature at the centerline is as follows, fi p H z z c W P q r r r r z q r max 0 max 0 max sec ln ln sin π π λ λ λ π

25 Pressure Distribution in hannels w/single Phase low () onsider mass and momentum balance in the following channel d w W+d W U+dU +d p+dp d g ϕ W U p g dz Slide No 5

26 Pressure Distribution in hannels w/single Phase low () he balance of forces projected on the channel axis is as follows: Pressure force p ( p + dp)( + d) pd dp Wall friction force dw PW dz τw ravity force dg -ρ dz g sinφ Momentum change -(W + dw)(u+ du) + W U -dw U - W du he resultant equation is: dp dz d dz ρ + p d dz + P τ w W + ρg sinϕ Slide No 6

27 Pressure Distribution in hannels w/single Phase low (3) or vertical channel with constant cross-section: dp dz PW τ w + ρg Integration along channel of length yields: dp dz dz P τ W w W w [ p( ) p(0) ] + ρg dz + ρg 0 0 P τ or P W w [ p( ) p(0 ] + ρg p ) tot τ Slide No 7

28 Pressure Distribution in hannels w/single Phase low (4) In a straight channel with constant cross-section area the only irreversible pressure loss is caused by friction: where p fric P W τ w thus τ w f ρ dp dz fric f P W ρ p fric f P W ρ Slide No 8

29 Pressure Distribution in hannels w/single Phase low (5) In a straight channel with constant cross-section area the only irreversible pressure loss is caused by friction: where p fric P W τ w thus τ w f ρ dp dz fric f P W ρ p fric f P W ρ Slide No 9

30 Pressure Distribution in hannels w/single Phase low (6) or local obstacles, additional irreversible pressure losses appear onsider sudden expansion of a channel Mass balance U U Momentum balance 0 ρ ( UU U U ) + p p Slide No 30

31 Pressure Distribution in hannels w/single Phase low (7) Solving mass and momentum equations for pressure difference yields ( ) p p ρu U U ρu pplying mechanical energy balance (Bernoulli equation): ρu + p ρu + p nd combining the two yields: p I ρu + p I Slide No 3

32 Pressure Distribution in hannels w/single Phase low (8) It is customary to express the local loss coefficient in the following form: ρu ploss ξloss ξ loss ρ hat is, for the sudden channel enlargement, the local pressure loss is p I ; ξenl ρ Slide No 3

33 Pressure Distribution in hannels w/single Phase low (9) In a similar manner, the local pressure loss for a sudden area contraction is found as: p I c ; ρ vena contracta ξ cont c c c ( ) c Slide No 33

34 Pressure Distribution in hannels w/single Phase low (0) he total pressure drop in a channel with length, constant cross-section area and local obstacles along the channel can be calculated as: p 4 f tot p fric ploc pelev + + D i ξi ρ ρ g sinϕ Slide No 34

35 Exercises () Exercise 8: pipe with the inner diameter 0 mm and the outer diameter mm is uniformly heated on the outer surface with heat flux q MW m- and cooled with water flowing inside the pipe. he inlet mass flux of water is 00 kg m - s -, the inlet enthalpy 5.6 kj kg - and the inlet pressure 70 bar. t what distance from the inlet the water will become saturated? Slide No 35

36 Exercises () Exercise 9: In a pipe as in Exercise 8 assume that the single-phase flow exists from the inlet to the pipe until the point where water becomes saturated. alculate the wall temperature on the outer surface of the pipe at the point where water becomes saturated if the wall conductivity is equal to 5 W m- K -. Slide No 36

37 Exercises (3) Exercise 0: or a fuel rod as in Example 7.3. (see ompendium, hapter 7, page 7-8) assume that the fuel heat conduction is the following function of temperature, [K], λ [W m- K-] What will be the total temperature drop difference when the heat transfer coefficient drops from 5000 to 300 W m - K -? Slide No 37

38 Exercises (4) Exercise : Water flows in a horizontal pipe with a sudden expansion. alculate the total pressure drop (reversible and irreversible) in a place where the pipe diameter suddenly increases from 0 cm to cm. ssume flow of water in direction from smaller to larger flow area with mass flow rate kg/s and density 740 kg m -3. Slide No 38

39 Exercises (5) Exercise : Solve the problem as in Exercise, but assume that the water flows in a direction from the larger to the smaller cross-section area (sudden contraction). Explain why the pressure drops in Exercises and are different. Slide No 39