SSC JE Mechanical Engineering 2014 Set 2 Paper

Question : 117 of 200

Marks: +1, -0

A single-stage impulse turbine with a diameter of 120 cm runs at 3000 rpm. If the blade speed ratio is 0.42, the inlet velocity of steam will be :

Solution:

Given D = 120 cm, speed N = 3000 rpm, speed ratio

\ϕ= 0.42.

\ϕ=µ/v={πDN}/{60V}

⇒

0.42 = {3.14 × 120× 3000}/{60 XV}

So V = 449.0 m /s

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