CN107784156A - The computational methods of nuclear power unit steam-dump system parameter - Google Patents

The computational methods of nuclear power unit steam-dump system parameter Download PDF

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CN107784156A
CN107784156A CN201710800214.0A CN201710800214A CN107784156A CN 107784156 A CN107784156 A CN 107784156A CN 201710800214 A CN201710800214 A CN 201710800214A CN 107784156 A CN107784156 A CN 107784156A
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mrow
msup
msub
steam
msubsup
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CN107784156B (en
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刘现星
罗小雨
孙海军
程堃
张靖
刘磊
王珺
唐征明
李广围
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719th Research Institute of CSIC
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Abstract

The present invention relates to the computational methods of nuclear power unit steam-dump system parameter, comprise the following steps：Initialization；Calculate steam flow, decompressor coefficient, discharge valve opening and pressure-drop in pipeline；Calculate steam flow, feedwater flow, circulating ratio, average heat transfer coefficient, reactor capability, Main Coolant mean temperature, recycling water-carrying capacity and the circulating water flow at current time；The vapor (steam) temperature at current time is calculated；Obtain the saturated vapour pressure at current time；The heat transfer zone for obtaining current time n goes out saliva enthalpy；Calculate the steam-water separation area temperature at current time；Calculate the decline area temperature at current time；Obtain the heat transfer zone entrance enthalpy at current time；Calculate the primary side of steam generator coefficient of heat transfer at current time；Calculate the coolant temperature leaving reactor vessel at current time；Step S2 is repeated to step S17；Determine the higher limit of coolant temperature leaving reactor vessel and steam pressure.The present invention is easy to use.

Description

The computational methods of nuclear power unit steam-dump system parameter
Technical field
The present invention relates to the computational methods of discharge of steam, specifically nuclear power unit steam-dump system parameter.
Background technology
Steam-dump system is an important system for ensureing nuclear power unit safe operation.It is big suddenly in steam turbine load After amplitude declines, reactor capability can not rapidly change as steam turbine generator load, and the effect of steam-dump system is Steam is controlledly directly discharged to condenser in the process, so as to by providing " artificial " load for reactor Method reduces the temperature transient of nuclear steam supply system reactor Main Coolant and pressure transient amplitude, while limits steam system System maximum working pressure, it is ensured that the safe operation of nuclear power plant.
The parameter designing of steam-dump system will take into full account the dynamic effect of system.The dynamic of steam-dump system operation Process is related to reactor, main coolant system, steam generator characteristic, and corresponding control strategy and control logic, due to The coupling effect of nuclear power unit operation, the design of parameter, which is difficult to calculate by the analysis of static state, to be completed.Usual nuclear power unit The calculation method of parameters of steam-dump system is：Discharge capacity and pressure are empirically first primarily determined that, then establishes nuclear power Device simulation system, the discharge capacity and pressure of initial designs are optimized by system emulation research, it is determined that specific row Parameter tuning value is put, is verified eventually through the debugging test of actual nuclear power unit.And nuclear power unit analogue system is contained Lid reactor physics, main coolant system, secondary coolant circuit system, power system and control system, the construction cycle is long, difficulty is big, Cost is high, the Iterative Design of steam-dump system and related system is influenceed, so as to influence the master-plan progress of nuclear power unit.
Therefore it provides under a kind of computable steam-dump system different designs discharge load easy to use steam pressure and The computational methods of the nuclear power unit steam-dump system parameter of primary side coolant temperature higher limit are very necessary.
The content of the invention
It is an object of the invention to provide under a kind of computable steam-dump system different designs discharge load easy to use The computational methods of the nuclear power unit steam-dump system parameter of steam pressure and primary side coolant temperature higher limit.
To achieve the above object, the present invention adopts the following technical scheme that：A kind of nuclear power unit steam-dump system parameter Computational methods, comprise the following steps：
S1. initialize：
S1.1. n=1 is made, obtains the reactor capability P at 0 moment0, steam generator average heat transfer coefficient K0, circulating ratio N0, the outlet stream amount G of steam generator secondary side heat transfer zoneof 0, go out saliva enthalpy h2f 0, heat transfer zone outlet vapor flow Gg 0、 Outlet vapor enthalpy h2g 0, heat transfer zone inlet flow rate G2i 0, entrance enthalpy h2i 0, the temperature T in steam-water separation area3 0, recycling water enthalpy h3f 0, feedwater flow Gf 0, decline the temperature T in area1 0；
S1.2. the Main Coolant mean temperature at 0 moment is obtained
The Main Coolant mean temperature at 0 moment is calculated by below equation
In formula：
--- reactor initial power P0Corresponding main coolant system mean temperature, DEG C；
△Tc--- reactor coolant controlling dead error, DEG C；
△Tm--- reactor coolant temperature measurement error, DEG C；
△TcProvided by Control System Design, △ TmBy temperature measuring instrument type selecting, producer provides；
S1.3. the vapor (steam) temperature T at 0 moment is obtaineds 0：
The steam pressure p at 0 moment is obtained by below equations 0：
ps 0=ps0+△ps(△Tc+△Tm)
In formula：
ps0--- the steam generator outlet steam pressure under Main Coolant design mean temperature, Pa；
△ps--- reactor capability P0When Main Coolant mean temperature compared with design load raise 1 DEG C when corresponding vapour pressure Power value added, Pa；
Vapor physical parameter table is looked into, is obtained and ps 0The vapor (steam) temperature T of corresponding saturated-steam temperature, i.e. 0 moments 0；
S2. steam flow, decompressor coefficient, last moment discharge valve opening and steam (vapor) outlet are calculated to drain valve front tube Road pressure drop：
According to steam-dump system reducing-and-cooling plant design capacity and pressure, and Main Coolant mean temperature deviation, Steam turbine target load value is calculated by below equation and corresponds to steam flowDecompressor coefficient k1, last moment n- 1 discharge valve opening k2 (n-1)With pressure-drop in pipeline before steam generator steam (vapor) outlet to drain valve
In formula：
--- steam turbine target load value corresponds to steam flow, kg/s；
P∞--- target load value, W；
G100--- 100% load corresponds to steam flow, kg/s；
In formula：
k1--- decompressor coefficient；
Gset--- the total design capacity of drain valve, kg/s；
pset--- drain valve design entry pressure, Pa；
In formula：
k2--- discharge valve opening；
--- the difference of Main Coolant actual average temperature mean temperature corresponding with target load；
△T100--- corresponding Main Coolant mean temperature Deviation Design value when discharge valve opening is 100%；
△T0--- corresponding Main Coolant mean temperature Deviation Design value when discharge valve opening is 0%；
IfLess than △ T0, then k2=0；IfMore than △ T0, then k2=1；
In formula：
--- pressure-drop in pipeline before steam generator steam (vapor) outlet to drain valve, Pa；
△pf--- steam generator steam (vapor) outlet to the journey friction pressure drop of drain valve forward position, Pa；
△pζ--- the form drag pressure drop of local resistance part, Pa before steam generator steam (vapor) outlet to drain valve；
Wherein △ pfWith △ pζComputational methods be if multistage parallel pipeline in pipeline be present, only in parallel pipeline One work branch is calculated；
In formula：
△Pf,i--- the i-th segment pipe friction pressure drop, Pa before drain valve；
△Pζ,j--- j-th of local resistance part form drag pressure drop, Pa before drain valve；
λ --- coefficient of pipe friction, dimensionless；
ζ --- local resistance part form drag coefficient；
L --- duct length, m；
ρ --- fluid density, kg/m3；
π --- pi；
D --- pipe diameter, m, Partial Resistance part, it is interface pipe diameter；
G --- steam flow, kg/s；
I --- pipeline section number；
J --- local resistance piece number；
Wherein：
In formula：
Wherein
The dynamic viscosity of η --- fluid, kg/ (ms)；
S3. the steam flow at current time is calculated：
The steam turbine target load value according to obtained by step S2 corresponds to steam flowDecompressor coefficient k1, and upper one Moment n-1 discharge valve opening k2 (n-1), pressure-drop in pipeline before steam generator steam (vapor) outlet to drain valveBy following Current time n steam flow is calculated in formula：
T --- time, s；
tg--- discharge valve events time delay, s；
--- steam turbine original negative charge values correspond to steam flow, kg/s；
ps--- steam generator outlet pressure, Pa；
t(n)=t(n-1)+△t
Wherein：
△ t --- time step, s；
S4. feedwater flow is calculated：
Feedwater flow is calculated by below equation：
In formula：
--- steam turbine original negative charge values correspond to feedwater flow, kg/s；
--- steam turbine target load value corresponds to feedwater flow, kg/s；
tf--- feedwater flow transformation period, s；
Wherein
--- feed-regulating valve valve position corresponding to steam turbine initial load；
--- feed-regulating valve valve position corresponding to steam turbine target load；
△tvp--- feed-regulating valve total travel time, s；
S5. the circulating ratio at current time is calculated：
According to steam generator static characteristic, current time n circulating ratio is calculated by below equation：
Wherein GgFor steam flow, f1() be static characteristic data are fitted obtained three ranks or Three rank above expression formulas；
S6. the average heat transfer coefficient at current time is calculated：
According to steam generator static characteristic, the steam generator that current time n is calculated by below equation is averaged The coefficient of heat transfer：
Wherein GgFor steam flow, f2() be static characteristic data are fitted obtained three ranks or Three rank above expression formulas；
S7. the reactor capability at current time is calculated：
The reactor capability at current time n is calculated by below equation：
In formula：
T --- time, s；
tP--- control rod acts time delay, s；
P0--- initial load, W；
c1--- design factor；
Wherein：
P100--- 100% load, W；
P90--- 90% load, W；
L100--- 100% load correspondingly controls stick position, mm；
L90--- 90% load correspondingly controls stick position, mm；
V --- control rod regular event speed, mm/s；
S8. the Main Coolant mean temperature at current time is calculated：
According to current time n reactor capability P(n)With steam generator mean heat transfer coefficient K(n), pass through below equation Calculate the Main Coolant mean temperature at current time n：
In formula：
P --- reactor capability, W；
△ t --- time step, s；
M --- the quality of primary Ioops cooling agent or structure, kg；
CP--- the specific heat capacity of primary Ioops cooling agent or structure, J/ (kg DEG C)；
--- Main Coolant mean temperature, DEG C；
Ts--- saturated-steam temperature, DEG C；
K --- steam generator average heat transfer coefficient, W/ (m2·℃)；
A --- steam generator heat transfer area, m2；
S9. the recycling water-carrying capacity and circulating water flow at current time are calculated：
According to current time n steam flow Gg (n), circulating ratio N(n)With feedwater flow Gf (n), calculated by below equation Current time n recycling water-carrying capacity Gof (n)With heat transfer zone inlet flow rate G2i (n)：
Gof (n)=(N(n)-1)Gg (n)
G2i (n)=Gof (n)+Gf (n)
Wherein：
Gof--- recycling water-carrying capacity, kg/s；
Gf--- feedwater flow, kg/s；
G2i--- heat transfer zone inlet flow rate, kg/s；
S10. the vapor (steam) temperature at current time is calculated：
By steam generator secondary side heat transfer zone energy-balance equation, current time n vapor (steam) temperature is calculated：
In formula：
m2--- the quality of secondary side heat transfer zone fluid or structure, kg；
cp2--- the specific heat capacity of secondary side heat transfer zone fluid or structure, J/ (kg DEG C)；
h2i--- heat transfer zone entrance enthalpy, J/kg；
h2g--- heat transfer zone outlet vapor enthalpy, J/kg；
h2f--- heat transfer zone goes out saliva enthalpy, J/kg；
S11. the saturated vapour pressure at current time is obtained：
According to current time n vapor (steam) temperature Ts (n), water and steam physical parameter database is consulted, obtains current time Saturated vapour pressure；
S12. obtain the heat transfer zone at current time n and go out saliva enthalpy：
According to current time n vapor (steam) temperature Ts (n), water and steam physical parameter database is consulted, obtains current time N heat transfer zone goes out saliva enthalpy h2of (n)；
S13. the steam-water separation area temperature at current time is calculated：
By steam-water separation area energy-balance equation, current time n steam-water separation area temperature is calculated：
In formula：
m3--- the quality of secondary side steam-water separation area's fluid or structure, kg；
cp3--- the specific heat capacity of secondary side steam-water separation area's fluid or structure, J/ (kg DEG C)；
T3--- the liquidus temperature in steam-water separation area, DEG C；
h3f--- recycling water enthalpy, J/kg；
S14. the decline area temperature at current time is calculated：
According to current time n steam-water separation area temperature T3 (n)With steam pressure Ps (n), consult vapor physical parameter data Storehouse, obtain the recycling water enthalpy h at current time n3f (n), and by declining area's energy-balance equation, current time n is calculated Decline area temperature：
In formula：
m1--- secondary side declines the quality of area's fluid or structure, kg；
cp1--- secondary side declines the specific heat capacity of area's fluid or structure, J/ (kg DEG C)；
T1--- decline the liquidus temperature in area, DEG C；
hf--- Enthalpy of Feed Water, J/kg；
S15. the heat transfer zone entrance enthalpy at current time is obtained：
According to current time n decline area temperature T1 (n)With steam pressure Ps (n), vapor physical parameter database is consulted, Obtain the heat transfer zone entrance enthalpy h at current time n2i (n)；
S16. the primary side of steam generator coefficient of heat transfer at current time is calculated：
According to steam generator static characteristic, the primary side of steam generator that current time n is calculated by below equation changes Hot coefficient；
Wherein GgFor steam flow, f3() be static characteristic data are fitted obtained three ranks or Three rank above expression formulas；
S17. the coolant temperature leaving reactor vessel at current time is calculated：
According to current time n Main Coolant mean temperature T(n), steam generator mean heat transfer coefficient K(n)Sent out with steam Raw device primary side coefficient of heat transfer α1 (n), current time n reactor export cooling agent is calculated by following equations group simultaneous Temperature
Reactor outlet temperature calculation expression：
In formula：
α1--- the primary side of steam generator coefficient of heat transfer, W/ (m2·℃)；
--- main coolant system hot arc temperature, DEG C；
--- cold section of temperature of main coolant system, DEG C；
S18. step S2 is repeated to step S17：
N=n+1 is made, repeats step S2 to step S17, until reactor capability is down to PnWhen untill；
Wherein：
S19. the higher limit of coolant temperature leaving reactor vessel and steam pressure is determined：
According to above result of calculation, the curve that coolant temperature and steam pressure change over time, above-mentioned two songs are obtained The maximum of line is respectively the higher limit of coolant temperature and steam pressure, if steam pressure and reactor coolant temperature Higher limit is no more than design permissible value, then design parameter meets to require, otherwise adjusts discharge capacity and recalculates.
The present invention is based on primary side of steam generator energy-balance equation, and the thermometer for obtaining primary side Main Coolant reaches Formula, by the vapor (steam) temperature and Main Coolant mean temperature of last moment, and reactor capability this moment, try to achieve master this moment Coolant average temperature；Secondary side in steam generator is divided into three regions, respectively declines area, heat transfer zone and Disengagement zone, The side heat transfer equation of steam generator first and second and each subregion energy conservation equation group of secondary side are established, Main Coolant is averaged temperature Degree is brought into, and carries out time discretization to these equations, is obtained on secondary side heat transfer zone, is declined the three of area and steam-water separation area Individual system of linear equations, simultaneous solution obtain steam-water separation area, decline the temperature of steam and water in area and heat transfer zone.
Using the present invention, steam pressure and primary side cooling agent under steam-dump system different designs discharge load can be calculated Temperature maximum, it can meet that main coolant system and secondary coolant circuit system are transported safely as differentiation steam-dump system parameter designing Capable foundation, it can also instruct the design of steam-dump system.The present invention is easy to use.
Embodiment
Below in conjunction with specific embodiment, the present invention is described in further detail, but the embodiment should not be construed as to this The limitation of invention.
The present embodiment be based on it is assumed hereinafter that：The temperature negative-feedback effect of moderator and fuel is not considered；Power and control rod Rod position is linearly related；Consider nuclear delay effect；Main Coolant is consistent with adjacent structure temperature；Using lumped parameter Calculate Main Coolant mean temperature；It it is one by the lump of steam generator quantity, lumped parameter includes flow, fluid and structure matter Amount, heat exchange area；Feedwater flow changes linearly over time after varying duty, keeps constant after reaching flow corresponding to target load； Consider that discharge valve signal postpones with action after varying duty, steam flow changes linearly over time in time delay；Steam generator Circulating ratio and average heat transfer coefficient are expressed as the function of steam flow under steam generator static characteristic.
In the present embodiment, the design permissible value 4.8MPa of steam pressure, the design permissible value of reactor coolant temperature 279.0 DEG C, the heat transfer area 856.82m of steam generator2, the total thermal capacitance 1.23 × 10 of primary side5KJ/ DEG C (including cooling agent, pressure Force container, main pipeline and primary side of steam generator structure), the total thermal capacitance 1.01 × 10 in steam generator secondary side heat transfer zone4kJ/ DEG C (including water, steam and steam generator secondary side structure), the total thermal capacitance 2.67 × 10 in Disengagement zone4KJ/ DEG C, decline area's total heat Hold 5.3 × 103KJ/ DEG C, control rod action time delay 1s, discharge valve events time delay 0.5s, feedwater flow transformation period 5s, 142 DEG C, feed pressure 4.2MPa of feed temperature, subsidiary engine steam flow 8.28kg/s.Time step chooses 0.1s, initial full Load operation reactor capability 100MW, steam generator outlet steam pressure 3.9MPa during oepration at full load, steam flow and give Water-carrying capacity is 45kg/s, 271 DEG C of coolant average temperature；Target load value is 55%.
Carried out according to step provided by the invention, the result of calculation that major parameter is inscribed when each is as shown in table 1：
Table 1
The present embodiment have chosen less time step, have higher computational accuracy, data volume is larger, from upper table When design discharge load is 55%, steam pressure higher limit can reach 4.62MPa, and reactor coolant temperature maximum is 274.54 DEG C, not less than design permissible value.
As it will be easily appreciated by one skilled in the art that the foregoing is merely illustrative of the preferred embodiments of the present invention, not to The limitation present invention, all any modification, equivalent and improvement made within the spirit and principles of the invention etc., all should be included Within protection scope of the present invention.
The content not being described in detail in this specification, belong to prior art known to those skilled in the art.

Claims (1)

1. a kind of computational methods of nuclear power unit steam-dump system parameter, comprise the following steps：
S1. initialize：
S1.1. n=1 is made, obtains the reactor capability P at 0 moment0, steam generator average heat transfer coefficient K0, circulating ratio N0, steam The outlet stream amount G of vapour generator secondary side heat transfer zoneof 0, go out saliva enthalpy h2f 0, heat transfer zone outlet vapor flow Gg 0, outlet steam Vapour enthalpy h2g 0, heat transfer zone inlet flow rate G2i 0, entrance enthalpy h2i 0, the temperature T in steam-water separation area3 0, recycling water enthalpy h3f 0, give Water-carrying capacity Gf 0, decline the temperature T in area1 0；
S1.2. the Main Coolant mean temperature at 0 moment is obtained
The Main Coolant mean temperature at 0 moment is calculated by below equation
<mrow> <msup> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mn>0</mn> </msup> <mo>=</mo> <msub> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mn>0</mn> </msub> <mo>+</mo> <msub> <mi>&amp;Delta;T</mi> <mi>c</mi> </msub> <mo>+</mo> <msub> <mi>&amp;Delta;T</mi> <mi>m</mi> </msub> </mrow>
In formula：
--- reactor initial power P0Corresponding main coolant system mean temperature, DEG C；
ΔTc--- reactor coolant controlling dead error, DEG C；
ΔTm--- reactor coolant temperature measurement error, DEG C；
ΔTcProvided by Control System Design, Δ TmBy temperature measuring instrument type selecting, producer provides；
S1.3. the vapor (steam) temperature T at 0 moment is obtaineds 0：
The steam pressure p at 0 moment is obtained by below equations 0：
ps 0=ps0+Δps(ΔTc+ΔTm)
In formula：
ps0--- the steam generator outlet steam pressure under Main Coolant design mean temperature, Pa；
Δps--- reactor capability P0When Main Coolant mean temperature when raising 1 DEG C compared with design load corresponding steam pressure increase It is value added, Pa；
Vapor physical parameter table is looked into, is obtained and ps 0The vapor (steam) temperature T of corresponding saturated-steam temperature, i.e. 0 moments 0；
S2. pipeline pressure before calculating steam flow, decompressor coefficient, last moment discharge valve opening and steam (vapor) outlet to drain valve Drop：
According to steam-dump system reducing-and-cooling plant design capacity and pressure, and Main Coolant mean temperature deviation, pass through Below equation is calculated steam turbine target load value and corresponds to steam flowDecompressor coefficient k1, last moment n-1 Discharge valve opening k2 (n-1)With pressure-drop in pipeline before steam generator steam (vapor) outlet to drain valve
<mrow> <msubsup> <mi>G</mi> <mi>g</mi> <mi>&amp;infin;</mi> </msubsup> <mo>=</mo> <mfrac> <msup> <mi>P</mi> <mi>&amp;infin;</mi> </msup> <msub> <mi>P</mi> <mn>100</mn> </msub> </mfrac> <msub> <mi>G</mi> <mn>100</mn> </msub> </mrow>
In formula：
--- steam turbine target load value corresponds to steam flow, kg/s；
P∞--- target load value, W；
G100--- 100% load corresponds to steam flow, kg/s；
<mrow> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>=</mo> <mfrac> <msub> <mi>G</mi> <mrow> <mi>s</mi> <mi>e</mi> <mi>t</mi> </mrow> </msub> <msub> <mi>p</mi> <mrow> <mi>s</mi> <mi>e</mi> <mi>t</mi> </mrow> </msub> </mfrac> </mrow>
In formula：
k1--- decompressor coefficient；
Gset--- the total design capacity of drain valve, kg/s；
pset--- drain valve design entry pressure, Pa；
<mrow> <msubsup> <mi>k</mi> <mn>2</mn> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> <mo>=</mo> <mfrac> <mrow> <mi>&amp;Delta;</mi> <msup> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msup> <mo>-</mo> <msub> <mi>&amp;Delta;T</mi> <mn>0</mn> </msub> </mrow> <mrow> <msub> <mi>&amp;Delta;T</mi> <mn>100</mn> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;T</mi> <mn>0</mn> </msub> </mrow> </mfrac> </mrow>
In formula：
k2--- discharge valve opening；
--- the difference of Main Coolant actual average temperature mean temperature corresponding with target load；
ΔT100--- corresponding Main Coolant mean temperature Deviation Design value when discharge valve opening is 100%；
ΔT0--- corresponding Main Coolant mean temperature Deviation Design value when discharge valve opening is 0%；
IfLess than Δ T0, then k2=0；IfMore than Δ T0, then k2=1；
<mrow> <mi>&amp;Delta;</mi> <msup> <mover> <mi>p</mi> <mo>~</mo> </mover> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msup> <mo>=</mo> <msubsup> <mi>&amp;Delta;p</mi> <mi>f</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> <mo>+</mo> <msubsup> <mi>&amp;Delta;p</mi> <mi>&amp;zeta;</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> </mrow>
In formula：
--- pressure-drop in pipeline before steam generator steam (vapor) outlet to drain valve, Pa；
Δpf--- steam generator steam (vapor) outlet to the journey friction pressure drop of drain valve forward position, Pa；
Δpζ--- the form drag pressure drop of local resistance part, Pa before steam generator steam (vapor) outlet to drain valve；
Wherein Δ pfWith Δ pζComputational methods be if multistage parallel pipeline in pipeline be present, only to one in parallel pipeline Work branch is calculated；
<mrow> <msubsup> <mi>&amp;Delta;p</mi> <mrow> <mi>f</mi> <mo>,</mo> <mi>i</mi> </mrow> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msubsup> <mo>=</mo> <mfrac> <mrow> <mn>8</mn> <msubsup> <mi>&amp;lambda;</mi> <mi>i</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> <msub> <mi>L</mi> <mi>i</mi> </msub> <msup> <mrow> <mo>(</mo> <msubsup> <mi>G</mi> <mi>i</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> <mrow> <msubsup> <mi>&amp;rho;</mi> <mi>i</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> <msup> <mi>&amp;pi;</mi> <mn>2</mn> </msup> <msubsup> <mi>d</mi> <mi>i</mi> <mn>5</mn> </msubsup> </mrow> </mfrac> </mrow>
<mrow> <msubsup> <mi>&amp;Delta;p</mi> <mrow> <mi>&amp;zeta;</mi> <mo>,</mo> <mi>j</mi> </mrow> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> <mo>=</mo> <mfrac> <mrow> <mn>8</mn> <msub> <mi>&amp;zeta;</mi> <mi>j</mi> </msub> <msup> <mrow> <mo>(</mo> <msubsup> <mi>G</mi> <mi>j</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> <mrow> <msubsup> <mi>&amp;rho;</mi> <mi>j</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> <msup> <mi>&amp;pi;</mi> <mn>2</mn> </msup> <msubsup> <mi>d</mi> <mi>j</mi> <mn>4</mn> </msubsup> </mrow> </mfrac> </mrow>
In formula：
ΔPf,i--- the i-th segment pipe friction pressure drop, Pa before drain valve；
ΔPζ,j--- j-th of local resistance part form drag pressure drop, Pa before drain valve；
λ --- coefficient of pipe friction, dimensionless；
ζ --- local resistance part form drag coefficient；
L --- duct length, m；
ρ --- fluid density, kg/m3；
π --- pi；
D --- pipe diameter, m, Partial Resistance part, it is interface pipe diameter；
G --- steam flow, kg/s；
I --- pipeline section number；
J --- local resistance piece number；
Wherein：
<mrow> <msup> <mi>&amp;lambda;</mi> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mfrac> <mn>64</mn> <msup> <mi>Re</mi> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> </mfrac> </mtd> <mtd> <mrow> <mn>0</mn> <mo>&lt;</mo> <msup> <mi>Re</mi> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>&lt;</mo> <mn>2320</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mfrac> <mn>0.3164</mn> <msup> <mrow> <mo>(</mo> <msup> <mi>Re</mi> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> <mn>0.25</mn> </msup> </mfrac> </mtd> <mtd> <mrow> <mn>2320</mn> <mo>&amp;le;</mo> <msup> <mi>Re</mi> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>&amp;le;</mo> <msup> <mn>10</mn> <mn>5</mn> </msup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>0.0032</mn> <mo>+</mo> <mfrac> <mn>0.221</mn> <msup> <mrow> <mo>(</mo> <msup> <mi>Re</mi> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> <mn>0.237</mn> </msup> </mfrac> </mrow> </mtd> <mtd> <mrow> <msup> <mn>10</mn> <mn>5</mn> </msup> <mo>&amp;le;</mo> <msup> <mi>Re</mi> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>&amp;le;</mo> <mn>3</mn> <mo>&amp;times;</mo> <msup> <mn>10</mn> <mn>6</mn> </msup> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
In formula：
<mrow> <msup> <mi>Re</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msup> <mo>=</mo> <mfrac> <mn>4</mn> <mrow> <mi>&amp;pi;</mi> <mi>d</mi> </mrow> </mfrac> <mfrac> <msup> <mi>G</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msup> <msup> <mi>&amp;eta;</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msup> </mfrac> </mrow>
Wherein
The dynamic viscosity of η --- fluid, kg/ (ms)；
S3. the steam flow at current time is calculated：
The steam turbine target load value according to obtained by step S2 corresponds to steam flowDecompressor coefficient k1, and last moment N-1 discharge valve opening k2 (n-1), pressure-drop in pipeline before steam generator steam (vapor) outlet to drain valvePass through below equation meter Calculation obtains current time n steam flow：
<mrow> <msubsup> <mi>G</mi> <mi>g</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msubsup> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msubsup> <mi>G</mi> <mi>g</mi> <mn>0</mn> </msubsup> <mo>-</mo> <mfrac> <mrow> <msubsup> <mi>G</mi> <mi>g</mi> <mn>0</mn> </msubsup> <mo>-</mo> <msubsup> <mi>G</mi> <mi>g</mi> <mi>&amp;infin;</mi> </msubsup> </mrow> <msub> <mi>t</mi> <mi>g</mi> </msub> </mfrac> <msup> <mi>t</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> </mrow> </mtd> <mtd> <mrow> <msup> <mi>t</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>&amp;le;</mo> <msub> <mi>t</mi> <mi>g</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>&amp;CenterDot;</mo> <msubsup> <mi>k</mi> <mn>2</mn> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> <mrow> <mo>(</mo> <msubsup> <mi>p</mi> <mi>s</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msubsup> <mo>-</mo> <mi>&amp;Delta;</mi> <msup> <mover> <mi>p</mi> <mo>~</mo> </mover> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>G</mi> <mi>g</mi> <mi>&amp;infin;</mi> </msubsup> </mrow> </mtd> <mtd> <mrow> <msup> <mi>t</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>&gt;</mo> <msub> <mi>t</mi> <mi>g</mi> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
In formula：
T --- time, s；
tg--- discharge valve events time delay, s；
--- steam turbine original negative charge values correspond to steam flow, kg/s；
ps--- steam generator outlet pressure, Pa；
t(n)=t(n-1)+Δt
Wherein：
Δ t --- time step, s；
S4. feedwater flow is calculated：
Current time n feedwater flow is calculated by below equation：
<mrow> <msubsup> <mi>G</mi> <mi>f</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msubsup> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msubsup> <mi>G</mi> <mi>f</mi> <mn>0</mn> </msubsup> <mo>-</mo> <mfrac> <mrow> <msubsup> <mi>G</mi> <mi>f</mi> <mn>0</mn> </msubsup> <mo>-</mo> <msubsup> <mi>G</mi> <mi>f</mi> <mi>&amp;infin;</mi> </msubsup> </mrow> <msub> <mi>t</mi> <mi>f</mi> </msub> </mfrac> <msup> <mi>t</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> </mrow> </mtd> <mtd> <mrow> <msup> <mi>t</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>&amp;le;</mo> <msub> <mi>t</mi> <mi>f</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>G</mi> <mi>f</mi> <mi>&amp;infin;</mi> </msubsup> </mtd> <mtd> <mrow> <msup> <mi>t</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>&gt;</mo> <msub> <mi>t</mi> <mi>f</mi> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
In formula：
--- steam turbine original negative charge values correspond to feedwater flow, kg/s；
--- steam turbine target load value corresponds to feedwater flow, kg/s；
tf--- feedwater flow transformation period, s；
<mrow> <msub> <mi>t</mi> <mi>f</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msubsup> <mi>V</mi> <mi>p</mi> <mn>0</mn> </msubsup> <mo>-</mo> <msubsup> <mi>V</mi> <mi>p</mi> <mi>&amp;infin;</mi> </msubsup> <mo>)</mo> </mrow> <msub> <mi>&amp;Delta;t</mi> <mrow> <mi>v</mi> <mi>p</mi> </mrow> </msub> </mrow>
Wherein
--- feed-regulating valve valve position corresponding to steam turbine initial load；
--- feed-regulating valve valve position corresponding to steam turbine target load；
Δtvp--- feed-regulating valve total travel time, s；
S5. the circulating ratio at current time is calculated：
According to steam generator static characteristic, current time n circulating ratio is calculated by below equation：
Wherein GgFor steam flow, f1() is that static characteristic data are fitted with obtained three ranks or three ranks Above expression formula；
S6. the average heat transfer coefficient at current time is calculated：
According to steam generator static characteristic, current time n steam generator Average heat transfer is calculated by below equation Coefficient：
Wherein GgFor steam flow, f2() is that static characteristic data are fitted with obtained three ranks or three ranks Above expression formula；
S7. the reactor capability at current time is calculated：
The reactor capability at current time n is calculated by below equation：
<mrow> <msup> <mi>P</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <msup> <mi>P</mi> <mn>0</mn> </msup> </mtd> <mtd> <mrow> <msup> <mi>t</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>&amp;le;</mo> <msub> <mi>t</mi> <mi>P</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msup> <mi>P</mi> <mn>0</mn> </msup> <mo>-</mo> <msub> <mi>c</mi> <mn>1</mn> </msub> <msup> <mi>t</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> </mrow> </mtd> <mtd> <mrow> <msup> <mi>t</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>&gt;</mo> <msub> <mi>t</mi> <mi>P</mi> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
In formula：
T --- time, s；
tP--- control rod acts time delay, s；
P0--- initial load, W；
c1--- design factor；
<mrow> <msub> <mi>c</mi> <mn>1</mn> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>P</mi> <mn>100</mn> </msub> <mo>-</mo> <msub> <mi>P</mi> <mn>90</mn> </msub> </mrow> <mrow> <msub> <mi>L</mi> <mn>100</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mn>90</mn> </msub> </mrow> </mfrac> <mi>v</mi> </mrow>
Wherein：
P100--- 100% load, W；
P90--- 90% load, W；
L100--- 100% load correspondingly controls stick position, mm；
L90--- 90% load correspondingly controls stick position, mm；
V --- control rod regular event speed, mm/s；
S8. the Main Coolant mean temperature at current time is calculated：
According to current time n reactor capability P(n)With steam generator mean heat transfer coefficient K(n), calculated by below equation Current time n Main Coolant mean temperature：
<mrow> <msup> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <msup> <mi>P</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>&amp;CenterDot;</mo> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>+</mo> <msub> <mi>&amp;Sigma;MC</mi> <mi>P</mi> </msub> <mo>&amp;CenterDot;</mo> <msup> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>+</mo> <msup> <mi>K</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mi>A</mi> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>&amp;CenterDot;</mo> <msup> <msub> <mi>T</mi> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> </mrow> <mrow> <msub> <mi>&amp;Sigma;MC</mi> <mi>P</mi> </msub> <mo>+</mo> <msup> <mi>K</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mi>A</mi> <mi>&amp;Delta;</mi> <mi>t</mi> </mrow> </mfrac> </mrow>
In formula：
P --- reactor capability, W；
Δ t --- time step, s；
M --- the quality of primary Ioops cooling agent or structure, kg；
CP--- the specific heat capacity of primary Ioops cooling agent or structure, J/ (kg DEG C)；
--- Main Coolant mean temperature, DEG C；
Ts--- saturated-steam temperature, DEG C；
K --- steam generator average heat transfer coefficient, W/ (m2·℃)；
A --- steam generator heat transfer area, m2；
S9. the recycling water-carrying capacity and circulating water flow at current time are calculated：
According to current time n steam flow Gg (n), circulating ratio N(n)With feedwater flow Gf (n), calculated by below equation current Moment n recycling water-carrying capacity Gof (n)With heat transfer zone inlet flow rate G2i (n)：
Gof (n)=(N(n)-1)Gg (n)
G2i (n)=Gof (n)+Gf (n)
Wherein：
Gof--- recycling water-carrying capacity, kg/s；
Gf--- feedwater flow, kg/s；
G2i--- heat transfer zone inlet flow rate, kg/s；
S10. the vapor (steam) temperature at current time is calculated：
By steam generator secondary side heat transfer zone energy-balance equation, current time n vapor (steam) temperature is calculated：
<mrow> <msup> <msub> <mi>T</mi> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <msup> <mi>K</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mi>A</mi> <mo>&amp;CenterDot;</mo> <mi>&amp;Delta;</mi> <mi>t</mi> <mrow> <mo>(</mo> <msup> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>-</mo> <msup> <msub> <mi>T</mi> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> <mo>-</mo> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>&amp;CenterDot;</mo> <mrow> <mo>(</mo> <msup> <msub> <mi>G</mi> <mrow> <mi>o</mi> <mi>f</mi> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <msup> <msub> <mi>h</mi> <mrow> <mn>2</mn> <mi>f</mi> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>+</mo> <msup> <msub> <mi>G</mi> <mi>g</mi> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <msup> <msub> <mi>h</mi> <mrow> <mn>2</mn> <mi>g</mi> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>-</mo> <msup> <msub> <mi>G</mi> <mrow> <mn>2</mn> <mi>i</mi> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <msup> <msub> <mi>h</mi> <mrow> <mn>2</mn> <mi>i</mi> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>&amp;Sigma;m</mi> <mn>2</mn> </msub> <msub> <mi>c</mi> <mrow> <mi>p</mi> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> <mo>+</mo> <msup> <msub> <mi>T</mi> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> </mrow>
In formula：
m2--- the quality of secondary side heat transfer zone fluid or structure, kg；
cp2--- the specific heat capacity of secondary side heat transfer zone fluid or structure, J/ (kg DEG C)；
h2i--- heat transfer zone entrance enthalpy, J/kg；
h2g--- heat transfer zone outlet vapor enthalpy, J/kg；
h2f--- heat transfer zone goes out saliva enthalpy, J/kg；
S11. the saturated vapour pressure at current time is obtained：
According to current time n vapor (steam) temperature Ts (n), water and steam physical parameter database is consulted, obtains the full of current time And steam pressure；
S12. obtain the heat transfer zone at current time n and go out saliva enthalpy：
According to current time n vapor (steam) temperature Ts (n), water and steam physical parameter database is consulted, obtains the biography at current time n Hot-zone goes out saliva enthalpy h2f (n)；
S13. the steam-water separation area temperature at current time is calculated：
By steam-water separation area energy-balance equation, current time n steam-water separation area temperature is calculated：
<mrow> <msup> <msub> <mi>T</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>&amp;CenterDot;</mo> <msup> <msub> <mi>G</mi> <mrow> <mi>o</mi> <mi>f</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mrow> <mo>(</mo> <msup> <msub> <mi>h</mi> <mrow> <mn>2</mn> <mi>f</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>-</mo> <msup> <msub> <mi>h</mi> <mrow> <mn>3</mn> <mi>f</mi> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>&amp;Sigma;m</mi> <mn>3</mn> </msub> <msub> <mi>c</mi> <mrow> <mi>p</mi> <mn>3</mn> </mrow> </msub> </mrow> </mfrac> <mo>+</mo> <msup> <msub> <mi>T</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> </mrow>
In formula：
m3--- the quality of secondary side steam-water separation area's fluid or structure, kg；
cp3--- the specific heat capacity of secondary side steam-water separation area's fluid or structure, J/ (kg DEG C)；
T3--- the liquidus temperature in steam-water separation area, DEG C；
h3f--- recycling water enthalpy, J/kg；
S14. the decline area temperature at current time is calculated：
According to current time n steam-water separation area temperature T3 (n)With steam pressure Ps (n), vapor physical parameter database is consulted, Obtain the recycling water enthalpy h at current time n3f (n), and by declining area's energy-balance equation, it is calculated current time n's Decline area's temperature：
<mrow> <msup> <msub> <mi>T</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>&amp;CenterDot;</mo> <mrow> <mo>(</mo> <msup> <msub> <mi>G</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <msub> <mi>h</mi> <mi>f</mi> </msub> <mo>+</mo> <msup> <msub> <mi>G</mi> <mrow> <mi>o</mi> <mi>f</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <msup> <msub> <mi>h</mi> <mrow> <mn>3</mn> <mi>f</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>-</mo> <msup> <msub> <mi>G</mi> <mrow> <mn>2</mn> <mi>i</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <msup> <msub> <mi>h</mi> <mrow> <mn>2</mn> <mi>i</mi> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>&amp;Sigma;m</mi> <mn>1</mn> </msub> <msub> <mi>c</mi> <mrow> <mi>p</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>+</mo> <msup> <msub> <mi>T</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> </mrow>
In formula：
m1--- secondary side declines the quality of area's fluid or structure, kg；
cp1--- secondary side declines the specific heat capacity of area's fluid or structure, J/ (kg DEG C)；
T1--- decline the liquidus temperature in area, DEG C；
hf--- Enthalpy of Feed Water, J/kg；
S15. the heat transfer zone entrance enthalpy at current time is obtained：
According to current time n decline area temperature T1 (n)With steam pressure Ps (n), vapor physical parameter database is consulted, is obtained Current time n heat transfer zone entrance enthalpy h2i (n)；
S16. the primary side of steam generator coefficient of heat transfer at current time is calculated：
According to steam generator static characteristic, the primary side of steam generator heat exchange that current time n is calculated by below equation is Number；
Wherein GgFor steam flow, f3() is that static characteristic data are fitted with obtained three ranks or three ranks Above expression formula；
S17. the coolant temperature leaving reactor vessel at current time is calculated：
According to current time n Main Coolant mean temperatureSteam generator mean heat transfer coefficient K(n)And steam generator Primary side coefficient of heat transfer α1 (n), current time n coolant temperature leaving reactor vessel is calculated by following equations group simultaneous
<mrow> <msup> <mi>K</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mrow> <mo>(</mo> <msup> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>-</mo> <msup> <msub> <mi>T</mi> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> <mo>=</mo> <msup> <msub> <mi>&amp;alpha;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mrow> <mo>(</mo> <msup> <msub> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>-</mo> <msup> <msub> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mi>o</mi> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> </mrow>
<mrow> <msup> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <msup> <msub> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mi>o</mi> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> </mrow> <mn>2</mn> </mfrac> </mrow>
Reactor outlet temperature calculation expression：
<mrow> <msup> <msub> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mi>o</mi> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>=</mo> <msup> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>-</mo> <mfrac> <msup> <mi>K</mi> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mrow> <mn>2</mn> <msup> <msub> <mi>&amp;alpha;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> </mrow> </mfrac> <mrow> <mo>(</mo> <msup> <mover> <mi>T</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mo>(</mo> <mi>n</mi> <mo>)</mo> </mrow> </msup> <mo>-</mo> <msup> <msub> <mi>T</mi> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> </mrow>
In formula：
α1--- the primary side of steam generator coefficient of heat transfer, W/ (m2·℃)；
--- main coolant system hot arc temperature, DEG C；
--- cold section of temperature of main coolant system, DEG C；
S18. step S2 is repeated to step S17：
N=n+1 is made, repeats step S2 to step S17, until reactor capability is down to PnWhen untill；
Wherein：
<mrow> <msub> <mi>P</mi> <mi>n</mi> </msub> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msup> <mi>P</mi> <mi>&amp;infin;</mi> </msup> <mo>-</mo> <msub> <mi>P</mi> <mn>100</mn> </msub> <mo>&amp;CenterDot;</mo> <mn>10</mn> <mi>%</mi> </mrow> </mtd> <mtd> <mrow> <mfrac> <msup> <mi>P</mi> <mi>&amp;infin;</mi> </msup> <msub> <mi>P</mi> <mn>100</mn> </msub> </mfrac> <mo>&amp;GreaterEqual;</mo> <mn>15</mn> <mi>%</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>P</mi> <mn>100</mn> </msub> <mo>&amp;CenterDot;</mo> <mn>5</mn> <mi>%</mi> </mrow> </mtd> <mtd> <mrow> <mfrac> <msup> <mi>P</mi> <mi>&amp;infin;</mi> </msup> <msub> <mi>P</mi> <mn>100</mn> </msub> </mfrac> <mo>&lt;</mo> <mn>15</mn> <mi>%</mi> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
S19. the higher limit of coolant temperature leaving reactor vessel and steam pressure is determined：
According to above result of calculation, the curve that coolant temperature and steam pressure change over time is obtained, above-mentioned two curves Maximum is respectively the higher limit of coolant temperature and steam pressure, if the upper limit of steam pressure and reactor coolant temperature Value is no more than design permissible value, then design parameter meets to require, otherwise adjusts discharge capacity and recalculates.
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CN109783874A (en) * 2018-12-18 2019-05-21 中国人民解放军海军工程大学 A kind of mixing adjusting core Steam Turbine coupling variable condition calculation model peculiar to vessel
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CN112417681A (en) * 2020-11-19 2021-02-26 上海交通大学 Estimation method for primary and secondary side convective heat transfer coefficient distribution of steam generator
CN112417782A (en) * 2020-11-19 2021-02-26 上海交通大学 Method for estimating circulation multiplying power of working medium in two loops of steam generator
CN112417681B (en) * 2020-11-19 2022-03-22 上海交通大学 Estimation method for primary and secondary side convective heat transfer coefficient distribution of steam generator
CN112380713B (en) * 2020-11-19 2022-03-22 上海交通大学 Method for estimating temperature distribution of metal wall of inverted U-shaped tube of steam generator
CN112417782B (en) * 2020-11-19 2022-04-26 上海交通大学 Method for estimating circulation multiplying power of working medium in two loops of steam generator
CN112699523B (en) * 2020-11-19 2022-04-26 上海交通大学 Method and device for estimating vaporization starting height of working medium in secondary loop of steam generator
CN112417780B (en) * 2020-11-19 2023-03-24 上海交通大学 Method and system for estimating mass flow of secondary loop recycled water of steam generator
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