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The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:

Serggg [28]

Answer:

a) 2622.903 N/m^3

b) 1.38233

c)4.878811765

Explanation:

Find the void ratio using the formula:

$y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1$

Here;

$G_{s}$ is specific gravity of soil solids

$y_{w}$ is unit weight of water = 998 kg/m^3

$w$ is the moisture content = 0.17

$e$ is the void ratio

$y$ is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

$w*G_{s} = S*e ..... Eq2$

S is saturation rate

Substitute Eq 2 into Eq 1

$y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w} }{1+e}$

$14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233$

Specific gravity of soil solids

$G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765$

Saturated Unit Weight

$y_{s} = \frac{(G_{s} + e)*y_{w} }{1+e} \\=\frac{(4.878811765 + 1.38233)*998 }{1+1.38233}\\\\= 2622.902571 N/m^3$

7 0

2 years ago

What are the three elementary parts of a vibrating system?

zhenek [66]

Answer:

the three part are mass, spring, damping

Explanation:

vibrating system consist of three elementary system namely

1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.

2) Spring - the part that has elasticity and help to hold mass

3) Damping - this part considered to have zero mass and zero elasticity.

7 0

2 years ago

What does this work for

Anastaziya [24]

Answer:

it allows your dash board to light up you MPH RPM and all the other numbers on the spadomter

Explanat

8 0

2 years ago

Which of these is the coarsest grit abrasive that may be used on aluminum?

natali 33 [55]

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).

5 0

3 years ago

Read 2 more answers

11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t

lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

$v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2} }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2} }{4}) } =0.729m/s$

It is necessary to get the Reynold's number:

$Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343$

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

$Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517$

The overall heat transfer coefficient:

$Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} } }$

Here

$h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C$

Substituting values:

$Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278} } =1855.8923W/m^{2} C$

5 0

3 years ago