the first law and other basic concepts.pdf

7/29/2019 The First Law and Other Basic Concepts.pdf

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MODULE 2

The First Law andOther Basic Concepts


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Relevant definitions

1. System:Portion of the universe that is under

consideration;

2. Surroundings:the rest of the universe;

3. Boundary:the invisible and infinitesimally

thin surface that separate the system from the

surroundings;

4. Isolated system: cannot exchange energy or

matter with the surroundings;


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Example for isolated system:

A perfect thermal flask would be a good

example, unfortunately does not exist.


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5. Closed system: can exchange energy, but not

matter, with the surroundings;

6. Open system: can exchange both energy and

matter with the surroundings;

7. Adiabatic system: can exchange work, but not

heat ormatter, with its surroundings


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Example for closed system (Refrigerant circuit)


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What system is this?


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8. Property of a system: a macroscopic characteristicof the system;

9. Extensive property: value is additive, and

proportional to quantity of matter (examples: totalmass, total volume, total energy);

10. Intensive property: value independent ofquantity of matter (examples: density, pressure,temperature, etc);

If you put two identical objects of temperature Ttogether, the assembly has temperature T insteadof 2T)


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First Law of Thermodynamics

It is an expression of the principle ofconservation of energy which statesthat:

Although energy assumes manyforms, the total quantity of energyis constant, and when energy

disappears in one form it appearssimultaneously in other forms.


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Energy balance for closed systems

All energy exchanged between a

closed system and its surroundings

appears as heat and work.

WQUt Q=Energy transferred to or from the system as heat;W=Energy transferred to or from the system as work;

U= Internal energy of the system

(2.1)


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Numerical values of Q and W are +

for transfer into the system fromthe surroundings


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Internal Energy

Internal energy refers to energy of the moleculesinternal to the system.

It includes:

- Kinetic energy due to the motion of particles(translational, rotational, vibrational);

- Potential energy associated with vibrational andelectric energy of atoms within molecules orcrystals.

It does not include energy that it may possess as aresult of its macroscopic position or movement


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For differential changes, the energy balance of a

closed system is written as:

(2.2)dWdQdUt

Using intensive properties, Eqt. (2.1) and (2.2) become:

WQUnnU )(

dWdQndUnUd )( (2.4)

(2.3)


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Note: The absolute value of internal

energy is unknown, only the changes in

internal energy can be determined


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Example 2.1

Water flows over a waterfall 100 m in height. Take 1kg of water as the system and assume that it doesnot exchange energy with its surroundings.

(a) What is the potential energy of the water at thetop of the falls with respect to the base of thefalls?;

(b) What is the kinetic energy of the water just

before it strikes bottom?(c) After the 1 kg of water enters the river below

the falls, what change has occurred in its state?

p22


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THERMODYNAMIC STATE AND STATE

FUNCTIONS

Thermodynamic state = Set of values ofproperties (state variables) of a thermodynamicsystem that must be specified to reproduce the

system. State variable/parameter = Measurable property

required to describe the state of system, forexample: T, P, V, mass, density, Energy, etc.

Once a minimum number of variables have beenspecified, values of all other properties of thesystem become automatically fixed


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Once a minimum number of variables have been

specified, values of all other properties of the

system become automatically fixed.For example N2 gas at a temperature of 300K

and a pressure of 1 atm has a fixed volume or

density or a fixed molar internal energy.


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State functions

A property whose values depend on only the

state of the system and not on the path bywhich the change from initial to final state isbrought about is called state function. Thechange in the value of the state functions

depends only upon the initial and final statethe system.

Some common state functions used in

thermodynamics are, pressure (P), volume (V),temperature (T), internal energy (U), enthalpy(H), entropy (S), free energy (G) etc.


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Example 2.2When a system is taken from state a to state b in figure below

along path acb, 100 J of heat flows into the system and thesystem does 40J of work.

(a) How much heat flows into the system along path aeb ifthe work done by the system is 20 J?;

(b) The system returns from b to a along path bda. If the

work done on the system is 30 J, does the system absorbor liberate heat? How much?


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Equilibrium

Equilibrium in thermo:

- Absence of change;

- Absence of any tendency toward change on amacroscopic scale.

We can have: thermal equilibrium, mechanical

equilibrium, chemical equilibrium etc.


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Equilibrium

Thermal equilibrium: Occurs when a systemsmacroscopic thermal observables have ceased tochange with time;

Mechanical equilibrium: Occurs when the sum of allforces on all particles of the system is zero, and also thesum of all torques on all particles of the system is zero;

Chemical equilibrium: occurs when chemical activitiesor concentrations of the reactants and products have

no net change over time. Usually, this would be thestate that results when the forward chemical processproceeds at the same rate as their reverse reaction.


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Phase ruleFor any system at equilibrium, the number of

independent variables that must be

arbitrarily fixed to establish its intensive state

is given by the Gibbs phase rule:

NF 2 (2.5)

= number of phases;N = number of chemical species;

F= degree of freedom


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The intensive state of system at equilibrium is

established when its temperature (T), Pressure

(P) and the compositions of all phases are

fixed.

F, the degree of freedom gives the minimum

number of variables from the above set which

must be arbitrarily specified to fix all

remaining phase-rule variables.

When F = 0 the system is invariant


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Example 2.3

Determine the degree of freedom for each of

the following systems

(a) Liquid water in equilibrium with its vapour;

(b) Liquid water in equilibrium with a mixture of

water vapor and nitrogen;

(c) A liquid solution of alcohol in water in

equilibrium with its vapor.


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Reversible process

A process is reversible when its direction can be

reversed at any point by an infinitesimal

change in external conditions without loss or

dissipation of energy.


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A reversible process

Is frictionless;

Is never more than differentially removed fromequilibrium;

Traverses a succession of equilibrium states; Is driven by forces whose imbalance is differential in

magnitude;

Can be reversed any point by differential change in

external conditions; When reversed, retraces its forward path, and restores

the initial state of system and surroundings.


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Exercise

Derive the expression of work of compression or

expansion of a gas caused by the differential

displacement of a piston in a cylinder.

tPdVdW

Solution

Refer to module 1

(2.6)


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Remarks

The reversible process is ideal in that it can

never be fully realized; it represents a limit to

the performance of actual processes.

Results for reversible processes in combination

with appropriate efficiencies yield reasonable

approximations of the work for actual

processes


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Example 2.4

A horizontal piston/cylinder arrangement is placed in aconstant-temperature bath. The piston slides in thecylinder with negligible friction, and an external forceholds it in place against an initial pressure of 14 bar.

The initial gas volume is 0.03 m

3

. The external force onthe piston is reduced gradually, and the gas expendsisothermally as its volume doubles. If the volume ofthe gas is related to its pressure so that the product PVis constant, what is the work done by the gas in moving

the external force?How much work would be done if the external force weresuddenly reduced to half value instead of beinggradually reduced?


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Constant V and constant P processes

The energy balance of a closed system of nmoles is

dWdQnUd )( (2.4)For a mechanically reversible closed system process

the infinitesimal work is given by:

)(nVPdPdVdW

t

(2.7)(2.7) in (2.4) yields:

)()( nVPddQnUd (2.8)


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(2.8) is the general First Law equation for aMechanically reversible, closed-system process

)()( nVPddQnUd (2.8)


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Constant V process

dQnUd )(

(2.8) becomes:

After integration we have:

UnQ (2.10)

For a mechanically reversible, constant volume,

Closed system process, the heat transferred is equal

to the internal energy change of the system

0)( tdVnVd

(2.9)


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Constant P process

)()()()( nPVdnUdnVPdnUddQ

HnQ (2.12)

For a mechanically reversible, constant pressure,

Closed system process, the heat transferred is equal

to enthalpy change of the system

)()( nVPddQnUd

)()]([ nHdPVUnddQ (2.11)


Where PVUH is the enthalpy of the system


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Enthalpy

The enthalpy H=U+PV is also a state function

as U, P and V are all state function;

As U, P and V, H is an intensive property of the

system;

The differential form of H is given by:

)(PVddUdH (2.13)


)(PVUH (2.14)


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Example 2.5

Calculate U and H for 1 kg of water when it is

vaporised at the constant temperature of

100C and the constant pressure of 101.33

kPa. The specific volumes of liquid and vaporwater at these conditions are 0.00104 and

1.673 m3kg-1. For this change heat in the

amount of 2256.9 kJ is added to the water


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Heat capacity

dTdQC

Heat capacity at constant volume, Cv

vvTUC )(

(2.15)

(2.16)

For a constant-volume process in a closed system,

(2.16) can be written as:

dTCdU v (2.17)


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(2.10) can now be written as:

(2.18)

(2.19)

(2.19) is only valid for a constant-volume process .

The integration of 2.17 yields:

2

1

T

TvdTCU

2

1

T

TvdTCnUnQ


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Notes:

State function like U in (2.18) areindependent of the process, i.e. it only

depends on the initial and final conditions but

heat and work depends on the process path.

For the calculation of property changes, an

actual process may be replaced by any other

process which accomplishes the same change

in state.


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Heat capacity

Heat capacity at constant pressure, Cp

PPT

HC )(

(2.20)

For a constant-pressure process in a closed system,(2.20) can be written as:

dTCdHP

(2.21)

The integration of (2.21) yields:

2

1

T

TPdTCH (2.22)


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(2.12) can now be written as:

2

1

T

TvdTCnHnQ (2.23)


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Example 2.6Air at 1 bar and 298.15K is compressed to 5 bar and 298.15K by

two different mechanically reversible processes:(a) Cooling at constant pressure followed by heating at

constant volume;

(b) Heating at constant volume followed by cooling at constantpressure.

Calculate the heat and work requirements and U and H ofthe air for each path.

The following heat capacities for air may be assumedindependent of temperature: Cv=20.78 and Cp=29.10 Jmol-1K-1.

Assume also for air that PV/T is a constant, regardless of thechanges it undergoes. At 298.15K and 1 bar the molarvolume of air is 0.02479 m3 mol-1


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M d E b l f


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Mass and Energy balances for open

systems

Measures of flow:

Mass flow rate:

Molar flow rate:

Volumetric flow rate: q

Velocity:u

m.

n.


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Relations of measures of flow

m=M. n.q=uA

m=uA.n=uA.

A = cross sectional area of a conduit;

M = molar mass;

= specific or molar density;

u = average speed of a stream in

the direction normal to A


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Mass balance for open systems

Start by defining the thermodynamic system for

which mass and energy balance are written.

For open systems this system is called control

volume which is separated from itssurrounding by a control surface.


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1mo

2m

o3

mo

Control surface

Control volume

321

)(mmm

ooo

CV

dt

md 0213

)( mmmooo

CV

dt

md

fs

o

m)(

0)()(

fso

CV mdt

md0)(

)( fs

CV uAdt

md

Continuity Equation

(2.24)


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Steady state conditions

0)()(

fsCV uA

dt

md

0)( fsuA (2.25)


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The General Energy balance

P1, V1, U1, H1, u1

1

2

P2, V2, U2, H2, u2Control volume

Q

SW

o

rateworkQmzguUmzguUdt

mUd oooCV _])2

1[(])

2

1[(

)(2

2

1

2

fs

o

mzguU ])2

1[( 2

rateworkQmzguUdt

mUd o

fs

oCV _])

2

1[(

)( 2 (2.26)


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P1, V1, U1, H1, u1

1

2


Q

SW

o

o

fs

o

WmPVratework ])[(_

fs

o

mPV ])[( Work associated with moving the flowing streamsthrough entrances and exits

o

W Include shaft work, work associated with expansion or

Contraction of the control volume and stirring work


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P1, V1, U1, H1, u1

1

2


Q

SW

o

o

fs

oo

fs

oCV WmPVQmzguU

dt

mUd ])[(])

2

1[(

)( 2

oo

fs

oCV WQmzguPVUdtmUd ])

21[()( 2

H

oo

fs

oCV WQmzguH

dt

mUd ])

2

1[(

)( 2 (2.27)

For applications where kinetic and potential energy changes


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oo

fs

oCV

WQmzguHdt

mUd

])2

1[(

)( 2

For applications where kinetic and potential energy changes

in the flowing streams are negligible we have:

oo

fs

oCV WQmH

dt

mUd ][(

)( (2.28)


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Example 2.7

An insulated, electrically heated tank for hotwater contains 190 kg of liquid water at 60oCwhen a power outage occurs. If water is

withdrawn from the tank at a steady rate of= 0.2 kg s-1, how long will it take for thetemperature of the water in the tank to dropfrom 60 to 35oC? Assume that cold water

enters the tank at 10oC, and that heat lossesfrom the tank are negligible. For water letCv=Cp = C, independent of T and P.

o

m

Energy Balances for Steady state Flow


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Energy Balances for Steady-state Flow

Processes

S

oo

fs

oCV WQmzguH

dt

mUd ])

2

1[(

)( 2

S

oo

fs

o

WQmzguH ])2

1[( 2 (2.29)

For single entrance stream and single exit stream we have:

S

ooo

WQmzguH ])2

1[( 2 (2.30)

so

o

o

o

WQSzguH

m

W

m

Q )]2

1[( 2

SWQzguH 2

2

1(2.31)

Energy Balances for Steady state Flow


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Energy Balances for Steady-state Flow

Processes

SWQzguH 2

2

1(2.31)

Equation 2.31= First law for a steady-state, steady-flow process

between one entrance and one exit.

In applications where kinetic and potential energy terms

are negligible we have:

SWQH (2.32)


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Homework

Read and understand the principle of a flow

calorimeter for Enthalpy Measurements


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Example 2.8

Air at 1 bar and 25oC enters a compressor at low

velocity, discharges at 3 bar, and enters a

nozzle in which it expands to a final velocity of

600 ms-1 at the initial conditions of pressureand temperature. If the work of compression

is 240 kJ per kilogram of air, how much heat

must be removed during compression?


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Example 2.9

Water at 200(F) is pumped from a storage tank

at the rate of 50 (gal)(min)-1. The motor for

the pump supplied work at the rate of 2(hp).

The water goes through a heat exchanger,giving up heat at the rate of 40000 (Btu)(min-

1), and is delivered to a second storage tank at

an elevation 50(ft) above the first tank. Whatis the temperature of the water delivered to

the second tank?

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