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250 litres/sec of water flowing in pipe having diameter of 300mm. If pipe is bent by $135^{\circ}$, Find magnitude, direction of resultant force on bend.The pressure of water flowing $39.24N/cm^{2}$
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written 5.6 years ago by |
Given P=$P_{1}=P_{2}=39.24N/cm^{2}=39.24\times 10^{4} N/m^{2}$
Q=250 d/s = $0.25m^{3}/s$
d=$d_{1}=d_{2}$=300mm=0.3 m
A=$A_{1}=A_{2}=\frac{\pi}{4}\times d^{2}=\frac{\pi}{4}\times (0.3)^{2}=0.67069m^{3}$
Q=AV
Velocity of water V=$V_{1}=v_{2}=\frac{Q}{A}=\frac{0.05}{0.077069}$=3.537m/s
From impulse-momentum principle, the force on x direction
F=$P_{1}A_{1}-(-P_{2}A_{2}cos \theta)+PQ[V_{1}-(-V_{2}Cos \theta)$
=PA$(1+cos\theta)+PQV(1+cos\theta)$
=$(39.24\times 10^{4})\times 0.07069(1+cos45)+1000\times 0.25\times 3.537(1+cos45)$
=48863 N
Similarity the force along y-direction using momentum equation can be calculated as
o-$P_{2}A_{2}sin\theta-Fy=PQ[V_{2}sin\theta-0]$ …
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