ME 63-Lecture 4-AY 20112012

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Lecture 4

Chapter 5: First Law of

Thermodynamics for a Control Mass

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THE 1ST LAW FOR A CONTROLMASS UNDERGOING A CYCLE

During any cycle a system undergoes, thecyclic integral of the heat is proportional tothe cyclic integral of work.

J Q W where

Q = net heat transfer for the cycleW

= net heat work for the cycle

The proportionality constant J is due to the difference in energy units for heat (Btu or calorie) and work (ft-lbf) in the English system; J=1 in SI Formulated from measurements of work and heat during a cycle for a wide variety of systems

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THE 1ST LAW FOR A CONTROLMASS UNDERGOING A CHANGE

OF STATEThe 1 st law for a change in state of a control mass or closed system can beestablished using the first law applied to a cycle.

Consider two arbitrary cycles, Cycle AB and Cycle AC , that go throughstates 1 and 2 as shown in Fig. 5.2

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OF STATE For Cycle AB:

(1) For Cycle CB:

(2) Subtracting (2) from (1),

or

AB ABQ W 2 1 2 11 2 1 2 A B A BQ Q W W

CB CBQ W 2 1 2 1

1 2 1 2C B C BQ Q W W

2 2 2 2

1 1 1 1 A C A C Q Q W W

2 2

1 1 A C Q W Q W

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OF STATE

2 2

1 1 A C Q W Q W

Since processes A and C are arbitrary between states 1 and 2,the quantity ( Q - W ) is the same for all processes between 1and 2 ( Q - W ) depends o n ly on t he in i t ia l and f ina l s t at es ,therefore a po int funct io n, and a d i fferent ia l of a pro per ty This leads to a def in i t ion of a new prop er ty ca l led to ta l energy

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The Property TOTAL ENERGY, E

The property called total energy is defined as,dE Q W

It represents all the energy of the system or control mass in any given state

This energy may be present in a variety of forms as system kinetic energy,potential energy, chemical energy, energy associated with molecular motion andposition, etc. Having identified and defined the property E , the 1 st Law for a control massor closed system undergoing a change of state is written as

Q dE W

12 2 1 12Q E E W For a contro l m ass (c losed sy s tem) und ergoing a change of s ta te , the net chang e in to ta l energy of the sys tem is equ al to the net energy tha t c rosses the

sys tem boun dary in the fo rm of hea t and /o r work .

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THE PROPERTY Internal Energy, U

The property internal energy U is defined as a resultof segregating the total energy E into threecomponents as

E = Internal Energy + Kinetic Energy + Potential Energy The reason for this segregation being

Both Kinetic Energy (KE) and Potential Energy (PE) are associated with thecoordinate frame selected for the system; can be specified by mass m , velocity v ,and elevation z.

212

KE mv PE m g z U represents all other forms of energy which the system contains and isassociated with the thermodynamic state of the system

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Example A Car of mass 1100 kg drives a velocity such

that it has a kinetic energy of 400 kJ.Find thevelocity of the car. If the car is raised with acrane how high should it be lifted in thestandard gravitational field to have a potentialenergy that equals the kinetic energy?

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THE PROPERTY Internal Energy, U

Because the internal energy U or u ( = U/m ) is associated only with thethermodynamic state of a substance, it is a property.

It can be used therefore to specify the state of a substance together withthe other properties P , V , T , and x .

In the saturated region: U = Uliq + Uvap mu = m liqu liq + mvap uvap = m liqu f + mvap ug

In terms of the quality x , u = (1-x) u f + x ug = u f + x u fg In the compressed liquid region: In the absence of tables, the compressed liquid internal energy is

approximated by the saturated liquid internal energy at the sametemperature,

uCL)T,P

u f )T

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Determine the missing Property(P,T or x) and also v

Water at A) T=300 C, u=2780 kJ/kg

B) P=2000kPa, u=2000 kJ/kg

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1st Law of Thermodynamics

Having identified the components of the totalenergy E, the 1 st Law for a control massundergoing a change of state is now written as

Q = dU + d(KE) + d(PE) + W = dU + d(mv 2 ) + d(mgz) + W

or in integrated form as Q12 = (U 2 - U 1 ) + m (v 22 v 12 ) + mg (z 2 z1 ) + W 12

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1st Law of Thermodynamics Q12 = (U 2 - U 1 ) + m (v 22 v 12 ) + mg (z 2 z1 ) + W 12 Note that the 1 st law gives only the changes in internal,

kinetic, and potential energies. To assign values to U, KE, and PE, reference states must be

assumed and values assigned to U, KE, and PE at thesereference states. The 1 st law tells us that as a control mass or closed system

changes state: energy may cross the boundary as heat and work the total energy of the system may change due to changes ininternal energy, kinetic energy, and potential energy the net change in total energy of the system is equal to the net

energy that crosses its boundaries

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Example A tank containing a fluid is stirred by a paddle

wheel. The work input to the paddle wheel is5090 kJ. The heat transfer from the tank is

1500 kJ. Consider the tank and the fluid insidea control surface and determine the change ininternal energy of this control mass.

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Example A vessel having a volume of 5 m contains 0.05

m of saturated liquid water and 4.95 m of saturated water vapor at 0.1 MPa. Heat is

transferred until the vessel is filled withsaturated vapor. Determine the heat transferfor this process.

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Enthalpy Let us introduce and define another property called enthalpy : H U + PV h u + Pv Units: H, (Joule) h, Joule/kg The reasons for so doing include

The combination u + Pv appears often in thermodynamic problems that itbecomes convenient to define a property to represent it Some thermodynamic tables only have entry for h so that u has to be

calculated from this definition (e.g., Tables B.4.2, B.5.2, B.6.2, etc.) When dealing with open systems or control volumes, h is not only a

property of the substances involved but also represents what is known asflow wor

Being a property, it can be used to specify the state of a substancetogether with the other properties P, v, T, x, & u.

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Enthalpy In the saturated region , the enthalpy of a substance is

given by h = (1-x) h f + h g = h f + x h fg In the compressed liquid region :

In the absence of tables, the compressed-liquidenthalpy is approximated by the following:

h CL )T h f )T + ( P CL P sat ) v f )T For small pressure differences, the pressure correction

is usually quite small so that h CL )T h f )T

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Example

A cylinder fitted with a piston has a volume of 0.1 m and contains 0.5 kg of steam at 0.4Mpa.Heat is transferred to the steam until thetemperature is 300C, while the pressureremains constant.

Determine the heat transfer and thework for this process.

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Specific Heat of a SimpleCompressible Substance

The specific heat of a simple compressible substance with ahomogeneous phase is

the amount of heat required per unit mass to raise thetemperature by one degree.

It is a measurable thermodynamic property typically usedto calculate internal energy or enthalpy (e.g., for propertytables)

The relationships among specific heat, internal energy, andenthalpy can be derived from

the specific heat definition the 1 st Law for a simple compressible system undergoing aquasiequilibrium process neglecting changes in KE and PE

Q = dU + W = dU + P dV

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Constant-Volume Specific Heat, C v The relationship between specific heat and internal

energy is determined by considering a constant-volume process for a closed system in which

V = const., dV = 0 Q = dU thus, Note that the resulting expression for the constant-

volume specific heat contains only thermodynamicproperties Cv is also a thermodynamic property

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Constant-Pressure Specific Heat, C p The relationship between specific heat and

enthalpy is determined by considering aconstant-pressure process for a closed system inwhich

P = const., Q = dH thus,

Note that the resulting expression for theconstant-pressure specific heat contains onlythermodynamic properties Cp is also athermodynamic property

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Change in Enthalpy and Internal Energy of Solids or Liquids When solids or liquids undergo a change of state without a change

of phase, a common approximation to the change in internalenergy and enthalpy is determined by considering that

h = u + Pv dh = du + d(Pv) = du + P dv + v dPsince a) these phases are nearly incompressible, dv 0 b) their specific volume is very small, v dP 0 so that

dh du C dT where C = Cv or C = Cp For cases where the specific may be assumed constant, then

h2 h1 u2 u1 C (T2 - T1)

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The Internal Energy, Enthalpy,and Specific Heats of IDEAL GASES

Internal Energy It can be shown analytically and experimentally that for an

ideal gas, the internal energy is a function of temperatureonly, i.e.,

u = f(T) The internal energy of an ideal gas can be determined from its

specific heat as follows:since u = f(T) only

Thendu = Cvo dT

dU = m Cvo dT

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Enthalpy For an ideal gas, the enthalpy is also a function of

temperature only

h = u + Pv = u + RT = h(T) The enthalpy of an ideal gas can be determined from its

specific heat as follows:since h = f(T) only

Thendh = Cpo dT

dH = m Cpo dT

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Specific HeatsCvo and C po are functions of temperature only since u = f(T)and h = f(T)

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Specific HeatsCvo and C po are functions of temperature only since u = f(T) andh = f(T)

The translational and rotational energies of molecules increase linearly with temperature contributions to thespecific heat are not temperature-dependent

Contributions from vibrational and electronic modes are temperature-dependent more vibrational modescauses higher increase in specific heat with temperature

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The difference of C vo(T) and Cpo(T) is always equal to the gasconstant

h = u + Pv = u + RT dh = du + R dTCpo dT = Cvo dT + R dT Cvo - Cpo = R

The specific heat ratio is defined as

Cpo / C vo R = ( - 1 ) Cvo

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Using Specific Heat to EvaluateIdeal Gas Enthalpy

In evaluating the enthalpy of gases using the specific heat,the accuracy depends upon the assumption used regardingthe specific heat.

I Constant Specific Heat

The specific heat is not a function of temperature but aconstant (see e.g. Table A.5)

Reasonable assumption for monatomic gases, smalltemperature range, or an average specific heat for the

temperature range is used.

h2 h1 = Cpo(T2 T1)

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II Specific Heat as a Function of Temperature The specific heat is expressed as an analytical

function of temperature which is an empirical

approximation to results of statisticalthermodynamic calculations (see e.g. Table A.6)

The change in enthalpy is a close empiricalapproximation and is given by:

2

12 1 ( )

T

poT h h C T dT

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III Specific Heat as Function of Temperature andIntegrated Directly from a Reference Temperature

The enthalpy of the gas at any temperature T

relative to a reference temp. T o is defined using thefunction

0

( )T

T poT h C T dT

where C po is calculated using statistical thermodynamics

Values of h T vs . T for some gases and air are tabulated in Table A.7 and Table A.8The enthalpy change is most accurate and is given by

h2

h1 = h T2

h T1

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The 1 st Law as a Rate Equation

The 1 st Law can be written in rate form toexpress the instantaneous or average rates of energy transfer through and energy change

within the system boundary. This departs from a strict classical point of

view but is useful in many applications.

d KE d PE dU Q W

dt dt dt

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Conservation of Mass The relationship between mass and energy

from relativistic considerations,E = m c2

will be disregarded in the present analysis of thermodynamic problems due to its negligibleeffects.

Thus, the conservation of mass andconservation of energy equations areconsidered as independent equations.

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Key Concepts and Formulas

12 2 1 12

2

1st Law of Thermodynamics:

Total Energy:

:

1

2

(

mass average:

1 )(1 )

C ;C

Heat:

f fg f g

f fg f g

v p

v p

Q E E W

E U KE PE mu mV mgZ

h u Pv

u u x

Enthalpy

Tw

u x u xuh h xh x h

o phase

Specific

xh

u u

T T

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Key Concepts and Formulas

2 1 2 1 2 1

2 1 2 1

2

[ A.3 and A.4]

h -h u

an

( )

( functions of T)

C ;

( )

d Liquids(Incompressible v=constant v and v are very small)

Gas

v p

v

f

p v

v v

Solid

Idea

C C C Tables

u C T T

h u Pv u RT only

du dhC C R

dT dT

u u C dT C T T

h

l

1 2 1( ) p ph C dT C T T

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ME 63-Lecture 4-AY 20112012