Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Thermodinamica > Lecture 8 - Entropy > Flashcards

Lecture 8 - Entropy Flashcards

1
Q

What is the Clausius inequality?

A

∮ δQ/T <= 0

where ∮ = cyclic or closed integral

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Entropy is derived from the Clausius inequality.

A

True

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What is the definition of entropy?

A

dS = (δQ/T)internally reversible

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Entropy is tabulated for steam and R134a. We can also calculate the entropy of ideal gases like we did for internal energy, u, and enthalpy, h.

A

True

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Entropy generation is …

A

An quantitative measure of the irreversibilities associated with a process

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

What is an isentropic process?

A

An adiabatic, internally reversible process where there is no entropy change of the system

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

If δQ = 0, we have a ….

A

an isentropic process, so dS = 0 as well

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Isentropic processes are idealizations for the best case scenarios for different types of processes.

A

True

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Our base/reference state for entropy, internal energy, or enthalpy is …

A

S, U, or H = 0 at 0K

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Entropy quantifies the transfer of the different forms of energy into random forms of energy, for example vibrational, kinetic energy of molecules, etc…

A

True

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Examples of entropy generation are …

A
  • mechanical friction
  • viscous dissipation
  • mixing of 2 fluids
  • chemical reactions
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

In mechanical friction, …

A

work is being converted into random thermal energy within the system

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

What is viscous dissipation?

A

The process whereby velocity shear/ differentials within a fluid are converted through viscosity and friction in the fluid into thermal energy

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

An example of entropy generation is viscous dissipation. An example of viscous dissipation is pressure drop in pipe flow. Explain.

A

Pressure is a form of mechanical energy. Since there is velocity shear within the pipe flow, the velocity shear becomes thermal energy, which causes the pressure to drop.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Entropy increases in the mixing of two fluids and in chemical reactions because …

A

it is difficult to take the resulting thermal energy or the final state and get back the original state

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

When calculating entropy, we can use a differential form of the first law applied to a closed stationary system.

The entropy of a closed stationary system is …

A

δQ - δW(int rev)out = dU + dKE + dPE

BUT if the system is stationary, δKE and δPE are 0.
THEREFORE:
δQ - δW = dU

if there is no other type of work, we have:

δQ - δWb = dU
TdS - PdV = dU
where the elements are not per unit mass

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

What is the first Gibb’s equation?

A

TdS - PdV = dU

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

Define enthalpy.

A

h = u + Pv

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Define flow work.

A

flow work = boundary work = pdv

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

h = u + Pv

Appropriately apply the product rule to the above.

A

dh = du + Pdv + vdP

21
Q

If we substitute dh = du + Pdv + vdP into the first gibb’s equation, we get …

A 
First Gibb's equation (per unit mass):
Tds - Pdv = du
Substituting:
Tds = du + Pdv
Tds = dh - vdP

WE GET THE SECOND GIBB’S EQUATION:
Tds = dh - vdP

22
Q

State Gibb’s Second Equation.

A

Tds = dh - vdP

23
Q

If we divide the first gibb’s equation or the second gibb’s equation by T, we get …

A

Gibb’s Second Equation:
Tds = dh - vdP / T
ds = dh/T - vdP/T

Gibb’s First Equation:
TdS - PdV = dU / T
dS - PdV/T = dU/T

24
Q

If we integrate Gibb’s First or Second law (divided by temperature:

ds = dh/T - vdP/T or ds - Pdv/T = du/T, between two end state, we get …

A

the change in entropy between two states irrespective or reversibility

25
Q

Determination of properties depend on …

A

the path to get from one state to another and not the end states

26
Q

Liquids and solids are in/compressible.

A

incompressible

27
Q

Because liquids and solids are incompressible, …

A

we can approximate the entropy.

28
Q

For liquids and solids, dv = …

A

dv = 0 because liquids ans solids are incompressible

29
Q

du =

A

CvdT

30
Q

For liquids and solids, the relation between Cv and Cp is …

A

Cv = Cp = cte

assume volume expansion/ compression is negligible for liquids

31
Q

For a liquid or a solid, ds =

A

ds = du/T + pdv/T
where pdv = 0, so

ds = du/T
ds = CdT/T
ds = C(T)dT/T

where C is a constant of temperature

INTEGRATING, WE GET

T(s2-s1) = C(T) * ln(T2-T1)

where we can approximate the specific heat as Cv(avg)

32
Q

When we have C(T), we can approximate it as the average specific heat. If we don’t we ..

A

need to know how the specific changes with the solid or liquid

33
Q

If we are dealing with an isentropic process, ds = 0, which implies that the temperature …

A

remains the same

T2 = T1

34
Q

Define enthalpy.

A

dh = Cp*dT

35
Q

If ds = du/T + Pdv/T, if we substitute du for the definition of internal energy and the definition of pressure, according to the ideal gas equation, we get …

A

ds = CvdT/T + RTdv/vT

ds = Cv*dt/T + Rdv/v

36
Q

If ds = dh/T - vdP/T, if we substitute dh for the definition of enthalpy and the definition of volume, according to the ideal gas equation, we get …

A

ds = Cp*dT/T - RTdP/PT

ds = Cp*dT/T - RdP/P

37
Q

How do find the entropy change between two states for an ideal gas?

A

By integrating

ds = Cvdt/T + Rdv/v or ds = CpdT/T - RdP/P

38
Q

If we integrate ds = CpdT/T - RdP/P or ds = Cvdt/T + Rdv/v, we get …

A

s2 - s1 = Cv(T) * ln(T2 - T1) + R * ln(v2 - v1)

s2 - s1 = Cp(T) * ln(T2 - T1) - R * ln(P2 - P1)

39
Q

Cv (T) = Cv(avg) and Cp(T) = Cp(avg) when …

A

we have constant specific heats like for a solid or liquid or for ideal gases like He, Ar, or N2

OR

when we assume the temperature change is relatively small (by about 200 K)

40
Q

For more accurate specific heat values, we use tables of …

A

s^0 = 0∫T Cp(T) dT/T

41
Q

s^0 means …

A

we integrate the entropy from 0 to some temperature in Kelvin

42
Q

To calculate an exact form of entropy change using the tables at the back of a thermodynamics textbook we do …

A

s2 - s1 = s2^0 - s1^0 - R*ln(P2-P1)

43
Q

In an isentropic process for an ideal gas, we can say about the entropy that …

A

s2 = s1

44
Q

Assuming constant specific heats for an isentropic process, we can say, from the equations of entropy change, that ln(T2/T1) = …

A

ln(T2/T1) = -R/Cv * ln(v2/v1)

Or

ln(T2/T1) = R/Cp * ln(P2/P1)

45
Q

k or ᴕ =

The gas constant, R =

A

ᴕ = Cp/Cv

R = Cp - Cv

46
Q

s^0 = 0∫T Cp(T) dT/T = Cp(T) * ln(T2 - T1)

A

True

47
Q

For isentropic processes of ideal gases, we can say that Tv^(k-1) = cte, TP^(1-k2/k1) = cte, and and Pv^k = cte

A

True

48
Q

We use Tv^(k-1) = cte, TP^(1-k2/k1) = cte, and and Pv^k = cte under the assumption that …

A

the process takes place under constant specific heat and so the change in temperature is not that high

49
Q

If you cannot assume constant specific heat, use …

A

the tables at the back of your textbook to find the relative pressure, Pr, and the relative specific volume, vr

Thermodinamica (10 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions