CN1118831C - Method for measuring power of pressurized-water reactor core for nuclear power station - Google Patents

Abstract

The present invention discloses a method for measuring the power of a reactor core of a pressurized water reactor in a nuclear power station. The measuring values of the temperature, the pressure, the flow capacity, etc. of a working medium of a secondary loop in a steam generator in the pressurized water reactor are measured to calculate enthalpy rise generated by the working medium of the secondary loop in the process of passing through the steam generator, and thus, energy transferred to the secondary loop from a primary loop of the reactor is obtained; energy obtained and lost by the first loop through other kinds of equipment is considered to calculate the power of the reactor core of the reactor by an energy balance principle. Thus, the present invention can greatly improve the speed and the precision of power measurement and improve the economic benefit of the operation of the machine set of the nuclear power station.

Description

A kind of method for measuring power of pressurized-water reactor core for nuclear power station

(1) technical field

The invention belongs to the measuring power of pressurized-water reactor core for nuclear power station technology.

Nuclear power station run duration, pressurized-water reactor core power are crucial parameters.Have only core power is measured accurately and rapidly, could guarantee reactor safety, economical operation.If the core power measured value is lower than actual value, reactor will move above under the condition of its operational factor, and reactor burns out easily, thereby initiation reaction heap accident jeopardizes equipment and personal safety.On the contrary, if the core power measured value is higher than actual value, reactor does not reach its specified service condition, and outside liberated heat of reactor and the generated energy that causes thus all are lower than design load, and the economy of unit will be affected.

(2) background technology

For pressurized-water reactor core power, useful in the world heat balance principle is measured.The heat balance method of that two the pressurized-water reactor core power in China GNPS adopt Electricite De France to provide is measured.There are two circulating water loops in presurized water reactor, one loop works medium absorption reaction is stacked heat, and by heat exchanger heat is passed to the water of secondary circuit, and the water of secondary circuit becomes the very high wet saturated steam of mass dryness fraction after absorbing heat, the pushing turbine work done finally becomes heat energy the electric energy of generator.Heat balance method of is exactly by measuring the heat that the secondary circuit actuating medium obtains, consider the internal heat loss simultaneously, calculating the reactor liberated heat, thereby obtain reactor core power.By actual measurement temperature, pressure and flow, can obtain reactor core power more exactly by heat balance method of, but in the heat Balance Calculation method that Electricite De France provides, what the water and steam thermodynamic properties calculated employing is simple binomial method, and its computational accuracy remains further to be improved.

(3) summary of the invention

The purpose of this invention is to provide a kind ofly, can improve the measuring method of pressurized-water reactor core power measurement precision utilizing on the basis of heat balance method of.

Technical scheme of the present invention is: a kind of method for measuring power of pressurized-water reactor core for nuclear power station, it is characterized in that: (1) will handle PC and be connected with the data acquisition PC, the data acquisition PC is connected with the isolation acquisition module, and the terminal cabinet of isolating acquisition module and power station unit is connected; (2) gather correlation parameter by isolating acquisition module, and the parameter that collects is input to acquisition PC machine, utilize the data of acquisition PC machine, the program by in the operation processing PC just can obtain reactor core power, and its concrete steps are:

1) the constant data that need in the time of will measuring are input to be handled in the PC, and these data comprise: 1. jet chimney diameter D, steam generator export to difference in height Δ z, coefficient of friction resistance λ between length L, pressure unit and the steam pipe axis of pressure unit interface under the reference temperature; 2. main feed system pipe diameter D under the reference temperature ₁, from the main feed system to the steam generator inlet length L ₁, pressure unit and feed pipe between centers difference in height Δ z ₁, coefficient of friction resistance λ 1,3. the feedwater orifice plate diameter d under the reference temperature _e, give water conduit circular section diameter D under the reference temperature _e, the 4. mudhole board diameter d under the reference temperature _p, reference temperature down blow conduit circular section diameter D _p

2) gather momentum parameter by isolating acquisition module, these data comprise: 1. steam generator top hole pressure P _VVP, 2. steam generator inlet pressure P of steam mean flow rate V, steam generator outlet steam quality x _Are, the differential pressure that 3. feeds water Δ p _Ei, 4. blowdown differential pressure Δ p _Pi

3) handling PC utilizes momentum parameter and constant parameter to calculate, the feedwater flow, blowdown flow, steam flow, feedwater enthalpy, blowdown enthalpy, outlet enthalpy of wet steam and other thermals source that calculate the steam generator secondary circuit are passed to the heat of reactor cooling system, and the concrete operation way is: 1. calculate each secondary circuit enthalpy of wet steam: according to formula $P=\mathrm{Pvvp}+{\mathrm{\ρ}}^{\′}\mathrm{g\Δz}+\frac{1}{2}{\left(\frac{4}{{\mathrm{\πD}}^2}\right)}^2\frac{Q^2}{\mathrm{\ρ}}+\mathrm{\λ}\frac{8L{Q}^2}{{\mathrm{\ρ\π}}^2{D}^5}$ , calculate the steam generator top hole pressure, according to this pressure value, calculate the enthalpy of saturation water and saturated dry steam under this pressure by the ASME formula, obtain the enthalpy of wet steam again according to the mass dryness fraction of wet steam, concrete computing formula is as follows: $\mathrm{\β}=\frac{P}{P_c}$ Be relative pressure, $\mathrm{\θ}=\frac{T}{T_C}$ Be relative temperature, $X=\frac{V}{V_C}$ Be relative specific volume, $\mathrm{\ϵ}=\frac{h}{p_C{V}_C}$ Be relative enthalpy, wherein P _c, T _c, V _cBe the value of critical point, when the pressure of known saturation water or saturated vapour, can iterate by following formula and try to achieve saturation temperature: $\mathrm{\β}\left(\mathrm{\θ}\right)=\exp (\frac{1}{\mathrm{\θ}}\frac{\underset{i=1}{\overset{5}{\mathrm{\Σ}}}{k}_i{(1-\mathrm{\θ})}^i}{1+k_6(1-\mathrm{\θ})+k_7{(1-\mathrm{\θ})}^2}-\frac{(1-\mathrm{\θ})}{k_8{(1-\mathrm{\θ})}^2+k_9})$ The computing formula of unsaturated water and saturation water is: x ₁=A ₁₁a ₅Z ^-5/17+ { A ₁₂+ A ₁₃θ+A ₁₄θ ²+ A ₁₅(a ₆-θ) ¹⁰+ A ₁₆(a ₇+ θ ¹⁹) ^-1}

-(a ₈+θ ¹¹) ^-1(A ₁₇+2A ₁₈β+3A ₁₉β ²)

-A ₂₀θ ¹⁸(a ₉+θ ²){-3(a ₁₀+β) ^-4+a ₁₁}+3A ₂₁(a ₁₂-θ)β ²

+ 4A ₂₂θ ^-20β ³Wherein:

Z＝Y+(a ₃Y ²-2a ₄θ+2a ₅β) ^1/2

Y＝1-a ₁θ ²-a ₂θ ^-6 ${\mathrm{\ϵ}}_1=a_0+A_0\mathrm{\θ}-\underset{v=1}{\overset{10}{\mathrm{\Σ}}}(v-2)A_v{\mathrm{\θ}}^{v-1}+A_{11}\[Z\{17(\frac{Z}{29}-\frac{Y}{12})+5\mathrm{\θ}\frac{Y^{\′}}{12}\}+a_4\mathrm{\θ}-(a_3-1)\mathrm{\θY}{Y}^{\′}\]Z^{-5/17}$ $+\{A_{12}-A_{14}{\mathrm{\θ}}^2+A_{15}(9\mathrm{\θ}+a_6){(a_6-\mathrm{\θ})}^9+A_{16}({20\mathrm{\θ}}^{19}+a_7){(a_7+{\mathrm{\θ}}^{19})}^{-2}\}\mathrm{\β}$ $-({12\mathrm{\θ}}^{11}+a_8){(a_8+{\mathrm{\θ}}^{11})}^{-2}(A_{17}\mathrm{\β}+A_{18}{\mathrm{\β}}^2+A_{19}{\mathrm{\β}}^3)+A_{20}{\mathrm{\θ}}^{18}({17a}_9+{19\mathrm{\θ}}^2)\{{(a_{10}+\mathrm{\β})}^{-3}+a_{11}\mathrm{\β}\}$ $A_{21}{a}_{12}{\mathrm{\β}}^3+21A_{22}{\mathrm{\θ}}^{-20}{\mathrm{\β}}^4$ The computing formula of saturated vapour is: $X_2=I_1\mathrm{\θ}/\mathrm{\β}-\underset{\mathrm{\μ}=1}{\overset{5}{\mathrm{\Σ}}}\mathrm{\μ}{\mathrm{\β}}^{\mathrm{\μ}-1}\underset{v=1}{\overset{n\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{B_{\mathrm{\μv}}X}^{z(\mathrm{\μ},v)}-\underset{\mathrm{\μ}=6}{\overset{8}{\mathrm{\Σ}}}\frac{(\mathrm{\μ}-2){\mathrm{\β}}^{1-\mathrm{\μ}}\underset{v=1}{\overset{n\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{\mathrm{\β}}_{\mathrm{\μv}}{X}^{z(\mathrm{\μ},v)}}{{\{{\mathrm{\β}}^{2-\mathrm{\μ}}+\underset{\mathrm{\λ}=1}{\overset{1\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{b}_{\mathrm{\μ\λ}}{X}^{x(\mathrm{\μ},\mathrm{\λ})}\}}^2}$ $+11\left(\frac{\mathrm{\β}}{{\mathrm{\β}}_L}\right)\underset{v=0}{\overset{5}{\mathrm{\Σ}}}{B}_{9v}{X}^v$

Wherein:

X＝exp{b(1-θ)} ${\mathrm{\ϵ}}_2=a_0+B_0\mathrm{\θ}-\underset{v=1}{\overset{5}{\mathrm{\Σ}}}{B}_{0v}(v-2){\mathrm{\θ}}^{v-1}-\underset{\mathrm{\μ}=1}{\overset{5}{\mathrm{\Σ}}}{\mathrm{\β}}^{\mathrm{\μ}}\underset{v=1}{\overset{n\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{B}_{\mathrm{\μv}}(1+z(\mathrm{\μ},v)\mathrm{b\θ})X^{z(\mathrm{\μ},v)}$

$+\mathrm{\β}{\left(\frac{\mathrm{\β}}{{\mathrm{\β}}_L}\right)}^{10}\underset{v=0}{\overset{6}{\mathrm{\Σ}}}\[\{1+\mathrm{\θ}(\frac{10{\mathrm{\β}}_L^{\′}}{{\mathrm{\β}}_L}+\mathrm{vb})\}{B}_{9v}{X}^v\]$ The assignment of each amount is as follows in above-mentioned each formula: I1=4.260321148A0=6824.687741, A1=-542.2063673, A2=-20966.66205, A3=39412.86787A4=-67332.77739, A5=99023.81028, A6=-109391.1774, A7=85908.41667A8=-45111.68742, A9=14181.38926, A10=-2017.271113, A11=7.982692717A12=-0.02616571843, A13=0.00152241179, A14=0.02284279054, A15=242.1647003, A16=1.269716088E-10, A17=2.074838328E-07A18=2.17402035E-08, A19=1.105710498E-09, A20=12.93441934A21=0.00001308119072, A22=6.047626338E-14a1=0.8438375405, a2=0.0005362162162, a3=1.72, a4=0.07342278489a5=0.0497585887, a6=0.65371543, a7=0.00000115, a8=0.000015108a9=0.14188, a10=7.002753165, a11=0.0002995284926, a12=0.204B0=16.83599274, B01=28.56067796, B02=-54.38923329, B03=0.4330662834, B04=-0.6547711697, B05=0.08565182058B11=0.06670375918, B12=1.388983801, B21=0.08390104328B22=0.02614670893, B23=-0.03373439453, B31=0.4520918904B32=0.1069036614, B41=-0.5975336707, B42=-0.08847535804B51=0.5958051609, B52=-0.5159303373, B53=0.2075021122B61=0.1190610271, B62=-0.09867174132, B71=0.1683998803B72=-0.05809438001, B81=0.006552390126, B82=0.0005710218649B90=193.6587558, B91=-1388.522425, B92=4126.607219B93=-6508.211677, B94=5745.984054, B95=-2693.088365B96=523.5718623b=0.7633333333, b61=0.4006073948, b71=0.08636081627b81=-0.8532322921, b82=0.3460208861n (1)=2, n (2)=3, n (3)=2, n (4)=2, n (5)=3, n (6)=2, n (7)=2n (8)=2z (1,1)=13, z (2,1)=18, z (3,1)=18, z (4,1)=25, z (5,1)=and 32z (6,1)=12, z (7,1)=24, z (8,1)=24, z (1,2)=3, z (2,2)=2z (3,2)=10, z (4,2)=14, z (5,2)=28, z (6,2)=11, z (7,2)=18z (8,2)=14, z (2,3)=1, z (5,3)=24l (6)=1, l (7)=1, l (8)=2x (6,1)=14, x (7,1)=19, x (8,1)=54x (8,2)=27k1=-7.691234564, k2=-26.08023696, k3=-168.1706546k4=64.23285504, k5=-118.9646225, k6=4.16711732k7=20.97506, k8=1.0E10, k9=9Tc=647.3K, Pc=22120000N/m ², Vc=0.00317m ³/ kgH _Esi=ε _1i* PcVc H _Vsi=ε _2i* PcVcH _Vi=x _i* H _Vsi+ (1-x _i) * H _Esi=x _i* ε _2i* PcVc+ (1-x _i) * ε _1i* 2. PcVc calculates the feedwater enthalpy of each secondary circuit: according to formula $P=P_{\mathrm{ARE}}+{\mathrm{\ρ}}^{\′}\mathrm{g\Δz}+\frac{1}{2}{\left(\frac{4}{{\mathrm{\πD}}^2}\right)}^2\frac{Q^2}{\mathrm{\ρ}}-\mathrm{\λ}\frac{{8\mathrm{LQ}}^2}{{\mathrm{\ρ\π}}^2{D}^5}$ Calculate the steam generator inlet pressure, revise on the basis of actual measurement temperature T 1 and draw the feed temperature value, calculate the feedwater enthalpy by the ASME formula then, concrete formula is: ${\mathrm{\ϵ}}_1=a_0+A_0\mathrm{\θ}-\underset{v=1}{\overset{10}{\mathrm{\Σ}}}(v-2)A_v{\mathrm{\θ}}^{v-1}+A_{11}\[Z\{17(\frac{Z}{29}-\frac{Y}{12})+5\mathrm{\θ}\frac{Y^{\′}}{12}\}+a_4\mathrm{\θ}-(a_3-1)\mathrm{\θY}{Y}^{\′}\]Z^{-5/17}$ $+\{A_{12}-A_{14}{\mathrm{\θ}}^2+A_{15}(9\mathrm{\θ}+a_6){(a_6-\mathrm{\θ})}^9+A_{16}({20\mathrm{\θ}}^{19}+a_7){(a_7+{\mathrm{\θ}}^{19})}^{-2}\}\mathrm{\β}$ $-({12\mathrm{\θ}}^{11}+a_8){(a_8+{\mathrm{\θ}}^{11})}^{-2}(A_{17}\mathrm{\β}+A_{18}{\mathrm{\β}}^2+A_{19}{\mathrm{\β}}^3)+A_{20}{\mathrm{\θ}}^{18}({17a}_9+{19\mathrm{\θ}}^2)\{{(a_{10}+\mathrm{\β})}^{-3}+a_{11}\mathrm{\β}\}$ $+A_{21}{a}_{12}{\mathrm{\β}}^3+{21A}_{22}{\mathrm{\θ}}^{-20}{\mathrm{\β}}^4$ Feedwater enthalpy Hei=ε _1i* 3. PcVc calculates each secondary circuit blowdown enthalpy: the blowdown enthalpy is got the enthalpy of steam generator exit saturation water, i.e. Hei=ε _1i* 4. PcVc calculates each secondary circuit feedwater flow, and concrete formula is as follows: $Q_{\mathrm{ei}}=C_{\mathrm{ei}}\·E_{\mathrm{ei}}\·{\mathrm{\ϵ}}_{\mathrm{ei}}{{\mathrm{\πd}}_{\mathrm{ei}}}^2/4\·\sqrt{2{\mathrm{\ρ}}_{\mathrm{ei}}\mathrm{\Δ}{P}_{\mathrm{ei}}}={\mathrm{\α}}_{\mathrm{ei}}\·{\mathrm{\ϵ}}_{\mathrm{ei}}\·{{\mathrm{\πd}}_{\mathrm{ei}}}^2/4\·\sqrt{2{\mathrm{\ρ}}_{\mathrm{ei}}\mathrm{\Δ}{P}_{\mathrm{ei}}}$ Wherein: Q _EiBe feedwater flow, C _EiBe efflux coefficient, E _Ei=1/ (1-β _Ei ⁴) ^1/2, β _Ei=d _{E '}/ D _{E '}, d _{E '}Being the orifice plate diameter under the operating condition, is d under the reference temperature _eModified value, D _{E '}For operating condition downcomer circular section diameter, be D under the reference temperature _eModified value, ε _eBe the expansion factor of fluid, ρ _eBe the device upstream fluid density, Δ P _eBe pressure reduction, α _eBe coefficient of flow, efflux coefficient C _EiObtain by the Stolz equation: $C_e=C_{1e}+C_{2e}{\left(\frac{{10}^6}{{\mathrm{Re}}_{\mathrm{De}}}\right)}^{0.75}$ Wherein: $C_{1e}=0.5959+0.0312{\mathrm{\βe}}^{2.1}-0.1840{{\mathrm{\β}}_e}^8+0.090L_{1e}\frac{{\mathrm{\βe}}^4}{1-{\mathrm{\βe}}^4}-0.0337L_{2e}^{\′}{\mathrm{\βe}}^3$

C _2e＝0.0029βe ^2.5 ${\mathrm{Re}}_{\mathrm{De}}=\frac{\mathrm{vDe}}{\mathrm{\μe}}=\frac{4\mathrm{Qe}}{\mathrm{\π\μDe}}$ Be Reynolds number μ _eBe the viscosity under the operating condition, β _Ei=d _{E '}/ D _{E '}Be diameter ratio, L _1eFor the orifice plate upstream face to the ratio L of the distance between the pressure port of upstream with pipe diameter _{2e '}For 5. orifice plate downstream face to the distance between the pressure port of downstream and the ratio of pipe diameter calculate the blowdown flow of each secondary circuit, the computing formula of blowdown flow is: $Q_{\mathrm{pi}}=C_{\mathrm{pi}}\·E_{\mathrm{pi}}\·{\mathrm{\ϵ}}_{\mathrm{pi}}{{\mathrm{\πd}}_{\mathrm{pi}}}^2/4\·\sqrt{2{\mathrm{\ρ}}_{\mathrm{pi}}\mathrm{\Δ}{P}_{\mathrm{pi}}}={\mathrm{\α}}_{\mathrm{pi}}\·{\mathrm{\ϵ}}_{\mathrm{pi}}{{\mathrm{\πd}}_{\mathrm{pi}}}^2/4\·\sqrt{2{\mathrm{\ρ}}_{\mathrm{pi}}\mathrm{\Δ}{P}_{\mathrm{pi}}}$ Wherein: Q _PiBe blowdown flow, C _PiBe efflux coefficient, E _Pi=1/ (1-β _Pi ⁴) ^1/2, β _Pi=d _{Pi '}/ D _Pi, d _{Pi '}, be the orifice plate diameter under the operating condition, be d under the reference temperature _PiModified value, D _{Pi '}For operating condition downcomer circular section diameter, be D under the reference temperature _PiModified value, ε _PiBe the expansion factor of fluid, ρ _PiBe the device upstream fluid density, Δ P _PiBe pressure reduction, α _PiBe coefficient of flow, efflux coefficient C _PiObtain by the Stolz equation: $C_p=C_{1p}+C_{2p}{\left(\frac{{10}^6}{{\mathrm{Re}}_{\mathrm{Dp}}}\right)}^{0.75}$ Wherein: $C_{1p}=0.5959+0.0312{{\mathrm{\β}}_p}^{2.1}-0.1840{{\mathrm{\β}}_p}^8+0.090L_{1p}\frac{{{\mathrm{\β}}_p}^4}{1-{{\mathrm{\β}}_p}^4}-0.0337L_{2p}^{\′}{{\mathrm{\β}}_p}^3$ C _2p＝0.0029β _p ^2.5 ${\mathrm{Re}}_{\mathrm{Dp}}=\frac{v{D}_p}{{\mathrm{\μ}}_p}=\frac{4Q_p}{\mathrm{\π}{\mathrm{\μ}}_p{D}_p}$ Be Reynolds number μ _pBe the viscosity under the operating condition, β p _i=d _{P '}/ D _{P '}Be diameter ratio, L _1pFor the orifice plate upstream face to the ratio L of the distance between the pressure port of upstream with pipe diameter _{2p '}For 6. orifice plate downstream face to the ratio of the distance between the pressure port of downstream and pipe diameter calculates the heat W Δ pr that other thermals source are passed to reactor cooling system, computing formula is: the heat that the heat that heat+replenishment pump that positive energy exchange=main pump is brought into is brought into+the voltage stabilizing well heater is brought into-(heat+system's thermal loss that heat+reactor cooling system that heat+the sealing water-to-water heat exchanger is taken away of taking away of non-regenerative heat exchanger is taken away), the numerical value change of W Δ pr is little under the nominal situation, so desirable constant, the total thermal loss value of reactor circuit is 11MW.4) utilize the operation values of above-mentioned each secondary circuit, calculate nuclear reactor power, computing formula is: $\mathrm{WR}=\underset{i=1}{\overset{3}{\mathrm{\Σ}}}\[(\mathrm{Hvi}-\mathrm{Hei})\mathrm{Qei}-(\mathrm{Hvi}-\mathrm{Hpi})\mathrm{Qpi}\]-\mathrm{W\ΔPr}$

Principle of work of the present invention is: the enthalpy liter that produces when calculating secondary circuit working medium by steam generator by measured values such as secondary circuit actuating medium temperature, pressure, flows in the steam generator in the pressurized water reactor, thereby obtain the energy that reactor-loop is passed to secondary circuit, and then the energy that a loop is obtained by other equipment and the energy consideration that loses enter, and obtains reactor core power by principle of energy balance.Specify as follows:

Heat balance method of is in measuring pressurized water reactor in the steam generator on the basis of physical parameters such as secondary circuit actuating medium temperature, pressure, flow, the enthalpy liter that produces when calculating secondary circuit working medium by steam generator, thereby obtain the energy that reactor-loop is passed to secondary circuit, and then the energy that a loop is obtained by miscellaneous equipment and the energy consideration that loses enter, and obtains the reactor core thermal power by principle of energy balance.Under steady state mode of operation, can obtain following energy-balance equation: $\mathrm{WR}=\underset{i=1}{\overset{3}{\mathrm{\Σ}}}\mathrm{WS}{G}_i-\mathrm{W\ΔPr}---\left(1\right)$ Wherein:

WR is reactor core (NSSS) thermal power (MW)

WSGi is the thermal power (MW) that the secondary circuit actuating medium obtains in i the steam generator

W Δ Pr is the heat that other thermal source is passed to reactor cooling system in the unit interval (MW)

Formula (1) is got by the thermal equilibrium relation, and promptly the heat of steam generator is provided by reactor and other thermal source, so the heat that the heat that steam generator obtains deducts other thermal source to be provided is the reactor core thermal power.

The heat that secondary circuit working medium obtains in a certain steam generator is tried to achieve by following computing formula:

WSG＝HvQv+HpQp-HeQe (2)

Qv＝Qe-Qp (3)

The implication of each symbol is in the formula: Qe is a steam generator secondary circuit feedwater flow (kg/s), Qp is a steam generator secondary circuit blowdown flow (kg/s), Qv is a steam generator secondary circuit steam flow (kg/s), He is steam generator secondary circuit feedwater enthalpy (kJ/kg), Hp is a steam generator secondary circuit blowdown enthalpy (kJ/kg), and Hv is steam generator secondary circuit outlet enthalpy of wet steam (kJ/kg).

The implication of formula (2), (3) for p.s. quality be that Qe, specific enthalpy are that the unsaturated water of He enters steam generator by steam generator secondary circuit inlet, after the steam generator heating, flow out by two outlets.One is the blowdown outlet, and the outflow mass rate is that Qp, specific enthalpy are the saturation water of Hp.Another is the wet steam outlet, and the outflow mass rate is that Qv, specific enthalpy are the wet steam of Hv.

(3) substitution (2) is obtained:

WSG＝(Hv-He)Qe-(Hv-Hp)Qp (4)

The enthalpy of wet steam can be expressed as:

Hv＝x Hvs+(1-x)Hes (5)

X is a steam generator outlet steam quality (dimensionless), and 1-x is the content (dimensionless) of water in the steam generator outlet steam, and Hes is the enthalpy (kJ/kg) of saturation water, and Hvs is the enthalpy (kJ/kg) of saturated vapour.The implication of formula (5) is: wet steam is made up of two parts, and a part is a saturated vapour, and the quality percentage composition is x; Another part is synthermal saturation water, and the quality percentage composition is 1-x.

(4) brought in (1) obtain: $\mathrm{WR}=\underset{i=1}{\overset{3}{\mathrm{\Σ}}}\[(\mathrm{Hvi}-\mathrm{Hei})\mathrm{Qei}-(\mathrm{Hvi}-\mathrm{Hpi})\mathrm{Hpi}\]-\mathrm{W\ΔPr}---\left(6\right)$

Wet steam enthalpy Hv is determined by the pressure of steam generator steam (vapor) outlet.Suppose that the steam quality of three steam generators is identical and be constant.Pressure P is obtained through correction by measured value Pvvp, the position pressure reduction Pz that pressure loss Δ P, pressure unit and the conduit axis difference in height that correction term has dynamic pressure Pd, steam generator to be exported to pressure unit causes. $\mathrm{Pd}=\frac{1}{2}{\mathrm{\ρV}}^2=\frac{1}{2}{\left(\frac{4}{{\mathrm{\πD}}^2}\right)}^2\frac{Q^2}{\mathrm{\ρ}}$

Pz＝ρ′gΔz $\mathrm{\ΔP}=\mathrm{\λ}\frac{L}{D}\frac{\mathrm{\ρ}}{2}{V}^2=\mathrm{\λ}\frac{{8\mathrm{LQ}}^2}{{\mathrm{\ρ\π}}^2{D}^5}$

Wherein: V is steam mean flow rate (m/s), and Q is steam mass flow (kg/s), and ρ is vapour density (kg/m ³), D is jet chimney diameter (m), and L is the length (m) that steam generator exports to the pressure unit interface, and ρ ' is the density (kg/m of water in the conduit ³), Δ z is the difference in height (m) between pressure unit and steam pipe axis, λ is the coefficient of friction resistance, and is relevant with the pipeline relative roughness with Reynolds number.Therefore the steam generator top hole pressure is: $P=\mathrm{Pvvp}+{\mathrm{\ρ}}^{\′}\mathrm{g\Δz}+\frac{1}{2}{\left(\frac{4}{{\mathrm{\πD}}^2}\right)}^2\frac{Q^2}{\mathrm{\ρ}}+\mathrm{\λ}\frac{{8\mathrm{LQ}}^2}{{\mathrm{\ρ\π}}^2{D}^5}---\left(7\right)$

Obtain the pressure of steam generator outlet wet steam by following formula after, can try to achieve the enthalpy of saturation water and dry saturated steam under this pressure, obtain the enthalpy of wet steam then by dryness of wet steam by water and steam thermodynamic properties computing formula.The ASME formula that the computing formula of water and steam thermodynamic properties provides for ASME, this formula obtains after being put in order according to experimental data separately by state scientists such as the U.S., the former Soviet Union, Germany, Japan, has bigger authority in the world.

Water and steam is divided into four zones by temperature, pressure, and to have provided respectively be the relative pressure of independent variable, the relative algebraic expression of specific volume, relative enthalpy and relative entropy with the relative temperature to the ASME formula in each district.Owing to only use wherein a part of formula in two zones in the KME system, so only list these formula here.Variable-definition: $\mathrm{\β}=\frac{P}{P_C}$ Be relative pressure, $\mathrm{\θ}=\frac{T}{T_C}$ Be relative temperature, $X=\frac{V}{V_C}$ Be relative specific volume $\mathrm{\ϵ}=\frac{h}{p_C{V}_C}$ Be relative enthalpy

P wherein _C, T _C, V _CBe the value of critical point.When the pressure of known saturation water or saturated vapour, can iterate by following formula and try to achieve saturation temperature: $\mathrm{\β}\left(\mathrm{\θ}\right)=\exp (\frac{1}{\mathrm{\θ}}\frac{\underset{i=1}{\overset{5}{\mathrm{\Σ}}}{k}_i{(1-\mathrm{\θ})}^i}{1+k_6(1-\mathrm{\θ})+k_7{(1-\mathrm{\θ})}^2}-\frac{(1-\mathrm{\θ})}{k_8{(1-\mathrm{\θ})}^2+k_9})---\left(10\right)$ The computing formula of unsaturated water and saturation water is: x ₁=A ₁₁a ₅Z ^-5/17+ { A ₁₂+ A ₁₃θ+A ₁₄θ ²+ A ₁₅(a ₆-θ) ¹⁰+ A ₁₆(a ₇+ θ ¹⁹) ^-1}

-(a ₈+θ ¹¹) ^-1(A ₁₇+2A ₁₈β+3A ₁₉β ²)

(11)

-A ₂₀θ ¹⁸(a ₉+θ ²){-3(a ₁₀+β) ^-4+a ₁₁}+3A ₂₁(a ₁₂-θ)β ²

+4A ₂₂θ ^-20β ³

Z＝Y+(a ₃Y ²-2a ₄θ+2a ₅β) ^1/2

Wherein:

Y＝1-a ₁θ ²-a ₂θ ^-6 ${\mathrm{\ϵ}}_1=a_0+A_0\mathrm{\θ}-\underset{v=1}{\overset{10}{\mathrm{\Σ}}}(v-2)A_v{\mathrm{\θ}}^{v-1}+A_{11}\[Z\{17(\frac{Z}{29}-\frac{Y}{12})+5\mathrm{\θ}\frac{Y^{\′}}{12}\}+a_4\mathrm{\θ}-(a_3-1)\mathrm{\θY}{Y}^{\′}\]Z^{-5/17}$ $+\{A_{12}-A_{14}{\mathrm{\θ}}^2+A_{15}(9\mathrm{\θ}+a_6){(a_6-\mathrm{\θ})}^9+A_{16}({20\mathrm{\θ}}^{19}+a_7){(a_7+{\mathrm{\θ}}^{19})}^{-2}\}\mathrm{\β}$ $-({12\mathrm{\θ}}^{11}+a_8){(a_8+{\mathrm{\θ}}^{11})}^{-2}(A_{17}\mathrm{\β}+A_{18}{\mathrm{\β}}^2+A_{19}{\mathrm{\β}}^3)+A_{20}{\mathrm{\θ}}^{18}({17a}_9+{19\mathrm{\θ}}^2)\{{(a_{10}+\mathrm{\β})}^{-3}+a_{11}\mathrm{\β}\}$ $+A_{21}{a}_{12}{\mathrm{\β}}^3+{21A}_{22}{\mathrm{\θ}}^{-20}{\mathrm{\β}}^4---\left(12\right)$

The computing formula of saturated vapour is: $X_2=I_1\mathrm{\θ}/\mathrm{\β}-\underset{\mathrm{\μ}=1}{\overset{5}{\mathrm{\Σ}}}{\mathrm{\μ\β}}^{\mathrm{\μ}-1}\underset{v=1}{\overset{n\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{B}_{\mathrm{\μv}}{X}^{z(\mathrm{\μ},v)}-\underset{\mathrm{\μ}=6}{\overset{8}{\mathrm{\Σ}}}\frac{(\mathrm{\μ}-2){\mathrm{\β}}^{1-\mathrm{\μ}}\underset{v=1}{\overset{n\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{\mathrm{\β}}_{\mathrm{\μv}}{X}^{z(\mathrm{\μ},v)}}{\{{\mathrm{\β}}^{2-\mathrm{\μ}}+\underset{\mathrm{\λ}=1}{\overset{1\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{b}_{\mathrm{\μ\λ}}{X}^{x(\mathrm{\μ},\mathrm{\λ})}{\}}^2}---\left(13\right)$ $+11\left(\frac{\mathrm{\β}}{{\mathrm{\β}}_L}\right)\underset{v=0}{\overset{5}{\mathrm{\Σ}}}{B}_{9v}{X}^v$

Wherein:

X＝exp{b(1-θ)} ${\mathrm{\ϵ}}_2=a_0+B_0\mathrm{\θ}-\underset{v=1}{\overset{5}{\mathrm{\Σ}}}{B}_{0v}(v-2){\mathrm{\θ}}^{v-1}-\underset{\mathrm{\μ}=1}{\overset{5}{\mathrm{\Σ}}}{\mathrm{\β}}^{\mathrm{\μ}}\underset{v=1}{\overset{n\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{B}_{\mathrm{\μv}}(1+z(\mathrm{\μ},v)\mathrm{b\θ})X^{z(\mathrm{\μ},v)}$

$+\mathrm{\β}{\left(\frac{\mathrm{\β}}{{\mathrm{\β}}_L}\right)}^{10}\underset{v=0}{\overset{6}{\mathrm{\Σ}}}\[\{1+\mathrm{\θ}(\frac{{10\mathrm{\β}}_L^{\′}}{{\mathrm{\β}}_L}+\mathrm{vb})\}{B}_{9v}{X}^x$

Can obtain the enthalpy that feeds water by the temperature and pressure value of measuring steam generator inflow point: when known steam generator inlet feed pressure and feed temperature, can try to achieve the enthalpy of water under this pressure and temperature by water and steam thermodynamic properties computing formula ASME formula.

Pressure can record static pressure PARE by the main feed system upstream and obtain through correction, and correction term has dynamic pressure Pd, pressure loss Δ P, position pressure reduction Pz. $\mathrm{Pd}=\frac{1}{2}{\mathrm{\ρV}}^2=\frac{1}{2}{\left(\frac{4}{{\mathrm{\πD}}^2}\right)}^2\frac{Q^2}{\mathrm{\ρ}},\mathrm{Pz}={\mathrm{\ρ}}^{\′}\mathrm{g\Δz},$ $\mathrm{\ΔP}=\mathrm{\λ}\frac{L}{D}\frac{\mathrm{\ρ}}{2}{V}^2=\mathrm{\λ}\frac{{8\mathrm{LQ}}^2}{{\mathrm{\ρ\π}}^2{D}^5}$

Wherein: V is feedwater mean flow rate (m/s), and Q is feed-water quality flow (kg/s), and D is main feed system pipe diameter (m), and L is the length (m) of inlet from the main feed system to the steam generator, and ρ ' is the density (kg/m of water in the transmitter conduit ³), Δ z is the difference in height (m) of pressure unit and feed pipe between centers, λ is the coefficient of friction resistance, and is relevant with the pipeline relative roughness with Reynolds number.

Therefore the steam generator inlet pressure is: $P=P_{\mathrm{ARE}}+{\mathrm{\ρ}}^{\′}\mathrm{g\Δz}+\frac{1}{2}{\left(\frac{4}{{\mathrm{\πD}}^2}\right)}^2\frac{Q^2}{\mathrm{\ρ}}-\mathrm{\λ}\frac{{8\mathrm{LQ}}^2}{{\mathrm{\ρ\π}}^2{D}^5}---\left(5\right)$

On the basis of measured temperature, consider that the main feed system pipeline to the thermal loss between the steam generator inlet, can obtain the feed temperature value.Thermal loss numerical value is amassed by the coefficient of heat transfer and pipe surface and calculates.

The blowdown enthalpy is got the enthalpy of steam generator exit saturation water.The computing method of enthalpy are with reference to the computing method of wet steam enthalpy, and pressure is the vapor pressure that measures.

Feedwater flow Qe is measured by orifice flowmeter, and according to the relevant regulations of international standard ISO5167-1-92 (alternate standard of NFX10-102), fluid pressure difference and fluid density are calculated feedwater flow by density and differential manometer before and after the measuring diaphragm.

Fluid density is obtained by pressure and temperature.Consider the orifice plate sectional area, can obtain the computing formula of fluid flow: $Q=C\·E\·\mathrm{\ϵ}\·\mathrm{\π}{d}^2/4\·\sqrt{2\mathrm{\ρ\ΔP}}=\mathrm{\α}\·\mathrm{\ϵ}\·\mathrm{\π}{d}^2/4\·\sqrt{2\mathrm{\ρ\ΔP}}---\left(16\right)$ Wherein: Q is flow (kg/s), and C is efflux coefficient (dimensionless), and this coefficient obtains according to the NFX10-102 criterion calculation.E is progressive velocity coefficient (dimensionless), E=1/ (1-β ⁴) ^1/2, wherein: β be diameter than (dimensionless), β=d/D, d are the orifice plate diameter (m) under the operating condition, D is an operating condition downcomer circular section diameter (m), notes: D and d should be revised by the expansion factor of sheet material and tubing.ε is the expansion factor (dimensionless) of fluid, incompressible fluid ε=1, and ρ is device upstream fluid density (kg/m ³), Δ P is pressure reduction (Pa), α is that coefficient of flow (dimensionless) efflux coefficient C is obtained by the Stolz equation: $C=C_1+C_2{\left(\frac{{10}^6}{{\mathrm{Re}}_D}\right)}^{0.75}$ Wherein: $C_1=0.5959+0.0312{\mathrm{\β}}^{2.1}-0.1840{\mathrm{\β}}^8+0.090L_1\frac{{\mathrm{\β}}^4}{1-{\mathrm{\β}}^4}-0.0337L_2^{\′}{\mathrm{\β}}^3$ $C_2=0.0029{\mathrm{\β}}^{2.5},{\mathrm{Re}}_D=\frac{\mathrm{vD}}{\mathrm{\μ}}=\frac{4Q}{\mathrm{\π\μD}}$ Be Reynolds number, μ is the viscosity (kg/s.m) under the operating condition, and β=d/D is the diameter ratio, and L1=e1/D is that the orifice plate upstream face is to the ratio of the distance between the pressure port of upstream with pipe diameter, L ' ₂=e2/D is that the orifice plate downstream face is to the ratio of the distance between the pressure port of downstream with pipe diameter.

The computing method of blowdown flow are identical with the computing method of feedwater flow, do not repeat them here.

Other thermal source is transferred to the thermal power (W Δ Pr) of reactor cooling system:

The heat of input has: the heat that the heat that the heat that main pump is brought into, replenishment pump are brought into, voltage stabilizing well heater are brought into.The heat of output has: the heat that non-regenerative heat exchanger is taken away, the heat that the sealing water-to-water heat exchanger is taken away, the heat that reactor cooling system is taken away, system's thermal loss.

More than every heat can try to achieve by energy budget method according to the ruuning situation of relevant devices.Concrete grammar is: obtained the intake of relevant devices and exported energy by pressure, temperature and flow measurements, the difference of the two is the energy that exchanges between this equipment and the steam generator.The numerical value change of W Δ Pr is little under the nominal situation, so desirable constant.Be taken as 11MW according in the past three total thermal loss in loop of operating experience.

Advantage of the present invention is: by measuring the enthalpy liter that produces when measured values such as secondary circuit actuating medium temperature, pressure, flow calculate secondary circuit working medium by steam generator in the steam generator in the pressurized water reactor, thereby obtain the energy that reactor-loop is passed to secondary circuit, and then the energy that a loop is obtained by other equipment and the energy consideration that loses enter, and calculates reactor core power by principle of energy balance.Therefore, can improve power measurement speed and accuracy greatly, improve the economic benefit of nuclear power station unit operation.

(4) description of drawings

The present invention is further illustrated below in conjunction with accompanying drawing and example.

Fig. 1 is pressurized water reactor of the present invention one loop and secondary circuit structural representation;

Fig. 2 is that measurements and calculations equipment of the present invention connects block diagram;

Fig. 3 is an executive routine logic diagram of the present invention.

Among Fig. 1, pressurized water reactor one loop and secondary circuit connect each other by steam generator.

Among Fig. 2, handle PC and be connected with the data acquisition PC, the data acquisition PC is connected with the isolation acquisition module, isolates acquisition module and is connected with ready-made measurement signal line.

Among Fig. 3, the constant parameter is input to the processing PC, after the processing PC powers up and brings into operation, handling PC utilizes momentum parameter and constant parameter to calculate, the feedwater flow, blowdown flow, steam flow, feedwater enthalpy, blowdown enthalpy, outlet enthalpy of wet steam and other thermals source that calculate each secondary circuit of steam generator are passed to the heat of reactor cooling system, calculate pressurized-water reactor core power at last.

(5) embodiment

In order to verify the measurement effect of above-mentioned measuring method, to calculate by the real time data of GNPS, result of calculation is as shown in the table:

Thermal power calculated value (MW)	260.5	509.8	1380.1	1727.0	2121.8
						Thermal power actual value (MW)	260.8	510.8	1381.4	1728.6	2123.6
The thermal power error of calculation (MW)	-0.3	-0.5	-1.3	-1.6	-1.8
						Thermal power calculated value (MW)	2359.4	2428.5	2544.0	2882.8	2899.1
Thermal power actual value (MW)	2361.4	2430.6	2546.2	2885.4	2901.7
						The thermal power error of calculation (MW)	-2.0	-2.1	-2.2	-2.6	-2.6

Last table result of calculation shows: nuclear reactor core power calculation value and actual value that this kind measuring method obtains differ very little, and maximum absolute error is 2,6MW, 1% of not enough full scale.

Claims (1)

1, a kind of method for measuring power of pressurized-water reactor core for nuclear power station is characterized in that:

(1) will handle PC and be connected with the data acquisition PC, the data acquisition PC is connected with the isolation acquisition module, isolates acquisition module and is connected with the in-site measurement signal wire;

(2) gather correlation parameter by isolating acquisition module, and the parameter that collects is input to acquisition PC machine, utilize the data of acquisition PC machine, the program by in the operation processing PC just can obtain reactor core power, and its concrete steps are:

1) the constant data that need in the time of will measuring are input to be handled in the PC, and these data comprise: 1. jet chimney diameter D, steam generator export to difference in height Δ z, coefficient of friction resistance λ between length L, pressure unit and the steam pipe axis of pressure unit interface under the reference temperature; 2. main feed system pipe diameter D under the reference temperature ₁, from the main feed system to the steam generator inlet length L ₁, pressure unit and feed pipe between centers difference in height Δ z ₁, coefficient of friction resistance λ 1,3. the feedwater orifice plate diameter d under the reference temperature _e, give water conduit circular section diameter D under the reference temperature _e, the 4. mudhole board diameter d under the reference temperature _p, reference temperature down blow conduit circular section diameter D _p

2) gather momentum parameter by isolating acquisition module, these data comprise: 1. steam generator top hole pressure P _VVP, 2. steam generator inlet pressure P of steam mean flow rate V, steam generator outlet steam quality x _ARE, the differential pressure that 3. feeds water Δ p _Ei, 4. blowdown differential pressure Δ p _Pi

3) handling PC utilizes momentum parameter and constant parameter to calculate, the feedwater flow, blowdown flow, steam flow, feedwater enthalpy, blowdown enthalpy, outlet enthalpy of wet steam and other thermals source that calculate the steam generator secondary circuit are passed to the heat of reactor cooling system, and the concrete operation way is:

1. calculate each secondary circuit enthalpy of wet steam: according to formula $P=\mathrm{Pvvp}+{\mathrm{\ρ}}^{\′}\mathrm{g\Δz}+\frac{1}{2}{\left(\frac{4}{\mathrm{\π}{D}^2}\right)}^2\frac{Q^2}{\mathrm{\ρ}}+\mathrm{\λ}\frac{8L{Q}^2}{\mathrm{\ρ}{\mathrm{\π}}^2{D}^5}$ , calculate the steam generator top hole pressure, according to this pressure value, calculate the enthalpy of saturation water and saturated dry steam under this pressure by the ASME formula, obtain the enthalpy of wet steam again according to the mass dryness fraction of wet steam, concrete computing formula is as follows: $\mathrm{\β}=\frac{P}{P_C}$ Be relative pressure, $\mathrm{\θ}=\frac{T}{T_C}$ Be relative temperature, $X=\frac{V}{V_C}$ Be relative specific volume, $\mathrm{\ϵ}=\frac{h}{p_C{V}_C}$ Be relative enthalpy, wherein P _c, T _c, V _cBe the value of critical point, when the pressure of known saturation water or saturated vapour, can iterate by following formula and try to achieve saturation temperature: $\mathrm{\β}\left(\mathrm{\θ}\right)=\exp (\frac{1}{\mathrm{\θ}}\frac{\underset{i=1}{\overset{5}{\mathrm{\Σ}}}{k}_i{(1-\mathrm{\θ})}^i}{1+k_6(1-\mathrm{\θ})+k_7{(1-\mathrm{\θ})}^2}-\frac{(1-\mathrm{\θ})}{k_8{(1-\mathrm{\θ})}^2+k_9})$ The computing formula of unsaturated water and saturation water is: x ₁=A ₁₁a ₅Z ^-5/17+ { A ₁₂+ A ₁₃θ+A ₁₄θ ²+ A ₁₅(a ₆-θ) ¹⁰+ A ₁₆(a ₇+ θ ¹⁹) ^-1}

-(a ₈+θ ¹¹) ^-1(A ₁₇+2A ₁₈β+3A ₁₉β ²)

-A ₂₀θ ¹⁸(a ₉+θ ²){-3(a ₁₀+β) ^-4+a ₁₁}+3A ₂₁(a ₁₂-θ)β ²

+ 4A ₂₂θ ^-20β ³Wherein:

Z＝Y+(a ₃Y ²-2a ₄θ+2a ₅β) ^1/2

Y＝1-a ₁θ ²-a ₂θ ^-6 ${\mathrm{\ϵ}}_1=a_0+A_0\mathrm{\θ}-\underset{v=1}{\overset{10}{\mathrm{\Σ}}}(v-2)A_v{\mathrm{\θ}}^{v-1}+A_{11}\[Z\left\{17\right(\frac{Z}{29}-\frac{Y}{12})+5\mathrm{\θ}\frac{Y^{\′}}{12}\}+a_4\mathrm{\θ}-(a_3-1)\mathrm{\θY}{Y}^{\′}\]Z^{-5/17}$ $+\{A_{12}-A_{14}{\mathrm{\θ}}^2+A_{15}(9\mathrm{\θ}+a_6){(a_6-\mathrm{\θ})}^9+A_{16}({20\mathrm{\θ}}^{19}+a_7){(a_7+{\mathrm{\θ}}^{19})}^{-2}\}\mathrm{\β}$ $-({12\mathrm{\θ}}^{11}+a_8){(a_8+{\mathrm{\θ}}^{11})}^{-2}(A_{17}\mathrm{\β}+{A_{18}\mathrm{\β}}^2+A_{19}{\mathrm{\β}}^3)+A_{20}{\mathrm{\θ}}^{18}({17a}_9+{19\mathrm{\θ}}^2)\{{(a_{10}+\mathrm{\β})}^{-3}+a_{11}\mathrm{\β}\}$ $+A_{21}{a}_{12}{\mathrm{\β}}^3+{21A}_{22}{\mathrm{\θ}}^{-20}{\mathrm{\β}}^4$ The computing formula of saturated vapour is: $X_2=I_1\mathrm{\θ}/\mathrm{\β}-\underset{\mathrm{\μ}=1}{\overset{5}{\mathrm{\Σ}}}\mathrm{\μ}{\mathrm{\β}}^{\mathrm{\μ}-1}\underset{v=1}{\overset{n\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{B}_{\mathrm{\μv}}{X}^{z(\mathrm{\μ},v)}-\underset{\mathrm{\μ}=6}{\overset{8}{\mathrm{\Σ}}}\frac{(\mathrm{\μ}-2){\mathrm{\β}}^{1-\mathrm{\μ}}\underset{v=1}{\overset{n\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{\mathrm{\β}}_{\mathrm{\μv}}{X}^{z(\mathrm{\μ},v)}}{{\{{\mathrm{\β}}^{2-\mathrm{\μ}}+\underset{\mathrm{\λ}=1}{\overset{1\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{b}_{\mathrm{\μ\λ}}{X}^{x(\mathrm{\μ},\mathrm{\λ})}\}}^2}$ $+11\left(\frac{\mathrm{\β}}{{\mathrm{\β}}_L}\right)\underset{v=0}{\overset{5}{\mathrm{\Σ}}}{B}_{9v}{X}^v$

Wherein:

X＝exp{b(1-θ)} ${\mathrm{\ϵ}}_2=a_0+B_0\mathrm{\θ}-\underset{v=1}{\overset{5}{\mathrm{\Σ}}}{B}_{0v}(v-2){\mathrm{\θ}}^{v-1}-\underset{\mathrm{\μ}=1}{\overset{5}{\mathrm{\Σ}}}{\mathrm{\β}}^{\mathrm{\μ}}\underset{v=1}{\overset{n\left(\mathrm{\μ}\right)}{\mathrm{\Σ}}}{B}_{\mathrm{\μv}}(1+z(\mathrm{\μ},v)\mathrm{b\θ})X^{z(\mathrm{\μ},v)}$

$+\mathrm{\β}{\left(\frac{\mathrm{\β}}{{\mathrm{\β}}_L}\right)}^{10}\underset{v=0}{\overset{6}{\mathrm{\Σ}}}\[\{1+\mathrm{\θ}(\frac{10{\mathrm{\β}}_L^{\′}}{{\mathrm{\β}}_L}+\mathrm{vb})\}{B}_{9v}{X}^v\]$ The assignment of each amount is as follows in above-mentioned each formula: I1=4.260321148A0=6824.687741, A1=-542.2063673, A2=-20966.66205, A3=39412.86787A4=-67332.77739, A5=99023.81028, A6=-109391.1774, A7=85908.41667A8=-45111.68742, A9=14181.38926, A10=-2017.271113, A11=7.982692717A12=-0.02616571843, A13=0.00152241179, A14=0.02284279054, A15=242.1647003, A16=1.269716088E-10, A17=2.074838328E-07A18=2.17402035E-08, A19=1.105710498E-09, A20=12.93441934A21=0.00001308119072, A22=6.047626338E-14a1=0.8438375405, a2=0.0005362162162, a3=1.72, a4=0.07342278489a5=0.0497585887, a6=0.65371543, a7=0.00000115, a8=0.000015108a9=0.14188, a10=7.002753165, a11=0.0002995284926, a12=0.204B0=16.83599274, B01=28.56067796, B02=-54.38923329, B03=0.4330662834, B04=-0.6547711697, B05=0.08565182058B11=0.06670375918, B12=1.388983801, B21=0.08390104328B22=0.02614670893, B23=-0.03373439453, B31=0.4520918904B32=0.1069036614, B41=-0.5975336707, B42=-0.08847535804B51=0.5958051609, B52=-0.5159303373, B53=0.2075021122B61=0.1190610271, B62=-0.09867174132, B71=0.1683998803B72=-0.05809438001, B81=0.006552390126, B82=0.0005710218649B90=193.6587558, B91=-1388.522425, B92=4126.607219B93=-6508.211677, B94=5745.984054, B95=-2693.088365B96=523.5718623b=0.7633333333, b61=0.4006073948, b71=0.08636081627b81=-0.853232292 1, b82=0.3460208861n (1)=2, n (2)=3, n (3)=2, n (4)=2, n (5)=3, n (6)=2, n (7)=2n (8)=2z (1,1)=13, z (2,1)=18, z (3,1)=18, z (4,1)=25, z (5,1)=and 32z (6,1)=12, z (7,1)=24, z (8,1)=24, z (1,2)=3, z (2,2)=2z (3,2)=10, z (4,2)=14, z (5,2)=28, z (6,2)=11, z (7,2)=18z (8,2)=14, z (2,3)=1, z (5,3)=241 (6)=1,1 (7)=1,1 (8)=2x (6,1)=14, x (7,1)=19, x (8,1)=54x (8,2)=27k1=-7.691234564, k2=-26.08023696, k3=-168.1706546k4=64.23285504, k5=-118.9646225, k6=4.16711732k7=20.97506, k8=1.0E10, k9=9Tc=647.3K, Pc=22120000N/m ², Vc=0.00317m ³/ kgH _Esi=ε _1i* PcVc H _Vsi=ε _2i* PcVcH _Vi=x _i* H _Vsi+ (1-x _i) * H _Esi=x _i* ε _2i* PcVc+ (1-x _i) * ε _1i* PcVc

2. calculate the feedwater enthalpy of each secondary circuit: according to formula $P=P_{\mathrm{ARE}}+{\mathrm{\ρ}}^{\′}\mathrm{g\Δz}+\frac{1}{2}{\left(\frac{4}{\mathrm{\π}{D}^2}\right)}^2\frac{Q^2}{\mathrm{\ρ}}-\mathrm{\λ}\frac{8L{Q}^2}{\mathrm{\ρ}{\mathrm{\π}}^2{D}^5}$ Calculate the steam generator inlet pressure, revise on the basis of actual measurement temperature T 1 and draw the feed temperature value, calculate the feedwater enthalpy by the ASME formula then, concrete formula is: ${\mathrm{\ϵ}}_1=a_0+A_0\mathrm{\θ}-\underset{v=1}{\overset{10}{\mathrm{\Σ}}}(v-2)A_v{\mathrm{\θ}}^{v-1}+A_{11}\[Z\{17(\frac{Z}{29}-\frac{Y}{12})+5\mathrm{\θ}\frac{Y^{\′}}{12}\}+a_4\mathrm{\θ}-(a_3-1)\mathrm{\θ}{\mathrm{YY}}^{\′}\]Z^{-5/17}$ $+\{A_{12}-A_{14}{\mathrm{\θ}}^2+A_{15}(9\mathrm{\θ}+a_6){(a_6-\mathrm{\θ})}^9+A_{16}({20\mathrm{\θ}}^{19}+a_7){(a_7+{\mathrm{\θ}}^{19})}^{-2}\}\mathrm{\β}$ $-({12\mathrm{\θ}}^{11}+a_8){(a_8+{\mathrm{\θ}}^{11})}^{-2}(A_{17}\mathrm{\β}+A_{18}{\mathrm{\β}}^2+A_{19}{\mathrm{\β}}^3)+A_{20}{\mathrm{\β}}^{18}({17a}_9+{19a}^2)\{{(a_{10}+\mathrm{\β})}^{-3}+a_{11}\mathrm{\β}\}$ $+A_{21}{a}_{12}{\mathrm{\β}}^3+21A_{22}{\mathrm{\θ}}^{-20}{\mathrm{\β}}^4$ Feedwater enthalpy Hei=ε _1i* PcVc

3. calculate each secondary circuit blowdown enthalpy: the blowdown enthalpy is got the enthalpy of steam generator exit saturation water, promptly

Hei＝ε _li*PcVc

4. calculate each secondary circuit feedwater flow, concrete formula is as follows: $Q_{\mathrm{ei}}=C_{\mathrm{ei}}\·E_{\mathrm{ei}}\·{\mathrm{\ϵ}}_{\mathrm{ei}}\·{{\mathrm{\πd}}_{\mathrm{ei}}}^2/4\·\sqrt{2{\mathrm{\ρ}}_{\mathrm{ei}}\mathrm{\Δ}{P}_{\mathrm{ei}}}={\mathrm{\α}}_{\mathrm{ei}}\·{\mathrm{\ϵ}}_{\mathrm{ei}}\·{{\mathrm{\πd}}_{\mathrm{ei}}}^2/4\·\sqrt{2{\mathrm{\ρ}}_{\mathrm{ei}}\mathrm{\Δ}{P}_{\mathrm{ei}}}$ Wherein: Q _EiBe feedwater flow, C _EiBe efflux coefficient, E _Ei=1/ (1-β _Ei ⁴) ^1/2, β _Ei=d _{E '}/ D _{E '}, d _{E '}Being the orifice plate diameter under the operating condition, is d under the reference temperature _eModified value, D _{E '}For operating condition downcomer circular section diameter, be D under the reference temperature _eModified value, ε _eBe the expansion factor of fluid, ρ _eBe the device upstream fluid density, Δ P _eBe pressure reduction, α _eBe coefficient of flow, efflux coefficient C _EiObtain by the Stolz equation: $C_e=C_{1e}+C_{2e}{\left(\frac{{10}^6}{{\mathrm{Re}}_{\mathrm{De}}}\right)}^{0.75}$ Wherein: $C_{1e}=0.5959+0.0312{\mathrm{\βe}}^{2.1}-0.1840{{\mathrm{\β}}_e}^8+0.090L_{1e}\frac{{\mathrm{\βe}}^4}{1-{\mathrm{\βe}}^4}-0.0337L_{2e}^{\′}{\mathrm{\βe}}^3$

C _2e＝0.0029βe ^2.5 ${\mathrm{Re}}_{\mathrm{De}}=\frac{\mathrm{vDe}}{\mathrm{\μe}}=\frac{4\mathrm{Qe}}{\mathrm{\π\μDe}}$ Be Reynolds number μ _eBe the viscosity under the operating condition, β _Ei=d _{E '}/ D _{E '}Be diameter ratio, L _1eFor the orifice plate upstream face to the ratio L` of the distance between the pressure port of upstream with pipe diameter _2eFor the orifice plate downstream face to the ratio of the distance between the pressure port of downstream with pipe diameter

5. calculate the blowdown flow of each secondary circuit, the computing formula of blowdown flow is: $Q_{\mathrm{ei}}=C_{\mathrm{pi}}\·E_{\mathrm{pi}}\·{\mathrm{\ϵ}}_{\mathrm{pi}}{{\mathrm{\πd}}_{\mathrm{pi}}}^2/4\·\sqrt{2{\mathrm{\ρ}}_{\mathrm{pi}}\mathrm{\Δ}{P}_{\mathrm{pi}}}={\mathrm{\α}}_{\mathrm{pi}}\·{\mathrm{\ϵ}}_{\mathrm{pi}}{{\mathrm{\πd}}_{\mathrm{pi}}}^2/4\·\sqrt{2{\mathrm{\ρ}}_{\mathrm{pi}}\mathrm{\Δ}{P}_{\mathrm{pi}}}$ Wherein: Q _PiBe blowdown flow, C _PiBe efflux coefficient, E _Pi=1/ (1-β _Pi ⁴) ^1/2, β _Pi=d _{Pi '}/ D _Pi, d _{Pi '}Being the orifice plate diameter under the operating condition, is d under the reference temperature _PiModified value, D _{Pi '}For operating condition downcomer circular section diameter, be D under the reference temperature _PiModified value, ε _PiBe the expansion factor of fluid, ρ _PiBe the device upstream fluid density, Δ P _PiBe pressure reduction, α _PiBe coefficient of flow, efflux coefficient C _PiObtain by the Stolz equation: $C_P=C_{1p}+C_{2p}{\left(\frac{{10}^6}{{\mathrm{Re}}_{\mathrm{Dp}}}\right)}^{0.75}$ Wherein: $C_{1p}=0.5959+0.0312{{\mathrm{\β}}_p}^{2.1}-0.1840{{\mathrm{\β}}_p}^8+0.090L_{1p}\frac{{{\mathrm{\β}}_p}^4}{1-{{\mathrm{\β}}_p}^4}-0.0337L_2^{\′}{{\mathrm{\β}}_p}^3$

C _2p＝0.0029β _p ^2.5 ${\mathrm{Re}}_{\mathrm{Dp}}=\frac{v{D}_p}{{\mathrm{\μ}}_p}=\frac{4Q_p}{\mathrm{\π}{\mathrm{\μ}}_p{D}_p}$ Be Reynolds number μ _pBe the viscosity under the operating condition, β p _i=d _{P '}/ D _{P '}Be diameter ratio, L _1pFor the orifice plate upstream face to the ratio L` of the distance between the pressure port of upstream with pipe diameter _2pFor the orifice plate downstream face to the ratio of the distance between the pressure port of downstream with pipe diameter

6. calculate other thermals source and pass to the heat W Δ pr of reactor cooling system, computing formula is: the heat that the heat that heat+replenishment pump that positive energy exchange=main pump is brought into is brought into+the voltage stabilizing well heater is brought into-(heat+system's thermal loss that heat+reactor cooling system that heat+the sealing water-to-water heat exchanger is taken away of taking away of non-regenerative heat exchanger is taken away), the numerical value change of W Δ pr is little under the nominal situation, so desirable constant, the total thermal loss value of reactor circuit is 11MW.

4) utilize the operation values of above-mentioned each secondary circuit, calculate nuclear reactor power, computing formula is: $\mathrm{WR}=\underset{i=1}{\overset{3}{\mathrm{\Σ}}}\[(\mathrm{Hvi}-\mathrm{Hei})\mathrm{Qei}-(\mathrm{Hvi}-\mathrm{Hpi})\mathrm{Qpi}\]-\mathrm{W\ΔPr}$

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