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Explanation:
Answer:
5984.67N
Explanation:
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?
from continuity equation
v1A1=v2A2
equation of continuity
v1=4ft /s=1.21m/s
d1=14 inch=.35m
d2=14-2=0.304m
A1=pi*d^2/4
0.096m^2
a2=0.0706m^2
from continuity once again
1.21*0.096=v2(0.07)
v2=1.65
force on the pipe
(p1A1- p2A2) + m(v2 – v1)
from bernoulli
p1 + ρv1^2/2 = p2 + ρv2^2/2
difference in pressure or pressure drop
p1-p2=2psi
13.789N/m^2=rho(1.65^2-1.21^2)/2
rho=21.91kg/m^3
since the pipe is cylindrical
pressure is egh
13.789=21.91*9.81*h
length of the pipe is
0.064m
AH=volume of the pipe(area *h)
the mass =rho*A*H
0.064*0.07*21.91
m=0.098kg
(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)
force =5984.67N
An o ring intended for use in a hydraulic system using MIL-H-5606 (mineral base) fluid will be marked with a blue stripe or dot.
Answer:
the answer how you analyzs the problwm
Answer:
fracture will occur as the value is less than E/10 (= 22.5)
Explanation:
If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.
![\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}](https://tex.z-dn.net/?f=%5Csigma_m%20%3D%202%5Csigma_o%20%5B%5Cfrac%7Ba%7D%7B%5Crho_t%7D%5D%5E%7B1%2F2%7D)
= 15 GPa
fracture will occur as the value is less than E/10 = 22.5
