International Journal of Engineering Research and Technology. ISSN 0974-3154, Volume 12, Number 9 (2019), pp. 1485-1493 © International Research Publication House. http://www.irphouse.com Steam Generator Overall Area Estimation for PWR NPP

Mai Quy Sang1 and Ihn Namgung2* 1Graduate Student, Dept. of NPP Engineering, KEPCO International Nuclear Graduate School, Kenya. 2Professor, Dept. of NPP Engineering, KEPCO International Nuclear Graduate School, South Korea.

Abstract: The rate of the heat transfer between the primary and the secondary side is dependent on the temperature gradient There are over 1,300 steam generators (SGs) installed in the between the average temperature and the saturation 450 nuclear power plants worldwide. The technology in SG temperature of the steam generator, the overall heat transfer design been continuously improved through experience and the area of the u-tubes, and the overall heat transfer coefficient continuous research activities in the industry. The heat transfer which is dependent on tube material, geometry, and the phenomena have continuously been studied to improve on the convective heat transfers of the tube inner and outer surfaces effectiveness and efficiency of SG around the world. In most [2]. To effectively remove the heat generated special PWR plant recirculating steam generators (RSG) are in use optimization of the overall heat transfer area of the u-tubes need basically shell and tube heat exchangers and mostly of the to be effectively deter-mined[2], the material selection is done vertical orientation besides the VVER models which are carefully to protect the tube from damage [1][4] while ensuring horizontal in orientation. The most advanced PWR plants use high rate of heat transfer[3], and standard geometries are being RSG with and without respectively. This forms the employed by many manufacturers of steam generators in the basis for this conceptual research problem to establish the industry. effectiveness of the economizer on the reduction of the overall heat transfer area (OHTA) of the PWR SG. The OHTA was Steam generators in operation in PWR’s are recirculating u- estimated in two steps where subcooled boiling region was tube type and once through types [1]. The recirculating types evaluated by the general heat transfer equation while the only partial evaporation of feedwater hence generating saturated boiling region used the modified Rohsenow equation saturated steam while the once through type allow for 100% by Westinghouse. From the results this conceptual study the evaporation of feedwater generating superheated steam[5]. estimated OHTA of the PWR SG design is 17.15% higher since They can also be either vertical like the APR1400 [3] and it is just a preliminary study. Optimization of the models is a AP1000[2] or horizontal like the Russian VVER models. topic for further studies through experiments. For the objectives In case of the Advanced Pressurized Reactor 1400 (APR1400) of the study the economizer reduced the OHTA by 3.11% from has two steam generators (SG) designed to transfer 4000MWt 18,459.77m2 to 17,906.41m2 which on optimization might be from the primary reactor coolant system (RCS) to the further increased secondary main steam system (MSS) and power conversion Keywords: Steam Generators, SG Economizer, Overall Heat system. The heat is transferred from the primary to secondary Transfer Area of SG, Heat Balance of SG system through SG tubes are Ni-Cr-Fe alloy 690 (Inconel 690); 3/4-inch OD and with a 0.042-inch nominal wall thickness

provided with a plugging margin of 10%. The reactor coolant I. INTRODUCTION enters the SG through a 42inch inlet nozzle, flows through the u-tubes, and leaves through two 30inch outlet nozzles both at Steam generators in pressurized water reactors (PWR) plant are the bottom spherical head. The SG has an integral economizer shell and tube heat exchangers typically two to four per reactor (Fig 1 and Fig 2) which is a separation of the cold side of the [1]. The steam generator is used to (1) produce dry saturated tube bundle from the hot side through vertical divider plates. steam for the turbine generator and auxiliary systems, (2) act as This allows feedwater to be preheated before its introduction to a heat sink for the reactor coolant system (RCS) during normal, the region of the SG [3] hence increasing efficiency abnormal, and emergency conditions, and (3) provide a reactor of heat transfer. The steam quality is improved to a limit of coolant pressure boundary avoiding release of the radioactive maximum moisture content of 0.25wt % by moisture separators primary coolant into the secondary system[2], [3]. To produce and steam driers in the shell side during normal operation at full steam the steam generator transfers the heat generated in the power. The pressure drop on the secondary side from the nuclear reactor core from the primary reactor coolant to the feedwater nozzles to the steam nozzles is 42psid secondary feedwater through the u-tube walls to generate steam (2.981kg/cm2). The steam generator design parameters are ensuring continuous removal of the steam from the core [2]. listed in Table 1 [3].

* Corresponding author

1485 International Journal of Engineering Research and Technology. ISSN 0974-3154, Volume 12, Number 9 (2019), pp. 1485-1493 © International Research Publication House. http://www.irphouse.com

Table 1. A typical steam generator main parameters. The APR1400 SG designed in compliance with all legal, regulatory requirements, codes, and standards [3] is one of the Parameter Value most recent advanced power reactors in the industry. One of the most distinguishing features is that it has an integral Primary Side economizer at the cold exit side of the u-tubes as described above. The economizer is a preheater that helps in the

Design pressure/temperature (psia/ 2,500/650 improvement of the thermal efficiency in the steam generator o 2 o F) (kg/cm / C) (175.76/343.33) by preheating the fluid before it enters the evaporator section

o o [5]. Other more recent high output models like the AP1000 do Coolant inlet temperature, F ( C) 615 (323.88) not possess such a feature but rather consist of just the evaporator region on the lower side of the steam generator [2]. Coolant outlet temperature, oF (oC) 555 (290.55) The AP1000 SG receives feedwater from the main feedwater 83.3 x 106 (37.78 nozzles through a feedwater ring which combines with the Coolant flow rate, each, lb/hr (kg/hr) x 106) recirculating water from the moisture separators and flows down the downcomer into the evaporator region [2]. The main objective of this project is to understand the concept in the design of SGs which would be used as an input in conceptual design of smaller capacity SGs while establishing the effect of the economizer provision on the APR1400 SG on OHTA

II. REVIEW OF HEAT TRANSFER IN PWR STEAM GENERATORS The main purpose of the steam generator is to produce steam for the turbine, using the heat generated by the core and carried by the primary coolant. This function is vital to safety of the reactor core as it ensure continuous removal of heat that is generated thus preventing core damage due to elevated temperatures. The basic modes of heat transfer are conduction, radiation, and . The temperature gradient, medium (solid, liquid, gas, or vacuum) determines the mode of heat transfer, where in some cases; all these forms of heat transfer may take place at the same time.

II-I. Conduction heat transfer Conduction is the transfer of energy from more energetic particles to less energetic particles due to interaction between the particles [7]. At molecular level, interactions of gas and liquid molecules or the lattice vibrations in solids enhance the

heat transfer in the media [7]. In the steam generators heat is Figure 1. A typical PWR SG major components generated on the primary side and transferred to the secondary side using the heat conduction mechanism in the wall of tubes in the bundle. While designing tubes, engineers select materials that have high thermal conductivity in order to improve heat transfer[8]. The conduction heat transfer process is quantified in terms of appropriate rate equations per unit time. Conduction follows the Fourier law where for one dimensional plane wall having a temperature distribution T(x) [7]. 푑푇 푞" = −퐾 (1) 푑푥 Where 푊 퐾 is the thermal conductivity of the medium ( ) 푚퐾

II-II. Convection heat transfer

Figure 2. A cross sectional view of SG economizer region Convection occurs by random molecular motion (diffusion)

1486 International Journal of Engineering Research and Technology. ISSN 0974-3154, Volume 12, Number 9 (2019), pp. 1485-1493 © International Research Publication House. http://www.irphouse.com and the bulk macroscopic motion of the fluid. In this study the Boiling occurs in distinct regions as indicated in Fig 4 below interest is in the convection heat transfer between a fluid in where in region I free convection is responsible for fluid motion motion and a bounding surface when the two are at different near the surface since the fluid near the surface is superheated temperatures. The fluid surface interaction leads to the slightly leading to its rise hence the region can be evaluated development of a hydrodynamic, velocity, or boundary layer using free convection correlations[8]. Bubbles form on the where the fluid velocity varies from zero (no slip condition) at surface and are dissipated in the fluid after breaking away from the surface to a finite value associated with the flow. The the surface in region II marking the onset of nucleate boiling. convection heat transfer is sustained by random molecular There is a rapid formation of bubbles which rise to the fluid motion and the bulk fluid motion within the boundary layer. surface on region III. Eventually the bubbles are too many that Therefore convection heat transfer can be classified by the they blanket the heating surface which leads to the increase of nature of flow as (1) Forced convection when the flow is thermal resistance and hence reduction of the heat flux marked caused by external means like , pumps or atmospheric winds by region IV. This is a very unstable region as it marks the (2) Free or when the flow is induced by transition between the nucleate boiling and the stable film buoyancy forces due to density differences caused by boiling in region V. The surface temperature required to temperature variations in the fluid [7]. Regardless of the nature maintain a stable film boiling are high, and once the condition of the heat transfer process, the appropriate rate equation for is reached most of the heat transfer is by as convection is of the form [7], [8]. indicated in region VI [8].

Figure 3. Boundary layer development in convection heat transfer

푞̈ = ℎ(푇 − 푇 ) (2) 푠 ∞ Figure 4. Boiling regimes from Farber and Scorah experiment 푊 Where ℎ is the convection heat transfer coefficient ( ), [7] 푚2퐾

푇푠 and 푇∞ are the surface and bulk temperatures (퐾). Rohsenow had correlated experimental data for nucleate pool boiling with the following relation [7][8], In the steam generator design, both forced and free convection 1 occur. Forced convection occurs since the liquid on the primary ⁄3 퐶푙∆푇푥 푞⁄ 퐴 푔푐휎 side of the SG is pumped by RCPs, through the reactor core, 푠 = 퐶푠푓 [ √ ] (3) before it enters the SG tube bundles. On the secondary side of ℎ푓푔푃푟푙 휇푙ℎ푓푔 푔(휌푙−휌푣) SG, both free and forced convection are driving heat transfer. where; In the economizer region, since the fluid is pumped by the feedwater pumps, forced convection heat transfer occurs. On 퐶푙 = Specific heat of saturated liquid, 퐵푇푈 퐽 the other hand, in the evaporator region, the heat transfer [ ] 표푟 [ ] mechanism can be described is free convection. 푙푏푚℉ 푘푔℃ ∆푇 = Excess temperature = 푇 − 푇 [℉] 표푟 [℃] 푥 푤 푠푎푡 퐵푇푈 퐽 ℎ = of vaporization, [ ] 표푟 [ ] II-III. Boiling heat transfer in Steam Generators 푓푔 푙푏푚 푘푔 푠 Boiling occurs when a surface is exposed to a liquid and 푃푟푙 = Prandtl number of saturated liquid maintained at a temperature above the saturation temperature 퐵푇푈 푊 of the liquid with the heat flux dependent on the temperature 푞⁄ 퐴 = heat flux per unit area, [ ] 표푟 [ ] 푓푡2ℎ 푚2 difference between the surface and the saturation temperature 푙푏푚 푘푔 [8]. Pool boiling occurs when the heated surface is submerged 휇푙 = liquid viscosity, [ ] 표푟 [ ] below the free surface of the liquid. Subcooled (local) boiling 푓푡 ℎ 푚 푠 is when temperature is below saturated temperature while bulk 𝜎 = surface tension of liquid-vapor 푙푏푓 푁 (saturated) boiling occurs if the temperature is maintained at interface,[ ] 표푟 [ ] saturated temperature [8]. 푓푡 푚

1487 International Journal of Engineering Research and Technology. ISSN 0974-3154, Volume 12, Number 9 (2019), pp. 1485-1493 © International Research Publication House. http://www.irphouse.com

푓푡 푚 푔 = gravitational acceleration, [ ] 표푟 [ ] Thus 푠2 푠2 푙푏푚 푘푔 푄푝 = 푚̇ 푝(ℎℎ표푡 − ℎ푐표푙푑) (2) 𝜌푙 = density of saturated liquid, [ 3 ] 표푟 [ 3] 푓푡 푚 The heat generated in the reactor is transferred to the secondary 푙푏푚 푘푔 𝜌 = density of saturated vapor,[ ] 표푟 [ ] side through the steam generator U-tubes. The heat transferred 푣 푓푡3 푚3 in these U-tubes is calculated as a function of the mass flow rate, the isobaric heat transfer rate at the average temperature, 퐶푠푓 = constant, determined from experimental data and the change in temperature at the entry and exit of the SG. 푠 = 1.0 for water and 1.7 for other liquids Heat transferred in SG from primary is expressed by the This equation can be rearranged to equation.

1⁄ 3 푄 = 푚̇ 푝푐푝 ∆T (3) ′′ 1⁄ 2 ℎ푓푔 푠 푞 휎 푇푤 − 푇푠푎푡 = 푃푟푙 퐶푠푓 [ ( ) ] (4) 푐푝,푙 휇푙ℎ푓푔 푔(휌푙−휌푣) Thus 푄 = 푚̇ 푐 ℎ (푇 − 푇 ) (4) 푝 푝푐 ℎ표푡 푐표푙푑 Also the heat transfer in the evaporator region can be ℎ where 푐푝 = isobaric heat capacity at the average of hot and expressed in terms of heat transfer coefficient as 푐 cold temperature 푞′′ = ℎ(푇 − 푇 ) (5) 푊 푠푎푡 The heat removed from the primary RCS is calculated as a Where the heat transfer coefficient is obtained after function of the mass flow rate in the secondary side ( 푚̇ 푠) and substituting eq (4) into (5) as the change in enthalpy between the feed-water and the steam. ′′ 2/3 ℎ = 퐾푅(푞 ) (6) 푄푠 = 푚̇ 푠(ℎ푠 − ℎ푓푤) (5) 2 푝 Figure 5 represents the fluids flow parameters with the left side 퐾푅 = 푓(푝푠푎푡) = 푎1 + 푎2푝 + 푎3푝 = 푏1 exp ( ) (7) 푏2 indicating the inlet and outlet temperature of coolant as it passes The constants in the above equation can be through the core as it enters through cold leg and leaves through determined through experimental work. An example is the hot leg. The central part represents the steam generator showing Westinghouse calculation where they determined the constant the primary coolant temperature reducing as it passes through 퐾푅 is determined by the equation [5]. the steam generator tubes. The secondary side shows different behavior since in the first section. There is a rise in temperature −4 퐾푅 = 4.1{1 + 1.4 × 10 (푝푠푎푡 − 800)} (8) of the feed-water as it passes through the economizer region and then it remains constant at the saturation temperature of Where the unit of 푝푠푎푡 is in psig. steam. Hence reducing the Rohsenow equation to [5]: The right side represents the turbine and the condenser. In the −4 ′′ 2/3 ℎ = 3.97[1 + 1.41 × 10 (푃푠푎푡 − 800)](푞 ) (9) turbine, the steam expands and losses heat as it drives the Other correlations that have been developed for the nucleate turbine leading to a temperature reduction. The steam from the pool boiling convective heat transfer coefficient include: turbine is discharged to the condenser where is converted back to feed-water by removing heat and use of water from the 103⁄ 2 푝 Jens and Lottes: ℎ = exp ( 푠푎푡) (푞′′)3⁄ 4 (10) ultimate heat sink. 60 900 Approximation of the total heat transfer area can be calculated 103⁄ 2 Weatherhead: ℎ = (푞′′)3⁄ 4 (11) by calculating the overall heat transfer in the steam generator 0.18(500−0.707 푇 ) 푠푎푡 which is evaluated from the equation Error! Reference source 1 푝 Thom: ℎ = exp ( 푠푎푡 ) (푞′′)1⁄ 2 (12) not found.. Heat flows in the steam generator from the hot 0.072 1260 coolant by convection, through the wall by conduction, and 4/3 푝 then by convection from the surface to the cold fluid. In the Levy: ℎ = ( 푠푎푡 ) (푞′′)2/3 (13) 495 three regions the heat flow is calculated as: Region I: hot liquid to solid convection

III. THEORETICAL DEVELOPMENT OF TOTAL 푞푥 = ℎℎ표푡(푇ℎ − 푇𝑖푤)푥 퐴𝑖 (6) HEAT TRANSFER AREA ESTIMATION III-I. Energy Balance and Thermal Sizing Heat generated by the nuclear fission in the reactor is equal to the mass flow rate through the reactor multiplied by the change in the enthalpy of the coolant for normal operating condition of reactor. Heat generated in the reactor is expressed by the equation.

푄 = 푚̇ (ℎ𝑖푛 − ℎ표푢푡) (1)

1488 International Journal of Engineering Research and Technology. ISSN 0974-3154, Volume 12, Number 9 (2019), pp. 1485-1493 © International Research Publication House. http://www.irphouse.com

푞 퐴 = 푥 (17) 표 푈 퐿푀푇퐷 A more precise calculation of the heat transfer are can be calculated by subdividing heat transfer area into evaporator region and economizer region.

a. Heat transfer area calculation for evaporator region Evaporator region of heat transfer is characterized by the secondary side saturated condition, hence, temperature maintains constant boiling temperature. A small segment of heat transfer tube is shown in Fig 6 below which shows heat flow from hot water of primary side to the secondary side cold water. This region is dictated by Rohsenow’s correlations. Figure 5. A depiction of the fluid flow

Region II: conduction across the wall,

2휋퐿푘(푇표푤− 푇𝑖푤) 푞푥 = 퐷 (7) 퐿푛 표 퐷𝑖

Region III: solid to cold liquid convection Figure 6. A differential tube element 푞 = ℎ (푇 − 푇 )푥 퐴 (8) 푥 푐표푙푑 표푤 푐 표 From the three equations above the overall heat transfer is determined as: A modified Rohsenow equation developed through experiment applicable to this case is 푞푥 = 푈퐴(푇ℎ − 푇푐) (9) Generally expressed as:

푞푥 = 푈퐴표LMTD (10) Where LMTD is calculated from the equation Error! Reference source not found.:

∆푇ℎ−∆푇푐 퐿푀푇퐷 = ∆푇 (11) In ℎ ∆푇푐 And U is the Universal Heat Transfer Coefficient calculated from: 1 푈 = (12) 푅𝑖+ 푅푤+ 푅표 Where:

푑표 푅𝑖 = (13) ℎ𝑖푑𝑖 Figure 1 Heat flux and Temperature against Average Tube 푑 푑 푅 = 표 푙푛 (14) length 푤 2푘 푑 푤 𝑖 2 푞 ⁄3 1 ℎ푠 = 퐾푠 ∙ ( ) (18) 푅표 = (15) 퐴 ℎ표 Where 퐾푠 is a constant that is a function of the saturation Where ℎ𝑖 is calculated from the Dittus-Boelter equation for heat transfer presented as follows: pressure and is calculated as, −4 푘 퐾 = 4.1{1 + 1.4 × 10 ∙ (푝 − 800)} (32) ℎ = 0.023 푅푒0.8푃푟푛 (16) 푠 푠푎푡 퐷ℎ In the differential element above, the heat transfer can be Where n is 0.3 for primary side (cooling) and 0.4 for secondary calculated as follows, side (heating) 푑푞 = 푚̇ 푝푐푝푑푇푝 (19) From the calculations above then the overall heat transfer area can be estimated as: But the total temperature difference through primary to secondary is

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2 ∆푇 = 푇 − 푇 (20) 퐴 1 −푆−2⁄ 3 푡 푝 푠 = [푅 푙푛 푆]2 + [( ) ] (45) 푚̇ 푐 푟 1 퐾 2 And since secondary side is saturated, the steam temperature, 푝 푝 푠 1 푇푠, is constant. Re-arranging the formula and replacing 푆, 푑(∆푇 ) = 푑푇 (21) 퐴 푑푞 푑푞 1 푡 푝 = 푅 [푙푛 ( ) − 푙푛 ( ) ] + ( ) ∙ 푚̇ 푐 푟 푑퐴 푑퐴 2퐾 Hence the equation of heat transfer from primary side above 푝 푝 2 1 푠 becomes: 푑푞 −2⁄ 3 푑푞 −2⁄ 3 [( ) − ( ) ] (26) 푑퐴 1 푑퐴 2 푑푞 = 푚̇ 푝푐푝푑(∆푇푡) (22) And finally the equations becomes, If 푅푟 is the summation of all the resistances, 푅푟 , except the 푑푞 boiling film resistance, 푅 , from (22), and it is rewritten in ( ) −2⁄ −2⁄ 표 푞 푑퐴 2 1 푑푞 3 푑푞 3 different form. 퐴 = ( ) {푅푟 푙푛 [ 푑푞 ] + ( ) [( ) − ( ) ]} (27) ∆푇푝 ( ) 2퐾푠 푑퐴 1 푑퐴 2 푑퐴 1 1 푞 = 푈퐴(푇ℎ − 푇푐) where 푈 = (22) 푅푟+푅표 Where the subscript 1 refers to the primary outlet and 2 to the primary inlet. 푅푟푞 + 푅표푞 = 퐴(푇푝 − 푇푠) = 퐴∆푇푡 This methodology is actualized by establishing the pinch point  푅푟푑푞 + 푅표푑푞 = 푑퐴∆푇푡 temperature and dividing the portion of the evaporator region 푑푞 푑푞 into small differential elements. With the finer the elements, we  ∆푇 = 푅 + 푅 푡 푟 푑퐴 표 푑퐴 can get more accurate result. Iterations for next sections are For boiling heat transfer, using Rohsenow correlation, done and finally the summation of all the elemental areas is done. The economizer region is also evaluated using the same 1 푞 2/3 1 푞 −2/3 methodology, but using the average temperatures for the 푅표 = and ℎ표 = 퐾푠 ( ) , and 푅표 = ( ) ℎ표 퐴 퐾푠 퐴 temperature differences. 푞 푑푞 Where for infinitesimal area = 퐴 푑퐴

푑푞 1 푑푞 푑푞 −2/3 1 푑푞 1/3 b. Heat transfer area calculation for economizer region  푅표 = ( )( )( ) = ( ) ( ) 푑퐴 퐾푠 푑퐴 푑퐴 퐾푠 푑퐴 Heat transfer in the economizer on the secondary side, Using the modified Rohsenow equation for the film boiling in 푄 = 푚̇ (ℎ − ℎ ) (28) above: 푒푐표 푠 푠 푓푤 Where 푚̇ is the flow rate of steam or feed water (secondary 푑푞 1 푑푞 1/3 푠 ∆푇푡 = 푅푟 ( ) + ( ) ( ) (37) side flow rate). 푑퐴 퐾푠 푑퐴 Differentiation of equation Error! Reference source not And the heat transfer in the primary side equals to the found.) by dq: secondary side, 1⁄ 3 푄 = 푚̇ 푐 (푇 − 푇 ) (29) d∆푇푡 푑 푑푞 1 푑 푑푞 푒푐표 푝 푝 푒 푐 = 푅푟 ( ) + ( ) ( ) (23) 푑푞 푑푞 푑퐴 퐾푠 푑푞 푑퐴 Where: 푚̇ 푠 is the coolant flow rate of steam generator, and 푇푒 d∆푇 1 Where 푡 = (24) is inlet temperature of primary side economizer and 푇푐 is outlet 푑푞 푚̇ 푝퐶푝 temperature of primary side (i.e. outlet of economizer). And replacing into the equation, Hence the pinch point temperature is the temperature of

1 푑 푑푞 1 푑 푑푞 1⁄ 3 economizer inlet, = 푅 ( ) + ( ) ( ) (25) 푟 푄 푚̇ 푝푐푝 푑푞 푑퐴 퐾푠 푑푞 푑퐴 푒푐표 + 푇 = 푇 (30) 푚̇ 퐶 푐 푒푐표 푑푞 푝 Replacing ( ) with S and multiplying the denominator 푑퐴 The 푇 is used for the calculation of heat transfer area 푑퐴 푒푐표 by( ) applying (47). 푑퐴

1 푑푆 1 푑푆1⁄ 3 = 푅푟 푑퐴 + ( ) 푑퐴 (41) 푚̇ 푝푐푝 푑푞( ) 퐾푠 푑푞( ) 푑퐴 푑퐴 IV. IMPLEMENTATION OF THEORY AND 1 푑푆 1 푑푆1⁄ 3 DETERMINATION OF THE HEAT TRANSFER AREA = 푅푟 푑푞 + ( ) 푑푞 (42) 푚̇ 푝푐푝 푑퐴( ) 퐾푠 푑퐴( ) 푑퐴 푑퐴 This section describes the implementation of the theory developed in the previous section. The procedure is divided in 푑퐴 푑푆 1 푑푆1⁄ 3 = 푅푟 + ( ) (43) to number of steps as follows. 푚̇ 푝푐푝 푆 퐾푠 푆

퐴 2 푑푆 1 2 푑푆1⁄ 3 1. Primary side heat transfer = 푅푟 ∫ + ( ) ∫ (44) 푚̇ 푝푐푝 1 푆 퐾푠 1 푆 2. Heat transfer through the U-tube wall 3. Secondary side heat transfer

1490 International Journal of Engineering Research and Technology. ISSN 0974-3154, Volume 12, Number 9 (2019), pp. 1485-1493 © International Research Publication House. http://www.irphouse.com

a. Bulk boiling area IV.III Secondary Side Heat Transfer b. Area of the economizer region In order to determine heat transfer in the secondary side of the Steam Generator, it was important to calculate first bulk 4. Estimation of total heat transfer area boiling area and in the next step to determine pinch temperature IV.I Primary Side Heat Transfer For the primary side of the Steam Generator, the coolant was a. Bulk boiling area assumed to be in subcooled state, thus all parameters were This region was divided to 10 segments in order to calculate calculated as for the subcooled liquid using Chemica Logic each segment temperature and heat transfer area, and Steam Tab Companion [6]. In order to determine heat transfer subsequently total heat transfer area in the bulk boiling region. in that region, it is important to calculate Nusselt number, In this evaluation Rohsenow modified equation by experiment which provides a measure of convection heat transfer occurring was used. at the inner surface of the U-tube. The general form of the −4 Nusselt number can be expressed by the equation: 퐾푠 = 4.1{1 + 1.4 × 10 (푝푠푎푡 − 800)} (35) ∗ 푁푢 = 푓(푥 , 푅푒퐿, 푃푟) (31) This equation is taking saturation pressure 푝푠푎푡 of steam in 푝푠𝑖푎 units. However, because 퐾 is a constant, it was not Where: 푠 required to convert the units. 푥∗ = spatial variable

푅푒 = Reynolds numbers, which specifies flow 퐿 b. Economizer region heat transfer area condition In economizer region, both inside and outside is saturated, 푃푟 = Prandtl number, characterizing heat transfer hence heat flows From the primary side convective heat between moving fluid and solid body transfer on the wall, conductive heat transfer through U-tube In the Steam Generator U-tube, the flow condition is described wall then convective heat transfer on the secondary side of the as fully developed turbulent flow in the smooth tube. Thus, the wall. In this region, the both side are the force convection. The correlation used for convective heat transfer coefficient can be heat transfer is calculated as following. expressed using Dittus-Boelter correlation [7]: 푄푒푐표 = 푈퐴∆푇 = 푚̇(ℎ푒 − ℎ푐) (36) 0.8 0.3 푁푢 = 0.023 푅푒퐿 푃푟 (32) 푚̇ (ℎ −ℎ ) 푄 퐽 And 푈퐴 = 푒 푐 = 푒푐표 = 9,593,017.04 [ ] (37) Temperature which was used for necessary parameters was ∆푇푚 ∆푇푚 푠℃ assumed to be average temperature of the fluid of the primary Where the rate of Heat transfer through Economizer region as, side and is expressed by the equation: ( ) 8 퐽 푇 +푇 푄푒푐표 = 푚̇ ℎ푒 − ℎ푐 = 2.75 × 10 [ ] (38) 푇 = 퐻 퐶 (33) 푠 푎푣푔 2 - Temperature difference between inlet and outlet Where: ∆푇 −∆푇 ∆푇 = 퐻 퐶 = 28.629 [℃] (39) 푚 ∆푇퐻 푇퐻 = temperature of the hot leg (input temperature of 푙푛 ∆푇퐶 the primary side) - ∆푇퐻 depends on the coolant outlet temperature and the feed- 푇퐶 = temperature of the cold leg (output temperature water temperature of the primary side) 훥푇ℎ = 푇푐 − 푇푓푤 = 290.55 − 232.22 = 58.33 [℃] (40)

- ∆푇퐶 depends on the pinch temperature and the saturated steam IV.II Heat Transfer through the U-Tube Wall temperature, Heat transfer through the U-tube wall is a simple calculation 훥푇 = 푇 − 푇 = 295.473 − 284.2 = 11.273[℃] (41) expressing thermal resistance for conduction through a solid 퐶 푒 푠푎푡 wall of the U-tube using Fourier’s Law [3], i) Convective heat transfer on the primary side of economizer region: 푑표 푑표 푅푘 = ln ( ) (34) 2푘 푑𝑖 Since primary coolant flow in tube and in single phase, Where: Dittus-Boelter correlation is used for the primary side of 푑표 the economizer: 푅𝑖 = 푘 = thermal conductivity of the Inconel 690 ℎ𝑖푑𝑖

ℎ 푑 퐺 푑 0.8 퐶 휇 0.3 푑표 = outer diameter of the U-tube 𝑖 𝑖 = 0.023 ( 𝑖 𝑖) ( 푝 ) (79) 푘 휇 푘 푑 = inner diameter of the U-tube 𝑖

1491 International Journal of Engineering Research and Technology. ISSN 0974-3154, Volume 12, Number 9 (2019), pp. 1485-1493 © International Research Publication House. http://www.irphouse.com

2 푃 √3 휋푑2 Mass flow rate per square area: 4( 푇 − 표) 4 8 푚̇ 푝 푚̇ 푝 푘푔 퐷푒 = = 0.018293344 [푚] (89) 휋푑표/2 퐺𝑖 = = 2 = 4669.32 [ 2 ] (80) 퐴푥 푛휋푑𝑖 푚 푠 4 The tube pitch: 푃푇 = 0.0254푚 = 1𝑖푛푐ℎ From equation Error! Reference source not found. 휇 is viscosity at the wall temperature convective Heat Transfer Coefficient (primary side): 푤 0.8 0.3 Substitute the values of parameter into Error! 푘 퐺𝑖푑𝑖 푐푝휇 푊 ℎ𝑖 = 0.023 ( ) ( ) = 42303.931 [ ] Reference source not found.: 푑𝑖 휇 푘 푚℃ 푊 (81) ℎ = 5397.111 [ ] (90) 표 푚℃ So the resistance coefficient is determined by (26) as following: Thus, the heat transfer resistance coefficient is determined as following: 푑표 −5 푅𝑖 = = 2.662 × 10 (82) 퐷표 ℎ𝑖푑𝑖 푅표 = = 0.000208654 (91) 퐷𝑖ℎ표 ii) Conductive heat transfer through the tube wall. Now with the known parameter value of 푅표, we can calculate 푑표 푑표 economizer region heat transfer area. From (82), (84), (91) into 푅푤 = 푙푛 (83) 2푘푤 푑𝑖 (25) 푑표 푑표 −5 푅푤 = ln ( ) = 6.54 × 10 (84) 1 푊 2푘푤 푑𝑖 푈 = = 3325.87 [ ] (92) 푅𝑖+ 푅푤+ 푅표 푚℃ With thermal conductivity of Inconel 690: From Error! Reference source not found. and Error! 푊 푘 = 17.3 [ ] Reference source not found. the area of the economizer region 푤 푚℃ 푈퐴 퐴 = = 1132.369 [푚2] (93) iii) Convective heat transfer on the secondary side of 푈 1 economizer: 푅표 = ℎ표 Using McAdams correlation for secondary side (force IV. DISCUSSIONS AND CONCLUSIONS convection region) as following: In a SG, thermal performance is a key critical metric and is the 0.55 1⁄ 3 0.14 main concept behind the conceptual design of the APR1400. ℎ표퐷푒 퐺푠퐷푒 푐푝휇 휇 푁푢 = = 0.36 ( ) ( ) ( ) (85) For the present project, thermal performance used the basic 푘 휇 푘 휇푤 assumptions in the heat balance equations for convection Where 퐷푒 is hydraulic diameter of wetted area. between fluid and wall in the primary side, between wall and 2 푃 √3 휋푑2 fluid in the secondary side, and conduction in the u-tube wall. 4( 푇 − 표) 4 8 The process started by establishing the heat balance in both For triangular pattern of tubes: 퐷푒 = 휋푑표/2 primary and secondary side. Calculated heat capacity on the

휋푑2 primary side from the Error! Reference source not found. is 4(푃2− 표) 푇 4 1,977.31푀푊푡. This value is less by 23푀푊푡 (approximately For square pattern of tubes: 퐷푒 = 휋푑표 1.15%) than the APR1400 rated capacity of 2000푀푊푡 per Shell side mass velocity: steam generator. On the other hand, using equation Error! Reference source not found., the secondary side heat rate was 푚̇ 푘푔 퐺 = 푠 = 500.626 [ ] (86) 2,007.86푀푊푡, which is 0.4% higher than the rated value. 푠 퐴 푚2푠 푠 These margins of error could be attributed to the human errors 푘푔 Mass flow rate of feedwater: 푚 = 1142.800852 [ ] like the use of average temperatures and discrepancies in the 푠 steam table values. Cross flow area at the center of the shell: From the calculations based on equation Error! Reference 휋퐷2 푛휋푑2 퐴 = 푆퐺 − 표 = 2.283 [푚2] (87) source not found., the economizer portion of the APR1400 푠 8 4 accounts for 292푀푊푡 which is about 14.6% of the total heat

Inside Diameter of Shroud: 퐷푆퐺 = 3.615푚 in the steam generator using an initial estimation of 10,000 tubes. The evaporator region of the steam generator accounts For the squared tube arrangement, thus the for the 1,715푀푊푡, which is 85.6% of the total heat transfer in hydraulic diameter is: the secondary side. To evaluate the heat transfer area, the 휋푑2 assumptions made in Error! Reference source not found.- 4(푃2− 표) 푇 4 퐷푒 = = 0.02407 [푚] (88) Error! Reference source not found. were used. In the 휋푑표 evaluation of the differential elements of the evaporator region, For the triangular tube arrangement, thus the it was noted that the heat flux was the highest at the entry region hydraulic diameter is: into the SG, hence a lower heat transfer area of 776.5m2 which increases gradually to 3,679.25푚2 at the economizer exit (pinch point) for 12,000 tubes. The heat transfer area

1492 International Journal of Engineering Research and Technology. ISSN 0974-3154, Volume 12, Number 9 (2019), pp. 1485-1493 © International Research Publication House. http://www.irphouse.com calculations for the economizer used the different way with for the evaporated region. The accuracy of the value of the heat transfer area was affected by the variable temperatures in both the primary and secondary side, hence fluid properties, which complicate the calculations. To simplify this, an assumption of the average temperatures for both the secondary and primary side was used in the estimation of the fluid properties. The economizer surface area was then calculated as 1,238.1푚2 representing a 7.1% of the total surface area for tube bundle.

Acknowledgements This work was supported by the Research fund of the KEPCO International Nuclear Graduate School (KINGS), Republic of Korea. The authors would like to express appreciation toward KINGS.

REFERENCES [1] W. Rohsenow, J. Hartnett, and Y. Cho, Heat transfer. 1998. [2] Q. Zhao, D. Deng, S. Nie, W. Chen, J. Wang, and D. Zhang, “The Research of Heat Transfer Area for 55/19 Steam Generator,” J. Power Energy Eng., vol. 3, no. 3, pp. 417–422, 2015. [3] T. Bergman, A. Lavine, F. Incropera, and D. Dewitt, Fundamentals of Heat and Mass Transfer. 2011. [4] “International Association for the Properties of Water and Steam.” [Online]. Available: http://www.iapws.org/index.html. [Accessed: 10-Mar- 2017]. [5] Wolfram Alpha LLC, “Wolfram|Alpha: Computational Knowledge Engine.” [Online]. Available: http://www.wolframalpha.com/. [Accessed: 10-Mar- 2017]. [6] SteamTabTM Companion, “Download SteamTab Companion.” [Online]. Available: http://www.chemicalogic.com/Pages/DownloadSteamT abCompanion.aspx. [7] K. A. A. Abdallah and I. Namgung, “Sensitivity study of APR-1400 steam generator primary head stay cylinder and tube sheet thickness,” Ann. Nucl. Energy, vol. 92, pp. 8–15, 2016. [8] C. Report, Crane Co., “Flow Of Fluids Through Valve, Fittings and Pipe.” p133. 1982. [9] Korea Hydro and Nuclear Power Co., APR1400 Design Control Document Tier 2, Chapters 3, 2014, [10] Crane Co., “Flow Of Fluids Through Valve, Fittings and Pipe.” p. 133, 1982. [11] E. Farber and Scorah R, Heat transfer to water boiling under pressure. Columbia: Univ. of Missouri, 1948.

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