Question

Solve the following problem using your choice of a software program The steps to solve the problem should be clearly outlined (can be handwritten or typed) The software program code should be printed out along with the results of the solution. DUE DATE - 8/3/2020 A steam power plant operates on the Rankine cycle with reheat, using the specified conditions below. Using stream nunbering shown in Figure 1, for each of the options given, determine: (a) The work output of each turbine per kg of steam; (6) The work input of the feedwater pump per kg of circulated water; (c) The flowrate of steam required; (d) The heat input required in the boiler/superheater and reheater; The thermal efficiency Option T, (C) P, (MPa) P. (MPa) Ts ("C) P. (MP) turbines pump 5 ii jii iv 450 400 400 500 600 0.4 0.6 0.5 0.8 1.2 4 4 + 400 450 500 500 600 0.01 0.01 0.01 0.01 0.01 0.8 0.8 0.9 0.85 0.85 Plant Capacity 80 MW 80 MW 75 MW 80 MW 80 MW 0.78 0.75 0.95 0.8 0.8 Figure 1 8 3 BOILER 7 SUPER HEATER REHEATER 7 5 CONDENSERİ 4. Please use MatLab, HYSYS, or Excel as a the software program for this assignment.

Answer 1

Dear student, kindly follow the steps.

Analytical solution for Option 1 is handwritten for a explanation.

Lo steam is @ 50 bae 13 3185.01 su Tsup Teat - = Sy ₂ - ha z option 1 @ obae, Tsat = 263.90: To = 450 °C Supechented. .. h3 = hg + Pamp & Temp -Teat) = 2794.2 +2:1 (450-263-9) kJ/kg for isentropic thekine, & hze S3 = 54 9. Sz = eg + Claup e sokae =5.974 + 211 In / 4504273 = 6.6 kJ/kg-k 263.9+273 @ 4 bar Sg = 6.894 kJ/kg-k Lo steam is wet . 'S4 = Sft Sp.x spg 1e4bae = 1.776 + 62 5.118) = 6.6 . i hp = hy * + x hfg lanne = 604.7 +(0.94% 2132.9) hy = 2609.63 ki kJ/kg Ideal Value Rio hz-hu hu hu = 27 24.71 XH = 0.94 Acthal Valve : 0.8 3185.ol hú 3185.ol 2609.63

Teat = 143.6 °C 4 bas, 4000 La steam is i ho '( Tsap cPane Teat) le a bear = 3276.04 ho=h 27 24.71KJ the lame are so 1. supecheated hg @ 4 bar .. h5 = 2737.6+ 2.1 (400 - 143-6) kJ/kg kJ/kg After turbine 2, point would be 6 ut since condition as 4, it is 4. Hence Calculations are also same no need to calculate again. There is am to et Table mentianas Po and the values are most likely different inconsistency in the table data values and figure from f4. A while fique & does not have any point is but is most the line 2 and but before Condenele. Hence calculating ho. m (Tsup Tsat ross = %= 6.894 +211 in (4007273 = 7.9 kJ/kg k 143.6 + 273 So= 8. @ for isentropic thaline turbine, ss = 56 @o. Il bae , Teat kJ/kyrk; Sf = 0.649 kJ/kg-k Lo steam is wet. 2.7.9 = 0.649 + (2x 8.151) x=0.89 ho = hata ta heg leoolbar ho 191.8 + (0.89 * 2392.9) = 2321.48 kJ h5-h6 .3276.04 - ho От hs-ho che = 2512.39 kJ 3/kg S5 56 = sg opene ola 8.151 ki S6 = Sf tasty J/kg 3276.04 - 2321.48

Py = P = 0.1 bal Wp 3 willy; Pe =P Vho = 0.0ololo = 0.001010 (50 - 0.1) Vfy (P& Py) :. We ANS. (6) 191.8 104 ) 79 at. KJ 191.81 81 10.75=191.879 - 191.8 -1918 Din lp = 5.04 x 104 ki KJ/kg Assung after condeneer, steam is entrented liquid hy = hf@ol kae kJ/kg . hg = hy + Wp = 191.8+(5.04 x 10 ks/kg ho-hy ho-hy hg' " he Qing=h₂ - hq = 3185.61 ? [SinB = 2993.2 KJ 2 kJ/kg ANS (d) Wri = hz - ha ANS (a) 1 = 191.813 kJ/kg 191.813 = 3185.01-27 24.71 WTA 460.3 kJ/kg O in Rh hs-hu = 3276.04-27 24.71 Q in Rh = 551.33 kJ/ky ).... Ans(d)

Wre hs-h6 ^ . WT2 Whet = 3276.04 - 2512.39 = 763.65 kJ ANS (a) kJ/kg] Wie + WT2 - Wp = 460.3 +763.65- (5.04 004) ?. Whet 1223.95 kJ .95 kJ/kg Wnet 80 MW = me x 1223.95 in ng = so x lor Ils mis = 65.36 kgle 122 3.95 x 103 103 J/kg ANS (C) = 65.36 * 2993:2 = 195.64 kW Sinen = 65.36 x 551.33 - 36.03 kW Qin B

| 223.95 Whet Qin 271302 + 55:33 4 أأ 3/4 . 53 دام ANS (e)

Only image files can be attached in the solution. Time exhausted so it took me a while to complete and update the solution. My sincere apologies for the same.

The excel sheet runs into 66 columns which was somehow not possible to add here in the form of a proper table but I have tried it below. But it appears it is going to be difficult to make any sense from the table below, hence screenshots from the excel sheet are also being attached.

Option 1 Dryness Fraction: x s4= (sf + x sfg] @ P4 Turbine efficiency n Dryness Fraction: x s6= [sf + x sfg] @ P6 Turbine efficiency n Cp sup hfg @ P4 hg @ P5 hfg @ P6 ASSUMPTION: Isentropic Turbine S4 kJ/kg-K 6.60 P3 bar 50 T3 °C 450 Tsat @ P3 °C 263.9 273+T3 K 723 273+Tsat @ P3 Compare T3 & Tsat @hg @ P3 K P3 kJ/kg 536.9 steam is superheated 2794.2 Wp = (vf7*(P8-P7)/100 h3= [hg + Cp sup (T3 - Tsat)) @ P3 kJ/kg 3185.01 kJ/kg-K 2.1 sg @ P3 kJ/kg-K 5.974 s3 = [sg + Cp sup (In(T3/Tsat)] @ P3 kJ/kg-K 6.60 P4 bar 4 sg @ P4 kJ/kg-K 6.894 Compare s4 & sg @ P4 steam is wet sf @ P4 kJ/kg-K 1.776 sfg @ P4 kJ/kg-K 5.118 hf @ P4 kJ/kg 604.7 Ideal Value: h4 h4[hf + x hfg) @P4 kJ/kg 2614.64 Actual Value: h4' n= (h3 - 44') / (h3-h4) | kJ/kg 2728.72 kJ/kg P5 bar 4 T5 °C 400 Tsat @ P5 C 143.6 273+T5 K 673 273+Tsat @ P5 Compare T5 & Tsat @ к P5 416.6 steam is superheated kJ/kg h5= [hg + Cp sup (T3 - Tsat) @ P5 kJ/kg 3276.04 sg @ P5 kJ/kg-K 6.894 ASSUMPTION: Isentropic Turbine s6 kJ/kg-K 7.90 s5 = [sg + Cp sup (In(T5/Tsat)] @ P5 kJ/kg-K 7.90 P6 bar 0.1 sg @ P6 Compare s6 kJ/kg-K & sg @ P6 8.151 steam is wet sf @ P6 kJ/kg-K 0.649 sfg @ P6 kJ/kg-K 8.151 hf @ P6 kJ/kg 191.8 Ideal Value: h6 h6= [hf + (x hfg)] @ P6 kJ/kg 2320.84 kJ/kg Actual Value: h6 n= (h5- i-h6') / (h5-h6) kJ/kg 2511.88 P7 = P6 bar 0.1 P8 = P3 bar 50 Vf7 @ P7 m^3/kg 0.00101 kJ/kg 0.94 2132.9 0.80 2737.6 0.89 2392.9 0.80 0.0005

ASSUMPTION: Qin B h3-h8 WT2 h5-h6 At condenser exit, steam Actual Value: h8 n= (h8 - 97) / (h8'-17) kJ/kg 191.80 17=hf @ P7 kJ/kg 191.8 Ideal Value: h8 Pump efficiency h8= h7 + Wp n kJ/kg 191.801 WT1 h3-h4' kJ/kg 456.29 kJ/kg Qin Rh h5-h4 kJ/kg 547.32 Thermal Efficiency Plant Capacity Mass flow rate of steam: m ŋ th w W = m* Wnet W net / Qin, % MW kg/s 34.47 80 65.55 W net WT1 + W T2-Wp kJ/kg 1220.46 kJ/kg saturated liquid 0.75 2993.21 764.16

File Edit View Insert Format Styles Sheet Data Tools Window Help Abc db 12 Liberation San: 10 a a a a - % 0.0 13 . .00 .00 0 BM16 ΑΣ A B с D E F G H J K Option 1 sg @ P3 P3 bar 50 T3 °C 450 Tsat @ P3 °C 263.9 273+T3 к 723 273+Tsat @ P3 Compare T3 & Tsat hg @ P3 K @ P3 kJ/kg 536.9 steam is superheated 2794.2 Cp sup kJ/kg-K 2.1 h3= [hg + Cp sup (T3 - Tsat)] @ P3 kJ/kg 3185.01 kJ/kg-K 5.974 s3 = [sg + Cp sup (In(T3/Tsat))] @ P3 kJ/kg-K 6.60 1 - -- 1 1 1 ! 1 1 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 - 1 1 + ! 1 1 - ! 1 1 1 - 1 1 1 1 - 1 1 1 1 1 22 + ! 1 1 T 1 1 1 1 1 23 24 25 26 27 28 29 30 31 - 1 1 1 1 1 1 KN + Option 1 Option 2 Sheet 1 of 2 Default English (India) * Average:; Sum: 0 + 100%

File Edit View Insert Format Styles Sheet Data Tools Window Help Abc 1 Last 12 Liberation San: 10 a a a 2 - % 0.01 0 .00 oo =0 BM16 fx = K L M N O P Q R S T U ASSUMPTION: Isentropic Turbine S4 kJ/kg-K 6.60 Dryness Fraction: x s4= (sf + x sfg @ P4 s3 = [sg + Cp sup (In(T3/Tsat))] @ P3 kJ/kg-K 6.60 P4 bar 4 sg @ P4 Compare s4 & sg @ P4 6.894 steam is wet sf @ P4 kJ/kg-K 1.776 kJ/kg-K sfg @ P4 kJ/kg-K 5.118 hf @ P4 kJ/kg 604.7 Ideal Valut h4= [hf + (x hf kJ/kg 2614.6 hfg @ P4 kJ/kg 2132.9 0.94 - 1 - - 1 1 - 1 1 - ! 1 1 1 1 1 ! - 1 1 - 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 1 1 1 1 1 ! - - 1 1 1 - - 1 1 1 1 1 1 1 1 1 1 1 1 1 --- 1 1 1 1 1 1 1 1 1 1 1 - ככ 1 Option 1 Option 2 KN + Sheet 1 of 2 Default English (India) Average:; Sum: 0 + 100%

File Edit View Insert Format Styles Sheet Data Tools Window Help RI Abc BE 12 Liberation San: 10 a a a a. % 0.0 . .00 .00 0 BM16 = T U V w X Y Z АА AB AC AD 1 - Ideal Value: h4 h4= [hf + (x hfg)] @ P4 kJ/kg 2614.64 Turbine efficiency n hfg @ P4 kJ/kg 2132.9 h5= [hg + C Actual Value: h4' n= (h3 - h4') / (h3-h4) kJ/kg 2728.72 P5 bar 4 T5 °C 400 Tsat @ P5 °C 143.6 273+T5 K 673 273+Tsat @ P5 Compare T5 & Tsat hg @ P5 K @ P5 kJ/kg 416.6 steam is superheated 2737.6 0.80 1 - 1 1 - 1 1 1 1 1 1 1 -- 1 1 1 1 - 1 1 - 1 1 - 1 - 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 1 1 1 1 1 - 1 1 - 1 1 1 1 1 1 + 1 1 - 1 1 1 - 1 ! 1 1 ! 1 1 - 1 1 1 1 1 1 1 1 1 1 1 1 1 1 - - 1 1 1 ! 1 1 1 1 1 - ככ 1 1 1 Option 1 Option 2 KN Sheet 1 of 2 Default English (India) Average; Sum: 0 + 100%

File Edit View Insert Format Styles Sheet Data Tools Window Help Abc 12 Just be in % 0.0 Liberation San: 10 a a = . .00 ood BM16 Ας Σ = AD AE AF AG AH Al AJ AK AL AM D s6 P6 hg @ P5 kJ/kg 2737.6 h5= [hg + Cp sup (T3 - Tsat)] @ P5 kJ/kg 3276.04 sg @ P5 kJ/kg-K 6.894 s5 = (sg + Cp sup (In(T5/Tsat)) @ P5 kJ/kg-K 7.90 ASSUMPTION: Isentropic Turbine s6 kJ/kg-K 7.90 bar 0.1 sg @ P6_Compare s6 kJ/kg-K & sg @ P6 8.151 steam is wet sf @ P6 kJ/kg-K 0.649 sfg @ P6 kJ/kg-K 8.151 1 1 1 - - - 1 1 1 1 1 ! - 1 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 1 - ! ! - - 1 1 1 1 1 - 1 - 1 1 - 1 1 1 1 22 1 1 1 - 1 1 1 - 1 - 1 1 1 1 23 24 25 26 27 28 29 30 31 22 KN 1 - - 1 Option 1 Option 2 Sheet 1 of 2 Default English (India) * Average:; Sum: 0 + 100%

File Edit View Insert Format Styles Sheet Data Tools Window Help M Og Abc OID 12 Liberation San 10 a a a a == % 0.0 bo .00 .00 BM16 fx = AM AN АО AP AQ AR AS AT AU AV AW AX 1 Dryness Fraction: x s6= (sf + x sfg @ P6 Turbine efficiency n sfg @ P6 kJ/kg-K 8.151 hf @ P6 kJ/kg 191.8 hfg @ P6 kJ/kg 2392.9 Ideal Value: h6 h6= [hf + (x hfg)] @ P6 kJ/kg 2320.84 Actual Value: h6' n= (h5-h6') / (h5-h6) kJ/kg 2511.88 P7 = P6 bar 0.1 P8 = P3 bar 50 ASSUMPTIOI Vf7 @ P7 Wp = (vf7"(P8-P7)7100 At condense m^3/kg kJ/kg 0.00101 0.0005 is saturated lid 0.89 0.80 - 1 1 1 1 - ! 1 1 1 1 1 - 1 - 1 - ! - 1 - 1 1 1 1 1 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - 1 1 - - 1 T 1 1 1 1 1 1 ! 1 1 1 1 - - 1 1 1 1 1 1 1 1 1 1 1 1 1 1 - ככ 1 1 Option 1 Option 2 Sheet 1 of 2 Default English (India) Average:; Sum: 0 + 100%

File Edit View Insert Format Styles Sheet Data Tools Window Help M Abc Y 12 Liberation San: 10 a a a a 를 % 0.0 ho .00 .00 0 BM16 6, Σ = AW AX AY AZ BA BB BC BD BE BF BG BH BI WT ASSUMPTION: Wp = (vf7*(P8-P7) 100 At condenser exit, steam kJ/kg 0.0005 is saturated liquid h7=hf @ P7 kJ/kg 191.8 Ideal Value : h8 Pump efficiency h8= h7 + Wp n kJ/kg 191.801 0.75 Actual Value: h8' n= (h8 - h7 / (h8'-17) kJ/kg 191.80 Qin B h3-h8' kJ/kg 2993.21 W T1 h3-h4' kJ/kg 456.29 Qin Rh h5-h4' kJ/kg 547.32 W T2 h5-h6' kJ/kg 764.16 -- - - --- 1 1 -- 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 22 -- --- -- - 1 ---- --- 1 1 --- + Option 1 Option 2 Sheet 1 of 2 Default English (India) Average: ; Sum: 0 + 100%

File Edit View Insert Format Styles Sheet Data Tools Window Help Abc Ω & Liberation San: 10 a a a % 0.0 . 00 00 =0 BM16 ΑΣ = BC BD BE BF BG BH B B.J BK BL BM BN - 1 Actual Value: h8' n= (h8 -h7) / (h8'-h7) kJ/kg 191.80 Qin B h3-h8' kJ/kg 2993.21 WT1 h3-h4' kJ/kg 456.29 Qin Rh h5-14 kJ/kg 547.32 WT2 h5-h6' kJ/kg 764.16 W net WT1 + W T2-Wp kJ/kg 1220.46 Thermal Efficiency Plant Capacity Mass flow rate of steam: m n th w W = m* Wnet W net / Qin, % MW kg/s 34.47 80 65.55 1 - 1 1 - 1 1 - 1 1 - 1 1 - - 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 1 1 1 -- 1 1 1 - - 1 1 1 1 1 I 1 1 1 1 1 1 1 1 Option 1 Option 2 KN + Sheet 1 of 2 Default English (India) Average:; Sum: 0 + 100%

I hope this helps. In case of any query, kindly message me.