82 TheRankine Cycle 273 EXAMPLE 8.1 | Analyzing an Ideal Rankine Cycle ‘Steam is the working fluid in an ideal Rankine cycle. Saturated ‘vapor enters the turbine at 8. MPa and saturated liguid exits the condenser ata pressure of 0.008 MPa, The ner power outpat of the eyele is 100 MW. Determine forthe eycle (a the thermal eff- ciency, (b) the back work ratio, (e) the mass low rate of the steam, inkg/h, @) the rate of heat transfer, Q, into the working fui ait passes through the bole, in MW, (e the rate of beat ans, Oy. from the condensing steam as it passes through the condenser in MW, (0 the mass low rate ofthe condenser cooling water in kg/h, if cooling water enters the condenser at 15°C and exits at 35°C, Solution Known An ideal Rankine eycle operates with steam as the ‘working Muid. The boiler and condenser pressures are specified, and the net power output is given, Find Determine the thermal efficiency, the back work ratio, the ‘mass flow rate of the team, in kg/h, the rate of heat transfer to the ‘working fluid as it passes through the boiler, in MW. the rate of ‘heat transfer from the condensing steam as it passes through the condenser, in MW, the mass flow rate of the condenser cooking hich enters at 15°C and exits at 35°C. La + Ove Coating + 3 Saturated Tigi t 0.008 Pa 1 1 4 Engineering Model 1, Each component ofthe cycle is analyzed asa contol volume at steady state. The contol volumes ar shown on the accon- ppanying sketch by dashed lines. 2. All processes ofthe working fluid are internally reversible. ‘3. The turbine and pump operate adishatically. 44. Kinetic and potential energy effects are negligible. ‘5. Saturated vapor enters the turbine, Condensate exits the con- sense as saturated liquid @Analysis To begin the analysis, we fix each of the principal ‘tates located on the accompanying schematic and T-s diagrams. Staring atthe inlet tothe turbine, the pressure is 8.0 MPa and the steam isa saturated vapor, so from Table A-3, fh, = 2758.0 iikg and 5) = 57432 Kk“ K ‘State 2 is fixed by ps = 0.008 MPa and the fat thatthe spe= entzopy is constant fore adiabatic, internally reversible ex- pansion through the turbine. Using saturate liquid and saturated ‘vapor data from Table A-3, we find thatthe quality at state is 5.7432 ~ 05926 7.6361 pow & 6745 The enthalpy is then a= he Sy = 1794.8 kg, 173.88 + (0.6745)2403.1 State 3s saturated liquid at 0.008 MPa, so hy = 173.88 kik State 4s xed by dhe boiler pressure p andthe specific enttpy 44.2.5, The specifi enthalpy canbe found by interpolation inthe ‘compressed liquid tables. However, because compressed liquid data ae relatively sparse, itis more convenient vo solve Bg. 83 for hy using Eg. 876 o approximate the pump work. With this approach ig = s+ Welt = hy + U(O4~ ps) By inserting property values from Table A-3 h 73.88 kITke + (1.0084 > 10° m'Fkg) to" Nf? || 14 Taba [fi0" N= 73 88-+ 8.06 = 181.94 kWkg ‘8. The net power developed by the cycle is Woe = We~ Wp Mass and energy rate balances for control volumes around the r= bine and pump give, respectively, % (8.0 ~ 0.008)MPa % where isthe mass flow rte ofthe steam. The rate of heat trans- fer to the working fuid as it passes through the boiler is deter- ‘mined using mass and energy rate balances as hh 7948) ~ (181.94 ~ 173.88) kk (27580 — 181.94) Kkg = 037137.1%) = (ari. The back work ratio is owe at SL 9 173 Wo yale (2788017948) Kke $06 = 508 5.37107 (sss) 9632 6 The mass flow rate of the steam can be obtained from the expression forthe net power given in part (a). Thus, Wo © Wah) a=) (2.00 MAW 10" KW/MAW} 3600s) (965.2 — 806) Kikg = 3.7710" kph 4. With the expression for @,, from part (a) and previously de- termined specitic enthalpy values Qn = Hit(hhy ~ hes) |3600 s/hj|10° kW/MW 269.7 MW «& Mass and energy rate balances applied to a control volume ‘enclosing the steam sie ofthe condenser give Qa = ithe ~ hy) (3.77 x 10° kh 794.8 — 173.88) ke [3600 sh10? kwW/MAW) 169.75 MW © Nove tat the ratio of Qa to is 0629 629%. Alternatively, can he determined from an energy rate balance onthe vera! vapor power plant. At steady state, the net power developed equals the ne rate of heat transfer to the plane Wee = Gis Qua Reatranging this expression and inserting values oa = Qin ~ Ways = 269.77 MW ~ 100 MW = 169.77 MW ‘The slight difference from the above value is due to roundoff f. Taking «control volume around the condenser, the mass and energy rate balances give a steady sate oo 0 = Bos Wee teal Mow in ~ Mow.na) + (hs ~ hy) Where rt. is the mass flow rate ofthe cooling water. Solving ft Fie rhs ~ fs) Goce aw ‘The numerator inthis expression is evaluated in part). For the cooling water = f(T), so with saturated liquid enthalpy values from Table A-2 at the entering and exiting temperatures ofthe cooling water (169.75 MW) |10" KW/MW}/3600 vb) fi. OSeEOeeeoeseseS = 73 10 gh (146.68 — 29) ATTkg © Nove that slihiy revised problem-solving methodology is used in this example problem: We begin with a systematic valuation ofthe specific enthalpy at each numbered stat. © Nove that the back work ratio is relatively low forthe Rat cycle. In the present case, the work required to operate the pump is less than 1% of the turbine ouput © in his example, 62.9% of the encray added to the working fd by hea aster is subsequemlly discharged to the cool ing water. though considerable energy is carried away by ‘the cooling water, its exergy is small because the water exis ata temperature only a few degrees greater than that of the surroundings See See. 8.6 for further discussion, SKILLS DEVELOPED Ability to. 1 shotch the T-r diagram ofthe basic Rankine cycle. + fix each ofthe principal states and retrieve necessary property data «apply mass and energy balances. + caleulate performance parameters for the eye Quick Quiz the mass low rate of steam were 150 kg/s, what would be the net power, in MW, and the thermal efficiency? Ans. 13.2 MW, mde.EXAMPLE 8.2 | Analyzing a Rankine Cycle with Irreversibilities [Reconsider the vapor power cycle of Example 8.1, but include in the analysis thatthe turbine and the pump each have an isentropic cliiency of 88%. Determine forthe motified cycle (a) the the- ‘mal efficiency (b) the mass flow rate of steam, in kp, for & net power output of 100 MW, (¢) the rate of heat transfer Q, into the ‘working fi sit passes through the bole, in MW, () the rate of heat transfer Q.» from the condensing steam as it passes through the condenser. in MW, (e) the mass flow rate of the condenser cooling wate, in ky if eooling water enters the condenser at 15°C and exits as 35°C. Solution ‘Known _A vapor power eyele operates with steam asthe work {ng Muid. The turbine and pump both have eficiencis of 85%, Find Determine the thermal efficiency, the mass flow rate in kph, the rate of heat transfer to the working fluid as it passes through the boiler in MW, the heat transfer rate from the con- ddensing steam asi passes through the condenser, in MW, and the ‘mass flow rate ofthe condenser cooing water in kg ‘Schematic and Given Data: r 4 4 3 Engineering Model 1. Each component ofthe eyele is analyzed as a control volume at steady state 2. The working fluid passes through the boiler and condenser at constant pressure, Saturated vapor enters the turbine. The condensate is saturated atthe condenser exit 3. The turbine and pump each operate adiabatically with an efficiency of 85% 44. Kinetic and potential energy effects are nezligible Analysis Owing to the presence of iereversibilities during the ‘expansion ofthe steam through te turbine, there isan inerease in specific entropy from turbine inlet to ext, as shown onthe accom- ‘panying T-s diagram, Similaly there isan inerease in specitic 1. With the net power expression of part (a, the mass ow rate of the steam is Woe Ga) AD (100 M1W)/3600 10" KW/MW (187 — 9.48) Ekg = A449 « 10" kg/h ‘e. With the expression for Q,, from pat (a) and previously de- termined specific enthalpy values 0. ih) — ho) (4.449 « 10" kg/h) 758 ~ 183.36) kirky [340010 KWAN] 182 MW 4d. The rate of heat transfer from the condensing steam to the cooling waters O° kg/h(1939.3 173.88) WWE _ 35 > apy [3600/10 wWiNW entropy from pump inlt to exit Let us begin the analysis by fixing ‘eh ofthe principal ithe same usin Example 8.1, o/h, = 2758.0 kikg and = 5.7432 kk» K ‘The specifi enthalpy atthe turbine exit, stat 2, canbe deter mined using the isentropic turbine efficiency, Eq, 8.9, aL hin, Iya, Where fyi the specific enthalpy at state 2s on the accompanying Ts digram, From the solution to Example 81, Solving for fs and inserting knowen valves ds I= mld 758 — O8S(27S8 ~ 1794.8) State 3s the same as in Example 81, so y= 173.88 kfkg. ‘To determine the specific enthalpy a the pump ext, state 4, reduce mass and energy fale balances for a contol volume around ‘the pump to obtain Wri = ha ~ fy. On rearrangement, the specific enthalpy a state 4 is 1939.3 kIrkg ga hy + Wy ‘To determine fi fom this expression requires the pump work. Pump work ean be evaluated using the isentropic pumpeficiency in the form of Fi. 8.10b: Solin for Wi ress in We le = 99 me “The numerator of this expression was determined inthe solution to Example 8.1. Accordingly, SOO 9.48 tak o8s “The specific enthalpy atthe pump exit then na f+ Wigs = (73.88 + 9.48 = 183.36 Karke 1 The net power developed by the cycle is W, ih — ha) ~ eg A fou = = Wy ‘The rate of heat wansfer to the working Mui as it passes through the boiler is Inserting values 2758 ~ 1939.) - 9.48 2758 — 8 0314 14%), The mass flow rate ofthe cooling water can be determined from _ ti = hy) ioe = Hon) (218.2 MW)|10" KWW/MW)/3600 sf) (146.68 = 62.99) ike 9.39 10" kg/h SKILLS DEVELOPED Ability to. ‘+ sketch the 7-s diagram ofthe Rankine cycle with turbine and ‘pump ireversibilites. ‘fix each ofthe principal states and retrieve necessary property © apply mas, energy, nd entropy principles ‘ caleulate performance parameters forthe cycle, Quick Quiz [the mass flow rate of steam were 150 kg/s, what would be the ‘pump power required, in kW, and the hack work ratio? Ans. 1422 kW, 0116.EXAMPLE 8.3 | Evaluating Performance of an Ideal Reheat Cycle stander Sum ems gee a AO MP {°C an expat? MPs then a oP foe euing th scone tine, whe xpi he ‘Sons pce 08 MP. The sponse MI, Detmine the thermal cio fhe ele (th mass Sofa gh ee tof ee ngincring Mode 1. Eactconpmne inde jl sanlze asa ce lane ying heh by de ines 2 Al reese fh working id ier eves 2 The tie and pump pert abba 4. Condens conden ate i 5 Kit and pt nr feta eg Anata To begin, we fx th of hpi at, Sing ‘helt he fit tine ag the prs 0 MPa {ictonprte 48° sea speed ape om ‘Tale Aah St84 gan 66886 Ug K ‘She 2s ely p07 MPa nd forthe sete expansion tgh he fststage tie Using strate tid ands sae dita foe Table A. he quay on BE I ages hehehe (722+ O9R9510663 = ELEM Stae 3 is superesad poe wih py = 07 Mand h=440°C fo Tale A, y= 3853 ga = 1757 hk To css 4, wep, = 1008 MP ands = fe sensope apasiondogh com tage tie With at fo Table ‘nS. the quay tse “Te specie is y= 17888 + (09842 408 = OS KIA Se iste igi 0008 MP = 1738 Fly he te a teppei Taal sok= IS1SGKIng. a, The ner poner develope he eles Mas and energy’ te alnces fete ine tage ad he pp rede ge epee, “ative Wt = y= Tee Wil yh Pimp Wyn "Tew at of eat tanerothe woring id ips ‘gh heb supethoweran het Oe, oy Ing Bt Opting press nena specif ahd ‘Seon poner et ee nt Dstemin he eek, te as Hw ‘etc it gh ath at ane or een ‘amar pe aight condense MW Ducane Schematic an Given ates sites expressions thermal eee it (hha haha) th hhh G44 27419) + 3883-2085) — OLS ‘GHG — 1 30) C353 2748) 640K NOH ISTIIAE 6 yoxsa 36) G65 610s” STT¥RIG ” xpesson fe et ome enn a Ton A (400 360 40 LWW) (6 MRE ROO 2.3610 gm The i of eat ser rom the condensing sta tthe ag tsp 23610 Ayn CD8 7A [6010 AWA sw ‘Tose the fects fk, we compare the prota ales wih ote in sane 1 Wih sper abd = Tet the hema eicieny ices ove tha of he ce of Example 1 Fora spiel set power opt 1D MM) er ttc eceny mons tha smal mass How rate of sear ire: Mrcover, with per thermal ein he ate (ced demand for cooling wate With cing. them gual fiyatthe bine ests abit eon veh ae forte ese of ample hoch he 7s dg of he ed Rankine ee witha 1 x each fhe prin sto and eee ecesry prope ta aula perfrmance parame freee vik Gute What i he rate of heat ato forthe rebeat process, in DMWWyand what erent that alo he tl heat aonEXAMPLE 8.4 | Evaluating Performance of a Reheat Cycle with Turbine Irreversibility Reconsider the reheat cycle of Example 8.3 but include in the analysis that each turbine stage has the same isentropic efficiency (@) In, = 85%, determine the thermal efficiency. (b) Plot the ther- ‘mal efficiency versus turbine stage isentropic efficiency ranging rom 85 10 100%. Solution [Known A reheat eyele operates with steam as tbe working uid. Operating pressures and temperatures ae specified. Each turbine sage has the same isentropic efficiency. Find If n= 85%, dotermine the thermal efiieney. Also plot the thermal efficiency versus turbine stage isentropic efficiency ranging ftom 85 to 100%. ‘Schematic and Given Data: ‘The thermal efficiency is then (hfs) + Us = fe) = ” (hha) +s ha) e (3348.4 ~ 2852.8) + G35 181.94 - 173.88) ~ (G38 4 — 181.94) + (33583 — 2832.8) 1293.6 ke 3515.18) 3687.0 ke b, The [Teode forthe solution follows, where etath is. etat2 is gs €ta isn, Wet = Wy i and Qin = Qn Engineering Model 1. Asin Example 83, each component i analyzed as a contol ‘volume at steady state 2, Except forthe two turbine stages, all processes ar internally reversible, 3. The turbine and pump operate adiatically 4. The condensate exis the condense as saturated liquid. ‘5. Kinede and potential energy effects are negligible. Analysis 2. From the solution to Example 8.3, the following specific en- thalpy values are known, in iMkg: hy = 3348.4 ha, = 27418, a, = 3353.3, y= 298 5, fg = 173.88, i= 181.94, ‘The specific enthalpy atthe exit of the fist-stage turbine, ‘i, am be determined by solving the expression forthe turbine ‘sentpic efficiency, Eq. 89, 1 obtain hy Mh ~ hs) 3348.4 ~ O8S(348.4 — 2741.8) = 28328RI/kg ‘The specific emhalpy atthe exit of te second-stage turbine ean be Found similarly ne = y= My ~ a) = 33533 ~085(3353.3 ~ 8.5) = 2567.2 ke Using the Explore button, sweep eta from 0.85 to 1.0 in steps of (0.01. Then, using the Graph button, obtain the following plot 092 03 L L as 090 095 700 Iseaopic turbine eficeney