Question

The high idle speed of engine is 2240 rpm. The peak torque of 180 Nm occurs, at 1450 engine rpm. If the lugging ability is 28 Nm. The engine power in KW at governor’s maximum position will be _________ (Governor regulation = 11.5%)

(a) 42.5 KW

(b) 51.49 KW

(c) 23.12 KW

(d) 31.3 KW

This question was posed to me in examination.

Origin of the question is Principles of Governor in chapter Governor of Farm Machinery

Answer 1

The correct option is (d) 31.3 KW

For explanation: Nmax = 2240 rpm

At 180 Nm = N = 1450 rpm

Lugging = 28 Nm

So, at 180 – 28 = 152 Nm, let speed be Nmin

GR = 11.5% = 0.115%

0.115 = \(\frac{[2(2240-Nmin)]}{2240+Nmin} \)

Nmin = 1966.40 rpm

P = 2πNτ = 2 * π * 1966.40 * 152 = 31.3 KW