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3 years ago

Explain what entropy is in relation to the second law of thermodynamics?

Engineering

1 answer:

Marat540 [252]3 years ago

5 0

Explanation:

Second law tell us that ,the entropy of reversible process is remain constant while the entropy of ir-reversible process increases.

Entropy of universe is the summation of entropy of system and entropy of surrounding.So we can say that

$\Delta S_{universe}=\Delta S_{system}+\Delta S_{surrounding}$

Entropy can be define as

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

Entropy is point function.This is property of system.Ir-reversibility cause the increase of entropy.

$\oint \dfrac{dQ}{T} =0 ,this\ is\ reversible\ process$

$\oint \dfrac{dQ}{T} >0 ,this\ is\ impossible\ process$

$\oint \dfrac{dQ}{T}$

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The specific volume of a system consisting of refrigerant-134a at 1.0 Mpa is 0.01 m /kg. The quality of the R-134a is: (a) 12.6%

Flura [38]

Answer:

option c is correct

47.2%

Explanation:

given data

consisting of refrigerant = 134 a

volume V = 0.01 m³/kg

pressure P = 1MPa = 1000 kPa

to find out

quality of the R 134a

solution

we will get here value of volume Vf and Vv from pressure table 60 kpa to 3 Mpa for 1 Mpa of R134 a

that is

Vf = 0.0008701 m³/kg

Vv = 0.0203 m³/kg

so we will apply here formula that is

quality = (V - Vf) / (Vv - Vf) ............1

put here value

quality = (0.01 - 0.0008701 ) / ( 0.0203 - 0.0008701 )

quality = 0.4698

so quality is 47 %

SO OPTION C IS CORRECT

4 0

3 years ago

For precipitation hardening, how many phases (in numeric form) exist (a) following the solution heat treatment, and (b) followin

Nitella [24]

The correct answer is B

4 0

2 years ago

(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows

maxonik [38]

Answer:

diameter of the sprue at the bottom is 1.603 cm

Explanation:

Given data;

Flow rate, Q = 400 cm³/s

cross section of sprue: Round

Diameter of sprue at the top $d_{top}$ = 3.4 cm

Height of sprue, h = 20 cm = 0.2 m

acceleration due to gravity g = 9.81 m/s²

Calculate the velocity at the sprue base

$V_{base}$ = √2gh

we substitute

$V_{base}$ = √(2 × 9.81 m/s² × 0.2 m )

$V_{base}$ = 1.98091 m/s

$V_{base}$ = 198.091 cm/s

diameter of the sprue at the bottom will be;

Q = AV = (π $d_{bottom}^2$ /4) × $V_{base}$

$d_{bottom}$ = √(4Q/π $V_{base}$ )

we substitute our values into the equation;

$d_{bottom}$ = √(4(400 cm³/s) / (π×198.091 cm/s))

$d_{bottom}$ = 1.603 cm

Therefore, diameter of the sprue at the bottom is 1.603 cm

6 0

2 years ago

A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici

Pavlova-9 [17]

Answer:

8.24μm

Explanation:

The theory of brittle fracture was used to solve this problem.

And if you follow through with the attachment made a the subject of the formula

Such that,

a = 2x(69x10⁹)x0.3/pi(40x10⁶)²

= 4.14x10¹⁰/5.024x10¹⁵

= 8.24x10^-06

= 8.24μm

This is the the maximum length of the surface flaw

4 0

2 years ago

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HACTEHA [7]

a quien alos israelitas

8 0

3 years ago