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3. First Law of Thermodynamics and Energy Equation 3.1 The First Law of Thermodynamics for a Control Mass Undergoing a Cycle The first law for a control mass undergoing a cycle can be written as W Q W Q ) cycle ( net ) cycle ( net This statement is experimentally observed by James Joule in 19 th century.
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3. First Law of Thermodynamics and Energy Equation

3.1 The First Law of Thermodynamics for a Control Mass Undergoing a Cycle

The first law for a control mass undergoing a cycle can be written as

WQ

WQ )cycle(net)cycle(net

This statement is experimentally observed by James Joule in 19th century.

3.2 The First Law of Thermodynamics for a Change in State of a Control Mass Consider the following cycle

W Q

)2((1)

(2) W W QQ

(1) W W QQ

1

2

B

2

1

C

1

2

B

2

1

C

1

2

B

2

1

A

1

2

B

2

1

A

C

2

1

2

1

A

2

1

C

2

1

A

2

1

C

2

1

A

)W Q( )WQ(

W W QQ

Because A and C is arbitrary processes between state 1 and 2, we conclude that Q W must be a point function and must be the differential of a property. This property is the energy (E) [J] of the mass: WQ dE The property E represents all forms of the energy. It is convenient to separate E in this manner: E = U + KE + PE

Internal energy (U) [J]: a property representing all microscopic forms of energy.

Kinetic energy (KE) and Potential energy (PE) are in the macro scale.

Thus, the first law of thermodynamics for a change of state

WQ d(PE) d(KE) dU For kinetic energy and potential energy

)( d m2

1 )m

2

1( d d(KE) 22 VV

For potential energy

dZmg (mgZ) d d(PE) The first law of thermodynamics is

WQ dZ mg )d( m2

1 dU 2 V

Integrating the above equation yields

W Q )Z(Z mg

)( m2

1 )U(U

212112

21

2212

VV

Notes: This law cannot be mathematically

proven. Another name of this law is the

conservation of energy: two ways to change the energy of a system are heat and work crossing the system boundary.

As energy quantities, heat and work are not different.

The first law gives only the changes in U, KE, and PE not the absolute values.

Since KE and PE are extrinsic properties, U is also an extrinsic property.

3.3 Definition of Work

Work (W) [N-m or J]: a force acting through a displacement x

dx F W dx F W2

1

Sign convention: Work done by a system (+) Work done on a system () Work is a form of energy being transferred across a system boundary.

Work associated with a rotating shaft:

dT drF dx F W Power ( ) [W], time rate of doing work

W

T VF t

W W or t WW

2

1

Specific work (w) [J/kg]: work per unit mass

m

W w

For work SI unit 1 N-m = 1 J other units 1 kWh = 3600 kJ For power, SI unit 1 J/s = 1 W other units 1 hp (mechanical) = 745.7 W 1 hp (metric) = 735.5 W For specific work, SI unit 1 J/kg = 1 (m/s)2

3.4 Work Done at the Moving Boundary of a Simple Compressible System Consider a gas in a piston/cylinder device undergoing a quasi-equilibrium process

If P is the pressure of the gas, the work done by the system is

dV P W

dLA P W

In case that the relationship between P and V is available as a diagram:

We conclude that

dV P W W2

1

2

1

21

= area under the curve 1-2

= area a-1-2-b-a

However, we can choose different quasi-equilibrium processes from state 1 to state 2 (process A, B or C)

Thus, work done during each process is different. As a result, work is not only a function of the initial and end states. work depends on the path of each process. work is a path function. W is an inexact differential.

On the other hand, thermodynamics properties are point functions. Notes: symbolic convention For a point function (such as V) the differentials are exact by using symbol dV

12

2

1

VV dV For a path function (such as W) the differentials are inexact by using symbol W

21

2

1

W W To determine the work from the area under P-V curve, the relationship between P and V must be obtained.

The most common relationship between P and V is a polytropic process: onstantC PVn

n1

VPVP PdV 1122

2

1

This result is valid for any values of n except 1. For n =1,

1

211

2

1V

VlnVP PdV

which is equivalent to the isothermal process of an ideal gas

3.5 General Systems that Involve Work Other types of work are Stretched wire

dL W

2

1

21 L W

Where is tension [N]

Wire

dL

Surface line

dA W

2

1

21 A W

Where is surface tension [N/m]

Liquid Film

dA

Electric

dt i dZ W

where is electric potential [V] Z is electric charge [C] i is electric current [A]

2

1

21 t i W

i t

W W

Battery

+ E

dZ

Combining all types of work gives

...dZdAdL dV P W

...iA V P W V

Notes:

The process above contributes no work due to nonequilibrium process.

(a) Work done on the system (b) No work done on the system

3.6 Definition of Heat

Heat (Q) [J]: a form of energy which is transferred by the difference of temperature

Heat is a path function

21

2

1

Q Q

Rate of heat transfer (heat rate) ( ) [W]: heat transferred per unit time

Q

t

Q Q

or t QQ2

1

21

Specific heat transfer (q) [J/kg]: heat per unit mass

m

Q q

Adiabatic process: a process, in which there is no heat transfer

3.7 Heat Transfer Modes

Three modes of heat transfer: Conduction: heat transmitted by

diffusion of energy through a medium without the bulk motion Fourier's law of conduction

dx

dTAk Q

where k is thermal conductivity [W/m-K]

Convection: heat transferred by the motion (or flow) of a medium Newton's law of cooling )TT(A h Q surs where h is convective heat transfer coefficient [W/m2-K]

Radiation: heat transmitted by electromagnetic waves in space Stefan-Boltzmann's law )TT (A Q 4

sur4s

where is emissivity, is Stefan-Boltzmann constant [W/m2-K4]

Notes: Comparison between heat and work

Heat and work are both transient phenomena. Systems never possess heat or work

Heat and work are boundary phenomena. Both represent energy crossing the boundary of the system.

Both heat and work are path function and inexact differential

(a) Heat transfers across the boundary. (b) Work transfers across the boundary.

3.8 Internal Energy - a Thermodynamic Property U is an extensive property. Thus, we can define

Specific internal energy (u) [J/kg]:

m

U u

In general, the value of the specific internal energy must be given relative to a reference state (which is arbitrary). For water, uf = 0 at 0.01oC Note: the values of the specific internal energy can be negative. For the two-phase mixture,

fgf

fgf

gf

u x u

)u (u x u

ux ux)1( u

u represents the sum of all microscopic forms of energy.

moleculeintntranslatiomoleculeext uuu u

originates from the

intermolecular force between molecules.

moleculeextu

originates from the translation of molecules

ntranslatiou

originates from many

sources such as molecular rotation, molecular vibration, electron translation, electron spinning, atomic bond, nucleus bond.

moleculeintu

Sensible energy: all kind to kinetic energies as a function of temperature

Latent energy: intermolecular force for phase change process

Chemical energy: atomic bond

Nuclear energy: nucleus bond

3.9 Problem Analysis and Solution Technique 1. What is the control mass or control

volume? draw a diagram to illustrate heat/work flows

2. What do we know about the initial state?

3. What do we know about the final state?

4. What kind of the process takes place? Is anything constant or zero?

5. Is it helpful to draw p-v diagram? 6. What is the thermodynamics model

we use (steam tables, ideal gas, etc)? 7. What is our analysis (find work, the

first law, the second law, etc)? 8. How do we proceed to find what we

need?

3.10 The Thermodynamics Property Enthalpy Consider a quasi-equilibrium constant-pressure process:

)VP(U)VP(U

)VPVP( U U

)VV( P U U

PdV U U

W U U Q

111222

112212

1212

2

1

12

211221

U + PV is a property because it is a combination of other properties.

Enthalpy (H) [J]: PV U H Specific enthalpy (h) [J/kg]:

vP u m

H h

Thus, for a quasi-equilibrium constant-pressure process

1221 HH Q Notes: 1) Some thermodynamics tables does not contain data of u. Therefore, we can find u from u = h P v. 2) Some substance use enthalpy as a reference state instead of internal energy (such as R-12, Ammonia).

For the two-phase mixture,

fgf

fgf

gf

h x h

)h (h x h

hx hx)1( h

3.11 The Constant-Volume and Constant-Pressure Specific Heats From the first law:

PdV dU W dU Q

1. Constant volume ( 0 PdV W ) Specific heat (at a constant volume) (Cv) [J/(kg-K)]:

vvvv T

u

T

U

m

1

T

Q

m

1 C

2. Constant pressure ( dH Q ) Specific heat (at a constant pressure) (Cp) [J/(kg-K)]:

PPPp T

h

T

H

m

1

T

Q

m

1 C

Cv and Cp are also thermodynamics properties

For solid and liquid (incompressible substance: v = constant)

C C C Pv

u and h can be written as

dTC du

dP vdTC dPvdu dh

If C is a constant, we have

1212 TTC uu

121212 PPvTTC hh

In most cases, if 12 PP is not large, we can neglect the second term of the right side. As a result,

1212 TTC hh

C for various solid and liquid are listed in Tables A.3 and A.4

Notes: For some substances, C could changes significantly at a very high temperature.

3.12 The Internal Energy, Enthalpy, and Specific Heat of Ideal Gases

For internal energy, u is generally a function of two independent properties.

For a low-density gas

u depends strongly on T, but not much on P. Thus for an ideal gas

Pv = RT and u = f(T) only

For Cv, we mathematically write

General cases : v

v T

u C

Ideal gases : Td

du Cvo

subscript "o" denotes for the specific heat of an ideal gas

Thus, we can write

dTCdu vo

dTCuu

2

1

vo12

For enthalpy, from the definition of h = u + P v, in case of an ideal gas :

only f(T) h

RT u vP u h

For Cp, we mathematically write

General cases : P

P T

h C

Ideal gases : Td

dh CPo

dCdh Po T

dTChh

2

1

Po12

Because u = f(T) and h = f(T), Cvo = f(T) and CPo = f(T) as well. The values of Cvo and CPo at standard condition are given in Table A.5. The variations of CPo as a function of T are given in Table A.6.

The main factor causing Cvo and CPo to vary with temperature is molecular vibration.

Monoatomic gas: Cvo and CPo are a very weak function of T Diatomic gas: Cvo and CPo are somewhat a function of T Polyatomic gas: Cvo and CPo are strongly a function of T For an ideal gas

R C C

RdT dTC dTC

RdT du dh

RT u vP u h

voPo

voPo

on a molar basis

R C C voPo

To utilize the specific heat, there are three possible ways to find the enthalpy difference of an ideal gas

Constant specific heat

)TT(C uu 12vo12 )TT(C hh 12Po12

Direct integration: an expression of CPo as a function of T (3rd degree polynomial) in Table A.6

Use Table A.7 and A.8, which are the integration results of the internal energy and the enthalpy over a reference point.


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