A body at
200oC undergoes an reversible
isothermal process. The heat energy removed in the process is 7875 J. Determine
the change in the entropy of the body.
System : Closed system
Known : T1 = T2
=
200oC
= 473 K
Qrejected = 7875 J
Process : Isothermal
To find : Ds
Diagram :
Analysis : S2 - S1 = for
an isothermal process
=
= - 16.65 J/K.
Comment : Entropy
decreases as heat is removed from the system.
A mass of 5
kg of liquid water is cooled from 100oC to 20oC.
Determine the change in entropy.
System : Closed
system
Known : Mass of water = 5kg
T1
= 100oC = 373 K
T2
= 20oC
= 293 K
Process : Constant pressure
To find : Change in entropy
Diagrams :
Assumptions : 1)
The process is reversible
2) The specific heat of liquid water is
constant
Analysis : S2 - S1 = m
For this problem
p2 = p1 & Cp = 4.186
\ S2 - S1 = 5
= -5.053
Comment : Entropy decreases as heat is removed from the
system.
Prob : 5.3 Air
is compressed isothermally from 100 kPa to 800 kPa. Determine the change in
specific entropy of the air.
System : Closed/Open
Known : p1 = 100 kPa
p2 = 800 kPa
To
find : DS - change
in Specific entropy
Diagram :
Analysis : DS =
= - R ln [Since the process
is isothermal]
= - 0.287 x ln
= - 0.597 kJ/kgK.
Prob : 5.4 A
mass of 5 kg of air is compressed from 90 kPa, 32oC
to 600 kPa in a polytropic process, pV1.3= constant. Determine the change entropy.
System : Closed
/ Open
Known : p1 = 90 kPa
T1 = 32oC = 305 K
p2 = 600 kPa
m
= 5 kg
Process : pV1.3 = Constant
To
find : DS - Change
in entropy
Diagram :
Analysis : S2 - S1 = m
Where
T2 = T1
= 305
= 473 K
\ S2 - S1 = 5
= - 0.517 kJ/K.
Comment : For
air the ratio of Cp and Cv is equal to 1.4. Therefore the polytropic index
n =
1.3(<1.4) indicates that some heat is removed from the system resulting
in negative entropy.
Prob : 5.5 A
rigid insulated container holds 5 kg of an ideal gas. The gas is stirred so
that its state changes from 5 kPa and 300 K to 15 kPa. Assuming Cp = 1.0 kJ/kgK and g = 1.4, determine the change of
entropy of the system.
System : Closed
Process : Constant volume since the
gas is stirred in an rigid container
Known : p1 = 5 kPa p2 = 15 kPa
m = 5 kg Cp = 1.0 kJ/kgK
T1 = 300 K g
= 1.4
Diagrams :
To find : Change in entropy
Analysis : S2 - S1 = m
By
applying the state equation.
Since V2 = V1
Also R = Cp - Cv
=
=
=
= 0.286 kJ/kgK
Substituting
these values we get
S2 - S1 = 5
= 3.922 kJ/K
Comment : Though
this process is adiabatic it is not isentropic since the process of stirring is
an irreversible process.
Prob : 5.6 An
insulated rigid vessel is divided into two chambers of equal volumes. One
chamber contains air at 500 K and 2 MPa. The other chamber is evacuated. If the
two chambers are connected d, what would be the entropy change ?
System : Closed
system
Process : Unresisted
expansion
Known : T1 = 500 K
p1 = 2 ´ 103 kPa
To
find : Entropy change
Diagrams :
Analysis : s2 - s1 =
s2 - s1 =
After
expansion air will occupy the entire volume of the container.
\ V2 = 2V1
Also
1W2 =
0 since it is an unresisted expansion
Q12 = 0 since the vessel is
insulated
Applying
the first law of thermodynamics
Q12 = DU + 1W2
Therefore
Du = 0
For
air
mcv(T2 - T1) = 0
i.e.
T2
= T1
Hence
s2 - s1 = Cvln + Rln
= 0.287
ln
= 0.199
kJ/kgK
Comment : Though the process is adiabatic entropy
increases as the process involving unresisted
expansion is an irreversible process. It also proves the fact that
Prob : 5.7 An
adiabatic chamber is partitioned into two equal compartments. On one side there
is oxygen at 860 kPa and 14oC. On the other side also, there
is oxygen, but at 100 kPa and 14oC. The chamber is insulated and
has a volume of 7500 cc. The partition is abruptly removed. Determine the final
pressure and the change in entropy of the universe.
System : Closed
Process : Adiabatic
Mixing
Known :
Subsystem I Subsystem
II
Fluid Oxygen Oxygen
Initial pressure 850 kPa 100 kPa
Initial Temperature
14oC 14oC
Initial volume
Diagrams :
Analysis : Here the energy interaction is taking place
only between the two fluids and therefore the energy lost by one fluid should
be equal to the energy gained by the other fluid. Taking tF as the final temperature we get
Since
the same fluid is stored in both the systems at the same temperature
C1 = C2 and
t1 = t2 = 14oC
Therefore
the final temperature will also be 14oC
After
removing partition total mass of oxygen is occupying the entire 7500cc at 14oC. Hence the
final pressure can be computed as given below :
= 0.0427 kg
= 0.00503 kg
To
find the final pressure
= m1 + m2
= 475 kPa
DSsystem = DS1 + DS2
DSsurroundings = 0
DSuniverse = 8.596
Prob : 5.8 Two
vessels, A and B each of volume 3 m3 may be connected by a tube of negligible
volume. Vessel A contains air at 0.7 MPa, 95oC
while vessel B contains air at 0.35 MPa, 205oC.
Find the change of entropy when A is connected to B by working from the first
principles and assuming the mixing to be complete and adiabatic.
System : Closed
Process : Adiabatic
mixing
Known : Properties Subsystem A Subsystem B
Fluid Air Air
pressure 0.7
MPa 0.35 MPa
volume 3
m3 3 m3
Temperature 95oC 205oC
Diagrams :
Analysis : Since
the energy interaction is taking place only between the two fluids energy lost
by one fluid is equal to the energy gained by the other fluid.
\ QA = QB
Taking t2 as the final temperature after mixing maCa (t2 -
t1a) = mbCb(t1b -t2) Since in both A and B the same fluid is stored, Ca = Cb
Also ma =
=
= 19.9 kg
mb =
=
=
7.65 kg
19.9 (t2 - 95) = 7.65
(205 - t2)
2.6 (t2 - 95) =
(205 - t2)
2.6t2 + t2 = 205 + 2.6 ´ 95
t2 = 125.6oC
Entropy change =
DSA + DSB
DSA = mA
DSB = mB
DSsys = 5.08 + 0.525
= 5.605
DSsurr = 0
\
DSuniverse = 5.605
Final pressure p2 =
=
= 525
kPa
Prob : 5.9 Air enters a turbine at 400oC,
30 bar and velocity 160 m/s. It leaves the turbine at 2 bar, 120oC
and velocity 100 m/s. At steady state it develops 200 kJ of work per kg of air.
Heat transfer occurs between the surroundings and the turbine at an average
temperature of 350 K. Determine the rate of entropy generation.
System : Open
Process : Steady
flow
Known : Properties Inlet Outlet
Pressure 30
bar 2 bar
Velocity 160
m/s 100 m/s
Temperature 400oC 120oC
Ambient temperature = 350 K
Work output = 200 kJ/kg
Diagram :
To
find : Rate
of entropy generation
Analysis :
For unit mass = Cp ln
= 1.005 ln
= 0.236 kJ/kgK
where Qsur =
=
Qsur = +89.2 kJ/kg
(DS)sur =
= 0.255 kJ/kgK
= 0.019 kJ/kgK.
Prob
: 5.10 A turbine operating at steady state receives
air at a pressure of p1 = 3.0 bar and temperature of 390 K.
Air exits the turbine at a pressure of p2 = 1.0 bar. The work developed is
measured as 74 kJ/kg of air flowing through the turbine. The
turbine operates adiabatically, and
changes in kinetic and potential energy between inlet and exit can be neglected. Using ideal gas model for
air, determine the turbine efficiency.
System : Open
Process : Steady
flow
Known :
p1 = 3.0 bar p2 = 1.0 bar
T1 = 390 K Wa =
74 kJ/kg
Diagrams :
Analysis : ht =
= for an ideal gas
Where
\
T2s =
\
T2s = 284.9 K
= Cp ( T1 - T2 ) = 74
\ T1 - T2 =
= 73.63 K
Hence ht =
=
= 0.7 (or 70%).
Prob : 5.11 A
closed system is taken through a cycle consisting of four reversible processes.
Details of the processes are listed below.
Determine the power developed if the system is executing 100 cycles per
minutes.
Process Q (kJ) Temperature
(K)
Initial
Final
1 -
2 0 300 1000
2 -
3 +1000 1000 1000
3 -
4 0 1000 300
4 -
1 - 300 300
System : Closed
Process : The
system is executing cyclic process.
Known : Heat
transfer in process 12, 23 and 34 and temperature change in all the process.
No
of cycles per minute.
To
find : Power developed.
Diagrams :
Analysis : To find the power developed Wnet per cycle must be known. From I Law Wnet = Qnet which can be computed from the following table
Process Q (kJ) Temperature
(K) DS
Initial
Final
1 -
2 0 300 1000 0
2 -
3 1000 1000 1000
3 -
4 0 1000 300 0
4 -
1 - 300 300 DS41
For a
cyclic process SDf = 0
where f is any property
\ SDS =
0
(i.e.,)
DS12 + DS23 + DS34 + DS41 = 0
0 + 1 + 0 + DS41 = 0
DS41 = -1
Since
the process 4-1 is isothermal
= -1
Q41 = -300 kJ
Therefore
Qnet = Q12 + Q23 + Q34 + Q41
= 0 + 1000 + 0 - 300
= 700
kJ per cycle
\ Wnet = Qnet = 700 kJ
and power
developed =
= 700
= 1166.7 kW
Prob : 5.12 Two kilogram
of air is heated from 200oC to 500oC at constant pressure. Determine the change in entropy.
System
: Open/closed
Working
: Air
fluid
Process : Constant pressure heating
Known : 1) t1 = 200oC
2) t2 = 500oC
Diagram
:
To
find : Change in
entropy DS
Analysis : DS =
=
=
= 0.987 kJ/K
Prob : 5.13 A
Carnot engine operated between 4oC and 280oC.
If the engine produces 300 kJ of work, determine the entropy change during heat
addition and heat rejection.
System : Open/closed
Process : The
working fluid is executing Carnot cycle
Known : 1) t1 =
280oC
2) t2 =
4oC
3) W = 300 kJ
Diagram :
To find : 1)
DS during
heat addition
2) DS during heat rejection
Analysis : 1) In carnot engine heat is added at constant
temperature
Therefore
DS =
Where
Qin =
h =
=
=
0.499
Therefore
Qin =
=
601.1 kJ
DS =
=
= 1.087 kJ/K
2) In
carnot engine heat rejection is also taking place at constant temperature
Therefore
DS =
Where
Qout =
Qin - W
=
601.1 -
300
=
301.1 kJ
DS =
= -1.087 kJ/K
Comment : In a carnot change two isothermal process and two isentropic
process. Therefore DS
during heat addition must be equal to DS during heating rejection so that
which obeys Clausius Inequality.
Prob : 5.14
Air flows through a perfectly insulated duct. At one section A the pressure and temperature are respectively
2 bar 200oC and at another section B further
along the duct the corresponding values are 1.5 bar and 150oC.
Which way the air flowing?
System : Open
Process : Steady
flow process
Known : 1) p1 = 2
bar
2) t1
= 200oC
3) p2
= 1.5
bar
4) t2
= 150oC
To find : To
know flow direction
Diagram :
Analysis :
This problem
cannot be solved by simple application of first law of thermodynamics. Because
there is nothing to tell us whether the fluid is expanding from A to B or being
compressed from B to A.
However, since the
duct is insulated the inference is that
there is no heat transfer to or from the environment and therefore there is no
change of entropy in the environment. But in any real process change of entropy
of the system plus the surroundings must be positive. In otherwords DSAB > 0
Thus
SA > SB and the flow is from B to A.
Even though entropy
cannot be measured directly it can still be used to find the sense of flow in a
well insulated duct given two salient states as above.
Prob 5.15 : A certain fluid undergoes
expansion in a nozzle reversibly and adiabatically from 500
kPa, 500 K to 100 kPa. What is the exit velocity? Take g = 1.4 and R = 0.287 .
System : Open
Process : Reversible
adiabatic expansion
Known : 1) Inlet pressure = 500 kPa
2) Inlet temperature
= 500 K
3) Exit pressure = 100 kPa
4) The ratio of
Specific heats = 1.4
5) Characteristic
Gas constant = 0.287
To find : Exit velocity
Diagram :
Analysis : Applying
Steady Flow Energy Equation
Therefore
where
Cp and T2 are unknowns.
To find CP
CP - CV = R
Substituting
g and R we
get
To find T2
It is
stated in the problem that the process of expansion is reversible. Therefore
Also
the process is given as adiabatic. That is
(or) ds = 0
S2 - S1 = 0
= 315.8 K
Substituting
numerical values for T2 and Cp, we get
Prob 5.16 : Show from the first principle
that, for a perfect gas with constant specific heat capacities
expanding polytropically (pvn = constant) in a non-flow
process, the change of entropy can be expressed by
Gaseous methane is
compressed polytropically by a piston from 25oC
and
0.8 bar to a pressure of 5.0 bar. Assuming an index of compression of
1.2, calculate the change of entropy and workdone, per unit mass of gas.
The relative molecular weight of methane is 16 and g = 1.3.
System : Closed
Process : Polytropic
(pVn = C)
Known : 1) T1 = 298 K
2) p1 = 80 kPa
3) p2 = 500 kPa
4) n = 1.2
5) M = 16
6) g = 1.3
To find :
1) 1W2 - Work done
2) DS - Change in entropy
Analysis : a) To prove S2 - S1 =
From First Law of Thermodynamics
Q12 = 1W2 + DU
In differential form
for a polytropic process
Therefore
Upon
integration we get
From the process
relation
Substituting for we get
We know that R = CP - CV
R = CV (g - 1)
Substituting for CV we get
(2) Workdone
= 404.45 K
Substituting numerical
values
(3) Change in entropy
Comment : The negative sign in
work indicates that work is given into the system. The
negative sign in entropy change indicates that there is a heat rejection by the
system to the ambient during compression.
Prob 5.17 : A closed system undergoes the
internally reversible process as shown below :
Compute the heat transfer.
System : Closed
Process : Defined
by a straight line on a T-S diagram.
Known : T1 = 200 K
T2 = 600 K
S1 = 1
kJ/K
S2 = 3
kJ/K
To find : Heat transfer
Analysis : Q = Area under the curve representing
the process in a T-S diagram
= 800 kJ
Prob 5.18 : In a refrigerant condenser
superheated vapour of ammonia enters steadily at 1.4 MPa, 70oC.
It leaves the condenser at 20oC. At 1.4 MPa condensation
begins and ends at 36.28oC. Cooling water enters the condenser
at 10oC and leaves 15oC.
Determine
(a) the
amount of heat rejected per kg of ammonia vapour condensed for the
given inlet and exit conditions.
(b) mass
of water to be supplied for each kg of ammonia vapour condensed
(c) the change in specific entropy of ammonia
(d) the
entropy generation per kg of ammonia
Take Cpvapour = 2.9 kJ/kgK, Cpliquid = 4.4 kJ/kgK and latent heat of evaporation
of ammonia at 1.4 MPa is 1118 kJ/kg. Also represent the process in a T-s diagram.
System : Open
Process : Steady
flow process
Known : T1 = 70oC
P1 = 1.4 MPa
T2 = 20oC
TW1 = 10oC
TW2 = 15oC
To find : (a) the amount of heat rejected per kg of
ammonia vapour condensed for
the given inlet and exit conditions.
(b) mass of water to be supplied for each kg of
ammonia vapour condensed
(c) the change in specific entropy of ammonia
(d) the
entropy generation per kg of ammonia
Diagrams :
Analysis : (a)
Heat rejected per kg of ammonia
Q1-2 = Q1 - 2a + Q2a - 2b + Q2b - 2
= 2.9 (70 -
36.28) + 1118 + 4.4 (36.28 - 20)
= 1287.42 kJ/kg
(b) Water flow rate
required per kg of ammonia
= 61.51
(c) Change in
Specific entropy of ammonia
= DS1 - 2a + DS2a - 2b + DS2b - 2
= -
4.153
(d) DSuniverse = DSWater + DSammonia
where DSWater = mCp ln
= 61.51 ´
4.186 ´ ln
= 4.509
Substituting
the values we get
DSuniverse = 4.509 + (- 4.153)
= 0.356
Comment : As heat is removed
from ammonia its entropy decreases whereas entropy of water
increases as it receives heat. But total entropy change will be
positive as heat is transferred through finite temperature difference.
Prob 5.19 : The specific heats of a gas are
given by CP = a + kT and CV = b + kT, where
a, b and k are constants and T is in K. Show that for an isentropic expansion
of this gas
Tb na-b ekT = constant
System : Closed
Process : Isentropic
Known : 1) CP = a + kT
2) CV = b + kT
To prove : Tb na - b ekT = constant for an isentropic process
Proof : For a gas
CP - CV = (a + kT) - (b + kT)
(or) R = a - b
For
an isentropic process
ds = 0
(or)
Substituting
for CV and R
Upon
integration
blnT + KT + (a - b) lnn = constant
Taking
antilog
Tb eKT na - b = constant
Prob 5.20 : Calculate the entropy change of
the universe as a result of the following process :
(a) A metal
block of 0.8 kg mass and specific heat capacity 250 J/kgK is
placed in a lake at 8oC
(b) The
same block, at 8oC, is dropped from a height of 100
m into the lake.
(c) Two
such blocks, at 100oC and 0oC,
are joined together.
Case (a)
System : A metal block
Process : Cooling
the metal block by dipping it in a lake.
Known : 1)
Initial temperature of the block (T1)=100 + 273 = 373 K
2)
Final temperature of the block (T2) = 8 +
273 = 281 K
3)
Mass of the metal block (m) = 0.8 Kg
4)
Specific heat capacity of the metal block (C) = 250
To find : Entropy
change of the universe
Diagram :
Analysis : DSuniverse = DSsystem + DSsurroundings
Where
Where Qsur = - Qsys
= - mC (T2 - T1)
= - 0.8 ´ 250 (281 - 373)
= 18400 J
Substituting the values we get
DSuniverse = -56.6
+ 65.48
= 8.84 J/K
Comment
: As discussed earlier the entropy
change of the universe is positive. The reason is the
irreversible heat transfer through finite temperature difference from the metal block
to the lake.
Case (b)
System : A
metal block
Process : Falling
of the metal block into the lake and reaching equilibrium
Known : 1)
Initial Temperature = 281 K
2)
Final Temperature = 281 K
3)
Initial height = 100 m
4)
mass of the metal block (m) = 0.8 kg
5)
Specific heat capacity of the metal block (C) = 250 J/kgK
Diagrams :
Analysis : DSuniverse = DSsystem + DSsurroundings
Where
DSsystem = 0, as the system is at the same temperature
at both the initial and final state.
Qsurroundings = DEsystem
= mgh
= 0.8 ´
9.81 ´ 100 = 784.8 J
Comment
: Increase in entropy of the universe indicates that there is a
irreversibility or
degradation of energy. That is the high grade potential energy is converted
low grade energy during this process.
Case (c)
Systems : Two
metal blocks
Process : Two
metal blocks are brought in thermal contact and allowed to reach equilibrium.
Known : Initial
temperatures of the blocks
T1a = 373 K
T1b = 273 K
To find : Entropy
change of the universe
Diagrams :
Analysis
: DSuniverse = DSa + DSb
Where
To find T2
Qa = - Qb
mc (T2 - T1a) = - mc (T2 - T1b)
Comment : In
this process also the heat transfer through finite temperature difference
makes the process irreversible which in turn results in increase
in entropy of the universe.
Prob 5.21 : Each of three identical bodies
satisfies the equation U = CT, where C is
the heat capacity of each of the bodies. Their initial temperatures are
200 K, 250 K and 540 K. If C = 8.4 kJ/K, what is the maximum amount
of work that can be extracted in a process in which these bodies
are brought to a final common temperature ?
System : Three identical bodies
Process : Extracting
work with heat transfer among the three bodies so that they reach
a common temperature.
Known : Initial
temperature of the three bodies
T1a = 540 K
T1b = 250 K
T1c = 200 K
Heat
capacity of all the three bodies = 8.4 T
To find : The
maximum amount of work that can be extracted.
Diagram :
Analysis : Let
us assume that the final temperature is greater than 250 K, so that heat
is transferred from body (1) to bodies (2) and (3).
The net work obtained
W = Q1 - Q2 - Q3
= (DU)1 - (DU)2 - (DU)3
= C [(540 - T2) - (T2 - 250) - (T2 - 200)]
= 8.4 [990 - 3T2]
This
work will be maximum if the process under consideration is reversible.
Therefore
DS1 + DS2 + DS3 = 0
Therefore
T2 = 300 K
This is the condition for the
process to be reversible. Hence the maximum work that can be obtained is
Wmax = 8.4 (990 - 3 ´ 300)
= 8.4 (90)
= 756 kJ
Prob 5.22 : A resistor of 50 ohm resistance
carries a constant current of 5A. The ambient temperature and
the temperature of the resistor remain to be 27oC
over a period of 10 seconds. What is the entropy change of the resistor,
ambient and the universe ?
System : A resistor
Process : Passing
of the electrical current through a resistor at constant temperature.
Known : 1)
Initial and final temperature of the resistor = 300 K
2) Ambient Temperature = 300 K
3) Duration (t) = 10 seconds
To find :
1) DSresistor
2) DSambient
3) DSuniverse
Analysis :
DSuniverse = DSresistor + DSambient
= 0 + 41.7
= 41.7
Comment : When current passes through a resistor it is
converted into heat. As the resistor is to be
maintained at the same temperature the heat is dissipated into the
ambient and hence the process is irreversible resulting in increase
of entropy of the universe.
Prob 5.23 : A closed system is assumed to have
a heat capacity at constant volume where a = 0.002 and T is the temperature in
K.
The
system is originally at 600 K and a thermal reservoir at 300 K is available.
What is the maximum amount of work that can be recovered as the system is cooled
down to the temperature of the reservoir ?
System : A closed system
Process : Obtaining
work with the help of the heat transfer from the system to the
reservoir.
Known : 1) Cv = e0.002T
2) T1 = 600 K
3) T2 = 300 K
To
find : Wmax
Diagram :
: Q1 = DU
If
the work is to be maximum, the process must be reversible. Therefore
(DS)universe = 0
DSsystem + DSreservoi
Neglecting
higher order terms,
Therefore
DSreservoir = -DS system
= 1.731
Thus maximum work = Q1 - Q2
= 911 - 519.3
= 391.7 kJ
