Page 344 - Steam Turbines Design, Applications, and Rerating

318 Chapter Fifteen

From Example 15.1, we found that a Curtis stage removes 121.6 Btu/lb
1459.6 − 121.6 = 1338.0 Btu/lb

Pressure = 590 psia
Assuming 70 percent stage efficiency:
121.6 × 0.70 = 85.1 Btu/lb
1459.6 − 85.1 = 1374.5 Btu/lb
From this end point to 165 psia, the isentropic heat drop for the Rateau stages is:

1374.5 − 1236.6 = 137.9 Btu/lb
From Example 15.2, the optimum isentropic heat drop per Rateau stage =
58.0 Btu/lb
137.9
Approximate stages required = = 2.4 (thermodynamically)
58.0
The turbine will therefore require one 25-in Curtis stage and two 35-in Rateau
stages.
Assuming an overall Rateau efficiency of 81 percent, the end point will be:
1374.5 − .81 (137.9) = 1262.8 Btu/lb

Here are the basic relationships in metric units:
πDN

V b = V j = 2 0 0 0 (h 1
− h 2 ) = 44.72 ΔH i
60,000
where V b = pitch line (blade) velocity, m/s
D = pitch diameter of wheel, mm (base diameter plus height of blade)
n = rotative speed, r/min
V j = steam jet velocity, m/s
h 1 = inlet steam enthalpy, kJ/kg
h 2 = isentropic exhaust steam enthalpy, kJ/kg
h 2e = stage exit steam enthalpy, kJ/kg
ΔH i = isentropic heat drop, kJ/kg (h 1 − h 2 )
V b /V j = velocity ratio, dimensionless
p a = absolute pressure, bar
p g = gauge pressure, bar = p a − 1.01
s = specific entropy, kJ/(kg K)

Example 15.5: Curtis stage performance Conditions: p g = 103.4 bar (p a = 104.4
bar), 510°C, 5000 r/min, 635-mm wheel diameter. Assume 25-mm blade height.
Find isentropic heat drop and end point. (See Fig. 15.5.)
πDN (3.14)(635 + 25)(5000)
V b = = = 173 m/s
60,000 60,000