# If the speed of the engine fluctuates between 990 rpm and 1010 rpm in a cycle operation, the coefficient of fluctuation of speed is:

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If the speed of the engine fluctuates between 990 rpm and 1010 rpm in a cycle operation, the coefficient of fluctuation of speed is:

1. 0.05
2. 0.01
3. 0.02
4. 0.04

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Correct Answer - Option 3 : 0.02

Concept:

Maximum Fluctuation of Speed:

The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed i.e. (N1 - N2).

Coefficient of Fluctuation of Speed (Cs):

The ratio of the maximum fluctuation of speed to the mean speed is called the coefficient of fluctuation of speed.

${C_s} = \;\frac{{{N_1} - \;{N_2}}}{N} = \;\frac{{2\left( {{N_1} - \;{N_2}} \right)}}{{{N_1} + \;{N_2}}}$

N = Mean speed in r.p.m. $= \frac{{{N_1} \;+ \;{N_2}}}{2}$

Coefficient of steadiness (m):

The reciprocal of the coefficient of fluctuation of speed is known as the coefficient of steadiness.

$\begin{array}{l} m = \frac{1}{{Coeff.\;of\;fluctuation\;of\;speed}}\\ = \frac{{{N_{}}}}{{{N_{max}} - {N_{min}}}} = \frac{{{N_1} + \;{N_2}}}{{2\left( {{N_1} - {N_2}} \right)}} \end{array}$

Calculation:

Given:

N1 = 990 rpm and N2 = 1010 rpm

N = Mean speed in r.p.m. $= \frac{{{N_1} + \;{N_2}}}{2}=\frac{{{990} + \;{1010}}}{2} =1000 \;rpm$

Maximum fluctuation of speed = 1010 - 990 = 20 rpm

and ${C_s} = \;\frac{{{N_1} - \;{N_2}}}{N} =\frac{20}{1000}$

∴ CS = 0.02

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