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
Answer:
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Explanation:
Answer:
Bending stress at point 3.96 is \sigma_b = 1.37 psi
Explanation:
Given data:
Bending Moment M is 4.176 ft-lb = 50.12 in- lb
moment of inertia I = 144 inc^4
y = 3.96 in

putting all value to get bending stress

Bending stress at point 3.96 is
= 1.37 psi
Answer:
0, 1, 2, 3, 6, 11, 20, 37, 68
Explanation:
Each number in the series, starting with 3, is the total of the three numbers before it. The following number is the total of the preceding three numbers, according to the pattern. The pattern's next number is 125.