Mechanical Engineering - Steam Nozzles and Turbines - Discussion

Discussion Forum : Steam Nozzles and Turbines - Section 2 (Q.No. 1)
1.
A single stage impulse turbine with a diameter of 1.2 m runs at 3000 r.p.m. If the blade speed ratio is 0.42, then the inlet velocity of steam will be
79 m/s
188 m/s
450 m/s
900 m/s
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
5 comments Page 1 of 1.

Parameshwar said:   3 years ago
D = 1.2M.
N =3000rpm,
π value= 22/7.
Velocity formule =πDN/60.
=(22/7 * 1.2 * 3000)/60.
=188.49m/s.

V = 188.49/0.42 = 488.80.
(1)

Phathutshedzo said:   8 years ago
The steam leaves the moving blades of the second stage at an angle 30 degree to the rotation of the blades, the inlet angle of the fixed blades is 15 degree, the average blade velocity is m/s, the velocity of flow at the inlet to the second stage is 90m/s, the velocity coefficient for all blades is 0.96, use the scale 1mm=4m/s and construct and the velocity diagram for the turbine in your answer book, indicate the length of all lines as well as the magnitude of the angles on the diagrams.
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Satish Hadiya said:   8 years ago
Blade speed ratio = blade speed / inlet velocity of steam.
Inlet velocity of steam = blade speed /blade speed ratio.
= (3.14*1.2)/0.47.
= 450 m/s.

Raj aadhavan said:   9 years ago
Blade speed ratio = (blade speed/absolute inlet velocity of steam).

= u/v1 = 0.42.

u = (piDN/60) = (3.14 * 1.2 * 3000)/60 = 188.5 m/s.

v1 = u/0.42 = 450m/s (Approximately).
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Suman said:   1 decade ago
v = (22/7*1.2*3000)/60 = 188.49m/s.

Then, u = (188.49/.42) = 448.78 = 450 m/s.
(1)

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