All categories

4 years ago

Explain the effect of heat transfer, reversibilities and irreversibilities on changes in entropy

Engineering

1 answer:

Natali5045456 [20]4 years ago

5 0

Answer is given below

Explanation:

The second law of thermodynamics states that in the reversible process, the entropy of the universe is constant, while in the unchanging process, such as the transfer of heat from a hot object to a cold object, the reflection of the universe increases.

ΔS = ΔQ ÷ T

The irreversible process increases the entropy of the universe. Since entropy is a state function, the change in entropy of the system is the same, the process is reversible or irreversible. The second law of thermodynamics can be used to determine whether a process is reversible.

for an irreversible process, ΔS is not same as reversible process, because more than 1 reversible process is required for 2 ΔS to be equal.

if the initial and final states are the same, ΔS for an irreversible process can be calculated as if it were a reversible process.

If the initial and final condition is same then ΔS can find out for irreversible process even if it were reversible process

You might be interested in

There are 22 people in the classroom 12 are we

Leni [432]

22 because all should wear safety glasses to be protected

6 0

3 years ago

Read 2 more answers

If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with

SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

$\sigma_b = \frac{M}{I} \times y$

putting all value to get bending stress

$\sigma_b = \frac{50.112}{144} \times 3.96$

$\sigma_b = 1.37 psi$

Bending stress at point 3.96 is $\sigma_b$ = 1.37 psi

3 0

4 years ago

Sea A una matriz 3x3 con la propiedad de que la transformada lineal x → Ax mapea R³ sobre R³.

skelet666 [1.2K]

Answer:

Explanation:

7 0

3 years ago

If the dry-bulb temperature is 95°F and the wet-bulb temperature is also 78°F, what is the relative humid- ity? What is the dew

Sidana [21]

Answer:

Relative humidity 48%.

Dew point 74°F

humidity ratio 118 g of moisture/pound of dry air

enthalpy 41,8 BTU per pound of dry air

Explanation:

You can get this information from a Psychrometric chart for water, like the one attached.

You enter the chart with dry-bulb and wet-bulb temperatures (red point in the attachment) and following the relative humidity curves you get approximately 48%.

To get the dew point you need to follow the horizontal lines to the left scale (marked with blue): 74°F

for the humidity ratio you need to follow the horizontal lines but to the rigth scale (marked with green): 118 g of moisture/pound of dry air

For enthalpy follow the diagonal lines to the far left scale (marked with yellow): 41,8 BTU per pound of dry air

5 0

3 years ago

Consider a cubic crystal with the lattice constant a. Complete the parts (a)-(c) below. (a) Sketch the crystallographic planes w

Anna [14]

Answer:

(a) See attachment

(b) The two planes are parallel because the intercepts for plane [220] are X = 0,5 and Y = 0,5 and for plane [110] are X = 1 and Y = 1. When the planes are drawn, they keep the same slope in a 2D plane.

Explanation:

(a) To determine the intercepts for an specific set of Miller indices, the reciprocal intercepts are taken as follows:

For [110]

$X = \frac{1}{1} = 1; Y = \frac{1}{1} = 1; Z = \frac{1}{0} = \inf.$

For [220]

$X = \frac{1}{2} = 0,5;Y = \frac{1}{2} = 0,5;Z = \frac{1}{0} = \inf.$

The drawn of the planes is shown in the attachments.

(b) Considering the planes as two sets of 2D straight lines with no intersection to Z axis, then the slope for these two sets are:

For (1,1):

$K_1 = \frac{1}{1} = 1$

For (0.5, 0.5):

$K_2 = \frac{0.5}{0.5} = 1$

As shown above, the slopes are exactly equal, then, the two straight lines are considered parallel and for instance, the two planes are parallel also.

The Miller indices are already given in the statement. Then, the distance is:

$\frac{1}{d^{2}} = \frac{h^{2} + k^{2} + l^{2}}{a^{2}}$

$d = \frac{a}{\sqrt{h^{2} + k^{2} + l^{2}}} = \frac{1}{\sqrt{2}} = 0,707$

7 0

3 years ago