Question

Figure 1 shows a number of components connected in series, all operating at steady state. Liquid water enters the boiler at 750 kPa. Steam exits the boiler at 750 kPa and 650 °C and undergoes a throttling process to 550 kPa before entering the turbine. Steam expands through the turbine to 65 kPa, 280 °C, and then undergoes a throttling process to 100 kPa before entering the condenser. (a) State the engineering model and assumptions that can be implemented for each process across each component. (b) Locate each of the states 2-5 and the corresponding processes directions on a sketch of the T-s diagram, (c) Determine the power developed by the turbine, in kJ per kg of steam flowing. (d) For the valves and the turbinc, evaluate the rate of entropy gencration, cach in kJ/K per kg of stcam flowing (e) Using the result of part (d), place the components in rank order, beginning with the component contributing the most to incfficient operation of the overall system. Comment. (f) What could be done in order to improve the performance of the overall system? Liquid water, p/= 750 kPa 2 p2 = 750 kPa T2 = 650°C Boiler Valve + 3 pz = 550 kPa Turbine +4 P4 = 65 kPa TL = 280 °C Valve Condensato, seconde em Condenser Condenser 5 ps = 100 kPa P6 = 100 kPa 4 Figure 1

Answer 1

Liquid water, p/= 750 kPa Liquid watering A B 2 oiler p2 = 750 kPa T2= 650 °C Boiler Valve +3 pz = 550 kPa Turbine P4 = 65 kPa +4 74 = 280 °C Valve Condensatc, po= 100 kPa Condenser 5 Ps = 100 kPa Figure 1 @ - Heating process in the Boiler can be considered as Constant pressure process. While practically the process definitely not fully isobaric - The Isen that pic valves placed before and after turbine are not actually I deal. There will be some small, negligible enthalpy

drop accross those values - The expansion in turbine is assumed isentropic for ease in calculations while practically Reversible adiabatic process is impossible to carry out - Again, it is assumed the process in condensor is an isobaric process. .. 750 kPa assokPa 7 cougoc Pooupa 91341°c) N 6) (230)

© oAut est @, P2 = 750 kPa, Tz=650°C From steam tables, (Super heated) ho = 3812.2191 kJ/kg Sz = 8.2899 kJ/xgik Aut Pt ③, - P3= 50 kPa, haaha From Steam tables, (superheated) Tg = 649.3997°C S3=8.4327 kJ / kg k a Aut Pt ① (superheated) Pu=65kpa, Tu=288°C haz 3035.3432 KJ/kg su=8.346 kJ /kg.

- Ast Put G, Ps : 100kPa, usaha From steam tables (super heated) ss=8.3554 kJ/kg k Ts=341.7513°C #Power Developed by turbine = hz.hu 3812.2191– 3035:3432 =776.876 kJ @ Rate of Entropy generation • For Valve before turbine = S3- S2

= 8.4327 - 8-2899 = 0.1428 kJ как • For Turbine, os= su-S3 = 8. 346-8-4327 2 -0.0867 kJ KIK e Entropy slightly reduced in Turbine) For Valve after Turbine: as= ss-su I 8. 3594-8. 346 z 0.0134 kJ/kg

k © Rank of Inefficiency os valve,> os valvez > Os turbine A value ® is most inefficient A Turbine is in good operating - condition. @ To improve performance, Remore valve ③. Instead add one high Pressure turbine and deat enthalpy doop accross it till Pressure reaches sso kpa, Then Reheat to 650°C ( ③ Point), Rest process keep same, what will happen because of this ? Neat irreversibility generated will

be same but the is more. sates for added to as in previous case, Neat work generated Hence, this compen - the irreversibility the system,