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An engine develops a power of 360 kw, when rotating at 30 revolutions per second. The Torque required to deliver this power is
  • a)
    191.08 Nm
  • b)
    19108 Nm
  • c)
    1910.8 Nm
  • d)
    19.108 Nm
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
An engine develops a power of 360 kw, when rotating at 30 revolutions ...
The power delivered by the torque τ exerted on rotating body is given by
P=τω or τ=P/ω
Here P=360KW=360000 Watt
ω=30 x 2π rad/sec,
ω=60π rad/sec
now,
τ=360000 /60×3.14Nm
τ= 1910.8 Nm
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Most Upvoted Answer
An engine develops a power of 360 kw, when rotating at 30 revolutions ...
The power delivered by the torque τ exerted on rotating body is given by
P=τω or τ=P/ω
Here P=360KW=360000 Watt
ω=30 x 2π rad/sec,
ω=60π rad/sec
now,
τ=360000 /60×3.14Nm
τ= 1910.8 Nm
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Community Answer
An engine develops a power of 360 kw, when rotating at 30 revolutions ...
Understanding Power and Torque
To calculate torque from power and rotational speed, we can use the formula:
Torque (T) = Power (P) / Angular Velocity (ω)
Converting Power
- Given power (P) = 360 kW = 360,000 W
Calculating Angular Velocity
- Rotational speed (N) = 30 revolutions per second
- Angular velocity (ω) in radians per second is calculated as:
  • ω = 2πN
  • ω = 2π × 30 = 60π rad/s
  • ω ≈ 188.5 rad/s


Calculating Torque
- Now, substituting the values into the torque formula:
  • T = P / ω
  • T = 360,000 W / (60π rad/s)
  • T ≈ 360,000 / 188.5
  • T ≈ 1910.8 Nm


Conclusion
Thus, the torque required to deliver this power is approximately 1910.8 Nm. Therefore, the correct answer is option 'C'.
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