Question

The torque (in Nm) exerted on the crankshaft of a two stroke engine can be described as $T=10000+1000sin2\theta -1200cos2\theta$, where $\theta$ 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

• A. 104.719

Answer 1

The correct option is A 104.719
$T=10000+1000sin2\theta -1200cos2\theta$

$N=100\text{rpm}$

$\omega =\frac{2\pi \times 100}{60}=10.472\text{rad/s}$

$T_{mean}=\frac{1}{2\pi }{\int }_0^{2\pi }[10,000+1000sin2\theta -1200cos2\theta ]d\theta$

$=10,000[2\pi -0]+\frac{1000}{2}\times [cos2\theta {]}_0^{2\pi }-\frac{1200}{2}[sin2\theta {]}_0^{2\pi }$

$=\frac{1}{2\pi }[10,000.2\pi +500.[cos4\pi -cos0]-600[sin4\pi -sin0]$

$=(10,000\times 2\pi )+500[1-1]-600[0-0]$

$=\frac{10000\times 2\pi }{2\pi }=10000\text{N-m}$

$\text{Power}=10000\times \frac{2\pi \times 100}{60}=104719.75W$

$=104.719\text{kW}$