Ask question

Ask question

All categories

3 months ago

12

A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

175 mm. In 90 s the discharge under a constant head of 38 em was 405 cm3 The sample had a dry mass of 4950 g and its Ps was 2710 kg!m3 Calculate (a) the coefficient of permeability, (b) the seepage velocity, and (c) the discharge velocity during the test.

1 answer:

Maru [3.3K]3 months ago

8 0

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Insert the values into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We will compute the coefficient of permeability

Applying the formula for permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross-sectional area

h=constant head that causes flow

Plugging the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability measures as $8.6\times10^{-3}\ cm/s$ .

(c). To ascertain the discharge velocity during the testing phase

Utilizing the discharge velocity formula

$v=ki$

$v=\dfrac{kh}{l}$

Substituting the values into the equation

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test measures 0.0187 cm/s.

Thus, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity computes at 0.0330 cm/s.

(c). The observed discharge velocity during the test equals 0.0187 cm/s.

You might be interested in

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha

Maru [3345]

Answer:

A) 5.1*10^10m B) 5.4*10^6m

Explanation:

Utilizing the formula for surface radiation P (energy per second in Watts) = emissivity constant * surface area * Stefan-Boltzmann constant * Temperature in Kelvin^4 *

2.7*10^31 = 1* 5.67*10^-8*A*11000^4

Rearranging to solve for A = 2.7*10^31 / (5.67*10^-8*1.46*10^16) = 0.3261*10^23m^2

Assuming the shape is spherical, the surface area is = 4πR^2 (radius of Rigel)

R = √(0.3261*10^23 / 4*π) = 5.1 * 10^10m

B) repeating the same calculation

2.1 *10^23 = 1*A*5.67*10^-8*10000^4 where A is the surface area of Procyon

Rearranging gives A = 2.1*10^23/(5.67*10^-8*10^16)

A = 0.37*10^15

Assuming the star is spherical;

A = 4πR^2 where R is Procyon's radius

R = √(0.37*10^15/4π) = 5.4*10^6m

4 0

2 months ago

Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g

Maru [3345]

Answer:

a) $X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

b) D does not influence the long-term results.

Explanation:

Given that

$\dfrac{dx}{dt}=-k(x-A)$

A = A0 cos(ωt)

$\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))$

$\dfrac{dx}{dt}+kx=kA_o cos(\omega t)$ This is a linear equation hence the integration factor, I

$I=e^{\int kdt}$

$I=e^{kt}$ Now using the characteristics of linear equations

$e^{kt} X=\int e^{kt} kA_o cos(\omega t) dt +C$

$e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C$

b) At t= 0

$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

Thus, the initial condition

does not affect the long-term outcome.

$X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C$

5 0

3 months ago

Albert uses as his unit of length (for walking to visit his neighbors or plowing his fields) the albert (a), the distance albert

Yuliya22 [3333]

To tackle this question, we know the following:

1 Albert equals 88 meters.

1 A = 88 m.

Initially, we square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since 1 acre equals 4,050 m^2, let’s divide both sides by 7,744 to find out how many acres match this value:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

<span>Thus, one acre is approximately 0.52 square alberts.</span>

7 0

3 months ago