Chapter 3 lecture

National Leading Research Lab. Nanoengineered Energy Conversion Devices Lab.

Evaluating Properties


Not a Pure Substance..?

Phase: a quantity of matter that is homogeneous throughout in both chemical composition and physical structure. !Pure substance: one that is uniform and invariable in chemical composition.

Gas Gas(1) (2)

(3) (4)


The State PrincipleTwo independent, intensive, thermodynamic properties are

required to fix the state of a simple compressible system.

For example:P and v

T and u

x and h

Intensive thermodynamic properties:

h – specific enthalpy

u – specific internal energy

x – quality (steam only)

s –specific entropy

P –absolute pressure

T – absolute temperature

v – specific volume

Less used: g - Gibbs free energy

a - Helmholz free energy

Coordinate

• Time: t

• Space: x, y, z

• Thermodynamics: P, T (or two independent properties)


P-v-T Relations

- Single phase regions- Two phase regions- Saturation states

- Triple line

- Critical Point


P-T (phase diagram)

National Leading Research Lab. Nanoengineered Energy Conversion Devices Lab.National Leading Research Lab. Nanoengineered Energy Conversion Devices Lab.National Leading Research Lab. Nanoengineered Energy Conversion Devices Lab.National Leading Research Lab. Nanoengineered Energy Conversion Devices Lab.

http://en.wikipedia.org/wiki/Image:Thermally_Agitated_Molecule.gif#file

http://socs.berkeley.edu/~murphy/Movies/movie.html

Solid Liquid Gas

http://en.wikipedia.org/wiki/Image:Thermally_Agitated_Molecule.gif#

Solid - Liquid - Gas


Expands or Contracts on Freezing

Expands on freezing (water) Contracts on freezing


P-T (phase diagram)

Triple point: solid/liquid/gas coexists

Critical point: no distinction between gas and liquid

• Vaporization line

• Melting line

• Sublimation line

Psat, Tsat Psat=f(Tsat)

Ideal gas


Saturated Vapor PressureSaturated vapor pressure

Therefore, usually Psat increases as Tsat increases.

# of molecules leaving liquid = # of molecules returning to liquid

Courtesy of V. Carey Psat = f Tsat( )


P-v Diagram

Isotherm

Pv = RT


Expands or Contracts on Freezing

Expands on freezing (water) Contracts on freezing

L+S


T-v Diagram

Subcooled Liquid

Pv = RT


Quality, x

• For Saturated Mixture (Liquid-Vapor) Region

– Quality; x; an intensive property

– x gives fraction that is vapor (gas)

– (1-x) gives Moisture Content

0 ≤ x ≤ 1; x = 0 → Saturated Liquid (subscript ‘f’) x = 1 → Saturated Vapor (subscript ‘g’) ‘fg’ → ‘g’-’f’ x ≡

mg

mf + mg


Quality Relations

LET b = ANY INTENSIVE PROPERTY – (b = v, u, h, s, etc.)

x =b − bfbg − bf

=b − bfbfg

b = bf + x ⋅bfgbfg = bg − bfb = x ⋅bg + 1− x( ) ⋅bf


Processes

• Adiabatic process: No thermal interaction with its surroundings !

• Isothermal process: A process that occurs at constant temperature !

• Isobaric process: A process that occurs at constant pressure !

• Isometric (isochoric) process: A process that occurs at constant volume !

• Isenthalpic process: A process that occurs at constant enthalpy !

• Isentropic process: A process that occurs at constant entropy


Find out whether you are interested in

• Compressed liquid

• Wet vapor

• Superheated vapor

• Ideal gas


Reference States

At triple point (water, 0.6113 kPa, 0.01°C), thermodynamic properties of saturated liquid are zero (as a standard)

• Compressed liquid

• Wet vapor

• Superheated vapor

• Ideal gas

Use saturated liquid data

or incompressible approximation

Ideal gas law & table

Look up the table

Look up the table


Enthalpy

Enthalpy is a property constructed due to the frequent occurrence of the above combination of properties.

Enthalpy of wet vapor

Class note

[J]

[J/kg]

H =U +PVh = u+Pv


Specific Heat (heat capacity)

Specific heat at constant volume

Specific heat at constant pressure

Specific heat ratio

cv =∂u∂T"

#$

%

&'v

cP =∂h∂T"

#$

%

&'P

k = cPcv


Approximation for Liquids

• Using Saturated Liquid Data (‘Compressed Liquid Rule’)

v T , P( ) ≈ v f T( )u T , P( ) ≈ u f T( )h T , P( ) ≈ hf T( )s T , P( ) ≈ s f T( )at fixed temperature (not pressure)Class note


Approximation for Liquids

• Using ‘Incompressible Substance Model’

Class note

cP = cv = c

u2 − u1 = c T2 − T1( )h2 − h1 = u2 − u1 + v P2 − P1( )h2 − h1 ≈ c T2 − T1( )


Ideal Gas

Class note

The Ideal Gas Model: When specific heats are assumed constant

Requirement:PV = Nk BT= nRuT= mRT

R = RuMW

P ≪ Pc or T ≫ Tc

u = u T( )h = u T( ) + Pv

= u T( ) + RT

= h T( )

cv (T ) =dudT

=R

k −1

cp (T ) =dhdT

=kR

k −1cp (T ) = cv (T ) + R

u2 − u1 = cv T2 − T1( )h2 − h1 ≈ cp T2 − T1( )

Pv= RT


Linear Interpolation

y − y1 =y2 − y1x2 − x1

x − x1( )


A vertical piston-cylinder assembly containing 0.05 kg of ammonia, initially a saturated vapor, is placed on a hot plate. Due to the weight of the piston and the surrounding atmospheric pressure, the pressure of the ammonia is 1.5 bars. Heating occurs slowly, and the ammonia expands at constant pressure until the final temperature is 25 ºC. Show the initial and final states on T-v and P-v diagrams, and determine

!(a)The volume occupied by the ammonia at each state, in m3.

(b)The work for the process, in kJ.


A closed, rigid container of volume 0.5 m3 is placed on a hot plate. Initially, the container holds a two-phase mixture of saturated liquid water and saturated vapor at P1 = 1 bar with a quality of 0.5. After heating, the pressure in the container is P2 = 1.5 bar. Indicate the initial and final states on a T-v diagram, and determine

!(a)The temperature, in ºC, at each state (b)The mass of vapor present at each state, in kg (c)If heating continues, determine the pressure, in bar,

when the container holds only saturated vapor.


A well-insulated rigid tank having a volume of 0.25 m3 contains saturated water vapor at 100 ºC. The water is rapidly stirred until the pressure is 1.5 bars. Determine the temperature at the final state, in ºC, and the work during the process, in kJ.


Water contained in a piston-cylinder assembly undergoes two processes in series from an initial state where the pressure is 10 bar and the temperature is 400 ºC.

!Process 1-2: The water is cooled as it is compressed at a constant pressure of

10 bar to the saturated vapor state.

Process 2-3: The water is cooled at constant volume to 150 ºC.

!(a)Sketch both processes on T-v and P-v diagrams.

(b)For the overall process determine the work, in kJ/kg

(c) For the overall process determine the heat transfer, in kJ/kg.


One pound of air undergoes a thermodynamic cycle consisting of three processes.

!Process 1-2: constant specific volume

Process 2-3: constant-temperature expansion

Process 3-1: constant-pressure compression

!At state 1, the temperature is 300 K, and the pressure is 1 bar. At state

2, the pressure is 2 bars. Employing the ideal gas equation of state,

(a)Sketch the cycle on P-v coordinates.

(b)Determine the temperature at state 2, in K

(c) Determine the specific volume at state 3, in m3/kg

Engineering

Chapter 3 lecture