Download basic thermodynamics

e-Notes by A.K.Bhatt, GIT, Belgaum
AVAILABILITY AND IRREVERSIBILITY
Available Energy
The sources of energy can be divided into two groups namely, high-grade energy
and low-grade energy. The conversion of high-grade energy to shaft work is
exempt from the limitations of the second law, while conversion of low grade
energy is subjected to them.
Example: High grade energy:
1) Mechanical work 2) electrical energy 3) water power 4) wind power 5) kinetic
energy of a jet 6) tidal power.
Example: Low grade energy: 1) Heat or thermal energy 2) heat derived from
nuclear fission or fusion. 3) Heat derived from combustion of fossil fuels. 4) Solar
energy.
The high-grade energy in the form of mechanical work or electrical energy is
obtained from sources of low-grade energy. The complete conversion of lowgrade energy, heat in to high-grade energy, shaft work is impossible. That part of
low-grade energy which is available for conversion is refereed to as available
energy, while the part which according to the second law must be rejected is
known as unavailable energy.
Fig 1: Heat transfer from a constant temperature energy source.
In the previous chapter the concept of efficiency of a device such as turbine,
nozzle and compressor are introduced and more correctly termed as first law
efficiency, since it is given as he ratio of two energy terms. This chapter gives
more meaningful definition of efficiency- second law analysis. Our main goal is to
use this analysis to manager our thermal resources and environment better.
1
Consider the simple situation shown in figure1 in which there is an energy source
Q in the form of heat transfer from a very large source and therefore constant
temperature reservoir at temperature T. what is the ultimate potential for
producing work?
To answer to this question we imagine that a cyclic heat engine is available as
shown in figure (b) to convert the maximum fraction of Q requires that the engine
be completely reversible, i.e. a Carnot cycle, and that the lower temperature
reservoir be at the lowest temperature possible, often but not necessarily at the
ambient temperature. From the first and second laws for the Carnot cycle and the
usual consideration of all the Q’s as positive quantities we find
W rev HE = Q –Qo
W rev HE = Q {1-( To / T ) }
Q / T = Qo / T o
The fraction of Q given by the right side of the equation is the available portion of
the total energy quantity Q.
Consider the situation shown on the T-S Diagram.
The total shaded diagram is Q.
The portion of Q that is below To, the environment temperature, can not be
converted into work by the heat engine and must instead be thrown away. This
portion is therefore the unavailable portion of the energy Q, and the portion lying
between the two temperatures T and To is the available energy.
Fig 2: T-S Diagram for a constant temperature energy source.
Let us consider the same situation except that the heat transfer Q is available
from a constant pressure source, for ex, a simple heat exchanger as shown in
the figure. The Carnot cycle must now be replaced by a sequence of such
engines, with the result shown in the figure B the only difference between the first
2
and the second example is that the second includes an integral, which
corresponds to S
S =  ( Q rev / T ) = Qo /To
W rev = Q – To * S
Note that this S quantity does not include the standard sign convention. It
corresponds to the change of entropy. The equation specifies the available
portion of the quantity Q. the portion unavailable for producing work in this
circumstance lies below To.
Thus the unavailable energy is the product of lowest temperature of heat
rejection ands the change of entropy of the system during the process of
supplying heat.
Fig 3: Changing temperature energy source.
Decrease in available energy when the heat is transferred through a finite
temperature difference:
Whenever heat id transferred through a finite temperature difference there is a
decrease in the availability of the energy so transferred. let us consider a
reversible heat engine operating between T1 and To as shown in the figure
3
Fig 4: Increase in unavailable energy due to heat transfer through a finite
temperature difference.
Then we have Q1 = T1 * S
Q2 = To * S
W = AE = (T1 – To) S
Let us now assume that q1 is transferred through a finite temperature difference
from the reservoir or source at T1 to the engine absorbing heat at T’1 lower than
T1 as shown in the figure
Fig 5: Constant Temperature energy source.
4
The availability of Q1 as received by the engine at T’1 and to receiving Q1 and
rejecting Q’2.
Q1 = T1 S = T ’1 S’
T1 > T ’1, Hence S’ > S
Q2 = Tc S and Q‘2 = To S’
Since S’ > S hence Q’2 > Q2
And hence W = Q1- Q’2 = T’1 S’ - To S’
And W = Q1 – Q2 = T1 S – To S
Hence W’ < W since Q’2 > Q2.
Available energy lost due to irreversible heat transfer through finite temperature
difference between source and working fluid during heat addition process is
given by
W- W’ = Q’2 - Q2 = To (S’ - S)
or decrease in AE = To (S’ - S)
Thus the decrease in available energy is the product of the lowest feasible
temperature of heat rejection and the additional entropy change in the system
while receiving heat irreversibly compared to the case of reversible heat transfer
from the same source. The greater is the temperature difference (T 1 – T’1) the
greater is the heat rejection Q’2 and greater will be the unavailable part of the
energy supplied. Energy is said to be degraded each time when it flows through
a finite temperature difference. That’s why the second law is some times called
the law of degradation of energy and the energy is said to run down hill.
Availability:
The availability of a given system is defined as the maximum useful work (total
work – pdV work) that is obtainable in a process in which the system comes to
equilibrium with its surroundings. Availability is thus a composite property
depending on the state of the system and surroundings.
Let U,S and V be the initial energy, entropy and volume of a system and Uo So,
Vo their final values when the system has come to equilibrium with its
environment. The system exchanges heat only with the environment. The
process may be either reversible or irreversible. The useful work obtained in the
process in the form of equation
W <= (U – ToS + po V) – (Uo – ToSo _ poVo)
Let 
= U – ToS + poV where  is called the availability function and a
composite property of both the system and its environment, with U,S,V being
properties of the system at some equilibrium state and To po are the temperature
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and pressure of the environment. The decrease in availability function in a
process in which the system comes to equilibrium with the environment is
 - o = (U – ToS + po V) – (Uo – ToSo _ poVo)
Hence W < =  - o
Thus the useful work is equal to or less than the decrease in availability function.
The availability A of a given system in a given environment is the maximum
useful work obtainable in reversible process.
A = W max =  - o .
This work is obtained in part from a decrease in the internal energy of the system
and in part from the heat withdrawn from the environment.
Let a system be taken from an equilibrium state 1, in which its availability is A1 to
a second equilibrium state 2 in which its availability is A 2. The end state 2 is not
in equilibrium with the environment. The maximum useful work that could be
obtained in the process
W max = A1 – A2 = (1 - o ) – (2 - o )
W max = (1 - 2)
Availability in steady flow:
The steady stet steady flow equation is given by
H1 + ½ (mv12 ) + mgz1 + Q = H2 + ½ (mv22 ) + mgz2 + W
The subscript 1 and 2 refer to the entrance and exit respectively. At the exit let
the system be in equilibrium with the environment at pressure po and
temperature To. Let symbols without subscripts refers to the entrance condition
of the system and changes in KE and PE are negligible. Then useful work is
given by
W = (H-Ho) + Q. the greater the value of Q larger will be the useful work. Thus
W will be maximum when Q is a maximum.
Let S and So be the entropies of the systems at the entrance and exit of the
device then
(S) system = So – S. and
(S)surr = - (Q / To)
From the entropy principle ( So –S ) – (Q - To) > = 0.
Therefore Q <= To (So- S) .
The useful work
W <= (H-Ho) + To(So-S)
<= (H-ToS) + (Ho - ToSo)
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Hence W <= B – Bo where B = H-ToS it is called availability function for
steady flow and is also called as Keenan function.
For entrance and exit condition the useful work is maximum when the heat
absorbed is a maximum. I.e when the internal irreversibility is zero.
The maximum work obtainable from a system at the entrance of a device
when the pressure and temperature at the exit are those of the environment is
called the available of the system in steady flow and is given by
A = W max = B- Bo.
The alternative names for availability and for unavailable quantity To S are
Exergy and Anergy respectively.
Reversible work in a non-flow process:
In a non-flow process dm1 = dm2 =0
The entropy equation for flow process reduces to
(dW)rev = - d (U1 - ToS + ½ (mv12 ) + mgZ)
Between two equilibrium end states 1 and 2
(W) rev1-2 = U1-U2 –To(S1-S2) + m( V12-V22)/2 + mg(Z1-Z2)
If the system doesn’t possesses any KE and PE
(W) rev1-2 = U1-U2 –To(S1-S2)
Irreversibility:
The actual work done by a system is always less than the idealized reversible
work, and the difference between the two is called the irreversibility of the
process.
I = W max – W
This is also some time referred to as degradation or dissipation. For a non flow
process between the equilibrium states, when the system exchanges heat only
with the environment
I = {(U1-U2) – To(S1-S2) } – {(U1-U2) + Q} = To (S2-S1) – Q
= To (S) system + To (S)surr
= To[ (S)system + (S)surr ]
Hence I > 0.
Similarly for the steady flow process I = W max – W
= To (S2-S1) – Q
= To (S) system + To (S)surr
= To[ (S)system + (S)surr ]
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Thus same expression for irreversibility applies to both flow and non-flow
process. The quantity To[ (S)system + (S)surr ] represents an increase in
unavailable energy or Anergy.
Second Law efficiency:
With the increased use of availability analysis in recent years a term called
second law efficiency has come into more common use. This term refers to
comparison of the desired output of a process with the cost or input in terms of
the thermodynamic availability. Thus the isentropic turbine efficiency defined by
the ratio of actual work output to the work for a hypothetical isentropic expansion.
From the same inlet state to the same exit pressure which is called first law
efficiency, in that it is a comparison of two energy quantities. The second law
efficiency as just described would be the actual work output of the turbine divided
by the decrease in availability from the same inlet state to the same exit state.
Thus the second law efficiency is
Second law = W / (1 - 2)
Where (1 - 2) is the decrease in availability for a steady state steady flow
process. Which is equal to the reversible work or maximum work obtainable.
In this sense this concept provides a rating or measure of the real process in
terms of the actual change of state and is simply another convenient way of
utilizing the concept of thermodynamic availability. In a similar manner the
second law efficiency of a pump or a compressor is the ratio of the increase in
availability to the work input to the device.
NUMERICAL EXAMPLES:
1) In a certain process a vapor while condensing at 420OC transfers heat to
water evaporating at 250OC. The resulting steam is used in a power cycle,
which rejects heat at 35OC. What is the fraction of the available energy in
the heat transfer from the process vapor at 420OC that is lost due to
irreversible heat transfer at 250OC.
Fig 6: Increase in unavailable energy
8
Solution:
ABCD would have been the power cycle if there was no temperature difference
between the vapor condensing and the water evaporating, and the area under
CD would have been the unavailable energy at 35 OC. EFGD is the power cycle
when the vapor condenses at 420OC and the water evaporates at 250OC. The
unavailable energy becomes the area under DG. Therefore the increase in
available energy due to irreversible heat transfer is represented by the area
under CG.
Q1= T1 S = T ’1 S’
= S’ / S = T1 /T ’1
W = work done in cycle ABCD = ( T1- To) S
W’ = work done in cycle EFGD = ( T’1- To) S’
The fraction of energy that becomes unavailable due to irreversible heat transfer
= (W-W’) / W = [ To(S’-S) ] / [( T1- To) S ]
= To [(S’/S) - 1 ] / ( T1- To) = To( T1- T’1) / T1( T1- To)
= [308( 693-523) ] / [ 523 (693-308) ] = 0.26
Ans
2) Air expands through a turbine from 500 kPa, 520OC to 100 kPa,. 300 OC.
during expansion 10KJ of heat is lost to the surroundings which is at 98
kPa, 20 OC. neglecting KE and PE changes determine per kg of air a) the
decrease in availability b) the maximum work c) the irreversibility.
Solution:
For air the change in entropy is given by S2-S1=[ mCp ln (T2/T1) - mR ln(P2/P1)]
For 1kg of air s2-s1 = Cp ln (T2/T1) - R ln(P2/P1)
The change in availability is given by, (1 - 2) = b1-b2 = (h1-Tos1) - (h2 –Tos2)
= (h1-h2) –To (s1-s2) = Cp(T1-T2) – To (Cp ln (T2/T1) - R ln(P2/P1))
= 1.005 (520-300) – 293(1.005 ln(573/793) -0.287 ln (1/5)
= 1.005 (220) – 293 (-0.4619 + 0.3267) = 221.1 + 39.6 = 260.7 kJ/kg
W max = Change in availability = (1 - 2) = 260.7 kJ/kg
From SSSF equation we have
Q + h1 = W + h2 or W = (h1-h2) + Q = 1.005(520-300) –10 =211.1 kJ/kg
Thus the irreversibility I = Wmax – W = 260.7 – 211.1 = 49.6 kJ/kg
Ans.
9
3) Consider a steam turbine that has throttle governor. The steam in the
pipeline flowing through the turbine has a pressure of 3 MPa and a
temperature of 360 OC. At certain load the steam is throttled in an adiabatic
process to 1.5 MPa. Calculate the availability per kg of steam before and
after the process and the reversible work and irreversibility per kg of steam
for this process. Assume To= 25OC po=1 bar = 0.1MPa
Solution:
The availability at the initial condition
= (h1-ho) –To(s1-so) = (3140.9 – 104.9) – 298 (6.7844 – 0.3664 )
= 3036-1912.38 = 1123.62 kJ/kg
Ans.
Similarly availability at 1.5MPa after adiabatic throttling = (h e-ho) – To(se-so)
since in adiabatic throttling he=hi=3140 kJ/kg and adiabatic Q = 0.
Therefore se from the steam table corresponding to he = 3140.9 kJ/kg and
1.5MPa = 7.0833 kJ/kg K.
Thus e = (3140.9 – 104.9) – 298 (7.0833-0.3664) = 3036.13 –2001.46 =
1034.67 kJ/kg
Hence W rev = i - e = 1123.62 – 1034.67 = 88.95 kJ/kg
Also irreversibility I= To –(se-si) = 298(7.0833-6.7844) = 89.07 as W actual is
zero.
4) Consider an air compressor that receives ambient air at 100 kPa, 25OC. It
compresses the air to a pressure of 1 MPa, where it exits at a temperature
of 540 OK. Since the sir and the compressor housing are hotter than
ambient it looses 50 kJ per kg air flowing through the compressor. Find the
reversible work, reversible heat transfer and irreversibility in the process.
-Q
-W
e
i
Fig 7
10
Solution:
It is a non-adiabatic compression with no change in KE and PE. Using SSSF
equation, which has single inlet and exit. From the ideal gas tables h 1= 298.62
kJ/kg and he = 544.69 kJ/kg, si = 6.8629 kJ/kgK and se = 7.4664 kJ/kgK.
The energy equation for the actual compressor is
W= h1-h2 + Q = 298.62 –544.69 –50 = -296.07 kJ/kg
The reversible work for the given change of state is given by
W rev = To (s2-s1) - (h2-h1) + Q[1-(To/Th)]
= 298.2 (7.4664 - 6.8629) – (0.287 ln 10)- (544.69-298.62) +0 = -263.17 kJ/kg
Thus irreversibility I =w rev-w = -263.17 – (-296.07) = 32.9 kJ/kg.
5) In a steam boiler, hot gases from a fire transfer heat to water which
vaporizes at constant temperature. In a certain case, the gases are cooled
from 1100oC to 550oC. The specific heat of gases is 1.005kJ/kg K and the
latent heat of water at 220oC is 1858.5kJ/kg. All the heat transferred from
the gases goes to the water. How much does the total entropy of the
combined system of gas and water increase a result of the reversible heat
transfer? Obtain the result on the basis of 1 kg of water evaporated. If the
temperature of the surroundings is 30oC, find the increase in unavailable
energy due to irreversible heat transfer.
Solution.
Fig 8 : h – s diagram
11
Gas is cooled from state 1 to state 2 (Fig). For reversible heat transfer, the
working fluid in the heat engine would have been heated along 2-1, so that at any
instant, the temperature difference between gas and the working fluid is zero.
Then 1-b would have been the expansion of the working fluid down to the lowest
possible temperature To, and the amount of heat rejection would have been
given by the area abcd.
When water evaporates at 220.o C as the gas gets cooled from 1100oC to 550
oC, the resulting power cycle has an unavailable energy represented by the area
aefd. The increase in unavailable energy due to irreversible heat transfer is thus
given by area befc.
(S) water = Latent heat absorbed /T = 1858.5/(273+220) = 3.77 kJ/kg-K
Q1 = Heat transferred from the gas
=Heat absorbed by water during evaporation
=mgCpg (1100-550)=1 X 1858.5 kJ
Hence, mgCpg = 1858.5/550 =3.38kJ/o C
S gas =  ( dQ /T ) =  mg Cpg dT/T
mg Cpg ln Tg2 /Tg1 = 3.38 ln 823 / 1373 = -3.38 * 0.51 = -1.725 kJ/K
hence S total = (S) water + (S) gas
= 3.77 – 1.725 = 2.045 kJ/K
Increase in unavailable energy = To (S) total = 303 * 2.045 = 620 kJ.
6) Calculate the available energy in 40 kg of water at 75oC with respect to
the surroundings at 5oC, the pressure of water being 1 atm.
Solution
If the water is cooled at a constant pressure of 1 atm from 75 oCto 5oC the
heat given up may be used as a source for a series of Carnot engines each using
the surroundings as a sink. It is assumed that the amount of energy received by
any engine is small relative to that in the source and the temperature of the
source does not change while heat is being exchanged with the engine.
Let us consider that the source has fallen to temperature T, at which level there
operates a Carnot engine which takes in heat at this temperature and rejects
heat at To = 278 K. If s is the entropy change of water, the work obtainable is
 W = - m (T-To ) s where s is negative.
 W – 40 (T-To) Cp  T/T =
-40Cp (1-To/T) T
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With a very great number of engines in the series the total work (Maximum)
obtainable when the water is cooled from 348K to 278 K would be
W (max) = AE = - lim  40 Cp (1- To/T) T
= -  40 Cp (1- To/T) T = 40 Cp [(348-278) – 278 ln 348/278]
= 1340 kJ.
Q1 = - 4 0 * 4.2 (348-278) = 11760 kJ.
Unavailable energy = Q1 – W
max =
11760 – 1340 = 10420 kJ.
7) A 5 kg copper block at a temperature of 200 oC is dropped into an
insulated tank containing 100 kg oil at a temperature of 30 oC. Find the
increase in entropy of the universe due to this process when copper block
and the oil reach thermal equilibrium. Assume that the specific heats of
copper and oil are respectively 0.4 kJ/kgK and 2.1 kJ /kgK.
Solution
Applying energy balance, Energy lost by copper block = Energy gained by the oil.
Mcu X Cp cu X Temperature difference = m oil X Cpw X Temperature difference.
5 X 0.4 X (200 – t ) = 100 X 2.1 (t – 30 ) 212t = 6700 , t = 31.6oC
Heat lost by Copper = 5 X 0.4 X (473 – 304.6) = 336 8 kJ, then associated
entropy change is dQ/T 336.8/473 0.7120 kJ/K.
Heat gained by oil = 100 X 2.1 X 1.6 = 1.1089 kJ/K.
Entropy of the universe = S cu +Soil = 0.712 +1.1089 = 1.8209 kJ/kg Ans
8) 0.5 kg of ice block at –10oC is brought into contact with 5 kg copper
block at
80oC in an insulated container. Determine the change in
entropy of
i)
ii)
iii)
Ice block.
Copper block.
The universe.
Given specific heat of ice=2kJ / kg K, specific heat of water = 4.2 kJ/kg K.
specific heat of copper =0.5 kJ /kg K, enthalpy of fusion of water at 0 oC= 334
kJ/kg.
Solution
Writing the energy balance equation for the system
Energy lost by the copper block = Energy gained by the ice
5 X 0.5 (80 - t) = 0.5 X 2.0 ( 0+10) + 0.5 X 334 + 0.5 X .4.2 (t - 0)
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Therefore the value of t is t = 5OC
Entropy change of copper block = dQ/T = 187.5/(273 +80) = 0.53116kJ/K
Entropy change of ice block= dQ/T = (10 +167+10.5)/263 = 0.7129kJ/K
Entropy change of the universe = ( S )Cu + ( S)w = 1.24406 Ans
9) Calculate the availability and unavailability of a system that absorbs
15000kJ of heat from a heat source at 500K temperatures while the
environment is at 290 K temperature.
Solution:
Entropy Change,
ds = Q/T=15000/500 =30 kJ/K
Unavailable work =T o ds =290 X 30 =8700 kJ
Available work = Q - To ds =15000 –290 X 30 =6300 kJ
10) 0.2 kg of air initially at 575 K temperature receives 300 kJ of heart
reversibly at constant pressure. Determine the available and unavailable
energies of the heat added.
Take Cp for air = 1.005 kJ / kg K and temperature of surroundings 300 K.
Solution:
Let T2 be the temperature of air after the addition of heat at constant pressure.
Then
300 = mCp (T2-T1) = 0.2 X 1.005 (T2 –557)
=0.201 T2 -115.57
T2 = (300+115.57)/0.201 =2067.5K
Entropy change, ds = mCp loge T2/T1 = 0.2 X 1.005 oge 2067.5/575 = 0.2572
kJ /K
Unavailable work = T o dS = 300 X 0.2572 = 77.16 kJ
Available work = Q- T o dS = 300-300 X 0.2572 = 222.84 kJ
11) A closed system contains 2 kg of air during an adiabatic expansion
process there occurs a change in its pressure from 500kPa to 100 kPa and
in its temperature from 350 K to 320 K. if the volume doubles during the
process make calculations for maximum work, the change in availability
and irreversibility. Take for air Cv = 0.718 kJ/kg K and R = 0.287 kJ/kgK. The
surrounding conditions may be assumed to be 100 kPa and 300 K.
Solution:
14
When the system undergoes a change from state 1 to 2 during a mass flow
process the maximum obtainable work is given by
W max = (U1-U2) –To(S1-S2). Now (U1-U2) =
mCv(T1-T2) = * 0.7185(350-220) = 43.08 kJ.
And
S1-S2 = m[Cv ln(T2/T1) – R ln V2/V1]
= 2*[0.718 l n 320 / 350 – 0.287 ln 2]
= 0.2693 kJ/K. then W max = 43.08 – 300 (-0.2693) = 123.87 kJ.
Change in availability is given by A1 – A2 = (U1-U2) –To(S1-S2) + Po (V1-V2)
V1 = mRT1 / P1 = 2 * 0.287 * 250 / 500. = 0.4018 m3.
V2 = 2 V1 = 2* 0.4018 + 0.8036 m3.
A1-A2 = 123.87 + 100(0.4018 – 0.8036) = 83.67 kJ
It is the measure of maximum useful work or newt work obtainable from the
system.
Irreversibility = max work – actual work
W actual = Q – (U2-U1)
Since the system – process is adiabatic Q = 0 and W actual = -(U2-U)1 = U1-U2
= n mCv(T1-T2) = 2* 0.718 (350-320) = 43.08 kJ.
Irreversibility = 123.87 - 43.08 = 80.79 kJ.
Problem from Availability.
12) A system at 500 K receives 7200kJ/min from a source at 1000 K. The
temperature of atmosphere is 300 K. assuming that the temperature of the
system and source remain constant during heat transfer find out
The entropy produced during heat transfer.
The decrease in available energy after heat transfer.
Temperature of the source T1 = 1000 K
Temperature of the system, T2 = 500 K
Temperature of atmosphere, To = 300 K
Heat received by the system, Q =7200kJ/min
Change in entropy of the source during heat transfer
= -Q/T1 = -7200/1000 = - 7.2 kJ/min K
Change in entropy of the system during heat transfer
= Q/T2 = -7200/500 =14.4kJ/min K
The net change of entropy, S -7.2 + 14.4 = 7.2kJ/min K
15
Decrease in available energy with source = (1000-300) 7.2 = 5040kJ/min
Decrease in available energy with the system = (500 –300) 14.4 = 2880 kJ/min.
13) 15 kg of water is heated in an insulated tank by a churning process
from 300 K to 340 K. If the surrounding temperature is 300 K, find the loss
in availability for the process
Solution
Work added during churning = increase in enthalpy of water
= 15* 4.187*(340 300) = 2512.2 kJ = Enthalpy in the water
Availability of this energy is given by
m[(u1-u0) – T0 s]
s= Cp loge (T1/T0)
s = 4.187 loge (340/300 )
s= 0.524 kJ/kgK
Available energy = m[Cv (T1 – T0)- T0 s]
=15[4.187 (340 – 300) – 300 X0.524]
= 2349.3kJ Ans
Loss of availability 2512.2 – 158.7
= 2353.5 kJ Ans
This shows that conversion of work into heat is highly irreversible process, since
out of 2512.5 kJ of work energy supplied to increase the temperature, only 158.7
kJ will be available again for conversion into work.
14) 5.0 kg of air at 550 K and 4 bar is enclosed in a closed system.
Determine,
i) The availability of the system if the surrounding pressure and temperature are
1 bar and 290 K respectively.
ii) If the air is cooled at constant pressure to the atmospheric temperature,
find availability and effectiveness.
Availability of the system is
= m[(u1 –u0)-T0(s1-s2)] = m[Cv(T1 – T0) - T0s]
s = Cp *loge (T1/T0) – R loge (p1/p0)
s = 1.005 * loge(550/290) – 0.287 * loge(4/1)
= 0.643 – 0.397 = 0.246 kJ/kg K
Availability of the system = m[Cv(T1 – T0) - T0s]
5[0.718(550 - 290) –290 X0.246] = 576.7 kJ
Ans
16
ii) Heat transferred during cooling
Q = m X Cp X (T1 –T0)
5.0 X1.005 X(550 - 290) = 1306.5 kJ Heat lost by the system
Change of entropy of the system during cooling
S = m X Cp X loge(T1/T0)
S = 5.0 X1005 X loge(550/290) = 3.216 kJ/K
Unavailable portion of this energy = T0 ()S = 290 X 3.216 = 932.64 kJ
Available energy = 1306.5 - 932.64 = 373.86 kJ
Ans
Effectiveness,
 = Available Energy/ Availability of the system
=(373.86/576.7) = 0.648 OR 64.8%

= 64.8% Ans
17
SESSION 1
Outcome of the Session:
 Microscopic and Macroscopic point of view
 Thermodynamics Definition
 Thermodynamic Systems
INTRODUCTION:
Thermodynamics is the science of energy transfer and its effect on the physical
properties of substances. The alternate definition is: thermodynamics is the science that
deals with work and heat and these properties of substances that bear a relation to heat
and work. Like all sciences, the basis of thermodynamics is experimental observation.
Thermodynamics (from the Greek therme, meaning "heat" and dynamis,
meaning "power") is a branch of physics that studies the effects of changes in
temperature, pressure, and volume on physical systems at the macroscopic scale by
analyzing the collective motion of their particles using statisticsThis subject was
developed mainly by
1) Carnot
2) Mayer
3) Clausius 4) Joule
5) Kelvin
6) Maxwell 7) Plank
8) Gibbs
A thermodynamic system is a device or combination of devices containing a
quantity of matter that is being studied. A typical thermodynamic system - heat moves
from hot (boiler) to cold (condenser), and work is extracted, in this case by a series of
pistons.
A typical thermodynamic system
The study of thermodynamics is the basis of such fields as steam power plants, IC
Engines, Gas dynamics and aerodynamics, fluid mechanics, Refrigeration and Air
conditioning and heat transfer.
18
A Steam Power Plant
Thermodynamics deals with four laws. Namely Zeroth law, first law, second law
and Third law of thermodynamics. Fortunately, there is no mathematical proof for any of
these laws of thermodynamics, like physical laws, but they are deduced from
experimental observations.
Refrigeration Cycle
Thermodynamics deals with three E’s, namely Energy, Equilibrium and
Entropy. Thermodynamics also talks about study of materials, chemical reactions,
plasmas and other biological reactions.
Macroscopic and Microscopic point of view:
This study deals with macroscopic, as opposed to microscopic or statistical
thermodynamics. In microscopic thermodynamics individual molecule is considered and
19
analysis of collective molecular action. In macroscopic thermodynamics, we concern
ourselves with the overall effect of the individual molecular interaction.
Macroscopic point of view: The macroscopic level is the level on which we live. We
measure most of the quantities on this level.
Ex: Temperature measurement, pressure measurement, total volume measurement,
specific volume measurement. Thus, microscopic point of view will be used only to
explain some phenomena that can’t be understood by macroscopic means.
Microscopic point of view: Considers a system containing a cube of 25 mm containing
monoatomic gas at atmospheric pressure and temperature. This volume contains approx.
1020 atoms. To describe the position of each atom in a coordinate system we require
three equations. To describe the velocity of each atom we have to specify 3 velocity
components. Thus to describe completely the behavior of the system from a microscopic
point of view we must deal atleast 6*1020 equations. It is a hopeless computational task.
The other approach that reduces number of variables to a few that can be handled
is the macroscopic point of view of Classical Thermodynamics. It concerns with the gross
or average effect of many molecules and can be measured by instruments. This
measurement is the time-averaged influence of many molecules.
In the present study, we concentrate on macroscopic point of view.
Statistical thermodynamics, classical thermodynamics deals with the significance
of microscopic point of approach. From the macroscopic point of view it is very clear that
continuum has to be there in the system because we are not concerned with the behavior
of the individual molecule.
Substance: What follows will be illustrations of the thermodynamics, one must be able
to solve problems and to do what the part of the problem must be enumerated. The first
consideration is that there must be something performing the energy transformations.
This something is called a substance. Ex: In case of IC engine gasoline and air mixture
constitutes the substance. In steam turbine the substance is steam.
The substance may be further divided into sub categories, namely pure substance
i.e. if it is homogeneous in nature- i.e. if it does not undergo chemical reaction and is not
a mechanical mixture of different spices. The other substance is a mixture substance
which is not a pure substance.
Thermodynamic System:
A substance does not exist alone. It must be contained. This brings us to the concept of a
system.
In thermodynamics a system is defined as any collection of matter or space of fixed
identity, the concept is one of the most important thermodynamics.
20
Concept of a Boundary
System boundary: When a system is defined, let us say, fluid in a cylinder, what
separates the fluid from the cylinder wall and the piston and every thing external to the
piston-cylinder? it is the system boundary. Everything not in the system is called the
surrounding. Note that piston can be raised or lowered, but the system, matter of fixed
identity is constant.
The system is further divided into closed system, open system and isolated system.
21
_____________________________________________________________
SESSION 2
Outcome of the Session
 Thermodynamic System
 Open and Closed and Isolated System
 Control Volume
 Units, Thermodynamic Properties
 Extensive and Intensive Property
Thermodynamic System
A Thermodynamic system is defined as a quantity of matter or a region in space
upon which attention is concentrated in the analysis of a problem. Everything external to
the system is called the surrounding or environment. The system is separated from the
surrounding by the system boundary. Boundary may be either fixed or moving. A system
and its surrounding together comprise a universe.
A System, Surroundings and Boundary
Open System: The open system is one in which matter crosses the boundary of the
system. There may be energy transfer also. Most of th4e engineering devices are
generally open systems. Ex: An air compressor in which air enters at low pressure and
leave at high pressure and there is energy transfer across the system boundary.
22
An Open System
Closed System: A closed system is a system of fixed mass. There is no mass transfer
across the system boundary. Ex: A certain quantity of fluid in a cylinder bounded by a
piston constitutes a closed system.
A Closed System
Isolated System: The isolated system is one in which there is no interaction between the
system and surrounding. It is of the fixed mass and energy and there is no mass or energy
transfer across the system boundary.
Control Volume and Control Surface:
In thermodynamic analysis of an open system such as air compressor, gas turbine
in which there is a flow of mass into and out of the system, attention is focused on a
certain volume in space surrounding the compressor known as control volume, bounded
by a surface called the control surface. Matter as well as energy can cross the control
surface.
Gas turbine
Homogeneous and Heterogeneous system:
A quantity of matter homogeneous throughout in chemical composition and
physical structure is called a phase. Every substance can exist in any one of the three
phases viz. Solid, Liquid or gas.
23
A system consisting of a single phase is called a homogeneous system while a
system consisting of more than one phase is known as a heterogeneous system.
THERMODYNAMIC PROPERTIES
Thermodynamic properties are taken from a macroscopic perspective. We are
dealing with quantities that can either directly or indirectly be measured and counted.
Therefore, the matter of units becomes an important consideration.
Mass, length and time are considered as fundamental physical quantities, they are related
by Newton's second law of motion, which states that the force acting on a body is
proportional to the product of mass and acceleration in the direction of force.
i.e. F= m * a
Mass – kg
Length – m
Time – s
This is adopted by CGPM- General Conference of Weights and Measures
In thermodynamics temperature is also considered as fundamental unit in Kelvin.
Energy: One of the very important concepts in a study of thermodynamics is the concept
of energy. It is defined as the capacity to do work. It is also defined as the capability to
produce an effect
When considered from molecular point of view, three general forms of energy
become important.
1) Intermolecular potential energy.
2) Molecular kinetic energy
3) Intermolecular energy.
The energy is the important concept which depends on the mass, velocity,
intermolecular attraction. In all intermolecular internal energy is most difficult to
evaluate.
Specific Volume: It is a macroscopic property and defined as the volume occupied by
unit mass. It is reciprocal of density and its unit is m3/ kg. The specific volume of a
system in a gravitational field may vary from point to point. Specific volume increases as
the elevation increases. Thus the definition of specific volume involves the specific
volume of a substance at a point in a system.
Pressure: The pressure in a fluid at rest at a given point is the same in all directions. We
define pressure as the normal component of force per unit area. Its unit is pascal or N/m2.
When dealing with liquids and gases we ordinarily speak of pressure. For solids we speak
of stresses.
Two other units not part of international system continue to be widely used are
Bar = 105 Pa = 0.1MPa and standard atmosphere is 1 atm = 101325 Pa.
In most thermodynamic investigations, we are concerned with absolute pressure.
Most pressure vacuum gauges however read the difference between the absolute pressure
and the atmospheric pressure at the gauge. This refers to as gauge pressure.
24
Pabs1
Ordinary Pressure gauge P = Pabs1 – P atm
Patm
Ordinary vacuum gauge
P = P atm – P abs2
Pabs2
Barometric reads atmospheric pressure
Illustration of terms used in pressure measurement.
Temperature: It is a fundamental property of thermodynamics and defined as the
hotness of the body. Temperature first of all as a sense of hotness or coldness when we
touch an object. We also learn that when a hot body and a cold body are brought into
contact the hot body becomes cooler and cold body becomes warmer. Because of these
difficulties in defining temperature we define equality of temperature.
Property: It is defined as any quantity that depends on the state of the system and is
independent of the path ( i.e. the prior history) by which the system arrived at the given
state. Conversely the state is specified or described by the properties and later we will
consider the number of independent properties a substance can have, i.e, the minimum
number of properties that must be specified to fix the state of a substance.
Thermodynamic properties can be divided into 2 general classes: intensive and extensive
properties:
Intensive property: An intensive property is independent of mass; thus intensive
property value remains same even if the matter is divided into two equal parts. Ex:
pressure, temperature, density etc.
Extensive Property: The value of an extensive property varies directly with the mass,
i.e. if a quantity of matter in given state is divided into 2 equal parts, the properties will
have the half the original values. Ex: mass, total volume, total enthalpy, total energy etc.
25
SESSION 3




Thermodynamic State Point
Process, Cycle
Cyclic and Non-cyclic Process
Quasi-static Process.
Thermodynamic State Point:The state may be identified or described by certain
observable macroscopic properties; some familiar one are temperature, pressure and
density. The state is specified or described by the properties. The state point can be
indicated on a thermodynamic coordinate system. Thermodynamic coordinate system
includes pressure volume diagram, temperature volume diagram, temperature entropy
diagram, enthalpy entropy diagram, pressure enthalpy diagram so on and so forth.
These co-ordinate system represents thermodynamic state of a substance.
Consider a system not undergoing any change. At this point all the properties can be
measured or calculated throughout the entire system, which gives us a set of properties
that completely describes the condition or the state of the system.
Thermodynamics deals with equilibrium state. When a system undergoes any
change then change of state will occur.
Weights
Piston
Gas
Cylinder
Piston Cylinder arrangement
Process: Whenever one or more of the properties of a system change we say that a
change in state has occurred. For ex: in a piston and cylinder arrangement, if weight is
removed from the piston rises and change in state occurs in which pressure decreases and
specific volume increases. The path of succession of states through which the system
passes is called process.
26
Path: Path is the complete series of states through which the system passes during a
change from one given state to other state. It is clear that the transformation of a system
from one fixed state to another state is called a process.
The thermodynamic processes that are commonly met within engineering practice are
1) Constant pressure process (Isobaric) 2) Constant volume process(Isochoric)
3)Constant temperature process( Isothermal) 4) Reversible adiabatic process
(Isentropic process) 5) Polytropic process 6) Throttling process.
If the system passes through a series of equilibrium states during the process its
called reversible process. On the other hand the system passes through a series of nonequilibrium states during a process it is called irreversible process. The state of the
processes can not be plotted on the co-ordinate systems since the path of the process is
not defined.
Generally the system is in equilibrium in the beginning and at the end of the process, the
reversible process can be plotted on the coordinate diagram by continuous line and an
irreversible process by a dotted line.
1
1
P
P
V
Representation of a reversible process
An irreversible process
2
2
V
Quasi – static process (very slow process) :
Quasi – meaning almost, static meaning infinite slowness. Thus quasi-static process is
infinitely slow transition of a system. Infinite slowness is the characteristic feature of a
quasi-static process. A quasi-static process is a succession of equilibrium states. It is a
reversible process.
Cycle: It is a process whose initial and final states are same. Thus at the end of a cycle
all the properties of a working fluid have the same values as they had in the initial states.
There are 2 types of cycles. viz. thermodynamic cycle and mechanical cycle.
27
Thermodynamic cycle:
It is one in which the working substance is recirculated. Ex: water that circulates
through steam power plant and refrigerant that passes through refrigeration plant are the
examples of thermodynamic cycle. There is change of phase during the process but the
end states do not change
Mechanical cycle:
In case of a mechanical cycle the working substance is not re circulated. In an IC
engine air and fuel are burnt in the engine, converted into the products of combustion and
are then exhausted into the atmosphere. Hence this type of cycle is called mechanical
cycle.
Thermodynamic Equilibrium:
The word equilibrium implies a state of balance. In an equilibrium state, there are
no unbalanced potentials within the system or driving forces. Thus, a system in
equilibrium experiences no changes when it is isolated from its surroundings.
There are many types of equilibrium. A system is not in thermodynamic
equilibrium unless the condition of all the relevant types of equilibrium are satisfied,
which includes 1) Thermal equilibrium 2) Mechanical equilibrium 3) Phase equilibrium
and 4) Chemical equilibrium.
Thermal Equilibrium:
If the temperature is the same throughout the entire system .i.e the system
involves no temperature differential which is the driving force or heat flow then we say
system is in thermal equilibrium.
Mechanical equilibrium:
It is related to pressure, velocity. A system is in mechanical equilibrium if there is
no change in pressure, velocity, specific volume at any point of the system with respect to
time. However the pressure may vary within the system with elevation as well as
resultant of gravitational effects. However there should not be any imbalance of forces.
Then we say the system is in mechanical equilibrium
Phase equilibrium:
If a system involves two phases it is in phase equilibrium when the mass of each
phase reaches equilibrium level and stays there.
Chemical equilibrium:
If the systems chemical composition does not change with time, i.e., no chemical
reaction occur then we say the system is in chemical equilibrium.
Thus if all thermal, mechanical, phase and chemical equilibrium exist for a system
then we say the system exist in thermodynamic equilibrium
28
Session 4
Outcome of the Session
 Diathermic wall
 Zeroth Law of Thermodynamics
 Temperature Concepts
 International Practical Temperature Scale.
Diathermic wall:
A wall which is impermeable to the flow of heat is an adiabatic wall, where as a
wall which permits the flow of heat is a diathermic wall. Thus heat flow takes place
through this wall.
Zeroth law of Thermodynamics:
It states that when two bodies have equality of temperature with the third body,
they in turn have equality of temperature with each other.
When a body A is in thermal equilibrium with body B and also separately with body C
then B and C will be in thermal equilibrium with each other.
Equality of temperature – Zeroth law of Thermodynamics.
Temperature scales:
Two scales are commonly used for measurement of temperature namely,
Fahrenheit after Gabriel Fahrenheit (1686-1736) and Celsius. The Celsius scale was
formerly called the centigrade scale but is now designated the Celsius scale after Anders
Celsius (1701-1744), the Swedish astronomer who devised this scale.
In SI units of temperature scale we use absolute temperature scale or absolute
scale of temperature which comes from second law of thermodynamics and its unit is
Kelvin.
29
Table: Thermometers and Thermometric Properties
Thermometer
Constant volume gas thermometer
Thermometric property
Symbol
Pressure
P
Volume
V
Constant pressure gas thermometer
Electrical Resistance thermometer
Thermocouple
Mercury in glass thermometer
Resistance
R
Thermal e.m.f.
Length
e
L
The absolute scale is related the Celsius scale is the Kelvin scale after William
Thompson, 1824-1907, who is also known as Lord Kelvin and is designated K. The
relation is K=oC + 273.15
In 1967 the CGPM defined the Kelvin as 1/273.16 of the temperature at the triple
point of water.
In order to obtain a quantitative measure of temperature a reference body is used,
and a certain physical characteristic of this body which changes with temperature is
selected. The changes in the selected characteristic may be taken as an indication of
change in temperature. The selected characteristic is called the thermometric property and
the reference body which is used in the determination of temperature is called the
thermometer.
A Very common thermometer consists of a small amount of Mercury in an
evacuated capillary tube. In this case the extension of the mercury in the tube is used as
the thermometric property.
A common Mercury thermometer
thermometer
Mercury-in-glass
Presently temperature of triple point of water which is an easily reproducible state is now
the standard fixed point of thermometry.
International practical temperature scale: an international temperature scale was
adopted at the seventh general conference on weights and measures held in 1927. It was
30
not to replace the Celsius or ideal gas scales, but to provide a scale that could be easily
and rapidly used to calibrate scientific and industrial instruments.
International practical scale agrees with Celsius scale at the defining fixed points
listed in following table.
Table: Temperature of Fixed Points
Normal boiling point of oxygen
Triple point of water (standard)
Normal boiling point of water
Normal boiling point of sulphur
Normal melting point antimony
Normal melting point of silver
Normal melting point of gold
Temperature in 0C
-182.97
+0.01
100.0
444.60
630.50
960.80
1063.0
31
Session 5
Outcome of the Session
Numerical Examples with Solutions:
1) A tank contains mixture of 20kg of nitrogen and 20 kg of carbon monoxide. The total
tank volume is 20m3. Determine the density and specific volume of the mixture.
Solution:
Total mass of the mixture: 20 kg N2 + 20 kg CO = 40 kg mixture
Specific volume = volume / mass = 20 m3 / 40kg = 0.5 m3/kg
Density of mixture = mass / volume = 1/sp. vol = 1/ 0.5 = 2 kg / m3.
Ans.
2) An automobile has a 1200 kg mass and is accelerated to 7m/s2. Determine the force
required to perform this acceleration.
Solution:
Force required F = m * a = 1200 * 7 =8400 kg m /s2 = 8400 N.
Ans.
3) The emf in a thermocouple with the test junction at t0C on gas thermometer scale and
reference junction at ice point is given by e = 0.20 t – 5 * 10-4 t2 m V. the millivoltmeter
is calibrated at ice and steam points. What will this thermometer read in a place where
gas thermometer reads 500C?
Solution:
At ice point, when t = 0oC, e= 0 mV.
At steam point, when t=100oC, e= 0.20 * 100 – 5* 10-4 * (100)2 = 15 mV
At t = 50oC, e = 0.20 * 50 – 5* 10-4 * (50)2 = 15 mV = 8.75 mV
When the gas thermometer reads 50oC, the thermocouple will read
t = 100/15 * 8.75 or 58.33 oC.
Ans.
4) A barometer to measure absolute pressure shows a mercury column height of 725mm.
The temperature is such that the density of the mercury is 13550 kg/m3. Find the ambient
pressure.
Solution:
Ambient pressure = ρ * g * h = 13550 * 9.81* 725 / 1000 = 96371Pa
or
P = 0.9637 bar.
Ans.
32
5) The temperature t on a Celsius thermometric scale is defined in terms of a property p
by the relation p = e(t-B)/A, where A and B are constants. Experiment gives values of p
of 1.86 and 6.81 at the ice and steam point respectively. Obtain relation for t and also find
the temperature t for the reading of p = 2.5
Solution:
At ice point,
t = 0oC, p= 1.86
Hence
1.86 = e –B/A or eB/A = 1/1.86
1.86 = e(0-B)/A
and ln eB/A = ln 1/1.86
B/A = - 0.62058 --------------- (1)
At steam point, t=100oC, p= 6.81
6.81 = e(100-B)/A
ln 6.81 = (100 - B) /A = 1.9184
1.9184 A + B =100--------------------------(2)
Solving equations (1) and (2)
A = 77.05 and B = -47.81 hence p = e [t – (-47.81) ] / 77.05
The temperature at p=2.5,
2.5 = e (t+47.81)/77.05
ln 2.5 =( t + 47.81) / 77.05 or t + 47.81 = 0.9163 77.05
or
t = 22.79 0C
Ans.
6) A hiker is carrying a barometer that measures 101.3 kPa at the base of the mountain.
The barometer reads 85 kPa at the top of the mountain. The average air density is
1.21kg/m3. Determine the height of the mountain.
Solution:
Pressure at the base of the mountain = ρ1 * g * h1.
h1= p / (ρ1 * g ) = 101.3 *1000 / (9.81 * 1.21) = 8534 m.
Also at the top of the mountain p = ρ2 * g * h2.
h2 = p / (ρ2 * g ) = 85 *1000 / (9.81 * 1.21) = 7161 m.
Hence the height of the mountain = h1- h2 = 8534 – 7161= 1373 m.
Ans.
7) A lunar excursion module (LEM) weighs 1500 kgf at sea level; o earth. What will be
its weight on the surface of the moon where g=1.7 m/s2? On the surface of the moon,
what will be the force in kgf required to accelerate the module at 10m/s2?
Solution:
33
The mass m of the LEM is given by W = mg/g0
1500 kgf = m * { (9.806 m/s2 ) / ( 9.806 kg/ kgf * m/s2 )}
i.e. m=1500 kg
The weight of the LEM on the moon would be
W = 1500 kg * {(1.7 m / s2) / (9.806 kg / kgf * m/s2 ) }
Ans.
The force required to accelerate the module at 10 m/s2
= [ 1500 kg / (9.806 kg/kgf * m/s2) ] * 10 m/s2 = 1530 kgf
Ans.
8) A cannon ball of 5 kg acts on a piston in a cylinder of 0.15m diameter. As the
gunpowder is burnt, a pressure of 7 MPa is created in the gas behind the ball. What is the
acceleration of the ball if the cylinder is pointing horizontally?
Solution:
Mass of the cannon ball = 5 kg
Diameter of the cylinder = 0.15 m
Force = Area * Pressure = (π / 4) * (0.15)2 *7* 106 = 123716.25 N
F = m*a
Hence a = F /m = 123716.25 / 5 = 24743.25 m / s2
Ans.
Ans.
34
Pure Substances
P-T and P-V diagrams
Triple point and critical point
Sub cooled liquid
Saturated liquid, mixture of saturated liquid and vapor
Saturated vapor and superheated vapor states of a pure substance with water as
example.
Enthalpy of change of phase, dryness fraction, T-S and H-S diagrams.
Representation of various processes on these diagrams.
Steam tables and its use.
Throttling Calorimeter, Separating Calorimeter.
Throttling and Separating Calorimeter.
Introduction:
A pure substance is one that has a homogeneous and in variable chemical
composition. It may exist in more than one phase, but the chemical composition
is the same in all phases. Thus liquid water, a mixture of liquid water and water
vapor, and a mixture of ice and liquid water are all pure substance: every phase
has the same chemical composition. On the other and mixture of liquid air and
gaseous air is not a pure substance. Because of the composition of the liquid
phase is different from that of the vapor phase.
Some times a mixture of gases such as air is considered a pure substance as
long as there is no change of phase. Strictly speaking this is not true.
Vapor liquid solid phase equilibrium in a pure substance:
Fig 1: Constant pressure change from liquid to vapor phase for a pure
substance
35
Consider a system one kg of water contained in a piston cylinder
arrangement as shown in the figure. Suppose that the piston and weight maintain
a pressure of 0.1 MPa in the cylinder and that initial temperature be 20 0C. As the
heat is transferred to the water the temperature increases appreciably, the
specific volume increases slightly under constant pressure. When the
temperature 99.60C, additional heat transfer results in a change of phase. I.e.
some of the liquid becomes vapor. During this process both temperature and
pressure remain constant where as sp. Volume increases considerably. When
the last drop of liquid has vaporized further transfer of heat results in an increase
in both temperature and sp. volume of the vapor.
The term saturation temperature designates the temperature at which
vaporization takes place at a given pressure. Or the pressure is called saturation
pressure corresponding to the saturation temperature. Thus for water 99.6 0C the
saturation pressure is 0.1MPa, and for water at 0.1MPa the saturation
temperature is 99.60C. Thus there is a definite relation between saturation
pressure and saturation temperature.
Pressure
Vapor-pressure curve
Temperature
Fig 2: Vapor – Pressure curve of a pure substance
If the substance exists as liquid at the saturation temperature and
pressure it is called saturated liquid. If the temperature of the liquid is lower than
the saturation temperature for the existing pressure it is called either a sub
cooled liquid or a compressed liquid.
When the substance exists as a part liquid and part vapor at the saturation
temperature a dryness fraction comes into picture. It is also called as quality and
it is defined as the ratio of mass of vapor to the total mass. It is denoted by the
symbol ’x’. Quality has meaning only when the substance is in saturated state. I.e
at the saturation pressure and temperature. The quality x is an intensive
property.
36
If a substance exists as vapor at the saturation temperature it is called
saturated vapor or dry saturated vapor with x=1. When the vapor is at a
temperature greater than the saturation temperature at the saturation pressure, it
is said to exist as superheated vapor. After that the temperature increases as
heat is added at constant pressure.
Temperature – Volume (T-V) diagram for water:
Fig 3: T-V diagram for water
It is clear from the figure that constant pressure lines are ABCD, EFGH,
IJKL etc. the peak point of the figure indicated by N is the critical point of water.
Thus the critical pressure 22.089 MPa and corresponding critical temperature is
374.14 0C.
A constant pressure process at a pressure greater than the critical
pressure is represented by curve PQ. Thus water at 40 MPa, 20 0C is heated in a
constant pressure process, there will be never be two phases present at the state
shown. Instead there will be continuously in density at all the times and there will
be only one phase present. The question arises is: when do we have a liquid and
when do we have a vapor? The answer is that this is not a valid question at
super critical pressures. We simply term the substance as fluid. However rather
arbitrarily at temperatures below the critical temperatures we usually refer to it as
a compressed liquid and at temperatures above the critical temperatures as
superheated vapor. It should be noted that however at pressures above the
critical pressures we never have a liquid and vapor phase of pure substance
existing in equilibrium.
37
Table1: Some Critical point data
Critical
Temperature, 0C
Water
374.14
Carbon dioxide
031.05
Oxygen
-118.35
Hydrogen
-239.85
Critical
Pressure, MPa
22.089
07.39
05.08
01.30
Critical Volume,
m3/kg
0.003155
0.002143
0.003438
0.032192
Consider another experiment with piston cylinder arrangement. Suppose that the
cylinder contains one kg of ice at –200C and one bar. When heat is transferred to
the ice the pressure remains constant the specific volume increases slightly and
the temperature increases until it reaches 00C, at which point the ice melts and
temperature remains constant. This state is called saturated solid state. For most
substances the specific volume increases during this melting process. But for
water specific volume of the liquid is less than the specific volume of the solid.
Sublimation: If the initial pressure of the ice at –200C is 0.26 kPa, heat
transferred to the ice results in an increase in the temperature to –100C. At this
point however the ice passes directly from the solid phase to the vapor phase.
This process is known as sublimation. Further heat transfer results in
superheating of the vapor
Triple Point: Consider the ice at 0.6113 kPa and temperature of –200C.
Through heat transfer let the temperature increase until it reaches 0.01C. At this
point however further heat transfer may cause some of the ice to become vapor
and some to become liquid. At this point it is possible to have three phases in
equilibrium. This point is called the triple point. Triple point is defined as the state
in which all three phases may be present in equilibrium. The pressure and
temperature at the triple point for a number of substance is given in following
table.
Table 2:Triple Point Data
Temperature, 0C
Hydrogen
-259
Oxygen
-219
Nitrogen
-210
Carbon Dioxide
-56.4
Mercury
-39
Water
0.01
Zinc
419
Silver
961
Copper
1083
Pressure, kPa
7.194
0.15
12.53
520.8
0.00000013
0.6113
5.066
0.01
6.000079
38
Consider a solid state as shown in the figure, when the temperature
increases with constant pressure the substance passes directly from solid to
vapor phase. Along the constant pressure line EF the substance passes from
solid to liquid phase at one temperature and then from liquid to vapor phase at
higher temperature.
Constant pressure line CD passes through the triple point and it is only at
the triple point the three phases exists together in equilibrium. At a pressure
above critical pressure such as GH line there is no sharp distinction between
liquid and vapor phases. The triple point temperature and critical temperature
vary greatly from substance to substance. For ex: critical temperature of helium
is 5.3K. Therefore absolute temperature of helium at ambient conditions is over
50 times greater than the critical temperature. On the other hand water has a
critical temperature 374.14 0C (647.29K) and at ambient conditions the
temperature of water is less than ½ the critical temperature.
Allotropic transformation
It should be pointed out that a pure substance can exist in a number of different
solid phases. A transition from one solid phase to another is called an allotropic
transformation. This can be well understood by the following figure.
Temperature
Fig 4: P-T Diagram for water.
39
Fig 5: P-T Diagram for iron
Independent properties of a pure substance:
One important reason for introducing the concept of pure substance is that
the state of a simple compressible pure substance is defined by two independent
properties. For ex: if the specific volume and temperature of a super heated
steam are specified the state of the steam is determined.
To understand the significance of the term independent property, consider
the saturate liquid and vapor state of a pure substance. These two states have
the same pressure and temperature but they are definitely not the same state. In
a saturation state therefore pressure and temperature are not independent
properties. Two independent properties such as pressure -specific volume,
pressure and quality, temperature-specific volumes are required to specify a
saturation state of a pure substance.
Equations of state for the vapor phase of a simple compressible
substance:
From the experimental observations it has been established that the
equation of state under low-density gases is given by PV=RuT. The ideal gas
equation of state and compressibility factor equation are good approximations at
low-density conditions. Therefore ideal gas equation of state is very convenient
to use in thermodynamic calculations. The question comes what is low density?
Or what range of density will the ideal gas equation of state hold with accuracy?
The analysis gives the pressure temperature deviations from ideal gas
behavior. To answer5 the question the concept of compressibility factor Z is
introduced and is defined by the relation Z= PV/R uT. For ideal gas Z=1 and the
deviation of Z from unity is a measure of the deviation of actual relation from the
ideal gas equation of state.
40
Fig 6: Compressibility of Nitrogen
It shows a skeleton compressibility charge for nitrogen. Three observations can
be made from this chart.
All the temperature Z
1 as P
0. i.e. as the pressure approaches zero the
P-V-T behavior closely approaches that predicted by the ideal gas equation of
state. Note also that at temperatures of 300K and above i.e. above room
temperature the compressibility factor is near unity up to a pressure of 10MPa.
This means that the ideal gas equation of state can be used for nitrogen over this
range with considerable accuracy.
Now suppose we reduce the temperature from 300K but keep the pressure
constant at 4MPa, the density will increase and we note a sharp decrease below
unity in the value of compressibility factor. Values of Z <1 mean that the actual
density is greater than would be predicted by ideal gas behavior. As the
temperature reduced from 300K and pressure remains constant at 400MPa the
molecules are brought closer together, there is an attractive force between the
molecules. The lower the temperature the greater is the intermolecular attractive
force. Thus more attractive force more density, which is greater than, would be
predicted by the ideal gas. (Ideal gas intermolecular attraction is negligible)
On comparison of various compressibility charts for other pure substances it is
observed that the diagrams are all similar in the characteristics described above
for nitrogen at least in qualitative sense. Quantitatively the diagram are all
different sine critical temperature, critical pressure of different substances vary
over a wide range. There is a way to put all these substances on a common
basis called reduced properties w.r.t. the values at critical point like
Reduced pressure = Pr = P / Pc and reduced temperature Tr = T / Tc
41
These equations state that the reduced property for a given state ids the value of
this property on this state divided by the value of the same property at the critical
point.
Tables of Thermodynamic properties:
Tables of thermodynamic properties of many substances are available and in
general these tables have the same form. The steam table is specially selected
because they are a vehicle for presenting thermodynamic tables and steam is
used extensively in power plants and industrial purposes. The table consists of
four separate tables namely:
1) Saturated steam table
2) Saturated water table
3) Temperature basis table and
4) Pressure basis table.
It contains values of enthalpy, internal energy, specific volume and entropy. If the
values of T and P are given other all values can be directly taken from the table.
Therefore we should first learn how to use the table. We note that all these are
independent variables.
To finding the correct table other nuisance of everyday use of the table in the
interpolation. I.e. when one of the stated values is not exactly equal to a value
listed in the table. Recently computerized tables are in use, which eliminates
these problems. But the students nevertheless must learn to understand the
significance, construction and limitations of the tables.
The actual steam table as presented in various books is a summary table based
on complicated curve fit to the behavior of water. Here we concentrate three
properties namely T,P and v and note that other properties listed namely u, h, s
are presented. It is further noted that the separation of phases in terms of values
of P and T is actually described by the relation illustrated in the pressuretemperature diagram.
The specific volume of a substance having given quality can be found out by the
definition of quality. The volume is the sum of volume of liquid and vapor.
V= Vliq + V vap.
In terms of masses: mv = mliq * vf + m vap* vg
By introducing the quality x we have v = (1-x) vf + x vg
We know that
vfg = vg - vf
We can write the specific volume equation for wet steam as v = vf + x vfg
42
In superheated region pressure and temperature are independent properties and
therefore for each pressure a large number of temperature is given and for each
temperature four thermodynamics properties are listed namely specific volume,
enthalpy, entropy and internal energy.
Temperature – specific volume diagram :
Figure shows a T-v plot for water with an indication of percent error in
assuming ideal gas behavior along the saturated vapor curve and also in several
area of superheated region. Generally a slight decrease and only a small error is
made if one same that the volume of a compressed liquid is equal to the specific
volume of the saturated liquid at the same temperature. It is generally accepted
procedure particularly when compressed liquid data are not available.
Fig 7: Temperature – specific volume diagram for water
In the modern scenario computerized table are available. The main
program operated with a visual interface in the window environment on a PC
which is generally user friendly. The program operates on DOS environment,
which covers not only the tables of water, but it covers most of the pure
substances used in engineering industries. The generalized chart with
compressibility factor is also included so it is possible to get the value of Z a
more little accurately than reading the graph. It is useful in the case of two-phase
mixture, like saturated liquid and vapor values are needed.
43
Thermodynamic Surfaces:
The matter discussed in this chapter can be well summarized by a consideration
of a pressure, specific volume, temperature surface (PVT surface). Two such
surfaces are shown in the figure for a substance such as water in which the
specific volume increases during freezing and the other in which a substance, its
specific decreases during freezing.
Fig 8 : PvT Surface -- Water
Fig 9 : PvT Surface – Water
In these diagrams the pressure, specific volume and temperature are plotted on
a mutually perpendicular coordinates and each possible equilibrium state is thus
44
represented by a point on the surface. This follows directly from the fact that a
pure substance has only two independent intensive properties. All points along a
quasi equilibrium surface lie on the PvT surface since such a process always
passes through equilibrium states.
The regions of the surfaces that represent a single surface- the solid, liquid and
vapor faces are indicated. These surfaces are curved. The two phase regions –
the solid-liquid, solid-vapor, liquid-vapor regions are ruled by surfaces. It is
understood that they are made up of straight lines parallel to the specific volume
axis. This of course follows from the fact that in the two-phase region lines of
constant pressure are also lines of constant temperature, although the specific
volume may change. The triple point actually appears as the triple line on the
PVT surface, since the pressure and temperature of the triple point are fixed but
the specific volume may vary depending upon the proportion of each phase.
It is also of interest to note that the pressure temperature and pressure volume
projections of these surfaces. We have already considered the P-T diagram for
water. It is on this diagram we observe the triple point. Various lines of constant
temperature are shown on the P-v diagram and corresponding constant
temperature sections are identically seen on the P-v-T surfaces. The critical
isotherm as a
point of
intersection at
the critical point.
Fig 10 : PvT Surface for a Substance which Contracts Upon Freezing
45
Fig 11 : PvT Surface for a Substance which Expands Upon Freezing
One notice that for a substance such as water which expands on freezing, the
freezing temperature decreases with an increase in pressure. For a substance
that contracts on freezing, the freezing temperature increases as the pressure
increases. Thus as the pressure of the vapor is increased along the constant
temperature line a substance that expands on freezing first becomes solid and
then liquid. For substance that contract on freezing, the corresponding constant
temperature line indicates that as the pressure of the vapor increases, it first
becomes liquid and then solid.
T-s Diagram for a pure substance:
Consider the heating of 1 kg of ice at –50C to steam at 2500C. The pressure
being maintained at 1 atm. It is observed that the entropy of steam increases in
different regimes of heating namely
1) Entropy increase of ice to saturated freezing temperature
2) Entropy increase of ice as it melts into water.
3) Entropy increase of water as it is heated from 0Oc to 100OC.
4) Entropy increase of water as it is vaporized at 100OC absorbing latent heat
of vaporization.
5) Entropy increase of vapor as it is heated from 100OC to 250OC.
46
Fig 12: T – s Diagram for water Constant pressure lines
These entropy changes are shown in T-S graph. It is a constant pressure
process. If during heating process the pressure had been maintained constant at
2 atm, a similar curve would be obtained. If these states for different pressures
are joined the phase equilibrium diagram of a pure substance on the T-s
coordinate would be obtained as shown below.
Fig 13: T –s Diagram for water
Most often liquid vapor transformation only are of interest, the following figure
the liquid, the vapor and the transition zones only at a particular pressure, sf –
specific entropy of saturated water and sg - specific entropy of saturated vapor.
The entropy change of the system during the phase change from a liquid to
vapor at that constant pressure is sfg. (= sg – sf). The value of sfg decreases as
pressure increases. And becomes zero at the critical point.
47
Fig 14: Phase equilibrium diagram –T – s Diagram
H-S diagram (Mollier diagram) for a pure substance:
From the first and second law of thermodynamics following property relations are
obtained.
Tds = dh – v dp
and ( h / s )p = T
These equations form the basis of h-s diagram of a pure substance. The slope of
an isobar on the h-s coordinates is equal to the absolute temperature. If the
temperature remains constant the slope will remain constant. If the temperature
increases the slope of the isobar will increase.
Consider once again the heating of ice at –5OC to steam at 250OC the pressure
being maintained constant at 1 atm. The slope of the isobar of 1 bar on the h-s
coordinates first increases as the temperature of the ice increases from –5OC to
0OC. the slope then remains constant as ice melts to water at 0OC. Te slope of
isobar again increases as the temperature of water rises from 0 OC to 100OC. the
slope again remains constant as water vaporizes at constant temperature. Finally
the slope of the isobar continues to increase as the temperature of steam
increases to 250OC and beyond (as shown in the figure below.
Fig 15: h –s diagram for water - constant pressure lines.
48
Similarly the isobars of different pressures can be drawn on h-s diagrams as
shown in the figure below.
Fig 16: h –s for water.
This figure shows the phase equilibrium diagram of a pure substance on the h-s
coordinate indicating the saturated solid line, saturated liquid line saturated vapor
line, the various phases and the transition zones.
Fig 17: h-s diagram for water:
This figure is the Mollier diagram indicating only the liquid and vapor phases. As
the pressure increases the saturation temperature increases, slope of the isobar
increases. Hence the constant pressure lines diverge from one another and the
critical isobar is at a tangent to the critical point. In the vapor region the states of
equal slopes at various pressures are joined by lines as shown, which are the
constant temperature lines. Here at a particular pressure hf is the specific
enthalpy of saturated water and hg is specific enthalpy of saturated vapor and h fg
(= hg - hf) is the latent heat of vaporization at that pressure. As the pressure
increases hfg decreases and at the critical pressure hfg becomes zero.
49
Dryness fraction and various equations:
Dryness fraction is defined as the ratio of mass of dry steam to total mass of
steam. It is denoted by x and is also called as quality of steam.
x = mv / (mv +ml ) where mv and ml are the masses of vapor and liquid
respectively.
Let V be the toal volume of a liquid vapor mixture of quality x in which Vf volume
of saturated liquid and Vg volume of saturated vapor, the corresponding being m,
mf, mg respectively.
We have m = mf + mg and V= Vf +Vg
Therefore mv = mf * vf + mg*vg
v= (1-x) vf + x vg.
Similarly s = (1-x) sf + x sg
h= (1-x) hf + x hg
u= (1-x) uf + x ug.
Same equations are written as
v = vf + x vfg,
h = hf + x hfg
u = uf + x ufg
s = sf + x sfg
If the condition of the steam is superheated then we have Degree of superheat,
which is difference between the superheated temperature to the saturation
temperature.
 super = Tsuper –Tsaturation
The other properties are calculated as v super = vsat * Tsuper / Tsat
hsuper = hg + Cp (Tsuper-Tsat)
ssuper = sg + Cp log(Tsuper/Tsat)
Measurement of Steam Quality:
The state of a pure substance gets fixed if two independent properties are given.
Thus the pure substance is said to have two degrees of freedom.
Fig 18: T – s and h – s diagram
50
Figure shows the values of pressure and temperature would fix up the state. But
when the substance is in the saturation state or two phase region the measured
values of pressure and temperature could apply equally well to saturated liquid
point f and saturated vapor point g. or two mixtures of any quality points x 1, x2 or
x3. of the two properties, P and T only one is independent; the other is a
dependent property. If the pressure is given the saturation temperature gets
automatically fixed for the substance. In order to fix up the state of the mixture
one more property such as specific volume, enthalpy or composition of mixture or
internal energy is required to be known. Since it is relatively difficult measure the
specific volume of the mixture devices such as calorimeters are used for
determining the quality or the enthalpy of the mixture.
In the measurement of quality, the object is always to bring the state of the
substance from the two phase region to the single phase region or superheated
region where, the pressure and temperature are independent and measured to
fix the state, either by adiabatic throttling or e3lectric heating.
There are four types of calorimeter are common in use namely
1) Separating Calorimeter
2) Throttling Calorimeter
3) Combined separating and throttling calorimeter.
4) Electrical calorimeter.
Separating Calorimeter: the steam whose dryness fraction is to be
determined is very wet then separating calorimeter gives the quality of the steam.
A known quantity of steam is passed through a separating calorimeter as shown.
The steam is made to change direction suddenly, the water being denser than
the dry steam is separated out. The quantity of water, which is separated out, is
measured at the separator. The dry steam coming out of the separator is sent
through a condenser where it is condensed separately.
The dryness fraction of the steam is calculated by weighing the mass of the
water and mass of dry steam after condensation separately.
Throttling Calorimeter: sample of wet steam of mass m at pressure p1 is
taken from the steam main through a perforated sampling tube as shown in the
figure.
51
Fig 19: Throttling calorimeter
Then it is throttled by the partially opened valve to a pressure p 2 measured by a
mercury manometer and temperature T 2 so that after throttling the steam is in the
superheated region. The process is shown on T-s and H-s diagram.
The steady flow energy equation gives the enthalpy after throttling as equal to
enthalpy before throttling. It is a irreversible process hence joined by a dotted
line. Thus the initial state of the steam is p1 , t1 and its dryness fraction is x1 and
the final state of the superheated steam is p2 x2.
Fig 20: T – s and h – s diagram for throttling calorimeter
Now h1 =h2
hf1+x1 hfg1 = h2
x1 =( h2 – hf1 ) / hfg1
52
With P2 and T2 being known, h2 can be found out from the superheated steam
table. The values of hf, hfg are taken from saturated steam table., thus quality of
the wet steam x1 can be calculated.
Combined separating and throttling calorimeter:
When the steam is very wet and the pressure after throttling is not low enough to
take the steam to the superheated region then a combined separating and
throttling calorimeter is used for the measurement of quality. Steam from the
main is first passed through a separator as shown in the figure, where some part
of the moisture separates due to sudden change in direction and falls by gravity
and partially dry vapor is then throttled and taken to the superheated region.
Fig 21: Schematic diagram for separating and throttling calorimeter.
As shown in the figure process 1-2 represents moisture separation from the wet
sample of steam at constant pressure P1 and process 2-3 represents throttling to
pressure P2 withP2 and T3 being measured, h3 can be found out from the
superheated steam table.
h3= h2 = hf1 + x2 hfg1
Therefore x2, the quality of the steam after partial moisture separation can be
evaluated. If m kg of steam is taken through the sampling tube in y seconds, m1
kg is separated and m2 kg is throttled and then condensed to water and
collected, we have
53
m=m1+m2.
The mass of dry vapor will be at state2 is x2m2.
Therefore the quality of the sample of the steam at state1 which ids x1 is given
by
x1= x2m2 / (m1 + m2)
There is one more method of measurement of quality of wet steam by using
electric calorimeter as shown in the figure. The sample of steam is passed in
steady flow through an electric heater. Electrical energy input Q should be
sufficient to taken the steam to the superheated region where pressure P 2 and
temperature T2 are measured. If I is the current flowing through the heater in
amperes and V the voltage across the coil at steady state Q=VI. If m is the mass
of steam taken in t seconds under steady flow condition then the steady flow
energy equation for heater is given by m1h1 + Q = m1h2
Fig 22: Electric calorimeter
Hence h1 + Q / m1 = h2. With h2,Q and m1 =being known h1 can be computed.
Thus h1= hf1 + x1 hfg1.
Hence x1 can be calculated.
54
1) A steam boiler initially contains 5 m3 of steam and 5 m3 of water at 1
MPa. Steam is taken out at constant pressure until 4 m3 of water is left.
What is the heat transferred during the process?
Solution:
Fig 23: Steam drum.
At 1 MPa vf = 0.001127 and vg= 0.1944 m3 / kg.
hg=2778.1 kJ/kg
uf = 761.68 , ufg = 1822 , ug = 2583.6 kJ/kg
The initial mass of saturated water and steam in the boiler = (Vf /vf) + (Vg / vg) =
[(5/0.001127) + (5/0.1944) ] = [(4.45 * 103) + (25.7) ] kg
Final mass of saturated water and steam = (4/0.001127) +(6/0.1944) = [ (3.55 *
103) + 30.8] kg
Mass of steam taken out of the boiler, ms = [4.45 * 103 + 24.7] - [ (3.55 * 103) +
30.8]
= 894.9 kg
Making an energy balance we have initial energy stored in saturated water and
steam + heat transfer from external source = final energy stored in saturated
water and steam + energy leaving the steam or
U1 + Q =U1 + ms*hg
Assuming that the steam taken out is dry.
Hence 4.45 * 103 * 761.68+27.7*2583.6+ Q = 3.55 * 103 *761.68 +30.8 *2583.6
+894.9 * 2778.1
Q = 2425000-685500+13176
Q = 1752676 kJ.
2) Steam flows in a pipeline at 1.5Mpa. After expanding to 0.1MPa in a
throttling calorimeter, the temperature is found to be 120 o C. Find the
quality of steam in the pipe line. What is the maximum moisture at1.5 MPa
55
that can be determined with this set-up if least 5
required after throttling for accurate readings?
o
C of superheat is
Fig 24 : h –s Diagram
Solution.
At state 2 when p = 0.1 MPa and t = 120 o C by interpolation,
h2 = 2716.2 kJ/kg and
p = 1.5 MPa hf = 844.89 and hfg = 1947.3 kJ/kg
and h1 = h2 hf1+ x1 hfg1 = h2
844.89 + x1 *1947.3 = 2716.2
x1 = 1871.3 / 1947.3 = 0.963
Ans.
When p = 0.1MPa and t = 99.63 + 5 = 104.63 oC,
h3=2685.5 kJ/kg
Since h3=h4 2685.5 = 844.89 + x4 *1947.3
x4 = 1840.6 / 1947.3 = 0.948
The maximum moisture that can be determined with this set up is only 5.2% Ans
3) The following data were obtained with a separating and throttling
calorimeter:
Pressure in pipeline :1.5 MPa
Condition after throttling:0.1 MPa.110o C
During 5 min moisture collected in the separator:0.150 litre at 70 oC,
Steam condensed after throttling during 5 min
Find the quality of steam in the pipeline
56
Fig 25: h –s Diagram
Solution :
AT 0.1 MPa, 110 oC, h3 = 2696.2 kJ/kg
Now h3=h2 = hf2 + x2 hfg2
2696.2=844.89 + x2 1947.3
x2 = 1851.31/1947.3 = 0.955
If m1 = mass of moisture collected in the separator in 5 min and
m2= mass of steam condensed after throttling in 5 min then
x1 = (x2 m2) / (m1 + m2)
at 70oC vf = 0.001023 m3/kg
m1= 0.1462 kg
and m2 = 3.24 kg
Hence x1= 0.955*3.24 / (0.1462 +3.24) = 0.915 Ans.
4) A vessel having a volume of .4 m3 contains 2.0 kg of liquid water and
vapor mixture in equilibrium at a pressure of 600 kPa. Calculate a) The
volume and mass of liquid b) The volume and mass of vapor.
Solution: The specific Volume is calculated first.
v =(0.4/2 ) = 0.2 m3/kg
The quality of steam can now be calculated
0.2 = 0.001101 + x * 0.3146, x = 0.6322
Therefore mass of liquid is 2.0*(1 - 0.6322) = 0.7356 kg
Mass of vapor is 2.0(0.6322) = 1.2644 kg
Volume of liquid is ml vf
= 0.7356(0.001101) =0.0008m3
Volume of vapor is mvvg = 1.2644(0.3157)=0.3992 kg
57
5) Steam at 1 bar and a dryness fraction of 0.523 is heated in a rigid vessel
until it becomes saturated vapor. Calculate the heat transferred per kg
steam.
Solution:
From the steam table at 1 bar pressure, ts = 99.62oC,
vf =0.001043m3/kg,vfg=1.69296m3/kg,
uf =417.33 kJ/kg, ufg =2088.72kJ/kg,
Volume of one kg of given state of vapor = vf + x vfg
= 0.001043 + 0.523 * 1.69296 = 0.8864 m3/kg
Enthalpy correspond to the state of steam,
u= uf + x ufg = 417.33 + 0.523 * 2088.72 =1509.73kJ/kg
v (0.8864 m3/kg ) correspond to the saturated condition of the steam, from steam
table, which is vg,
The pressure found to be 2 bar and ug =2529.49kJ/kg
Heat added = 2529.49– 1509.73 = 1019.76 kJ/kg Ans
.
6) A rigid vessel contains one kg of mixture of saturated water and
saturated steam at a pressure of 0.15MPa. When mixture is heated the state
passes trough the critical point. Determine,
a) Volume of the vessel
b) The mass of the liquid and vapor
c) The temperature of the mixture when the pressure rises to 3 MPa.
d) The heat transfer required to produce the final state.
Solution. Vc: Critical volume = 0.003155 m3/kg (From the steam table.)
v = vf + x X vfg
0.003155 = 0.001035 + x1 * 1.15828 ,
x1 = 0.00183
Hence, mass of vapor = x = 0.00183 kg
Mass of liquid = (1-x1) = 0.998 kg.
u1 = uf + x1 X ufg ;
u1 = 466.92 + 0.00183 X 2052072 = 470.68 kJ/kg
Saturation temperature correspond to 3 MPa, is 233.9 OC, the temperature of the
mixture
x2 = (0.003155 – 0.001216)/0.06546 = 0.02962
Heat transfer in constant volume process = u2 – u1,
u2 = 1004.76 + 0.02962 X 1599.34 = 1052.13 kJ/kg
58
Heat transfer = 1052.13 – 470.68 = 581.45 kJ/kg.
7) Steam initially at 0.3MPa, 250OC is cooled at constant volume. Find
i) At what temperature will the steam become saturated vapor?
ii) What is the quality at 80OC? What is the heat transferred per kg of steam
in cooling from 250 OC to 80OC?
Solution:
At 300kPa, 250 OC, from the steam table , it is a superheated condition.
v = 0.79636 m3/kg , u = 2728.69 kJ/kg
vg = 0.79636 m3/kg , Ps = 225 kPa, ts = 124 OC
at 80 OC
0.79636 = 0.001029 + x 3.45612
x = (0.79636 - 0.001029) /3.45612 = 0.23
u2 = 520.45+0.23(2013.1) = 483.463 kJ/kg
Heat transfer = change in internal energy as the process is const. volume,
Heat transfer = u1 – u2 = 2728.69 – 983.463 = 1745.227 kJ/kg
8) State whether the following samples of steam are wet, dry or
superheated: Justify your answer.
I)
Pressure = 1 MPa absolute enthalpy = 2880 kJ/kg
II)
Pressure=500kPa absolute, volume =0.35m3 /kg
III)
Temperature = 200oC Pressure = 1.2 MPa.
IV)
Temperature = 100oC, entropy =6.88kJ/kg K.
V)
Pressure= 10 kPa, enthalpy = 2584 kJ/kg.
Try as homework.
59