Pure Substance

1. Pure substances
B.Prabhu, T.Suresh, P.Selvan
Assistant Professor – Mechanical Engineering
Kamaraj College of Engineering & Technology,
Virudhunagar

2. UNIT III
PROPERTIES OF PURE SUBSTANCE
AND STEAM POWER CYCLE

3. Syllabus Topics
 Formation of steam and its thermodynamic properties
 p-v, p-T, T-v, T-s, h-s diagrams
 p-v-T surface
 Use of Steam Table and Mollier Chart
 Determination of dryness fraction
 Application of I and II law for pure substances
 Ideal and actual Rankine cycles
 Reheat and Regenerative cycles
 Economiser, Preheater
 Binary and Combined cycles

4. What is a Pure Substance?
• A substance that has a fixed
chemical composition throughout
• A mixture of various chemical
elements or compounds also
qualifies as a pure substance as
long as the mixture is
homogeneous. Eg. Air
• A mixture of oil and water is not a
pure substance

5. The Three Phases

6. Formation of steam and thermodynamic properties of
Pure Substances
• Compressed Liquid
Consider a piston-cylinder device containing liquid water at
20°C and 1 atm pressure (state 1). Under these conditions,
water exists in the liquid phase, and it is called a
compressed liquid, or a subcooled liquid, meaning that it is
not about to vaporize

7. Phase Change Process of Pure Substances
• Saturated Liquid
As more heat is transferred, the temperature will rise till it
reaches 100°C (state 2). At this point, water is still liquid,
but any heat addition will cause some liquid to vaporize. A
liquid that is about to vaporize is called saturated liquid.
State 2 is saturated liquid state.

8. Phase Change Process of Pure Substances
• Saturated Liquid – Vapour mixture
Once boiling starts, the temperature will stop rising until
the liquid is completely vaporized. That is, the temperature
will remain constant during the entire phase-change
process if the pressure is held constant. Midway about the
vaporization line (state 3), the cylinder contains equal
amounts of liquid and vapour.

9. Phase Change Process of Pure Substances
• Saturated Vapour
As we continue heat transfer, the vaporization process will
continue until the last drop of liquid is vaporized (state 4).
At this point, the entire cylinder is filled with vapour. Any
heat loss from this vapour will cause some of the vapour to
condense. A vapour that is about to condense is saturated
vapour. Therefore, state 4 is a saturated vapour state.

10. Phase Change Process of Pure Substances
• Super heated Vapour
Further transfer of heat to saturated vapour will result in
an increase in both temperature and the specific volume.
At state 5, if we transfer some heat from the vapor, the
temperature may drop somewhat but no condensation will
take place as long as the temperature remains above 100°C
(for P=1 atm). A vapor that is not about to condense is
called a superheated vapor

11. Phase Change Process of Pure Substances
• Saturated Temperature & Pressure
The statement “water boils at 100°C’’ is incorrect. The
correct statement is “water boils at 100°C at 1 atm
pressure.’’ The only reason the water started boiling at
100°C was because we held the pressure constant at 1 atm.
If the pressure is raised to 5 atm, the water would start
boiling at 151.9°C. That is, the temperature at which water
boils depend on pressure. At a given pressure, the
temperature at which a pure substance changes phase is
called the saturation temperature Tsat. Likewise, at a given
temperature, the pressure at which a pure substance
changes phase is called saturation pressure Psat.

12. Property diagrams for phase-change processes
• T-v diagram

13. Property diagrams for phase-change processes
• P-v diagram

14. Property diagrams for phase-change processes
• P-T diagram

15. h-s diagram / Mollier Diagram

16. Determination of dryness fraction
• Tank or Bucket Calorimeter

17. Tank or Bucket Calorimeter - continued
• Tank or Bucket Calorimeter

18. Determination of dryness fraction
• Throttling Calorimeter

19. Throttling Calorimeter - continued
• The steam to be sampled is taken from the pipe by means
of suitable positioned and dimensioned sampling tube.
• It passes into an insulated container and is throttled
through an orifice to atmospheric pressure.
• Here the temperature is taken and the steam ideally should
have about 5.5 K of superheat.
• The throttling process is shown on h-s diagram in Fig. by
the line 1-2.
• If steam initially wet is throttled through a sufficiently large
pressure drop, then the steam at state 2 will become
superheated.
• State 2 can then be defined by the measured pressure and
temperature. The enthalpy, h2 can then be found and hence
h2 = h1 = (hf1 + x1hfg1 ) at p1 where h2 = hf2 + hfg2 + cps (Tsup2 -Ts2)], x1 =
(h2-hf1)/hfg1

20. IDEAL RANKINE CYCLE PLANT

21. Processes in Simple Steam Power plant
Working fluid: Water
• Water Source →Boiler: Water is pumped from
source to boiler
• Furnace ↔ Boiler: Heat is transferred to boiler
water from furnace, where fuel is burnt
• Boiler →Turbine: High pressure & high temperature
steam expands in turbine to produce work
• Turbine →Condenser: Steam leaving the turbine is
condensed in condenser
• Condenser →Boiler: The condensate is circulated
back to boiler, by pump

22. P-v-T diagram for Pure substance

23. P-v, T-s & h-s diagrams of Ideal Cycle

24. Application of I law to Steam formation
[Steady Flow Energy Equation]

25. SFEE Continued
Considering mass of fluid as 1 kg
• Process in boiler : 4 → 1
qb = h1 – h4
• Process in turbine : 1 → 2
WT = h1 – h2(Work is delivered by turbine, so W is +ve
)
• Process in condenser : 2 → 3
qc = h2 – h3
• Process in pump : 3 → 4
WP = h4 – h3(Work is given to pump, so W is -ve
)
h4 – h3 = v(p1-p2) (from TΔs = dh-vΔp & from slide 7, Δs=0)

26. Thermodynamic Analysis
• Fluid undergoes a cyclic process
• So no net change in internal energy over the cycle
• ΔQ to&fromfluid= ΔWon&byfluid
Qtofluid – Qfromfluid = W byfluid – W onfluid
• Cycle efficiency, ηcycle = ΔWon & by fluid ÷ Qto fluid
= ΔQ to & from fluid ÷ Qto fluid
Therefore, ηcycle = [Qtofluid – Qfromfluid] ÷ Qtofluid
= 1 – [Qfromfluid ÷ Qtofluid]
Where, W byfluid = Turbine work = WT
W onfluid = Pump work = WP
Qtofluid (Q1) = Boiler heat = qb

27. Analysis continued
• Net work = WT-WP
• Positive work = WT
• Heat input = Q1
• Work Ratio = Net work ÷ Positive work
• Cycle Efficiency = Net work ÷ Heat input
• Steam Rate = 3600 ÷ (WT-WP), kJ/kWh
• Heat Rate = 3600Q1 ÷ (WT-WP)

28. Losses that leads to actual cycle
Piping Losses
• Δp due to friction in pipes from boiler to turbine
• Δp due to friction in pipes from pump to boiler
• Heat loss to the surroundings
Turbine losses
• Frictional effects
• Heat loss to the surroundings
Condenser Losses (considered as negligible losses)
• Δp in turbine makes fluid cool below the saturation line
• So additional heat energy needed for liquid to heat liquid
Pump losses
• Frictional effects

29. ACTUAL RANKINE CYCLE

30. Application of II law in the form of Carnot cycle
• Although Carnot cycle has maximum possible efficiency, it is
unsuitable in steam power plant.
• Carnot cycle is analogous with Rankine cycle in the
processes as follows:
Adiabatic compression Adiabatic compression
Isothermal heat addition Isobaric heat addition
Adiabatic expansion Adiabatic expansion
Isothermal heat rejection Isothermal heat rejection

31. Continued….
• η = (Q1 - Q2) ÷ Q1
• Q1C > Q1R
Therefore,
ηC > ηR
• For eg.,
Q1C= 100 Q1R= 50
Q2C = 40 Q2R =40
ηC = 60% ηR = 20%
Q1 – Heat addition
Q2 – Heat rejection

32. Continued….
• η = (Q1 - Q2) ÷ Q1
• Q1C < Q1R & Q2C < Q2R
• For eg.,
Q1C= 30 Q1R= 50
Q2C = 12 Q2R =40
ηC = 60% ηR = 20%
Therefore,
ηC > ηR
• We had already proved in
previous slide that
ηC > ηR

33. Continued….
• η = (Q1 - Q2) ÷ Q1
• Q1C > Q1R
Therefore,
ηC > ηR
• This is similar to that of
fig. (a)

34. Mean Temperature of Heat Addition
• To create an area
under 4s-1, equivalent
to area under 5-6, we
consider the process
4s-1
• Q1 = h1-h4s
= Tm1(s1-s4s)
Therefore,
Tm1 = (h1-h4s)÷(s1-s4s)
η = 1 – (T2s÷Tm1)

35. Effect of increase of pressure
• For fixed Tmax and rise in
pressure from P1 to P2, Tm
increases and hence the η.
i.e., Tm(7s→5) > Tm(4s→1)
• But the expansion process
shifts steam quality from 2s
to 6s and results in turbine
blade erosion.
• Limit of dryness fraction
after expansion must be
0.85 to avoid abrupt
erosion.

36. Choosing an optimal Pmax.
• Choose the Tmax. based on material that subjects
to high pressure and high temperature
• Fix dryness fraction at turbine exit, x2s = 0.85
• From the T-s diagram, intersection of Tmax. and
x2s helps us fix the optimized maximum
Pressure, Pmax.

37. Choosing an optimal Pmax.

38. Reheat Cycle
• If steam pressure more than optimized Pmax is used, but we are
in need to fix x=0.85, reheat cycle can be adopted.

39. Contd.
• The expansion of steam to the condenser pressure is
done in 2 stages.
• Partially expanded steam from HPT is taken inside
boiler for reheating, at constant pressure. (Partial
expansion means expansion done till the steam
reaches saturated vapor nearly)
• The reheated steam is again expanded in LPT till it
reaches condenser pressure as usual.

40. Reheat cycle analysis
Considering mass of fluid as 1 kg

41. Reheat – Merits
• Net work output increases.
• Steam rate decreases.
• Practically there is slight increase in η.
• Keeps quality of steam under permissible limit even
at higher pressures.
=============================================
• Increased number of reheats would give rise in
pressure, but mechanical stress would rise
proportionally.
• Cost of reheat fabrication is high.

42. Regenerative Cycle – why?
• Till now, Tm was increased by increasing heat supply at high
temperatures, such as superheat, reheat, high pressure
• But Tm can also be increased by decreasing the amount of heat
added at low temperatures.

43. Regenerative Cycle

44. Regenerative Cycle Analysis

45. How to evaluate m1 & m2?
Apply energy balance equation on Heaters
Heater 1: Heater 2:

46. Reheat-Regenerative Rankine cycle

47. Binary Vapor Cycles (BVC)
• Though water is better than any working fluid, at higher
temperature range of applications, there are better fluids like
Diphenyl ether, Al2Br6, Hg, Na, K, etc.
• Diphenyl ether not used because it decomposes gradually at
higher temperature.
• Al2Br6 is a possibility and is on verge of research
• At p=12 bar, saturation temperature of water, Al2Br6 and Hg
are 1870
C, 482.50
C and 5600
C respectively.
• Therefore Hg is actually used in practice, especially at high
temperature range.
• But at low temperature, the saturation pressure of Hg is too
low and it is impractical to maintain such a vacuum.
• For this reason, Hg is combined with steam to form a BVC.

50. Combined Cycle – Brayton & Rankine

51. FAQ under Part - A
• What is flow and non flow processes?
• Write the methods for improving performance of the Rankine cycle.
• Draw P-T diagram for a pure substance (water).
• Draw P-v, T-v diagram of a pure substance.
• Define exergy.
• What is meant by dead state?
• What is a pure substance? Give examples.
• Define degree of superheat, Dryness fraction.
• Explain comparison between Rankine cycle and Carnot cycle.
• What are the four processes that make up the simple ideal Rankine
cycle?
• Define the term critical pressure, critical temperature and critical
volume of water.

52. FAQ under Part - B
Theory Questions:
1. Explain steam formation with relevant sketch and label all
salient points
2. Explain the P-v, P-T, T-v and h-s plots for pure substance
such as water.
3. Explain with a neat sketch the Rankine cycle and perform
the thermodynamic analysis on the same using T-s
diagram.
4. Explain with a neat sketch the layout of Binary Vapour
Cycle along with T-s diagram
5. Explain anyone Combined power Cycles in detail. Also draw
T-s diagram and mention the processes.

53. FAQ under Part - B
Problems on Properties of steam:
1. One kg of steam at 10 bar exists at the following conditions
(i) wet and 0.8 dry (ii) dry and saturated (iii) at a
temperature of 199.90
C. Determine the enthalpy, specific
volume, density, internal energy and entropy in each case.
Take Cps = 2.25 kJ/kg.
2. Steam initially at 0.3 MPa, 2500
C is cooled at constant
volume. At what temperature will the steam become
saturated vapor? What is the steam quality at 800
C. Also
find what is the heat transferred per kg of steam in cooling
from 2500
C to 800
C.
3. A vessel of volume 0.04m3
contains a mixture of saturated
water and saturated steam at a temperature of 2500
C. The
mass of the liquid present is 9 kg. Find the pressure, the
mass, the specific volume, the enthalpy, the entropy and
internal energy.

54. FAQ under Part - B
Problems on Simple Rankine Cycle:
1. In a Rankine cycle, the steam at inlet to turbine is saturated
at a pressure of 35 bar and the exhaust pressure is 0.2 bar.
Determine the pump work, the turbine work, Rankine
efficiency, the condenser heat flow, the dryness at end of
expansion. Assume flow rate of 9.5 kg/s
2. Steam at 50 bar, 4000
C expands in a Rankine cycle to 0.34
bar. For a mass flow rate of 150 kg/sec of steam,
determine, (i) power developed, (ii) thermal efficiency, (iii)
specific steam consumption.

55. FAQ under Part - B
Problems on Reheat Rankine Cycle:
1. A steam power plant operates on a theoretical reheat
cycle. Steam at boiler at 150 bar, 5500
C expands through
the high pressure turbine. It is reheated at a constant
pressure of 40 bar to 5500
C and expands through the low
pressure turbine to condenser at 0.1 bar. Draw T-s and h-s
diagrams. Find, (i) Quality of steam at turbine exhaust, (ii)
Cycle efficiency, (iii) Steam rate, in kg/kWh
2. In a reheat cycle, the initial steam pressure and the
maximum temperature are 150 bar and 5500
C respectively.
If the condenser pressure is 0.1 bar and the moisture at the
condenser inlet is 5%, and assuming ideal processes,
determine, (i) the reheat pressure, (ii) the cycle efficiency,
(iii) the steam rate