Answer:
L= 312.75 mm
Explanation:
given data
elastic modulus E = 106 GPa
cross-sectional diameter d = 3.9 mm
tensile load F = 1660 N
maximum allowable elongation ΔL = 0.41 mm
to find out
maximum length of the specimen before deformation
solution
we will apply here allowable elongation equation that is express as
ΔL = ....................1
put here value and we get L
L =
solve it we get
L = 0.312752 m
L= 312.75 mm
Answer:
Explanation:
First we should recall how Newton's laws relates shear stress to a fluid's velocity profile:
where tau is the shear stress, mu is viscosity, v is the fluid's velocity and y is the direction perpendicular to flow.
Now, in this case we have a parabolic velocity profile, and also we know that the fluid's velocity is zero at the boundary (no-slip condition) and that the vertex (maximum) is at and the velocity at that point is
We can put that in mathematical terms as:
From the no-slip condition, we can deduce that and so we are left with just two terms:
We know that the vertex is at and so we can rewrite the last equation as:
where k and h are constants to be determined. First we check that :
So we found that h was the maximum velocity for the fluid, now we have to determine k, for that we need to make use of the no-slip condition.
And thus we find that the final expression for the fluid's velocity is:
where v is in mm/s and y is in mm.
In SI units it would be:
To calculate the shear stress, we need to take the derivative of this expression and multiply by the fluid's viscosity:
for we have:
Which is our final result
Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa
Explanation:
Given that;
patm = 79 kPa, h = 13 in of H₂O,
A sketch of the problem is uploaded along this answer.
Now
pA = patm + 13 in of H₂O ( h × density × g )
pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)
pA = 82.23596 kPa
the absolute static pressure in the gas cylinder is 82.23596 kPa
