An axial flow impulse steam turbine has a mean rotor diameter of 0.55m & run's at 3300rpm. The speed ratio is 0.45 & the nozzle at the rotor inlet is 20°. Such that Vu1 is positive. The blade velocity co-efficient is 0.91.
Find:-
1)Rotor blade angle assuming axial exit
2) Power developed
3) Axial thrust Assume mass flow rate of 10kg/s.
Answers
Answered by
1
Answer:
see in google
Step-by-step explanation:
please mark me as brainliest
Answered by
0
Step-by-step explanation:
given:
D=.55m
N=3300 rpm
speed ratio=0.45
nozzle angle=20°
Blade velocity coefficient=0.91
Blade inlet angle=Blade outlet angle
m=10kg/s
Cbl=πDN/60 =>Cbl=(π×0.55×3300)/60
Cbl=95.033m/s
speed ratio =Cbl/C1
0.45=95.033/C1
C1=211.18m/s
Blade inlet and outlet angle=35°
Cf1=75.07
Cf2=67.47
Cw1=223.32
Cw2=3.04
CW=220.28
Power=(m×Cw×Cbl)/1000
P=209.33kw
Axial trust=m(Cf1-Cf2)
Axial trust=76N
Similar questions