Answer:
(a) dynamic viscosity = ![1.812\times 10^{-5}Pa-sec](https://tex.z-dn.net/?f=1.812%5Ctimes%2010%5E%7B-5%7DPa-sec)
(b) kinematic viscosity = ![1.4732\times 10^{-5}m^2/sec](https://tex.z-dn.net/?f=1.4732%5Ctimes%2010%5E%7B-5%7Dm%5E2%2Fsec)
Explanation:
We have given temperature T = 288.15 K
Density ![d=1.23kg/m^3](https://tex.z-dn.net/?f=d%3D1.23kg%2Fm%5E3)
According to Sutherland's Formula dynamic viscosity is given by
, here
μ = dynamic viscosity in (Pa·s) at input temperature T,
= reference viscosity in(Pa·s) at reference temperature T0,
T = input temperature in kelvin,
= reference temperature in kelvin,
C = Sutherland's constant for the gaseous material in question here C =120
![\mu _0=4\pi \times 10^{-7}](https://tex.z-dn.net/?f=%5Cmu%20_0%3D4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D)
= 291.15
when T = 288.15 K
For kinematic viscosity :
![\nu = \frac {\mu} {\rho}](https://tex.z-dn.net/?f=%5Cnu%20%3D%20%5Cfrac%20%7B%5Cmu%7D%20%7B%5Crho%7D)
![kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec](https://tex.z-dn.net/?f=kinemic%5C%20viscosity%3D%5Cfrac%7B1.812%5Ctimes%2010%5E%7B-5%7D%7D%7B1.23%7D%3D1.4732%5Ctimes%2010%5E%7B-5%7Dm%5E2%2Fsec)
Answer:
a. 0.28
Explanation:
Given that
porosity =30%
hydraulic gradient = 0.0014
hydraulic conductivity = 6.9 x 10⁻4 m/s
We know that average linear velocity given as
![v=\dfrac{K}{n_e}\dfrac{dh}{dl}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7BK%7D%7Bn_e%7D%5Cdfrac%7Bdh%7D%7Bdl%7D)
![v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B6.9%5Ctimes%2010%5E%7B-4%7D%7D%7B0.3%7D%5Ctimes0.0014%5C%20m%2Fs)
![v=3.22\times 10^{-6}\ m/s](https://tex.z-dn.net/?f=v%3D3.22%5Ctimes%2010%5E%7B-6%7D%5C%20m%2Fs)
The velocity in m/d ( 1 m/s =86400 m/d)
v= 0.27 m/d
So the nearest answer is 'a'.
a. 0.28
Explanation:
Instantaneous center:
It is the center about a body moves in planer motion.The velocity of Instantaneous center is zero and Instantaneous center can be lie out side or inside the body.About this center every particle of a body rotates.
From the diagram
Where these two lines will cut then it will the I-Center.Point A and B is moving perpendicular to the point I.
If we take three link link1,link2 and link3 then I center of these three link will be in one straight line It means that they will be co-linear.
![I_{12},I_{23},I_{31} all\ are\ co-linear.](https://tex.z-dn.net/?f=I_%7B12%7D%2CI_%7B23%7D%2CI_%7B31%7D%20all%5C%20are%5C%20co-linear.)
Answer:
Relative density = 0.545
Degree of saturation = 24.77%
Explanation:
Data provided in the question:
Water content, w = 5%
Bulk unit weight = 18.0 kN/m³
Void ratio in the densest state,
= 0.51
Void ratio in the loosest state,
= 0.87
Now,
Dry density, ![\gamma_d=\frac{\gamma_t}{1+w}](https://tex.z-dn.net/?f=%5Cgamma_d%3D%5Cfrac%7B%5Cgamma_t%7D%7B1%2Bw%7D)
![=\frac{18}{1+0.05}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B18%7D%7B1%2B0.05%7D)
= 17.14 kN/m³
Also,
![\gamma_d=\frac{G\gamma_w}{1+e}](https://tex.z-dn.net/?f=%5Cgamma_d%3D%5Cfrac%7BG%5Cgamma_w%7D%7B1%2Be%7D)
here, G = Specific gravity = 2.7 for sand
![17.14=\frac{2.7\times9.81}{1+e}](https://tex.z-dn.net/?f=17.14%3D%5Cfrac%7B2.7%5Ctimes9.81%7D%7B1%2Be%7D)
or
e = 0.545
Relative density = ![\frac{e_{max}-e}{e_{max}-e_{min}}](https://tex.z-dn.net/?f=%5Cfrac%7Be_%7Bmax%7D-e%7D%7Be_%7Bmax%7D-e_%7Bmin%7D%7D)
= ![\frac{0.87-0.545}{0.87-0.51}](https://tex.z-dn.net/?f=%5Cfrac%7B0.87-0.545%7D%7B0.87-0.51%7D)
= 0.902
Also,
Se = wG
here,
S is the degree of saturation
therefore,
S(0.545) = (0.05)()2.7
or
S = 0.2477
or
S = 0.2477 × 100% = 24.77%