Question

(2) Non-ideal Rankine Recognizing that the Rankine cycle described in question (1) cannot be ideal in reality, your supervisor asks you to perform a more accurate Thermodynamic analysis. She tells you that the States 1, 3 and the mass flowrate are identical to those given above. Additionally, the pump manufacturer specifies its isentropic efficiency, np = 0.8. Through experimental measurement techniques (and the mathematical relations in Chapter 12), the specific entropy of water at State 4 is found to be s4 = 7.8302 kJ/kg.K. Given this data, what is the isentropic efficiency of the turbine, nt? Also calculate Wnet, Oh, and 9th- How do these values compare to the results you found in question (1)? Do these seem reasonable to you?

Answer 1

solution, mp = 0.8 Sya 7-8302 bo A 2 4 ballagk. r = 20kgls at state, P = 25 kPa. hiz 222 kJ/kg. 920.00 lo2 mol kg 080102 (5600-25) Merry Wpump = 6.343 kJ /kg h2= ht Wapcomp 222 + 6.343 = 278.34 kg/kg. Wpump = 0.8 State 3 P₂ = 5MPa, Tz = 50°C h = 340 kJ/kg S₂ = 6-98 kJ / bg k , Sate 4 Sy= S4= 7.8302 blak Set & Sg-so) 0.893 + xy (2 83 -0.893) 7-8362 → Ky Point'u' lies on satunted vapour condition

Now, 3 According to ist law of thermodynamics. Es= {W. 932 Wizy มัน - Byl t&2 Y Wzy & W21 941. 25 hz to = 3430 - 228.34 = 315166 kg/kg. NOW! Sul= hyrhi hy= hyt xu hey 272+ ( 2620 - 272) = 2620 kolky. hy-hi = 2620-222 = 2348 kJ/kg Was is Pump work. = 6-343 kJ /kg Sula Wzy + Wall .. Now using Sylt 823 = » (-2348) + 3151-2 W34 + (-6.343) Way = 809. 34 kJ/kg M p = by Way 809.31 -0.99g. 3430-2620 =99.91%

Whet Way -Wplamp = 809.34- 6.343 = 802.99 kalkg Sr = harha= - 3430 - 228.34 = 3151.66 kolky tha Whet 25.47% si