Question

22 6 points Dry saturated steam at 7 bar is expanded in turbine nozzles to a pressure of 5.5 bar 0.98 dry. Find (1) The velocity at exit. If the area at exit is 11.2cm, find (ii) the mass flow of the steam in kg/s. Use values from excerpt of the Steam Table below: he Sy 9 Srg [kJ/kg K] р bar] 3.5 4.0 4.5 5.0 5.5 0 2732 2739 T ve [°C] [mP/kg] 138.9 0.5241 143.6 0.4623 147.9 0.4139 0.3748 155.5 0.3427 158.8 0.3156 165.0 0.2728 170.4 0.2403 175.4 0.2149 151.8 и. ug [kJ/kg] 584 2549 605 2554 623 2558 639 2562 The 655 2568 696 2573 720 2577 742 2581 2744 hi hry [kJ/kg] 584 2148 605 2134 623 2121 640 2109 656 2097 670 2087 697 2067 721 2048 743 2031 1 2565 1.727 1.776 1.820 1.860 1.897 1.931 1.992 2.046 2.094 6 2749 2753 2757 2764 2769 2774 5.214 5.121 5.037 4.962 4.893 4.830 4.717 4.617 4.529 669 6.941 6.897 6.857 6.822 6.790 6.761 6.709 6.663 6.623 8 9 Dragn Drop here on Browse

Answer 1

7 Turbine 5.5 bar 0.98 quality so from the Steam table, At 7 bar and dry saturated steam h = at 7bar hi= 275 2764 KJ kg - And Ast 5.5 bar, quality = 0.98 hf = 656 KJ hg 2753 KJ kg kg ... h2 = 0.98 x 2753 +(1-0.98) 656 KJ kg kg energy balance actors the Husbino : (Neglecting potential energy, Heat loss, work done and Initial kinetič energy get, hi = hz + 1 v2 - 2711.06 KJ Doing the we so 2764 KT 2711:06 KJ + Kg 12v2 kg

/v² = (2764-2711.06) KS kg :v²= 105.88 V = √105.88 = 10.29 .: The velocity at exit = 10:29 m . Area at the exit = 11.2 cm2 11.2 x 10-4 m2 At outlet condition, Vg = = 0.34 27 m3 kg. do, volumetric flow rate at outlet velocity x Area 10:29 u 112x10-4 m2 = 0.01152 m2 S. in Mass flow rate = mong = 0.0336 kg 0-01152 mm 0.3427 m : Mass flow of the steam = 0.0336 kg 3