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 alcoholwatersystem
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)
PT Diagrams
Sublimation
Fusion or melting
Vapourization
Critical point
For H2O, TC = 374 C &
PC = 218 atm Triple point
TV Diagram
Figure: TV 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
PV Diagram

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Shape of the PV diagram, much like the TV diagram, T=
constant lines are downwards
PV line for compressed liquid region is almost vertical.
PV 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. liquidvapour mixture,
h = ( 1 X ) hf+ Xhg

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Specific entropy of a liquidvapour 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