Question

Problem 1. (40) Water is the working fluid in an ideal Rankine cycle with reheat. Superheated vapor enters the turbine at 12 MPa, 480 °C and the pressure at the exit of the second stage turbine is 8 kPa. Steam expands through the first stage turbine to 1 MPa and then is reheated to 440 °C. Saturated liquid water leaves the condenser. After the pump, pressure goes back to 12 MPa. Find: (1) Sketch the process on a T-s diagram and justify the location of each point. (10) (2) Find specific enthalpy at cach point from hi to ho in kJ/kg. (18) W cycle (3) The net work per unit of steam flow in kJ/kg. (2) (4) Total heat addition per unit of steam flow in the boiler Rin, in kJ/kg. (2) (5) Heat loss per unit of steam flow in the condenser Route , in kJ/kg. (2) (6) Thermal efficiency n. (2') (7) Knowing that liquid in a steam turbine causes corrosion, based on the phase of water at the end of each turbine, is it a good design and why? (4') mh th

Answer 1

Solutions from Steam table (super heated) at 1st stage forbine inlet Stage (1) - P,= 12 mea, Ti = 480c SE : 6.4186 IST hia 3295.3 Koleg kgk testine first stage " exit State (22 P2 = 1 mpa How since Turbine is isenbopic son S = f = 6.4186 kolegia from Steam tablel Besare based) Now at P= impa Sf = 2, 1381 kJ lk gk i Sg = 5.5850 kalagie hg=2771 kilag ha= 762.52 koliko Sol = (Sf) impa + " ( Sg - Sf) impa 604180 = 2.1381 4 318058 50 - 7.1381) 22=09625 (state is in mixture zone)

Similadly :- (halimpa + şily- hall n h2= Impa has 762.527.962512777.1-762.33 h = 2701.715 kolag The heated State) (3 ); - Pz = P2 = Impa T3 = 440°c from steam tablel Superheated) at Ind stage tistine inlet :- this state!! hz 3349.9 kolkg S3 7.5902 kulk gk 2nd stage tistine exit) Conden ser inlet State (4) Pu= 8 kpq = .008 mpa for isentropic testine; Su=ş = 7.5902 kg kga from steam table (Bessure based) at Pu 2008 mea

St=.5924 g KS kgk Sfg = 726348 Kilk gk hf = 113.84 KB/Kg hag = 2402.4 kJ/kg Sy = (SF).col + Hy ( 58g)..os 008 mea ме * 7.5902 = 59249 + au l 7.6348) Hy = .91655 (State is in lig. rapong mixture zone) also! hy (hadook taol hag) MPa mpa 173.84* .91653 (2402.4) hu= 2375, 70 koleg Condensor exit / Pump Inlet Inlet State (5):. Ps = Pu= .008 mea (Salisated liquid) hs = (hfloor 173.84 kilug mpa

Vs = (f) .00100848 m kg 008 mpa For isentropic pump :- Reume exit state (69!. ho= hs + (We) isentropic hst Ys (dp) hst Vs (PG-Ps) ho=113.84+ 100 100848 ( 12000-8) (P= P,= 12 mpa ho = 185.933 1657 = 12000 kPa) b Tas Diagram! 480 umo T 2 6 (°C) Wp 8kPa s c S(K5149%)

(3) Net work per unit steam flow). (utustine), + (Werline); - w sume (haha) + (hz - hu)-(ho-hs) - 13295.3 - 2701,7157+ (3349.9 349.9 -2375.70 -(185.933 - 173.84) Weycle 1555.622 kol kg mi 14) Total Heat addition per unit steam flow + Q Reheater = boiler ta = Chi-ha) & Chacha) a (3295.3 - 185.933) + (3349.9.2701.715) & in 3757.552 kJ/kg)

(5) Heat loss per unit steam flow Scondenser is Chuths) (2375.77 - 173.84) " a condensor 2201.93 kolkg mi (6) Thermal efficiency: M= wi vet Sadded n 15550622 =.4139 3957, 552 n = 41.39/ (7) According to above analysis in both stage tubine exit their is some moisture this design is Not So safee My = .91655 Ha a og625, So moisture present at both state