Question

A single cylinder oil engine has a stroke of 38 cm and crank rotates at 360 rpm. The reciprocating parts weigh 670 N and revolving parts are equivalent to 800 N at crank radius 'r'. A revolving balance weight is introduced at a radius of 15 cm to balance the whole of revolving and one half of reciprocating parts. The magnitude of balancing weight required and the residual unbalanced force on crank shaft are

• A. $1487.66\text{N,}9221.3\text{N}$
• B. $1437.66\text{N,}9221.3\cos \theta \text{N}$
• C. $1437.66\text{N,}9221.3\text{N}$
• D. $1492.66\text{N,}9221.3\sin \theta \text{N}$

Answer 1

The correct option is C $1437.66\text{N,}9221.3\text{N}$


Given:
$2r=38\text{cm}$
$r=19\text{cm}$

$m_{\text{rec}}=\frac{670}{9.81}=68.3\text{kg}$

$m_{\text{rec}}=\frac{800}{9.81}=81.55\text{kg}$

$\omega =\frac{2\pi N}{60}=\frac{2\pi \times 360}{60}=37.68\text{rad/s}$

$m_b.\left(0.15\right)\left({\omega }^2\right)=(m_{\text{rev}}+\frac{1}{2}{m}_{\text{rec}})\times 0.19{\omega }^2$

$m_b=(81.55+34.15)\times \frac{19}{15}=146.55\text{kg}$
$W_b=m_b\times 9.81=1437.66\text{N}$

Residual unbalanced force is

$=\sqrt{{(\frac{1}{2}{m}_{\text{rec}}.(0.196\left){\omega }^2\cos \theta \right)}^2+{\left(\frac{1}{2}{m}_{\text{rec}}\right(0.19\left){\omega }^2\sin \theta \right)}^2}$

$(\because m_{\text{rev}}\text{is completely balanced})$

$=\frac{1}{2}{m}_{\text{rec}}\left(0.19\right){\omega }^2$

$=9221.3\text{N}$