Answer: Because MM's CEO, Crosscut Sal, is a stickler for keeping machinery running, the company stocks quick-change replacement modules for the two most common ..
Explanation:
Answer:
Fluids
Explanation:
Fluids has special properties that allow forces and pressure to be distributed evenly within them.
- Fluids are gases and liquids whose intermolecular forces of attraction are generally weak or non-existence.
- Therefore, when pressure is applied to them, it permeates evenly on all parts.
- Their ability to tend to randomness makes liquids and gases very viable for distributing pressure.
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
In places with cold winters, space heating systems have a fundamental role in buildings. Without them, indoor temperatures would quickly become unsuitable for human occupancy. The local weather is one of the most important factors when designing a heating system; if two identical buildings are developed in Miami FL and New York City, the heating load will be much higher for the NYC property.
Answer:
w = 10.437 kips
deflection at 1/4 span 20.83\E ft
at mid span = 1.23\E ft
shear stress 7.3629 psi
Explanation:
area of cross section = 18*76
length of span = 32 ft
moment = 334 kips-ft
we know that
moment = load *eccentricity
334 = w * 32
w = 10.437 kips
deflection at 1/4 span
= 20.83\E ft
at mid span
shear stress
