Question 6.34: A 50% reaction turbine (with symmetrical velocity triangles)...
A 50% reaction turbine (with symmetrical velocity triangles) running at 400 r.p.m. has the exit angle of the blades as 20° and the velocity of steam relative to the blades at the exit is 1.35 times the mean blade speed. The steam flow rate is 8.33 kg/s and at a particular stage the specific volume is 1.381 m³/kg. Calculate for this stage :
(i) A suitable blade height, assuming the rotor mean diameter 12 times the blade height, and
(ii) The diagram work. (N.U.)
Speed, N = 400 r.p.m. ; α = 20°
C_{r_0} = C_{1} = 1.35 C_{bl} ; \dot{m}_{s} = 8.33 kg/s
v = 1.381 m³/kg ; D = 12 h
(i) Blade height, h :
Refer Fig. 49.
Axial flow velocity, C_{f_1} = C_{f_0} = C_{f} = C_{1} \sin α
= 1.35 C_{bl} \sin 20°
= 0.4617 C_{bl}
Area of flow, A = πDh = πD × \frac{D}{12} = \frac{πD²}{12}
Mass flow rate, \dot{m} = \frac{A C_{f}}{v} or 8.33 = \frac{A × 0.4617 C_{bl}}{1.381}
or \frac{8.33 × 1.381 }{0.4917} = A × C_{bl} = \frac{πD²}{12} × \frac{πDN}{60}
or 24.916 = \frac{π²D³ × 400}{720} or D³ = \frac{24.919 × 720}{π² × 400 } or D = 1.656 m
∴ Blade height, h = \frac{D}{12} = \frac{1.656}{12} = 0.138 m or 138 mm.
(ii) The diagram work :
Diagram work = \dot{m} × C_{bl} (C_{w_1} + C_{w_0})
= \dot{m} × C_{bl} (2 C_{1} \cos α – C_{bl})
= 8.33 × C_{bl} (2 × 1.35 C_{bl} \cos 20° – C_{bl})
= 8.33 × C_{bl}² (2 × 1.35 \cos 20° – 1 )
= 8.33 × ( \frac{πDN}{60} )² × 1.537 = 8.33 × ( \frac{π × 1.656 × 400}{60} )² × 1.537
= 15401.2 W or 15.4 kW.
