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Precept 2: Mass/energy balances of a Steam Cycle Mechanical Engineering - Energy Systems Authors: •Group xx: xxxxx xxxxx (xxxxxxxx), Fabio Santoro (xxxxxxxx) Lecturer:Prof. Stefano Consonni Teaching Assistants:Riccardo Cremona, Hamidreza Heydari and Nima Razmjoo Academic year:2025-2026Contents Description2 1 Thermodynamic conditions4 2 Mass Flow Rates10 3 Relevant performances10 4 RegeneratorrbT−˙ Qdiagram11 5 Results13 References13 1 Energy Systems- Report Precept 2 - Group xx Description This precept covers the design of a Rankine steam cycle plant, where the input thermal power is provided by a firing boiler using the combustion of natural gas, as represented in the scheme of Figure 1on page3. All the relevant data are collected in Table1on page4. Let us briefly describe the combustion process: the ambient air(a)is preheated in the Ljungstrom heat exchanger by the hot flue gases to state(b). At this point, the preheated air(b)is mixed with the fuel(h)in the combustion chamber(c), where the combustion reaction takes place. The resulting hot flue gases evaporate the water to state(d)and then heat the steam in the superheater and reheater to state(e). Finally, the flue gases preheat the feedwater to near-saturation conditions at state(f)and warm the inlet air in the Ljungstrom heat exchanger before exiting the boiler at state (g)to go to the stack. The plant is a subcritical Rankine steam plant, where electrical power is generated by the rotation of three turbines operating at different pressure levels (high, intermediate, and low). The three turbines are mounted on the same shaft, which is connected to the alternator. The steam generated in the boiler by the economizer(eco), the evaporator(eva), and the superheater (sh)enters the first turbine(hp)under superheated steam conditions. When the fluid leaves the first turbine, it is split into two streams: the first returns to the boiler through the reheater(rh)to reach the same inlet temperature as the high-pressure turbine and then enters the second turbine(ip), while the other part of the flow(2extr)is used to heat the feedwater through the regenerator(rb), passing from superheated steam(2extr)to subcooled water(2dr)conditions. The regenerator(rb)is a non-contact heat exchanger (assumed single-pass and counterflow). Since the two streams are at different pressures, they cannot be mixed. The pressure of the subcooled water is reduced by an isenthalpic expansion valve to allow mixing with another fluid stream(2drvalve). The superheated steam exiting the(ip)turbine is again split into two streams: part of it(4)enters the last low-pressure turbine(lp), while the other part(4extr)enters the deaerator (regenerator) (ra). Most steam power plants are characterized by the last stages of the(lp)turbine and the condenser operating at a pressure below atmospheric, which makes them susceptible to air infiltration, particularly through sealing defects in the condenser. The low-pressure condensate mixed with non-condensable gases (NCGs) enters an extraction pump to increase its pressure to match that of the deaerator(ra). In a steam plant of this type, the presence of the deaerator is essential to ensure the evacuation of the NCGs that leak into the condenser. The deaerator(ra)is essentially a direct-contact heat exchanger containing pressurized saturated liquid and is fed by three streams: the one from the extraction pump(7), the one expanded in the isenthalpic valve(2dr), and the steam extracted from the outlet of the(ip)turbine(4extr). We assume negligible steam evacuation from the deaerator; therefore, it is not necessary to introduce a make-up water stream into the plant. The saturated liquid(8)enters a feedwater pump, which increases its pressure(9)before entering the regenerator(rb), where it is heated(10)prior to entering the boiler through the economizer(eco)and restarting the cycle. All mechanical components are considered non-ideal machines, and some pressure drops occur in the piping system. 2 Energy Systems- Report Precept 2 - Group xx Figure 1:Process Flow Diagram 3 Energy Systems- Report Precept 2 - Group xx Table 1:Precept Data SymbolValueu.o.m.Description p E V A140 bar Evaporation pressureT in,H P550◦ C HP turbine inlet temperature (superheat outlet)˙ Qin1000 MW thThermal power input to the steam cyclep out,H P42 bar HP turbine discharge pressureT in,I P550◦ C IP turbine inlet temperature (reheater outlet)p out,I P6 bar IP turbine discharge pressurep cond0.07 bar Condensation pressurep 2,extr42 bar Pressure of1st regenerative bleedingp 4,extr6 bar Pressure of2nd regenerative bleeding (deaerator)η H P0.87 Isoentropic efficiency of HP turbine sectionη I P0.915 Isoentropic efficiency of IP turbine sectionη LP0.86 Isoentropic efficiency of LP turbine sectionη or,T0.98 Turbine mechanical efficiencyη el,T0.985 Alternator electrical efficiencyη hyd,P10.84 Hydraulic efficicncy of condenser extraction pumpη hyd,P20.84 Hydraulic efficicncy of feedwater pumpη me,P1,P20.9 Mechanica/electrical efficiency of pumps(∆p/p) boiler0.2 Economizer pressure drop (water/steam side)(∆p/p) S H/RH0.08 SH and RH pressure drop (steam side)T in,E C O255◦ C Economizer inlet temperature∆T S C5◦ C Subcooling at the inlet of the evaporator∆Tat the cold end of the surface feedwater heater ∆TRB,c.e.5◦ C (Th,out−T c,in)η boiler0.949 Steam generator thermal efficiencyW aux0.032·W elS T ,grossOther auxiliary electric power consumption1.Thermodynamic conditions In this section, we are required to determine the thermodynamic properties at each point of the steam cycle. All results are collected in Ta- ble2(the red values are given in the text) at page13; here, we will show only the calculation steps. The remaining thermodynamic properties are evaluated using an Excel add-in developed by a research team [1].•Point 1 In point(1)(superheated steam) we need to consider the pressure drop in the(sh), the pres- sure is: p1=p E V A· 1− ∆pp  S H The temperature isT1=T in,H Pand the other propertiesh1,s 1andv 1are evaluated with Excel add-in.•Point 2is This is the results of the isoentropic expan- sion of an ideal(hp)turbine. The pressure is p2is=p out,H P. The enthalpyh 2isis obtained withp2isand the entropys 2is=s 1. The other thermodynamic propertiesT2isandv 2isare ob- tain with Excel add-in. The point(2is)is at superheated steam conditions. •Point 2 The real expansion depends on the isoentropic efficiencyηH P. h2=h 1−η H P(h 1−h 2is) The pressure is the turbine outputp2=p out,H P. The other propertiesT2,s 2andv 2are evaluated with Excel add-in. 4 Energy Systems- Report Precept 2 - Group xx •Point 2extr The point(2extr)(superheated steam) has the same thermodynamic properties of the point (2)but, it has different mass flow rate. •Point 2extrSV The point(2extrSV)is the saturated vapor conditions at pressurep2extrS V=p 2. The other thermodynamic propertiesT2extrS V,h 2extrS V, s2extrS Vandv 2extrS Vare computed with Excel add-in.•Point 2extrSL The point(2extrSL)is the saturated liquid conditions at pressurep2extrS L=p 2. The other thermodynamic propertiesT2extrS L,h 2extrS L, s2extrS Landv 2extrS Lare computed with Excel add-in.•Point 3 The pressure drops during(rh): p3=p 2· 1− ∆pp  RH The stream is reheated toT3=T in,H P(super- heated steam) and the other propertiesh3,s 3 andv3are evaluated with Excel add-in. •Point 4is This is the results of the isoentropic expan- sion of an ideal(ip)turbine. The pressure is p4is=p out,I P. The enthalpyh 4isis obtained withp4isand the entropys 4is=s 3. The other thermodynamic propertiesT4isandv 4isare ob- tain with Excel add-in. The point(4is)is at superheated steam conditions.•Point 4 The real expansion depends on the isoentropic efficiencyηI P. h4=h 3−η I P(h 3−h 4is) The pressure is the turbine outputp4=p out,I P. The other propertiesT4,s 4andv 4are evaluated with Excel add-in.•Point 4extr The point(4extr)(superheated steam) has the same thermodynamic properties of the point (4)but, it has different mass flow rate.•Point 5is This is the results of the isoentropic expan- sion of an ideal(lp)turbine. The pressure is p5is=p cond. The enthalpyh 5isis obtained with p5isand the entropys 5is=s 4. The point(5 is) falls in the two phase region, so the tempera- tureT5,isis only function of the pressurep 5is.The other thermodynamic propertiesx 5isand v5isare obtain with Excel add-in. •Point 5 The real expansion depends on the isoentropic efficiencyηLP. h5=h 4−η LP(h 4−h 5is) Also the point(5)belongs in the two-phase re- gion becausehS L(p 5)< h 5< h S V(p 5). Again, the pressure is again the condensation pressure p5=p cond. The The pressure is the turbine outputp4=p out,I P. The other thermodynamic propertiesx5,s 5andv 5are obtain with Excel add-in.•Point 6 At point(6)the flow is full condensed, so we are in saturated liquid conditionsx6= 0at p6=p cond. All the other thermodynamic prop- ertiesT6,s 6andv 6are obtain with Excel add-in. •Point 7 The flow exiting the condenser(6)enters the extraction pump which compresses the fluid to sub-cooled conditions(7). The pump is not an ideal device but, it is affected by the hydraulic efficiencyηhyd,P1. The final compressionp 7is given. The enthalpy is computed as: h7=h 6+v 6(p 7−p 6)η hyd,P1 The other thermodynamic propertiesT7,s 7and v7are obtain with Excel add-in. •Point 8 The stream(8)exiting the deareator(ra)is imposed at saturated conditions at the same out- let pressurep8=p 7of the extraction pump. So given the pressure and the vapor titlex8= 0, we can evaluate the other thermodynamic proper- tiesT8,h 8,s 8andv 8with the Excel add-in. The stream(8)is the result of the mixing of three streams:(4extr),(7)and(2drvalve). The last will be evaluated later.•Point 9 After(8), the fluid enters the feedwater pump that compresses the stream at sub-cooled con- ditions(9)at a pressurep9which technically should have been the boiler pressure but, due to pressure drop, it is at higher pressure: p9=p E V A1−  ∆pp  boiler 5 Energy Systems- Report Precept 2 - Group xx The enthalpyh 9is evaluated as as the one at (7): h9=h 8+v 8(p 9−p 8)η hyd,P2 The other thermodynamic propertiesT9,s 9and v9are computed with the Excel add-in. •Point 10 At point(10), the fluid is at sub-cooled con- ditions and corresponds with the exiting of the regenerator(rb). The temperatureT10= Tin,E C Oof at which the fluid enters the econ- omizer(eco)is given by the text. The fluid is at the same pressure of(9)p10=p 9. The other thermodynamic propertiesT10,s 10andv 10are computed with the Excel add-in. •Point 2dr We neglect any pressure drop in the regenerator (rb), so the pressure of the hot fluid remains constant:p2dr=p 2extrbut, the temperature is constraint by the conditions(9)in order to maintain a minimum value of temperature dif- ference: T2dr=T 9+ ∆T RB,c.e. The other thermodynamic propertiesh2dr,s 2dr andv2drare computed with the Excel add-in. The value of the enthalpyh2dris lower that the saturated liquid corresponding at that pressure, so(2dr)is at sub-cooled liquid conditions. Be- tween(2extr)and(2dr), the fluid changes phase, so it is necessary to evaluate the points at saturated liquid and vapor conditions •Point 2extrsv The stream(2extrsv)enters the two phase region without any pressure drop in the regen- erator(rb)p2extrS V=p 2extrand it is at satu- rated vapor conditionx2extrS V= 1. The other thermodynamic propertiesT2extrS V,h 2extrS V, s2extrS Vandv 2extrS Vare evaluated with the Ex- cel add-in. •Point 2extrsl The stream(2extrsl)exits the two phase re- gion without any pressure drop in the regenera- tor(rb)p2extrS L=p 2extrS Vand it is at satu- rated liquid conditionx2extrS L= 0. The other thermodynamic propertiesT2extrS L,h 2extrS L, s2extrS Landv 2extrS Lare evaluated with the Ex- cel add-in.•Point 2drvalve The stream(2dr)enters an isoenthalpic valve (2drvalve)h2drvalve=h 2drto reduce the pressure to the deareator(ra)onep2drvalve= pdea=p 7to allow the streams mixing. The other thermodynamic propertiesT2drvalve,h 2drvalve, s2drvalve,x 2drvalveandv 2drvalveare evaluated with the Excel add-in. We notice that the point (2drvalve)belongs in the two-phase region. •Point 11 At point(11)the fluid exits the economizer (eco)at sub-cooled conditions at a tempera- ture inferior to the evaporation: T11=T E V A−∆T S C In the economizer, the pressure drops to the evaporation onep11=p E V A. The other ther- modynamic propertiesh11,s 11andv 11are com- puted with the Excel add-in.•Point 12 The exit of the evaporator(12)is at saturated vapor conditionsx12= 1at evaporation pressure p12=p E V A. The other thermodynamic condi- tionsT12,h 12,s 12andv 12are computed with the Excel add-in.•Point 11sl In the evaporator(eva), the fluid(11)goes from sub-cooled conditions to complete evapora- tion(12)so, in the device the fluid reaches the saturated liquidx11S L= 0conditions(11sl). The pressure isp11S L=p E V A. The other ther- modynamic conditionsT11sl,h 11sl,s 11slandv 11sl are computed with the Excel add-in. In Figure2, we present the steam cycle on the T−sdiagram, while Figure3highlights specific details. In the left-hand detail, we observe:•the feedwater compression from(8)to (9), •the heating in the first section of the re-generator(rb), from(9a)to(9b), •the isoenthalpic expansion from(2dr)to(2drvalve). In the right-hand detail, we illustrate the superheating process, the high-pressure tur- bine expansion from(1)→(2), the reheating phase from(2)→(3), and the onset of the intermediate-pressure turbine expansion. It is clearly visible that both the superheating and reheating stages are accompanied by a pressure drop. 6 Energy Systems- Report Precept 2 - Group xx Figure 2:Steam CycleT−sdiagram Figure 3:Steam CycleT−sdiagram details 7`´`ˆ`˜`¨`˝`˚`ˇ`˘`¯ `˙`¸`˛`‚`‹`›`“ `”`„`«`»`–`—`«`‌` `ı`ȷ `˚`´ `ˆ`´`´ `ˆ`˚`´ `˜`´`´ `˜`˚`´ `¨`´`´ `¨`˚`´ `˝`´`´ `˝`˚`´ `˚`´`´ `˚`˚`´ `ff`fi`fl `›`fi`‚`ffi`˛`ffl`‚`fi `ff `„`´`␣`ȷ `!`´ `ˆ`˜`¨`˝ `ˆ `˜ `!`´ `ˆ`˜`¨`˚ `ˆ `˜`´ `˜ `ˆ `˜ `ˆ `˜ `ˆ `˜ `ˆ `˜ `ˆ `˜ `ˆ`´ `!`´ `˜`ˇ`˘ `!`´ `˜`¯`˝ `!`´ `ˇ`˘`˙`¸ `!`´ `ˆ`˜`¨`˚ `!`´ `ˆ`˜`¨`˝ `!`´ `˛`‚`‹`˛`›`˜ `ˆ `˜ `ˆ `!`“ `”`˜ `ˆ`˜ `ˆ `!`“ `”`ˆ `ˆ `ˇ`˘`¯`"`"`ffi`"`#`ˆ`´`ˆ`ˆ `ˆ`ˆ`”`$ `˜`%`” `˝`%`” `˚`%`”`ˆ`˜`¨`˝`˚ `˜`fi`&`˛`‚`’`( `˜`fi`&`˛`‚`’`) `˝`fi`&`˛`‚`’`( `˜`fi`&`˛`‚ `˝`fi`&`˛`‚ `!`´ `„`« `!`´ `ˆ`« `!`“ `«`” `!`“ `»`” `!`“ `–`” `˜`*`‚ `˜`*`‚`+`ffi`$`+`fi `ˆ`˜ `´`ˆ`˜`´`ˆ`¨`˝`˝`ˆ`´ `˚`ˇ`˘`¯`˙`¸`˛ `‚`‹`›`“`”`„`›`«`»`–`— `´`‌` `´`‌`‌ `´`ı` `´`ı`‌ `´`ȷ` `´`ȷ`‌ `´`˜` `ff`fi`fl `¸`fi`¯`ffi`˘`ffl`¯`fi`ff `‹`´`␣`— `!`´ `ˆ`˝ `´ `˜`¨`¨`ffi`˝`"`¯ `˝`"`¯`#`ffi`$`#`fi `ˆ `˜ `!`˜ `¨`˝`˚`¨`ˇ `´`ˆ`˜`¨`¨`ˆ`˜ `˝`˚`ˇ`˘`¯`˙`¸ `˛`‚`‹`›`“`”`‹`„`«`»`– `—`‌`‌ `—`˜`‌ `˜`‌`‌ `˜`˜`‌ ``ı`ȷ `˙`ı`˘`ff`ˇ`fi`˘`ı` `‚`´`fl`– `ˆ `´ `ˆ `´ `ˆ `´ `ˆ `´ `ffi`ffl`˛`␣`ffi`!`ffi`ı`"`ˇ`˘ `#`´ `ˆ`˜ `#`´ `¨`˜ `#`˝ `˜`˚ `#`˝ `ˇ`˚ Energy Systems- Report Precept 2 - Group xx Figure 4:Steam Cycleh−s(Mollier) diagram 8`´`ˆ`˜`¨`˝`˚`ˇ`˘`¯ `˙`¸`˛`‚`‹`›`“ `”`„`«`»`–`—`«`‌``ı`ȷ `´ `˚`´`´ `ˆ`´`´`´ `ˆ`˚`´`´ `˜`´`´`´ `˜`˚`´`´ `¨`´`´`´ `¨`˚`´`´ `˙`¸`˛`ff`fi`fl`›`“ `ff `„`«`»`–`«`‌`ȷ `´ `ˆ `ffi`´ `ˆ`˜`¨`˝ `´ `ˆ `ffi`´ `ˆ`˜`¨`˚ `ˆ `´ `ffi`´ `ˇ`˘`¯`˙ `ffi`´ `˜`¸`˝ `ˆ `´ `ffi`´ `˛`‚ `˜`˜`‹`›`“`”`˛`¸ `˝`‹`›`“`”`˛`¸ `´ `ffi`´ `ˆ`‚ `´`´ `ffi`„ `«`» `´ `ffi`„ `–`» `ffi`„ `‚`» `´ `´ `ffi`´ `˜`ˇ`˘ `´ `ffi`´ `ˆ`˜`¨`˚ `´ `ffi`´ `ˆ`˜`¨`˝ `´ `ffi`´ `—`‌``—`‹ `˜`‹`›`“`”`˛`– `˜`ı`” `˜`ı`”`—`‌``—`‹ `´ `ffi`„ `»`˜ `´ `ffi`„ `»`ˆ`ˇ`˘`¯`ffl`ffl`fi`ffl`␣`ˆ`´`ˆ`ˆ `ˆ`ˆ`”`fl`ˆ`˜`ˆ`¨ `˝`˚`˚`!`” `˝`!`” `˜`!`” Energy Systems- Report Precept 2 - Group xx Figure 5:Steam Cycleh−s(Mollier) diagram detail 9`´`ˆ`˜ `´`ˆ`˜`¨ `´`ˆ`˝ `´`ˆ`˝`¨ `´`ˆ`˚ `´`ˆ`˚`¨`ˇ`ˇ`ˆ`˘`¨ `ˇ`ˆ`´ `ˇ`ˆ`´`¨ `ˇ`ˆ`ˇ `¯`˙`¸`˛`‚`‹`› `“`”`„`«`»`–`„`—`‌``ı `ȷ`¨`˘ `ȷ`ȷ`˘ `ȷ`˜`˘ `ȷ`˝`˘ `ȷ`˚`˘ `˜`˘`˘ `˜`´`˘ `˜`ˇ`˘ `˜`ff`˘ `¯`˙`¸`fi`fl`ffi`‹`› `fi `”`„`«`»`„`—`ı `´ `ffl`´ `ˆ`˜`¨`ˆ`˝ `ˇ`˚`ˇ `ˇ`˚`ˇ`ˆ`˜`¨`ˆ`˝ `ˆ `´ `ffl`˘ `¯`ˇ`˝`˚ `´`ˆ`˜`´`ˆ`´`´`ˆ`¨`´`ˆ`˝`´`ˆ`˚`¨`¨`ˆ`ˇ`¨`ˆ`˘`¨`ˆ`¯`¨`ˆ`˙ `¸`˛`‚`‹`›`“`” `„`«`»`–`—`‌`»``ı`ȷ`ff `˘`˝`fi`fi `˘`˚`fi`fi `¯`fi`fi`fi `¯`ˇ`fi`fi `¯`˘`fi`fi `¯`¯`fi`fi `¯`˙`fi`fi `¯`˜`fi`fi `¯`´`fi`fi `¸`˛`‚`fl`ffi`ffl`“`” `fl `«`»`–`—`»``ff `˘ `␣`´ `ˆ`˜ `´ `´ `´ `´ `␣`¨ `˝`˚ `␣`¨ `˜`˚ `´ `´`ˇ`¯ `˙ `˙`!`"`‚`‹ `˘`!`"`‚`‹ `˙`#`„ `˘`#`„ Energy Systems- Report Precept 2 - Group xx 2.Mass Flow Rates In order to compute all the mass flow rates we need to combine mass conservation and energy conser- vation equations. First let’s write all the mass conservations equations: ˙m1= ˙m 8= ˙m 9= ˙m 10= ˙m 11= ˙m 11S L= ˙m 12˙m 2= ˙m 1−˙m 2extr ˙m2dr= ˙m 2dr,valve= ˙m 2extr= ˙m 2extr,S L= ˙m 2extr,S V= ˙m 2dr,valve˙m 3= ˙m 2 ˙m4= ˙m 3−˙m 4extr˙m 4= ˙m 5= ˙m 6= ˙m 7 ˙m8= ˙m 4extr+ ˙m 7+ ˙m 2dr,valve We can write the energy conservation equation of the control volume that comprehend the boiler and the(hp)turbine and the energy conservation equation of the regenerator(rb). If we combine them in a system we can extract the values of the mass flow rates:    ˙ Qin= ˙m 1(h 1−h 10) + ( ˙m 1−˙m 2extr)(h 3−h 2) ˙m2extr(h 2extr−h 2dr) = ˙m 1(h 10−h 9) ↓                 ˙m 1=˙ Qin(h 1−h 10) + 1−h 10−h 9h 2extr−h 2dr (h3−h 2) ˙m2extr= ˙m 1·h 10−h 9h 2extr−h 2dr We these results we know:˙m8,˙m 9,˙m 10,˙m 11,˙m 11S L,˙m 12,˙m 2,˙m 2extr,S L,˙m 2extr,S V,˙m 2dr,˙m 2dr,valve and˙m3. Now, we must write the energy balance of the deareator(ra)and combined with the energy conser- vation equations we obtain: ˙m8h 8= ˙m 4extrh 4extr+ ˙m 2dr,valveh 2dr,valve+ ˙m 7h 7 ↓ ˙m4extr=( ˙m 2−˙m 2extr)h 7+ ˙m 2dr,valveh 2dr,valve−˙m 8h 8h 7−h 4extr With this, all the remaining mass flow rates are found. All the numerical results are collected in the Table2at page13. 3.Relevant performances In this section we are required to compute the relevant efficiencies of the steam plant. First we can compute and verify the energy balance of the steam cycle so, we compute the modulus of the blade output power of the turbines (thermodynamic power): ˙ WT ,blade=HP turbine z}|{ ˙m1(h 1−h 2) +IP turbine z}|{ ˙m2(h 3−h 4) +LP turbine z}|{ ˙m4(h 4−h 5)≈440.39MWth The blade power modulus of the condensate extraction pump:˙ WP1,blade= ˙m 4(h 7−h 6)≈0.182MWth The blade power modulus of the feedwater pump:˙ WP2,blade= ˙m 8(h 9−h 8)≈8.239MWth 10 Energy Systems- Report Precept 2 - Group xx The thermal power that enters the cycle through the boiler is equivalent to the data given ˙ Qin: ˙ Qth,in=˙ QE C O z}|{ ˙m1(  h11−h 10) +˙ QE V A z}|{ ˙m1(  h12−  h11) +˙ QS H z}|{ ˙m1(h 1−  h12) +˙ QRH z}|{ ˙m2(h 3−h 2) = ˙m1(h 1−h 10) + ˙m 2(h 3−h 2) = 1000MWth The modulus of the thermal power discharged by the condenser: ˙ Qcond= ˙m 4(h 5−h 6)≈568.04MWth over than 50% of the thermal power entering the steam plant is discharged to the environment. We can verify the overall energy balance (we use the natural sign convention, positive entering): ˙ Qth,in−˙ WT ,blade−˙ Qcond+˙ WP1,blade+˙ WP2,blade= 0 The gross electrical power generated by the turbines taking into account mechanical and electrical efficiencies is:˙ WT ,el,gross=˙ W·ηor,T·η el,T≈425.10MWe The net electrical power considering the inefficiencies of the hydraulic pumps: ˙ Wel,cycle,net=˙ WT ,el,gross−˙ WP1,bladeη me,P1−˙ WP2,bladeη me,P2≈415.75MWe The thermodynamic efficiency of the steam cycle: ηT DN,sc=˙ Wth,net˙ Qth,in= ˙ WT ,blade−˙ WP1,blade−˙ WP2,blade˙ Qth,in≈43.196% The net electric power considering the auxiliary power consumptions:˙ Wel,plant,net=˙ Wel,cycle,net−˙ Waux=˙ Wel,cycle,net−0.032·˙ WT ,el,gross≈402.14MWe The net electric power of the plant is approximately 40.21% of the input thermal power. The total thermal power generated by the fuel can be evaluated from the definition of boiler efficiency: ˙ Qf uel= ˙m f uel·LH V=˙ Qinη boiler≈1053.74MWth The net electric efficiency based on the fuel consumption is: ηel,plant,net,f uel=˙ Wel,plant,net˙ Qf uel≈38.16% 4.RegeneratorrbT−˙ Qdiagram In this section we are required to draw theT−˙ Qof the flows inside the regenerator(rb). We consider it a non-contact counter flow single pass shell-tube heat exchanger in which, the hot fluid that comes from part of the outlet stream(2extr)of the(hp)turbine decreases its temperature till it reaches the sub-cooled conditions(2dr)before going in the deareator. In order to guarantee the heat transfer from the hot source to the cold stream, it is necessary to verify there would always be a temperature difference so, it is better if we compute the thermodynamic properties of the two intermediate points of the stream from(9)to(10)inside the regenerator. We define these points(9a)and(9b)in the order of the cold stream direction. We can compute the enthalpies with energy balance: h9a=h 9+˙m 2,extr˙m 9(h 2,extr−h 2,extrS V)h 9b=h 10+˙m 2,extr˙m 9(h 2,extrS L−h 2,dr) 11 Energy Systems- Report Precept 2 - Group xx The pressure at these points does not change because there is no pressure drop in the heat exchanger piping:p9a=p 9b=p 9. The other thermodynamic properties for both points were computed using the Excel add-in. All numerical results are reported in Table2on page13. In Figure6, we show the T−˙ Qdiagram. The diagram is plotted using the actual values of temperatures and enthalpies, not only at the extreme points but also at the intermediate ones. This allows a clearer appreciation of the fact that the lines are not straight due to the variation of the specific heat capacity. In particular, between points(2extr) and(2extrsv)of the hot stream, where the fluid is superheated steam, the slope changes noticeably (in the diagram, we plotted a dashed straight line in region(rb iii)to show the difference). Even if it is difficult to see clearly, the slope of the hot steam in section(rb i)is almost half that of section(rb iii). This is due to the fact that the specific heat capacity of liquid water is almost double that of the vapor. The feedwater line is nearly straight because the variation of heat capacity in liquid water is very small. Its slope is lower compared to the slope of the hot stream in(rb i)because the mass flow rate is almost six times higher, which increases the effective heat capacity of the fluid. The overall exchanged power in the regenerator is approximately 155.27 MW of which≈68.92%is transferred during evaporation(rb ii), while≈15.04%and≈16.04%respectively for(rb i)and (rb iii).Figure 6:Regenerator(rb)T− ˙ Qdiagram 12`´`ˆ`´`˜`´`¨`´`˝`´`˚`´`´ `˚`ˆ`´ `˚`˜`´ `˚`¨`´ `ˇ`˘`¯`˙`¸`˛`¯`‚`‹`›`˘`˛`“`”`„`˘`˛ `«`´ `»`– `— `‌ `˚``´ `ˆ`´`´ `ˆ``´ `ı`´`´ `ı``´ `˜`´`´ `¸`˘`ȷ `ff`˘`˛`¯`˙`fi`˛`˘`¸ `»`´`fl`‌ `´`´`´`´`ˆ`ˆ`˜`ˆ`¨`˝`˚`˜`ˇ`˘ `ˆ`¨`˝`˚`˜`ˇ`¯`ˆ`¨`˝`˚`˜`ffi`ffi`˙`ffi`¸ `˚`´ `´ `ˆ `˜ `´ `ˆ `˜`˜ `´ `ˆ `˜`˜`˜ `ffl`␣`˘`˘`!`˘`!`‹`˙`˘`¯`ȷ `"`˘`˘`!`„`¯`˙`˘`˛ Energy Systems- Report Precept 2 - Group xx 5.Results Table 2:Thermodynamics properties and mass flows resultsip iT ih is iv ix V ,istatusm ibar°CkJ/kgkJ/(kgK)m 3 /kgkg/s 1 128.85503472.63 6.614 0.0271 - steam 372.09 2 is42 364.69 3125.32 6.614 0.0650 - steam 309.08242383.16 3170.47 6.684 0.0675 - steam 309.08 2 extr42383.16 3170.47 6.684 0.0675 - steam 63.01 2 extr,S V42 253.27 2799.85 6.049 0.0473 1 SV 63.012 extr,S L42 253.27 1101.63 2.823 0.0013 0 SL 63.012 dr42166.65 706.51 2.005 0.0011 - sub-cool 63.01 2 dr,valve6 158.83 706.51 2.014 0.0065 0.0173 L-V 63.013 38.645503561.50 7.252 0.0960 - steam 309.08 4 is6 267.51 2994.36 7.252 0.4082 - steam 255.4946290.63 3042.56 7.340 0.4269 - steam 255.49 4 extr6290.63 3042.56 7.340 0.4269 - steam 53.59 4 extr,S V6 158.83 2756.14 6.759 0.3156 1 SV 53.594 extr,S L6 158.83 670.50 1.931 0.0011 0 SL 53.595 is0.07 39.00 2279.94 7.340 18.0383 0.8788 L-V 255.4950.0739.00 2386.71 7.682 18.9482 0.9232 L-V 255.49 60.0739.00 163.37 0.559 0.0010 0 SL 255.49 7 is6 39.03 163.95 0.559 0.0010 - sub-cool 255.497 6 39.06 164.08 0.560 0.0010 - sub-cool 255.49 8 6 158.83 670.50 1.931 0.0011 0 SL 372.09 9 175 161.65 692.64 1.940 0.0011 - sub-cool 372.09 9 a175 177.19 759.56 2.091 0.0011 - sub-cool 372.099 b175 241.62 1047.16 2.687 0.0012 - sub-cool 372.0910 1752551109.92 2.807 0.0012 - sub-cool 372.09 11 140 331.669 1533.59 3.562 0.0016 - sub-cool 372.09 11 S L140 336.669 1570.88 3.623 0.0016 0 SL 372.0912 140 336.669 2638.09 5.373 0.0115 1 SV 372.09 References [1]Keith Woodbury, Bob Taylor, Joseph Chappell, Kenny Mahan, Jesse Huguet, and Troy Dent. Ther-modynamics.https://excelinme.net/thermodynamics/, 2025. Excel in Mechanical Engineering Research Team. 13