Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S = 
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M
Answer:
a) 
b)
Explanation:
Given that:
diameter d = 12 in
thickness t = 0.25 in
the radius = d/2 = 12 / 2 = 6 in
r/t = 6/0.25 = 24
24 > 10
Using the thin wall cylinder formula;
The valve A is opened and the flowing water has a pressure P of 200 psi.
So;
b)The valve A is closed and the water pressure P is 250 psi.
where P = 250 psi
The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below
Answer:
80grit
Explanation:
80 grit is coarsest grit that may be used on aluminum
The lowest grit sizes range from 40 to 60. From the given options 80 grit is practically available grit.
What is a sandpaper used for?
They are essentially used for surface preparation. Sandpaper is produced in a range of grit sizes and is used to remove material from surfaces, either to make them smoother (for example, in painting and wood finishing), to remove a layer of material (such as old paint), or sometimes to make the surface rougher (for example, as a preparation for gluing).
Answer:
a) 0.3
b) 3.6 mm
Explanation:
Given
Length of the pads, l = 200 mm = 0.2 m
Width of the pads, b = 150 mm = 0.15 m
Thickness of the pads, t = 12 mm = 0.012 m
Force on the rubber, P = 15 kN
Shear modulus on the rubber, G = 830 GPa
The average shear strain can be gotten by
τ(average) = (P/2) / bl
τ(average) = (15/2) / (0.15 * 0.2)
τ(average) = 7.5 / 0.03
τ(average) = 250 kPa
γ(average) = τ(average) / G
γ(average) = 250 kPa / 830 kPa
γ(average) = 0.3
horizontal displacement,
δ = γ(average) * t
δ = 0.3 * 12
δ = 3.6 mm
Answer:
E= 15 GPa.
Explanation:
Given that
Length ,L = 0.5 m
Tensile stress ,σ = 10.2 MPa
Elongation ,ΔL = 0.34 mm
lets take young modulus = E
We know that strain ε given as
We know that
Therefore the young's modulus will be 15 GPa.
