Question

The torque (in N-m) exerted on the crank shaft of a two stroke engine can be described as T = 10000 + 1000 sin 2θ – 1200 cos 2θ, where θ is the crank angle as measured from inner dead center position. Assuming the resisting torque to be constant, the power (in kW) developed by the engine at 100 rpm is __________.

Answer 1

Concept:

Power (P) = T_mean × ω

T_mean = Mean torque

ω = Angular velocity = 2πN/60

P = 2πNT/60

N = RPM of engine 

For a two-stroke engine,

\({T_{mean}} = \frac{1}{π }\mathop \smallint \nolimits_0^π Tdθ \)

For a four-stroke engine,

\({T_{mean}} = \frac{1}{{2π }}\mathop \smallint \nolimits_0^{2π } Tdθ \)

Calculation:

T = (10000 + 1000 sin2θ - 1200 cos2θ) N-m

N = 100 rpm

\(\begin{array}{l} {T_{mean}} = \frac{1}{π }\mathop \smallint \limits_0^π Tdθ = \frac{1}{π }\mathop \smallint \limits_0^π \left( {10000 + 1000sin2θ - 1200cos2θ } \right)dθ \\ \Rightarrow {T_{mean}} = \frac{1}{π }\left[ {10000θ - 500cos2θ - 600sin2θ } \right]\begin{array}{*{20}{c}} π \\ 0 \end{array} = 10000\ N - m \end{array}\)

and Power,

\(Power = \frac{{2π NT}}{{60}} = \frac{{2 × π × 100 × 10000}}{{60}} = 104719.75\ W\ or\ 104.719\ kW\)