Page 345 - Steam Turbines Design, Applications, and Rerating

Elliott Shortcut Selection Method for Steam Turbines 319

From Fig. 15.4, velocity ratio for optimum Curtis stage efficiency = 0.23
173
V b
V b /V j = 0.23; V j = = = 752 m/s
0.23 0.23
V j = 44.72 Δ H j = 752 m/s

ΔH i = 282.8 kJ/kg

h 1 (from Mollier chart or steam table) = 3394.1 kJ/kg
h 2 = 3394.1 − 282.8 = 3111.3 kJ/kg
Exhaust pressure (from Mollier chart) = 40.6 bar
Assuming a stage efficiency of 70 percent, the stage exit conditions are:
exhaust pressure p a 40.6 bar
h 2e = 3394.1 − (.7) (282.8) 3196.2 kJ/kg
Example 15.6: Rateau stage performance Conditions: p g = 27.6 bar (p a = 28.61
bar), 315°C, 5000 r/min, 890-mm wheel diameter. Assume 25-mm blade height.
Find isentropic heat drop and end point. (See Fig. 15.6.)
πDN (3.14)(890 + 25)(5000)
V b = = = 239 m/s
60,000 60,000
From Fig. 15.6, velocity ratio for optimum Rateau stage efficiency = 0.46
239
V b /V j = 0.46; V j = V b /.46 = = 520 m/s
0.46
V j = 44.72 Δ H i = 520 m/s
2
2
ΔH i = (V j /2000) = (520 /2000) = 135.2 kJ/kg
h 1 (from Mollier chart or steam tables) = 3034.9 kJ/kg
h 2 = 3034.9 − 135.2 = 2899.7 kJ/kg
Exhaust pressure (from Mollier chart) = 15.9 bar

Assuming a stage efficiency of 80%, the stage exit conditions are:
exhaust pressure 15.9 bar;
h 2e = 3034.9 − (.8) (135.2) 2926.7 kJ/kg.
Example 15.7: Straight Rateau staging Conditions: p g = 27.6 bar (p a = 28.61 bar),
315°C, exhaust p g = 7 bar (p a = 8.01 bar). Find the number of 890-mm-diameter
Rateau stages required, at 5000 r/min, assuming optimum stage efficiency. (See
Fig. 15.7.)
From Mollier chart, or steam table, the isentropic heat available is:

3034.9 − 2754.9 = 280.0 kJ/kg

Alternately, using a TSR table:
TSR = 12.8 kg/(kWh)