Summary Slides • Isentropic processes 3-7 NTEC Module: Water Reactor Performance and Safety • Propagation of small amplitude Lecture 9: Critical flow pressure wave 8-12 • Isentropic flow 13-20 G. F. Hewitt • Critical flow 21-27 Imperial College London • Example of critical flow of steam 28-31 • Homogeneous equilibrium model (HEM) for two-phase flow 32-38

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Isentropic processes I Isentropic processes II

Ideal gas law Definition of specific enthalpy

Ru hupvuRT  (ideal gas) pv RT R ' M u  specific internal energy (J/kg)

1 3 v specific volume  (m /kg)   Definition of ccvp and p  pressure (Pa) dh cp dT R  gas constant (Nm/kg.K) du cv dT T  temperature (K)

Ru  Universal gas constant Thus:  8314 (Nm/kmole.K) dh cpv dT du RdT c dT RdT ' M  Molecular weight (kg/kmole) ccRpv 3 4

1 Isentropic processes III Isentropic processes IV Define c Work done given by p  k ( often also used) cv dw pdv kR R Thus ccpv , kk11 du dq pdv Define: Chenge in internal energy du given by dq du dq dw ds  dq  heat added per kg (J/kg) T dw  work done by system (J/kg) Thus: Tds du pdv 5 6

Isentropic processes V Isentropic processes VI But For an adiabatic, reversible ("isentropic") process ds  0. Thus: dT R dv dh d ( u pv ) du pdv vdp  0 --- (1) Tcvv Tds dh vdp dT R dp  0 --- (2) Tcpp Subtracting (2) from (1) and dividing by R : Thus cv dpc dv du p dT dv p 0 ds dv  cv  R pcv TT T v v Integrating dh v dT dp ds dp  c  R c TTp T p lnpvp ln constant cv

k 7 pv  constant 8

2 Propagation of small amplitude pressure wave I Propagation of small amplitude pressure wave II

Consider frame of reference moving with wave

p + dp V = 0 a - dV a dV ρ + dρ V = dV p, ρ, T p + dp T + dT ρ + dρ p, ρ, T a T + dT

• Long cylinder filled initially with motionless fluid Control volume moving with wave Continuity equation on control volume • Piston at one end given infinitesmal velocity dV aadVd ( )(  ) • Pressure wave moves at velocity a separating fluid moving at velocity dV and fluid which a a ad   dV  dVd  remains stationary ad dV --- (3)

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Propagation of small amplitude pressure wave III Propagation of small amplitude pressure wave IV

But from equation 3 (continuity), dV ad Momentum equation for control volume: dp 2 ad22  ad pA ( p dpA )  aA22 ( d )( a  dV ) A dp a2  Dividing through by A and expanding: d 2 For thin region around wave with a disturbance of p p dp  a22  (  d )( a  2 adV  dV ) infinitesmal amplitude, the process is essentially a22  a  2  adV   dV2  a 2 d   2 adVd  ddV 2 ADIABATIC AND REVERSIBLE, hence ISENTROPIC Thus 2 2 dp dp 2 adV  a d a   d S

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3 Propagation of small amplitude pressure wave V Isentropic flow I For ideal gas, isentropic process gives p pvk  constant ( X )  k

kk11p dp Xk dk k d  Procedure: Derive continuity p V and momentum equations for  kd element of length dx.  ρ pkp kRT S

x akRT 13 14

Isentropic flow II Isentropic flow III

Momentum equation (steady flow without body or shear forces) Continuity equation (steady flow) d 11ppdA VA  0 pdxdApdxdx dx 22x xdx  or VA M const pA pA dx pA x Differentiating 2 22 ρV A VA dx VA VAd VdA  AdV  0 dx x ddAdV or   0  AV A A dA Momentum flux out – momentum flux in = sum of (pressure) forces on element 15 16

4 Isentropic flow IV Isentropic flow V

222 VA dx VA VA  p x dx V2 A Adx 1 pA   xx pA p dx dx  pA  dx pA  p 2 xx  x  or VA2  A 0  xx VA222 dx VA VA  p x or VVAA 0 ApAAp1 2 xx pA pdx  dx  pA pdx  Adx x 2 xx x x Vp  p VA V VA A 0 --- (4) dx V2 A Adx x xx xx

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Isentropic flow VI Isentropic flow VII

But from the continuity equation dV also follows from the continuity expression  V VA  0 x dV dA d   VA Thus equation 4 becomes (eliminating A ) and introducing this into equation 5 we have: Vp dA d dp V  0   xx ApV 2 or dp VdV dA d dp dp V 2    1  dV dp 22 or  --- (5) ApVpVdpd /  VV 2

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5 Isentropic flow VIII Critical flow I But adpd2222 / and VaM / where M is the MACH NUMBER. Thus * * po p pd po > p > pd pdAA/ v  in 22 When V = M at throat of VM1 V = a nozzle, further decrease in Thus: M = 1 downstream pressure Subsonic flow MdpdA 1: positive if positive (a) Nozzle does not change velocity (or mass flux) at throat or vena contracta or p*. dp negative if dA negative p* p p o d Hence MAXIMUM or vin Supersonic flow Mdp 1: negative if dA positive CRITICAL flow observed in these conditions. V = a M = 1 dp positive if dA negative (vena contracta) 21 (b) Orifice 22

Critical flow II Critical flow III

Kinetic energy per unit mass of fluid  V 2 / 2. kRT V 2 kRT  o First law of thermodynamics: kk12 1 V 2 hho  constant 2 If V a kRT*2* then V  kRT . Thus:

ho  STAGNATION ENTHALPY ** kRT kRT kRTo  enthalpy of fluid brought to rest ( upstream enthalpy if Va )  in kk12 1 kRT hcTp (Slide 5) k 1 * Dividing through by kRT and multiplying through by k  1 Thus:

2 kk11To  kRT V kRTo 1  * kk12 1 22T

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6 Critical flow IV Critical flow V * To obtain ppo / we use k Maximum (critical) mass rate of flow pv constant = c1 (isentropic expansion) aA* pv RT (ideal gas law) MaAAkRT **  ** * max v* RT vck constant  ** v 1 AVaACAC is flow area at which  ( , coefficient of discharge) c Tvk 1  constant 1 c 2 ** R We express TpT and  in terms of (known) oo and 1 kk1/ k  cc11k 1  * 2To v  v  T  pp  k 1 T p*  constant  *  (ideal gas law) pkk1/ RT * Thus kk/1   kk/1   pToo 1 k **  pT 2 25 26

Critical flow VI Critical flow VII 2T T *  o k 1 1 * po 11k  * MA  kRT * p crit1/k  1 o    * (ideal gas law) RTo k 1 2 RT  p 2 p*  o kk/1 11 1 k 1/2   k12  * kk1 2  Apo  RTo 2 * po 1   kk/1 1  k 1 1 k 1 k 1/2   RTo  kk1 21k  2 2 *   Apo  p 1 RTo 2 o 1/k  1 RTo 1 k  2 27 28

7 Example of critical flow of steam I Example of critical flow of steam II

• A 1100 MW(e) reactor operating at 150 bar is shut down due to a smqll break corresponding to an area of 0.011m2. A short time after For steam the shutdown, the rate of power generation has reduced to 200MW(t). Ru 8314 • The reactor has remained pressurised and the steam generators are R '  461.9 not operating. Heat removal is by “feed and bleed” with water being M 18 injected by the HPIS and the heat being removed in the form of 7 steam which flows through the break. po  150 bar 1.5 10 Pa • Assuming that the specific heat ratio k (= cp/cv) is 1.3 for steam and that the steam is saturated at 343 C, calculate the maximum flow To  342 273 615 K rate of steam which could occur. Also, assuming a latent heat of vaporisation of 1000 kJ/kg, calculate the maximum amount of heat ACA*  1 0.011 0.011 which could be removed. Is this sufficient? • The molecular weight of water should be taken as 18 kg/kmole and the break treated as a nozzle with a discharge coefficient of unity. The Universal Gas Constant is Ru = 8314 J/kmole K

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Example of critical flow of steam III Example of critical flow of steam IV

k1 1/2  21k  * kk1 Maximum rate of heat removal MAp crit o  RTo 2 3 Mh max LG 206 1000 10 1/2 1.3 1 /2 1.3  1 7 1.3 1.3 1  0.011 1.5 10    206 MW 461.9 615 2  0.011 1.5 107 0.002139  1.15 3.833 (i.e. just sufficient to remove decay heat)  206 kg/s

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8 Homogeneous equilibrium model Homogeneous equilibrium model (HEM) for two phase flow I (HEM) for two phase flow II

Gas Temperature Tg

Velocity ug Density of fluid ( ) given by Composition yi Mole fraction of th component 1 i xx/1/  gl Liquid Temperature Tl (lg and and liquid and gas densities) Velocity ul x  quality (vapour mass flow as a fraction of total flow) Composition xi  Mole fraction of i th component Assumptions:

TTgl Applied along uu path of flow gl from upstream to throat yxii 's in equilibrium with 's 33 34

Homogeneous equilibrium model Homogeneous equilibrium model (HEM) for two phase flow III (HEM) for two phase flow IV

Method general and applies to both single component and multi-component fluids

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9 Homogeneous equilibrium model (HEM) for two phase flow V

Thermodynamic Equilibrium

Tg = Tl

Yes No

Mechanical Equilibrium

ul = ug

Yes No Yes No

Phasic Equilibrium

xyii's in equilibrium with 's

Yes No Yes NoYes No Yes No

HEM Example: Dispersed Flow Model (DSM) 37 (Lemonier and Selmer Olsen)

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