Question 6.37: A twenty-stage Parson turbine receives steam at 15 bar at 30...
A twenty-stage Parson turbine receives steam at 15 bar at 300°C. The steam leaves the turbine at 0.1 bar pressure. The turbine has a stage efficiency of 80% and the reheat factor 1.06. The total power developed by the turbine is 10665 kW. Find the steam flow rate through the turbine assuming all stages develop equal power.
The pressure of steam, at certain stage of the turbine is 1 bar abs., and is dry and saturated. The blade exit angle is 25° and the blade speed ratio is 0.75. Find the mean diameter of the rotor of this stage and also the rotor speed. Take blade height as 1/12th of the mean diameter. The thickness of the blades may be neglected.
Number of stage = 20
Steam supply pressure = 15 bar, 300°C
Exhaust pressure = 0.1 bar
Stage efficiency of turbine, η_{stage } = 80%
Reheat factor = 1.06
Total power developed = 10665 kW
Steam pressure at a certain stage = 1 bar abs., x = 1
Blade exit angle = 25°
Blade speed ratio, ρ = \frac{C_{bl}}{C_{1}} = 0.75
Height of the blade, h = \frac{1}{12} D (mean dia. of rotor)
(i) Steam flow rate, \dot{m}_{s} :
Refer Fig. 50.
Isentropic drop, (Δ h)_{isentropic } = h_{1} – h_{2} = 3040 – 2195 = 845 kJ/kg
η_{overall } = η_{stage } × Reheat factor = 0.8 × 1.06 = 0.848
Work done = Actual enthalpy drop
= (Δ h)_{isentropic } × η_{overall }
= 845 × 0.848 = 716.56 kJ/kg
Work done per stage per kg = \frac{716.56}{20} = 35.83 kJ …(i)
Also, total power = No. of stages × \dot{m}_{s} × work done/kg stage
∴ 10665 = 20 × \dot{m}_{s} × 35.83
∴ \dot{m}_{s} = \frac{10665}{20 × 35.83} = 14.88 kg/s.
(ii) Mean diameter of rotor, D :
Rotor speed, N :
Refer Fig. 51.
Work done per kg per stage = C_{bl} × C_{w} = C_{bl} (2C_{1} cos 25° – C_{bl})
Also, \frac{C_{bl}}{C_{1}} = 0.75 …(Given)
∴ C_{1} = \frac{C_{bl}}{0.75} = 1.33 C_{bl}
i.e., Work done per kg per stage
= C_{bl} (2 × 1.33 C_{bl} × 0.906 – C_{bl})
= 1.41 C_{bl}² Nm …(ii)
Equating (i) and (ii), we get
1.41 C_{bl}² = 35.83 × 1000
∴ C_{bl}² = \frac{35.83 × 1000}{1.41} or C_{bl} = 159.41 m/s
∴ C_{1} = = 1.33 × 159.41 = 212 m/s
From Fig. 51
C_{f_1} = C_{1} \sin α = 212 \sin 25° = 89.59 m/sv_{g} = Specific volume at 1 bar when steam is dry and saturated
= 1.694 m³/kg (from steam tables)
Mass flow rate, \dot{m}_{s} = \frac{πDhC_{f_1} }{v}
∴ 14.88 = \frac{π × D × (\frac{D}{12}) 89.59}{1.694} or D² = \frac{14.88 × 1.694 × 12}{π × 89.59}
∴ D = 1.036 m.
Now, h = \frac{D}{12} = \frac{1.036}{12} = 0.086 m = 8.6 cm.
Also, C_{bl} = \frac{πDN}{60}
∴ N = \frac{ C_{bl} × 60}{πD} = \frac{159.41 × 60}{π × 1.036} = 2938.7 r.p.m.
