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3 years ago

Do you understand entropy? Why the concept of entropy is difficult to engineering students?

Engineering

1 answer:

Darya [45]3 years ago

4 0

Answer:

Entropy:

Entropy is the measure of randomness of system.In other words the entropy is the measurement of tendency of system towards the disorder.

The concept of entropy arise from second law of thermodynamics.It is given as follows

$ds=\int \dfrac{dQ}{T}$

Entropy is a extensive property of system .

Entropy of universe = Entropy of system + Entropy of surrounding.

The entropy of the system can be zero,positive and negative.But entropy of the surrounding can not be negative,but it can be zero or positive.

Actually the concept of entropy is difficult to understand because we can not visualize because it is not like beam and like rods.Only we have to realize that there is entropy.

You might be interested in

A hollow, spherical shell with mass 2.00kg rolls without slipping down a slope angled at 38.0?.

mezya [45]

Answer:

$\mu = 0.31$

Explanation:

Given data:

mass = 2.00 kg

slope angle = 38.0

From figure

balancing force

$mgsin\theta - f = ma$ .....1

Balancing torque

$F_R = \frac{2}{3} mR^2 \alpha$ ......2

for pure rolling

$\alpha = \frac{a}{R}$

$F = \frac{2}{3} ma$

from 1 and 2nd equation

$mgsin\theta - \frac{2}{3}ma = ma$

$mgsin\theta = \frac{5}{3} ma$

$a = \frac{3}{5} g sin\theta$

$= \frac{3\theta 9.8 sin 38}{5} = 3.62 m/s^2$

$F =\mu N$

$= \frac{2}{3} ma$

$= \frac{2}{3} 2\times 3.62 = 4.83 N$

N =normal force $= mgsin\theta = 2 \times 9.8 sin 38 = 15.44 N$

$\mu \times 15.44 = 4.83$

solving for coefficent of friction we get

$\mu = 0.31$

4 0

3 years ago

The current through a 10-mH inductor is 10e−t∕2 A. Find the voltage and the power at t = 8 s.

NNADVOKAT [17]

Answer:

voltage = -0.01116V

power = -0.0249W

Explanation:

The voltage v(t) across an inductor is given by;

v(t) = L $\frac{di(t)}{dt}$ -----------(i)

Where;

L = inductance of the inductor

i(t) = current through the inductor at a given time

t = time for the flow of current

From the question:

i(t) = $10e^{-t/2}$ A

L = 10mH = 10 x 10⁻³H

Substitute these values into equation (i) as follows;

v(t) = $(10*10^{-3})\frac{d(10e^{-t/2})}{dt}$

Solve the differential

v(t) = $(10*10^{-3})\frac{-1*10}{2} (e^{-t/2})$

v(t) = -0.05 $e^{-t/2}$

At t = 8s

v(t) = v(8) = -0.05 $e^{-8/2}$

v(t) = v(8) = -0.05 $e^{-4}$

v(t) = -0.05 x 0.223

v(t) = -0.01116V

(b) To get the power, we use the following relation:

p(t) = i(t) x v(t)

Power at t = 8

p(8) = i(8) x v(8)

i(8) = i(t = 8) = $10e^{-8/2}$

i(8) = $10e^{-4}$

i(8) = 10 x 0.223

i(8) = 2.23

Therefore,

p(8) = 2.23 x -0.01116

p(8) = -0.0249W

7 0

3 years ago

Read 2 more answers

Explain the function of both of the organelles pictured below.

just olya [345]

Answer:

The mitochondria's job is to produce the majority of the chemical energy required to power a cell's metabolic reactions.

The chloroplast's job is to create food for a cell through the process of photosynthesis.

6 0

3 years ago

). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially

Delicious77 [7]

Answer:

$\frac{e'_z}{e_z} = 0.87142$

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

εx = - [ σx - ν( σy + σz ) ] / E

εy = - [ σy - ν( σx + σz ) ] / E

εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

εx = [ ν*σz ] / E

εy = [ ν*σz ] / E

εz = - [ σz ] / E .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

εx' = - [ σx' - ν( σy' + σz ) ] / E = [ ν*σz ] / 2E

εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

εx' = - [ σy' - ν( σx' + σz ) ] / E = [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note: σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

- [ σx' - ν( σx'+ σz ) ] = [ ν*σz ] / 2

σx' ( 1 - v ) = [ ν*σz ] / 2

σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

εz' = - [ σz - ν( σy' + σx' ) ] / E

εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

$\frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ] }{-\frac{s_z}{E}} \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142$ ... Answer

5 0

3 years ago

(TCO 1) Name one disadvantage of fixed-configuration switches over modular switches. a. Ease of management b. Port security b. F

frutty [35]

Answer:

The answer is A.

Explanation:

Two main types of network switches, modular and fixed configuration switches, are used for connecting the devices with one another provided they are on the same network.

As the name suggests, modular switches can be configured according to your needs and specific situations where you need a different setup.

The one advantage fixed-configuration switches have over the modular switches is that they are easier to operate. You can't change anything for a different application but they are simpler to setup and use, you can just plug them in and start using. They are usually for the more casual end-user and home networks etc.

I hope this answer helps.

5 0

3 years ago