Evaluating Properties For mechanical and Industrial Engineering

Lesson 3:

Evaluating Properties

1 By Meng Chamnan GIM

State Principle • State principle has been developed as a guide in

determining the number of independent properties

required to fix the state of a system.

• the state principle indicates that the number of

independent intensive properties of a simple

compressible substance is two.


Diagram p-v-T


Fig. 1: p-v-T surface and

projections for water. (a)

Three-dimensional view. (b)

Phase diagram. (c) p-v

diagram.

T–v Diagram

By Meng Chamnan GIM 4

Phase Change


- Liquid State

-Two-phase liquid–vapor mixture, the ratio of the

mass of vapor present to the total mass of the

mixture is its quality, x. In symbols,

The value of the quality ranges from zero to unity:

at saturated liquid states, x = 0, and at saturated

vapor states, x = 1

- Vapor state


vapor

liquid vapor

mx

m m

Vapor and Liquid Tables


a critical point, also called a critical state, specifies the conditions (temperature, pressure

and sometimes composition) at which a phase boundary ceases to exist

http://en.wikipedia.org/wiki/Critical_temperature

http://en.wikipedia.org/wiki/Phase_%28matter%29

Example1:

To gain more experience with Tables T-4 and T-5

verify the following: Table T-4 gives the specific

volume of water vapor at 10.0 MPa and 600oC as

0.03837 m3/kg. At 10.0 MPa and 100oC, Table T-5

gives the specific volume of liquid water as

1.0385×10-3 m3/kg. Table T-4E gives the specific

volume of water vapor at 500 lbf/in.2 and 600oF as

1.158 ft3/lb. At 500 lbf/in.2 and 100oF, Table T-5E

gives the specific volume of liquid water as

0.016106 ft3/lb.


Example2: Let us determine the specific volume of water vapor at a state

where p = 10 bar and T = 215oC. Shown in this Figure is a

sampling of data from Table T-4. At a pressure of 10 bar, the

specified temperature of 215oC falls between the table values

of 200 and 240oC, which are shown in bold face. The

corresponding specific volume values are also shown in bold

face. Determine the specific volume v corresponding to

215oC.


Evaluating Specific Volume


1 f g f g fv x v x v v x v v

liq vapV V V

liq vapV VVv

m m m

liq vap

f g

m mv v v

m m

liq liq fV m v

vap vap gV m v

/ 1liqm m x /vapx m m

Example3:

Consider a system consisting of a two-phase

liquid–vapor mixture of water at 100oC and a quality

of 0.9. What is the specific volume of the mixture?


Example4: Let us determine the pressure of water at each of three states

defined by a temperature of 100oC and specific volumes,

respectively, of v1 = 2.434 m3/kg, v2 = 1.0 m3/kg, and v3 =

1.0423 × 10-3 m3/kg. And try to find the vapor quality, x, at

state 2?


Example5: 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 water 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 continued, determine the pressure, in bar, when the

container holds only saturated vapor.


Evaluating Pressure, Specific Volume,

and Temperature inside the dome


1 f g f g fu x u x u u x u u

1 f g f g fh x h x h h x h h

h u pv h u pv

H U pV

Example6:

We determine the specific enthalpy of Refrigerant

134a when its temperature is 12oC and its specific

internal energy is 132.95 kJ/kg. Referring to Table

T-6, the given internal energy value falls between uf

and ug at 12oC, so the state is a two-phase liquid–

vapor mixture.


Example7:

A well-insulated rigid tank having a volume of 10 ft3

contains saturated water vapor at 212oF. The water

is rapidly stirred until the pressure is 20 lbf/in.2.

Determine the temperature at the final state, in oF,

and the work during the process, in Btu.


Example8:

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 400oC.

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 150oC.

(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.


Evaluating Specific Heats cv and cp


v

v

uc

T

p

p

hc

T

p

v

ck

c property k, called the specific heat ratio

kJ/kg ∙ K or kJ/kmol ∙ K or Btu/lb ∙ oR or Btu/lbmol ∙ oR

Universal Gas Constant


8.314 kJ/kmol K

1.986 Btu/lbmol R

1545 ft lbf/lbmol R

R

Evaluating Properties Using the

Ideal Gas Model


pV mRT pV nRT

Ideal Gas Model

pv RT

( ) ( )h h T u T RT

Example9:

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 540oR, and the

pressure is 1 atm. At state 2, the pressure is 2 atm.

Employing the ideal gas equation of state,

(a) sketch the cycle on p–v coordinates.

(b) determine the temperature at state 2, in oR.

(c) determine the specific volume at state 3, in ft3/lb.


Polytropic Process for Ideal Gas


constantnpV 1 1 2 2

2 1

1 2

n n

n

pV p V

p V

p V

When n= 0, the process is an isobaric (constant-pressure) process, and when n=

±∞ , the process is an isometric (constant-volume) process

n is a constant

Ideal gas equation of state


Example10: Air undergoes a polytropic compression in a piston–

cylinder assembly from p1 = 1 atm, T1 = 70oF to p2 = 5

atm. Employing the ideal gas model, determine the work

and heat transfer per unit mass, in Btu/lb, if n = 1.3.


Evaluating Properties For mechanical and Industrial Engineering

Engineering

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