SSC JE Mechanical Engineering 11 Oct 2020 Shift 1 Papers

Question : 115 of 200

Marks: +1, -0

A turbine rotates at 600 rpm. Of moment of momentum of working fluid is reduced by 15.915 kNm, then power developed in MW is:

Solution:

N = 600 rpm; T = 15.915 kN-m ⇒ 15.915 × 10

^{3}

Nm ;

{P}={2 \×\π\×600 \×15.915 \×10^{3}}/{60}

P = 0.999 MW = 1 MW.

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