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Chapter 6

Q. 6.35

300 kg/min of steam (2 bar, 0.98 dry) flows through a given stage of a reaction turbine. The exit angle of fixed blades as well as moving blades is 20° and 3.68 kW of power is developed. If the rotor speed is 360 r.p.m. and tip leakage is 5 per cent, calculate the mean drum diameter and the blade height. The axial flow velocity is 0.8 times the blade velocity.                                                                                                                                                                     (Roorkee University)

Step-by-Step

Verified Solution

Rate of flow of steam through the turbine,    \dot{m}_{s}   =  \frac{300}{60}   =  5  kg/s
Pressure and condition of steam, p = 2 bar, x = 0.98.
The exit angles of fixed blades as well as moving blades, α = φ = 20°
Power developed,                             P = 3.68 kW
Speed of the rotor,                            N = 360 r.p.m.
Tip leakage                                          = 5 per cent
Axial flow velocity,                              C_{f}   =   0.8   C_{bl}   (blade velocity)
Refer Fig. 49.

Mean drum diameter, D :

Mean blade velocity,                                C_{bl}   =  \frac{πDN}{60}  =  \frac{πD    ×     360}{60}  = 18.85 D m/s

Power developed,                             P =  \frac{\dot{m}_{s} C_{bl} C_{w}}{1000}

or                                            3.67 = \frac{(5    ×    0.95)   ×    18.85   D    ×    C_{w} }{1000}

∴                                                            C_{w} =  \frac{3.67      ×  1000}{(5    ×    0.95)   ×    18.85   D}  =  \frac{40.988}{D}

Assuming Parson’s reaction turbine, we have

C_{f_1}    =   C_{1}     sin    α        or    C_{1}  =  \frac{C_{f_1}  }{sin    α}  =  \frac{0.8    C_{bl}  }{sin    α}  =  \frac{0.8   ×    18.85    D}{sin     20°} = 44.091 D
(C_{f_1}   =  C_{f_0}  =    C_{f})

Also,                                                C_{w}   =   2  C_{1}    cos   α   –  C_{bl}    or          \frac{40.988}{D} = 2 × 44.091  D  \cos  20° – 18.85  D = 64.01  D

or                         40.988 = 64.01 D²    or    D = 0.8 m or 800 mm.

Blade height, h :

Mean steam flow rate,      \dot{m}_{s}   =  \frac{πDhC_{f}}{x v_{g}} 

or                            (5 × 0.95) =  \frac{π    ×   0.8    ×   h    ×   C_{1}     sin    α}{0.98   ×    0.885}  =  \frac{π    ×   0.8    ×   h    ×   (44.091  D      ×    sin       20°)}{0.98   ×    0.885}

(At 2 bar :   v_{g} = 0.885 kg/m³)

or                                h =  \frac{(5     ×    0.95)    ×    0.98   ×    0.885}{π    ×   0.8    ×   (44.091    ×    0.80      ×    0.3420)}    = 0.1359 m        or        135.9 mm.

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