Download - 3 Properties of Fluids

8/11/2019 3 Properties of Fluids

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Properties

of

Fluids

To evaluate heat and work interactions of a

thermodynamic system, a knowledge of properties isessential.

As well as the relationship between properties

(P, v, T, U and H etc.)

Pure substance:

A substance which has single chemical species

(Air is not a pure substance)

Phase:

Commonly physical states of aggregation of anysubstance

(a) solid (b) liquid (c) gas


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Phase: Different states in which a pure substance canexists

Generally, A system which is uniform in chemicalcomposition and physical state, is called a phase.

System consisting different phases, all phases areseparated by phase boundaries.

Thermodynamic properties of substance change abruptlyat phase boundary even though the temperature andpressures are same.

Mixtures also exist in different phases.

Example: liquid and vapour phases of alcohol-watersystem

Phase Change of Water

Consider a piston cylinder device containing

liquid water at 20C and 1 atm pressure

water exists in the liquid phase, is called a

compressed liquid, or asubcooled liquid.

As temperature rises, liquid water expands

slightly, specific volume increases.

As more heat is transferred, the temperature

keeps rising until it reaches 100C. At this

point water is about to vaporize, called a

saturated liquid.


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Once boiling starts, the temperature

stops rising until the liquid is

completely vaporized.

,

saturated liquid and saturated

vapour exist in equillibrium.

At end, the entire cylinder is filled with

.

A vapor that is about to condense is

called a saturated vapor.

Further heating to saturated vapor,rise of temperature and specific

volume, superheated vapour

Again, cooling results in same

reversible process (superheated

vapour to subcooled liquid)

Saturation pressure


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Thermodynamic diagrams

(Phase change of pure substance)

P-T Diagrams

Sublimation

Fusion or melting

Vapourization

Critical point

For H2O, TC = 374 C &

PC = 218 atm Triple point

T-V Diagram

Figure: T-V diagram for pure substance


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boiling at higher temperature at higher pressure

Above the critical state, no line that separates thecompressed liquid and superheated vapor.

Saturated liquid line and saturated vapour line meet at

the critical point

Com ressed li uid re ion, su erheated va or re ion,

saturated liquidvapor mixture region, or the wet

region

P-V Diagram


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Shape of the P-V diagram, much like the T-V diagram, T=

constant lines are downwards

P-V line for compressed liquid region is almost vertical.

P-V line for superheated vapour region is less steep.

For a mixture of saturated liquid and saturated vapour

Quality or dryness fraction =

.

X = mg

/m

Total volume of mixture is given by,

m v = (m mg )vf+ mgvg . (1)

Where, vg = specific volume of saturated vapour and

vf= specific volume of saturated liquid

v = specific volume of the mixture

Divide b total mass m

v = (1 X ) vf+ X vg

P u t m = mf+ mg in equation (1),

(mf+ mg)v = mfvf+ mgvg

Rearranging,

(vg v) / (v vf) = mf/mg (lever rule)

Similarly, specific enthalpy of a Sat. liquid-vapour mixture,

h = ( 1 X ) hf+ Xhg


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Specific entropy of a liquid-vapour mixture,

s = (1 X)sf+ Xsg

Above the critical temperature, no amount of pressure can

condense the va or to a li uid.

abc


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Table2

Table3


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Table4

Table5


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Example (Use of Sat. Temp. table)

A vessel having a capacity of 0.05 m3 contains a mixture of

saturated water and saturated steam at a temperature of 245C.

The mass of the liquid present is 10 kg. Find the following:

(i) The pressure (ii) The mass (iii) The specific volume (iv) The

spec c ent a py v) e spec c entropy an v ) e spec c

internal energy.

Solution:

Given, V = 0.05 m3, Sat. liq. vap. mixture, T = 245 C, mf= 10 kg

From steam tables, corresponding to 245C :

Psat = 36.523 bar, vf = 0.001240 m3/kg, vg = 0.0546 m3/kg, hf =1061.6 kJ/kg, hg = 2801.6 kJ/kg, sf = 2.748 kJ/kg K, sg = 6.106

kJ/kg K

(i) Pressure = Sat. pressure = 36.5 bar (3.65 MPa)

(ii) Mass (m):

We have, Volume of liquid = Vf= mfvf

= 10 0.00124 = 0.0124 m3

So, Volume of vapour, Vg = 0.05 0.0124 = 0.0376 m3

and mass of vapour = mg = V/vg = 0.0376/0.0546 = 0.689 kg

So, total mass of mixture = mf+ mg = 10 + 0.688 = 10.689 kg

(iii) The specific volume (v) :

Here, Quality of the mixture x = mg / (mg + mf) = 0.064

Now, v = x.vg + (1 - x)vf

= 0.00356 m3/kg


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(iv) Specific enthalpy, h :

h = x.hg + (1 x).hf= 1172.96 kJ/kg

(v) Specific entropy, s :

s = x.s + 1 x .s = 2.9629 kJ/k K

(vi) Specific internal energy, u :

u = h pv = 1172.96 3650 0.00356 = 1159.97 kJ/kg

Example (Use of superheated tables)

Steam at 120 bar has a specific volume of 0.01721 m3

/kg, find thetemperature, enthalpy and the internal energy.

Solution:

Given: Pressure of steam (p) = 120 bar and specific volume (v) =0.01721 m3/kg

(i) Temperature:

With use of Sat. pressure steam table,

At 120 bar, vf= 0.001527 and vg = 0.014285 m3/kg

ere, vactual . m g > vg . m gSo, steam is superheated.

From the superheat steam volume tables at 120 bar, the specific volumeis 0.01721 m3/kg at a temperature of 350C.

So, steam temperature = 350C


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Criticalisothermmustshowapointofinflectionat

criticalpoint.

a=27R2TC2/64PC and b=RTC/8PC

Compressibility Factor

Gases deviate from idealgas behavior significantly

(e.g. near the saturation region and critical point)

Accounting of deviation from idealgas behavior by

use of compressibility factor (Z)

Defined as:

=Ideal Volume/ Actual Volume


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Z=1foridealgas

Fornon

ideal

gas

behaviour,

Z=f(Pr,

Tr)

Where,Tr=Reducedtemperature= T/Tc

andPr=Reducedpressure=P/Pc

Tc=Criticaltemperatureofgas

Pc=Criticalpressureofthegas

Compressibility Charts


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At very low pressures (Pr 2), idealgas behavior can

be assumed with good accuracy regardless of

pressure (except when Pr>> 1).

e ev a on o a gas rom eagas e av or s

greatest in the vicinity of the critical point.

Example: One kmol of ammonia is filled in a 0.1 m3 vessel at a

temperature of 200 C. Using the generalized compressibilitychart determine the pressure which ammonia exerts.

(Given: TC = 405.5 K, PC = 112.77 bar)

Solution:

Here, Tr= 473/405.5 = 1.17

Pr= P / PC = Pideal (assume) / PC = (RT/v) / PC

= (8.314*473.15/0.1) / 11277 = 3.5

Use, generalized compressibility charts,

Z = 0.57 (for Pr= 3.5 and Tr= 1.17)

So, New P = ZRT/v = (0.57* 8.314*473.15/0.1) = 22423 kPa


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New Pr= P/PC = 22423 / 11277 = 2

Again, use of generalized compressibility charts,Z = 0.52 (for Pr= 2 and Tr= 1.17)

So, New P = ZRT/v = (0.52* 8.314*473.15/0.1) = 20455 kPa

And New Pr= P/PC = 20455 / 11277 = 1.81

Again, use of generalized compressibility charts,

= . or r= . an r= .

So, New P = ZRT/v = (0.53* 8.314*473.15/0.1) = 20.8 MPa