All categories

3 years ago

6) A deep underground cavern Contains 980 cuft

Engineering

1 answer:

Elza [17]3 years ago

3 0

Answer:

15625 moles of methane is present in this gas deposit

Explanation:

As we know,

PV = nRT

P = Pressure = 230 psia = 1585.79 kPA

V = Volume = 980 cuft = 27750.5 Liters

n = number of moles

R = ideal gas constant = 8.315

T = Temperature = 150°F = 338.706 Kelvin

Substituting the given values, we get -

1585.79 kPA * 27750.5 Liters = n * 8.315 * 338.706 Kelvin

n = (1585.79*27750.5)/(8.315 * 338.706) = 15625

You might be interested in

In almost all cases, touching power lines or coming into contact with energized sources will result in what?

Fynjy0 [20]

Answer:

electrocution

Explanation:

You will end up getting shoked because the electricity is attracted to the water in your body.

6 0

3 years ago

A thin plastic membrane is used to separate helium from a gas stream. Under steady-state conditions the concentration of helium

Juli2301 [7.4K]

Answer:

N_A=1.5*10^-8 kmol/s.m^2

Explanation:

KNOWN:

Molar concentration of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness.

FIND:

Molar diffusion flux.

ASSUMPTIONS:

(1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = C_A + C_B.

ANALYSIS: The molar flux may be obtained from

N_A=D_AB/L(C_A,1-C_A,2)

=10^-9 m^2/s/0.001 m(0.02-0.005)kmol/m^3

N_A=1.5*10^-8 kmol/s.m^2

COMMENTS: The mass flux is:

n_A,x=M_a*N_A,x

n_A,x=6*10^-8 kg/s m^2

5 0

3 years ago

Sexual Harassment falls under Title IX. What are the two types of Sexual Harassment, not examples, but there are only two kinds.

Anit [1.1K]

Answer:

According to the Equal Employment Opportunity Commission (EEOC), there are two types of sexual harassment claims: "quid pro quo" and "hostile work environment." The EEOC provides guidance on defining sexual harassment and establishing employer liability. hope that helps

Explanation:

4 0

3 years ago

In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images

Zarrin [17]

Answer:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

$A= \frac{r^5 \rho}{t^2}$

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant $\rho_{air}$ .

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= $\frac{ML^2}{T^2}$ A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as $\frac{L}{T}$

[t]=T represent the time

$[\rho]=\frac{M}{L^3}$ represent the density.

Solution to the problem

And if we analyze the function for r we got this:

$[r]=L=[A]^x [\rho]^y [t]^z$

And if we replpace the formulas for each on we got:

$[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z$

And using algebra properties we can express this like that:

$[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}$

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

$x+y =0 , 2x-3y=1, -2x+z=0$

We can solve for x like this x=-y and replacing into quation 2 we got:

$2(-y)-3y = 1$

$-5y = 1$

$y= -\frac{1}{5}$

And then we can solve for x and we got:

$x = -y = -(-\frac{1}{5})=\frac{1}{5}$

And if we solve for z we got:

$z=2x =2 \frac{1}{5}=\frac{2}{5}$

And now we can express the radius in terms of the dimensional analysis like this:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

$A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}$

$A= \frac{r^5 \rho}{t^2}$

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed $\rho =1.2 kg/m^2$ , and replacing we got:

$A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}$

And we can convert this into ergs we got:

$A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs$

And then we know that 1 g of TNT have $4x10^4 erg$

And we got:

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

3 0

3 years ago

A technique for resolving complex repetitive waveforms into sine or cosine waves and a DC component is known as:

tatiyna

Answer:

(A) Fourier Analysis

Explanation:

Fourier Analysis

It is the form of study of the way a general functions can be represented via the sum of the simple trigonometric functions .

It is named after Joseph Fourier , who represented a function as a sum of its trigonometric functions and it simplifies the study of the heat transfer .

Hence ,

The technique for resolving the complex repetitive waveforms into the sine or the cosine waves and the DC component is known as the Fourier Analysis .

7 0

4 years ago