A 150 mm diameter valve, against which a steam pressure of 2 MN / m ^{2} is acting, is closed by means of a square threaded screw 50 mm in external diameter with 6 mm pitch. If the coefficient of friction is 0.12 ; find the torque required to turn the handle.
Question : A 150 mm diameter valve, against which a steam pressure of 2...
Given : D = 150 mm = 0.15 mm = 0.15 m ; P s=2 MN / m ^{2}=2 \times 10^{6} N / m ^{2} ; d_{0}=50 mm ; p=6 mm ; \mu=\tan \phi=0.12
We know that load on the valve,
W=\text { Pressure } \times \text { Area }=p_{ S } \times \frac{\pi}{4} D^{2}=2 \times 10^{6} \times \frac{\pi}{4}(0.15)^{2} N
= 35 400 N
Mean diameter of the screw,
d=d_{0}-p / 2=50-6 / 2=47 mm =0.047 m
\therefore \tan \alpha=\frac{p}{\pi d}=\frac{6}{\pi \times 47}=0.0406
We know that force required to turn the handle,
P=W \tan (\alpha+\phi)=W\left[\frac{\tan \alpha+\tan \phi}{1-\tan \alpha \cdot \tan \phi}\right]=35400\left[\frac{0.0406+12}{1-0.0406 \times 0.12}\right]=5713 N
\therefore Torque required to turn the handle,
T=P \times d / 2=5713 \times 0.047 / 2=134.2 N – m