ENTROPHYERT 206

THERMODYNAMICS

WAN KHAIRUNNISA WAN RAMLI

OBJECTIVEApply the second law of thermodynamics to processes.

Define a new property called entropy to quantify the second-law effects.

Calculate the entropy changes that take place during

processes for pure substances, incompressible substances, and ideal gases.Examine a special class of idealized processes,

called isentropic processes, and develop the property relations for these processes.Derive the reversible steady-flow work relations Introduce and apply the entropy balance to

various systems.

OBJECTIVE

2ND LAW OF THERMODYNAMI

CS

Involve many inequalities: Ŋ irrev < ŋrev

COP HP/ R < COP rev

CLAUSIUS INEQUALITIESVALID FOR ALL CYCLES

Differential over entire cycle

Sum of all differential amounts of heat transfer per T at boundary

Energy balance on combined system,

Consider the cyclic device reversible,

Eliminates δQR,

The combine system undergo cycle,

So, we have CLAUSIUS INEQUALITY

If no irreversibilities & cyclic device is reversible,The combined system is internally reversible

THE CLAUSIUS INEQUALITY

The system considered in the development of Clausius Inequality

Violates Kevin-Planck of 2nd Law, so Wc cannot be work output (Wc≠+ve)

CLAUSIUS INEQUALITYThe EQUALITY for totally or just internally reversible cycles The INEQUALITY for the irreversible ones.

Definition of Entrophy

The net change in a property (i.e. volume) during a cycle is always zero

SPECIAL CASE: INTERNALLY REVERSIBLE ISOTHERMAL HEAT TRANSFER PROCESSES

RELATION TO ENTROPHY?

Property A quantity which its cyclic integral = 0

The entropy change between two specified states is the same whether the process is reversible or irreversible

ENTROPYPROPERTY

1. Processes can occur in a certain direction only, not in any direction. A process must proceed in the direction that complies with the increase of entropy principle, that is, S gen ≥ 0. A process that violates this principle is impossible.

2. Entropy is a nonconserved property, and there is no such thing as the conservation of entropy principle. Entropy is conserved during the idealized reversible processes only and increases during all actual processes. (Entropy is generated or created during irreversible process due to the presence of irreversibilities).

3. The performance of engineering systems is degraded

by the presence of irreversibilities, and entropy generation is a measure of the magnitudes of the irreversibilities during that process. It is also used to establish criteria for the performance of engineering devices.

REMARKS ON ENTROPHY

Entropy is a property: The value of entropy of a system is fixed once the state of the system is fixed.

Determination of S value = determination of other property (i.e. h)

The entropy of a pure substance is determined from the tables (like other properties).

ENTROPHY CHANGE OF PURE SUBSTANCES

Entrophy change

S used as coordinate on T-s diagram

Compressed Liq & Superheated Vapor straight from data

For saturated mixture, given quality, x

EXAMPLE 1

A rigid tank contains 5kg of R-134a initially at 20°C and 140 kPa. The refrigerant is cooled while being stirred until its pressure drops to 100 kPa. Determine the entrophy change of R-134a during this process.

SOLUTION:ASSUMPTIONS: the volume of the tank is constant, so v2 = v1ANALYSIS:1. Closed system, no mass crosses the system boundary2. Change in entrophy = s2-s1, state 1 is fully specified3. Specific volumes remains constant

State 1: at P1=140 kPa, T1 = 20°C s1 = 1.0624 kJ/kg.K, v1 = 0.16544 m3/kg

State 2: P2 = 100 kPa, v2 = v1 = 0.16544 m3/kg saturated mixture since vf<v2<vg

Determine the quality for state 2 x = (v2-vf)/vfg = 0.859Thus s2=sf+xsfg = 0.8278 kJ/kg.KEntrophy change, ΔS = m(s2-s1)= -1.173 kJ/K

ISENTROPIC PROCESS A PROCESS WHICH THE ENTROPHY REMAINS CONSTANT.

During an internally reversible, adiabatic (isentropic) process, the entropy remains constant.

ISENTROPIC PROCESSES

The isentropic process appears as a vertical line segment on a T-s diagram.

ISENTROPIC PROCESS: NO HEAT TRANSFER, AREA = 0

ENTROPHY A MEASURE OF MOLECULAR DISORDERS OR MOLECULAR RANDOMNESSAs a system becomes more disordered, the entrophy will increase.

3RD LAW OF THERMODYNAMICS

A pure crystalline substance at 0 temperature is in perfect order &

its entrophy is 0

WHAT IS ENTROPHY?

The level of molecular disorder (entropy) of a substance increases as it melts or evaporates.

S is related to the thermodynamic probability, p (molecular probability)

Derived from Eq. 7-23 Gibbs equation

Liquids & solids can be approximated as

incompressible substances since

their specific volumes remain nearly constant during a process

Liquids, solids:

For an isentropic process of an incompressible substance

ISENTROPIC

ENTROPHY CHANGES OF LIQUIDS & SOLIDS

ISOTHERMAL

ENTROPHY CHANGES OF IDEAL GASESIDEAL GAS PROPERTIES

From the first T ds relation (Eq 7-

25)

From the second T ds relation (Eq

7-26)

Entropy change of an ideal gas on a unit–mole basis multiplying by molar mass

Under the constant-specific-heat assumption, the specific heat is assumed to be constant at some average value.

Constant Specific Heats (Approximation Analysis)

For gases whose C vary linearly with

T range

Absolute zero is chosen as the reference temperature & define a function s° as

Thus, entrophy changes between T1 & T2

In unit-mole basis

On a unit–mole basisThe entropy of an ideal gas depends on both T and P. The function s° represents only the temperature-dependent part of entropy.

Variable Specific Heats (Exact Analysis)

For gases whose C vary NONlinearly with T range

EXAMPLE 2Air is compressed from an initial state of 100kPa and 17°C to a final state of 600 kPa and 57°C. determine the entrophy change of air during this compression process by using (a) property values from the air table (b) average specific heats

SOLUTION: Air is compressed between 2 specified states. The entrophy change is to be determined by both methodASSUMPTIONS: Air is an ideal gasANALYSIS: the initial & final states are fully specified(a) Property table for s0 (Table A-17) & substituting into eq. for exact analysis [-

0.3844 kJ/kg.K](b) Find Cp values to determine the Cpave at Tave of 37°C (Table A-2b) and solve

using approximation analysis [-0.3842 kJ/kg.K]

Constant Specific Heats (Approximate Analysis)

For Isentropic (Δs=0) process, setting the above Eqs. equal to zero to get:

The isentropic relations of ideal gases are valid for the isentropic processes of ideal gases only.

Isentropic Processes for Ideal Gases

Pv=RT

Variable Specific Heats (Exact Analysis)

Isentropic Processes for Ideal Gases

Relative Pressure & Relative Specific Volume

The use of Pr data for calculating the final temperature during an isentropic process

T/Pr is the relative specific volume vr.

EXAMPLE 3Helium gas is compressed by an adiabatic compressor from an initial state of 100 kPa and 10°C to a final temperature of 160°C in a reversible manner. Determine the exit pressure of helium.

SOLUTION:ASSUMPTION: At specified conditions, helium can be treated as ideal gas. ANALYSIS:1. Find the k value for helium (Table A-2)2. Calculate the final pressure of helium using eq.

Energy balance for a steady-flow device (internally reversible, +ve for Wout) but Yield

For work input,

For the steady flow of incompressible liquid (v constant) through a device that involves no work interactions (such as a pipe section), the work term is zero.

REVERSIBLE STEADY-FLOW WORK

BERNOULLI EQUATION

v , W

Energy balance for 2 (irrev & rev) steady-flow devices (+ve for Qin, Wout)

Both operate between the same end states,

however,

Gives,

T is absolute T

Work-producing devices (turbines )deliver more work, and work-consuming devices (pumps, compressors ) require less work when they operate reversibly

A reversible turbine delivers more work than an irreversible one if both operate between the same end states.

Proof that Steady-Flow Devices Deliver the Most and Consume the Least Work when the Process Is

Reversible IRREVERSIBL

E

REVERSIBLE

Entrophy transfer + Entrophy generation = Entrophy change

Entropy Change of a System, ∆S system

When the properties of the system are not uniform

Energy and entropy balances for a system

ENTROPHY BALANCE

HEAT TRANSFER

Entropy transfer by heat transfer:

Entropy transfer by work:

Heat transfer is always accompanied by entropy transfer in the amount of Q/T, where T is the boundary temperature.

No entropy accompanies work as it crosses the system boundary. But entropy may be generated within the system as work is dissipated into a less useful form of energy.

Mechanisms of Entrophy Transfer, Sin & Sout

T ≠ constant

MASS FLOW

Entropy transfer by mass:

When the properties of the mass change during the process

Mass contains entropy as well as energy, and thus mass flow into or out of system is always accompanied by energy and entropy transfer.

s = specific

entrophy

Mechanisms of Entrophy Transfer, Sin & Sout

Mechanisms of entropy transfer for a general system.

Entropy generation outside system boundaries can be accounted for by writing an entropy balance on an extended system that includes the system and its immediate surroundings.

Entrophy Generation, Sgen

The entropy change of a closed system during a process is equal to the sum of the net entropy transferred through the system boundary by heat transfer and the entropy generated within the system boundaries.

ADIABATIC CLOSED SYSTEM

SYSTEM + SURROUNDINGS

Closed Systems

STEADY FLOW SINGLE STREAM ADIABATIC

STEADY FLOW

STEADY FLOW SINGLE STREAM

The entropy of a substance always increases (or remains constant in the case of a reversible process) as it flows through a single-stream, adiabatic, steady-flow device.

The entropy of a control volume changes as a result of mass flow as well as heat transfer.

Control Volumes

THANK YOU

TUTORIAL IIA horizontal cylinder is separated into two compartments by an adiabatic, frictionless piston. One side contains 0.2 m3 of nitrogen and the other side contains 0.1kg of helium.. Both initially at 20°C and 95 kPa. The sides of the cylinder and the helium end are insulated. Now heat is added to the nitrogen side from a reservoir at 500°C until the pressure of the helium rises to 120 kPa. Determine (a) the final temperature of helium, (b) the final volume of the nitrogen, (c) the heat transferred to the nitrogen and (d) the entrophy generation during this process.ANS: (a) 321.7K, (b) 0.2838 m3, (c) 46.6287kJ, (d) 0.057 kJ/K