Notes prepared by M.Sundaresan.
AERO-ENGINEERING THERMODYNAMICS
A.
Two mark question and answers.
1.0.
State the Kelvin-Planck statement of second law of thermodynamics.
Ans. Kelvin-Planck states that it is impossible to construct a heat engine working on a cyclic
process, whose only purpose is to convert all the heat energy given to it in an equal amount of
work.
It essentially means that while the heat is getting converted to work you need a second body
which is at a lower temperature through which part of the heat should also pass. That is why heat
is considered a lower grade of energy compared to work.
2.0
State Clausius statement of the second law of thermodynamics.
Ans. It states that heat can flow from a cold body to a hot body only with external aid or work
added in a cyclic process. But heat can easily pass from a hot body to a cold body without any
external aid and in the process does work
3.0
State Carnot’s theorem.
Ans. No heat engine operating in a cyclic process between two fixed temperatures can be more
efficient than that of a reversible engine operating between the same temperature limits.
4.0
What are the corollaries of Carnot theorem ?
Ans.
i) All the reversible engines operating between the two given thermal reservoirs with
the fixed temperatures have the same efficiency.
.ii) The efficiency of any reversible heat engine operating between two reservoirs is
independent of the nature of the working fluid and depends only on the temperature of the
reservoirs.
5.0
Define PMM of the second kind.
Ans. Perpetual motion machine of the second kind draws heat continuously from a single
reservoir and converts it into equivalent amount of work. Thus it gives 100% efficiency. This is
impossible.
6.0
What is the difference between a heat pump and refrigerator ?
Ans. Heat pump is a device which operating in a cyclic process maintains the temperature of a
hot body at a temperature higher than that of the surroundings
A refrigerator is a device operating in a cyclic process maintains the temperature of a
cold body at a temperature lower than the temperature of the surroundings.
7.
What is meant by a heat engine.?
Ans. A heat engine is a device, which is used to convert the thermal energy int mechanical
energy.
8.
Define the term COP ?
Ans. Co-efficient of performance is defined as the ratio of heat extracted or rejected to work
input.
COP = Heat extracted /rejected
Work input.
9.
Write the expansion for COP of a heat pump and refregirator.
Ans
COP for a heat pump
1
COPHP =
Heat...rejected
work ...input
=
T1
(T1 − T2 )
COP for a refrigerator
COPref =
Heat...extracted
work ...input
=
T2
(T1 − T2 )
10
Why Carnot cycle cannot be realized in practice ?
Ans. (i) In a Carnot cycle, all the four processes are reversible but in actual practice there is no
process that is reversible.
(ii) There are two processes to be carried out during compression and expansion. For
isothermal process, the piston moves slowly and for adiabatic process the piston moves as fast
as possible. This speed variation during the same stroke of the engine is not possible.
(iii) It is not possible to avoid friction between moving parts completely.
11.
Name two distinct methods by which the efficiency of a Carnot cycle can be
increased.
Ans (i) Efficiency can be increased as the higher temperature T2 increases.
(ii) Efficiency can be increased as the lower temperature T1 decreases.
12.
Why a heat engine cannot have 100 % efficiency ?
Ans. For all the heat engines there will be a heat loss between system and the surroundings.
Therefore we cannot convert all the heat input into useful work.
13.
When the Carnot cycle efficiency will be the maximum?.
Ans. Carnot cycle efficiency is maximum when the initial temperature is 0K.
14.
What is the process involved in Carnot cycle ?
Ans. Carnot cycle consists of the following processes.
(i)
reversible adiabatic compression.
(ii)
Reversible isothermal heat addition.
(iii) Reversible adiabatic compression.
(iv)
Reversible isothermal heat rejection.
15.
Sketch the p-v and T-s diagram for Carnot cycle.
T
p
w
V
s
16.
Is the second law independent of the first law ?
Ans. Yes , the second law of thermodynamics independent of the first law. The second law
speaks about quality of energy.
17.
Define Entropy ?
2
Ans. The entropy of an assembly in a state of thermal equilibrium is a thermodynamic
property, in principle related through equations of state to other thermodynamic properties. The
change of entropy is defined only for reversible processes stated as :
2
dq
∫T
= ds .
1
Hence in a closed isolated system the entropy change ds ≥ 0 . This aspect relates to the second
law of thermodynamics in dictating the direction and flow of heat in a workable engine or heat
pump. The property of the entropy namely specific entropy has been arrived on the basis that
equilibrium values are easily and quickly achieved though the various molecules are in random
and disorderly motion and are in combination as various meta states in a system. Once the
assembly are system is in a highly probable macro-state though disorderly, the probability of
returning to a low probable macro-state is very small.
18.
Define the terms source sink and heat reservoir.
Ans. Source.
The part from where heat is rejected to a colder body for doing work, or
the part from where heat is generated for doing work is called as the source,
Sink. The part which receives heat from work absorbing or work developing device is called
sink.
Reservoir. The part which supplies or receives heat continuously without change in its
temperature is called a reservoir.
19.
Why is the performance of refrigerator and heat pump are given in terms of COP
and not in terms of efficiency ?
Ans. The performance of any device is expressed in terms of efficiency for work developing
machines.. But for Heat pump and refrigerator are work absorbing machines. So the performance
of these devices are based on COP only.
20.
What is meant by principle of increase of entropy ?
Ans. For any infinitismal process undergone by a system, change in entropy,
(ds ) ≥ dQ
T
For a reversible, dQ = 0 and hence dS = 0
For irreversible, dS > 0
So the entropy of a system would never decrease, it will always increase and remains constant if
the pressure is reversible and is called as principle increase of entropy.
21.
What is meant by Clausius inequality?
Ans. It is impossible for a self acting machine working in a cyclic process unaided by any
external agency to convey heat from a body at a lower temperature to a body at a higher
temperature,
22.
Explain briefly the Clausius inequality.
Ans.
dQ
. ∫
≤0
is known as inequality of Clausius
T
3
If
∫
∫
dQ
= 0 , the cycle is reversible
T
dQ
< 0 , the cycle is irreversible and possible
T
dQ
> 0 , the cycle is impossible. (Violation of second law)
T
23.
For the compressible process between the same two states, which work will be more
reversible or irreversible ?
Ans. Irreversible work will be more in the compressor. Generally for compression, the actual
work given will be higher than the calculated work (Wrev).
24. A heat pump pumps 10MJ/KWhr to the high temperature reservoir. What is the COP
?
Ans.
COP= Heat supplied / work output.
10 × 10 3
= 2.78
3600
25.
Find the entropy of the universe where 1000kj of heat is transferred from800K to
500K.
Ans.
− Q Q − 1000 1000
+
=
+
= 0.75 KJ / K
Entropy change of universe =
T1 T2
800
500
26.
Give the expressions for change in entropy during constant pressure and polytropic
process. Show on T-s diagram
2s
Ans. For the constant pressure process,
∫
∆S = S 2 − S1 = mC P ln
T2
.
T1
For polytropic process,
∆S = S 2 − S1 = m[C P ln
T2
p
− R ln 2 ]
T1
p1
2n
T
2T
or
T2
v
s
+ R ln 2 ] (adiabatic ∆S =0 ).
T1
v1
27.
Explain the term reversibility.
Ans If the process traces the same path during the process when reversed is called as
reversibility. And the entropy change is zero.
28.
Can the entropy of the universe decrease? why?
Ans Entropy of the universe cannot decrease. It will be constant or will increase due to
irreversibility.
∆S = S 2 − S1 = m[C v ln
4
29.
Ans.
What is the essence of the second law of thermodynamics?
1.0
To know the feasibility of process.
2.0
To know about the quality of energy.
30.
If the Carnot engine efficiency is 50%. Find COP of Carnot refrigerator working
between same temperatures
Ans.
Efficiency (HP) =
T1 − T2
=0.5
T1
T1
=2
T2
T2
1
1
=
=
= 1 Ans.
T1 − T2 T1
2 −1
−1
T2
31.
Define the term absolute entropy.
Ans. The change in entropy of the system with respect to ambient conditions or any other
standard reference condition is known as absolute entropy.
COP of refrigerator
32.
Define a system.
Ans. A THERMODYNAMIC SYSTEM IS DEFINED AS A DEFINITE SPACE OR AREA WHICH THE STUDYOF
ENERGY TRANSFER AND ENERGY CONVERSIONS IS MADE.
BOUNDARY
SYSTEM
SURROUNDINGS
33.
What is a closed and open system.? and an isolated system
Ans. In open systems both the mass and energy transfer take place. The open system is often
specified within a control volume. e.g. air compressor. The system boundary is determined by
the space that the matter occupies. In a closed system it does not permit mass transfer, but only
the energy transfer takes place. A closed system is also referred to control mass.
The isolated system is not affected by the surroundings. Simply there is no heat, work and mass
transfer. e.g Universe.
33.
Write the two property equations involving entropy.
Ans.
Tds = du + pdv
Tds = dh − vdp
34.
Define the steam rate in using the Rankine cycle
5
Ans.
3600 Kg
where Wt and Wp are the work done by
WT − WP kW .h
Steam rate =
turbine and pump respectively.
35.
Define the heat rate in using the Rankine cycle
Ans.
3600 × Qi
3600kJ
=
where Qi heat supplied.
heat rate =
WT − WP η CYCLE kW .h
36.
Ans.
Define dryness fraction of steam, and total volume of steam.
mf
X =
.
m f + mg
v = (1 − x) + v f + xv g where v f and vg are specific liquid and gas volume respectively.
37.
Give the formulae relating temperature , pressure and value for an adiabatic
process.
ANS .
γ −1
γ −1
V 
T2  P2  γ
=   =  1 
T1  P1 
 V2 
38. What is meant by one tonne of refrigeration ?
Ans. One tonne of refrigeration is defined as the rate of heat removal from the surroundings
equivalent to the heat required to melt 1 tonne of ice in oe day. If the latent heat of fusion of ice
is taken as 336kj /kg then 1 tonne is equivalent to heat removal at the rate of
91000x336)/24=14000kJ/hr.
Therefore capacity of the refrigeration plant is=
w × (h1 − h 4) × 3600
1400
C O N STA N T VO L UM E P R O C ESS
( F O R STEA M )
P1
P
1
1
T
P2
T 1
2
V 2=V 1
2
T 2
V 2=V 1
S
V
dq = du
q = u 2 − u 1 = C v o (T 2 − T1 )
2
S 2 − S1 =
q = u 2 − u1
2
S 2 − S1 =
∫
1
2
dq
=
T
∫
1
∫
1
du
T
du
T
= C v o ln 2
T
T1
P1
P
= 2
T1
T2
6
REVERSIBLE ISOTHERMAL PROCESS ( STEAM )
P2
P
T2 = T1
2
P1
T
2
P2
1
T2
1
P1
W
Q
S
V
V2=V1
T ( S 2 − S1 ) = q = u 2 − u1 + w, and ...dq = du + dw
u 2 = u1, P2V2 = P1V1, and ....q = w =
∫
v
pdv = RT ln 2
v1
2
S 2 − S1 =
∫
1
dq
v
== R ln 2 ; and .... p1v1 = p2v2
T
v1
P
P2
=P
1
REVERSIBLE CONSTANT PRESSURE PROCESS ( FOR STEAM )
T
T1
2
P2=P1 1
2
1
W
Q
T2
S
V
2
WORK
.. W =
∫
= p ( v 2 − v 1 ) = R ( T 2 − T1 )
pdv
1
dq = du + dw = dh = C po ( T 2 − T 1 )
q = h 2 − h1
2
S 2 − S1 =
∫
1
2
dq
T
=
∫
1
dh
T
= C po ln
T2
T1
v1
v
= 2
T2
T1
7
REVERSIBLE ADIABATIC PROCESS
P1
P
T
1
P1
P2
1
2
P2
2
S2=S1
S
V
FOR
STEAM
P1
P2
T
T1
P1
P
1
S2=S1
T2
T1
2
P2
T2
S
V
FOR
FOR
AIR
AIR
CONSTANT PRESSURE
T
P
T1
P2=P1
2
1
2
1
P2=P1
T2
Q
W
S
V
CONSTANT TEMPERATURE
P2
P1
T
P
P2
T2=T1
2
1
P1
W
T2=T1
Q
V
S
8
First law and second law
dw = −du
Q = Tds = 0; and ..W = −(u2 − u1 )
S 2 = S1
W = −(u2 − u1 ) = −Cvo (T2 − T1 )
T2
T1
k −1
P  k
V
=  2 
=  1
 P1 
 V2




k −1
PHASE RULE.
The number of independent variables with a multicomponent, multiple system is
given by the phase rule. It is also called the Gibbs phase rule. It is expressed by the
equation as:
N= C – Φ +2
Where,
N= the number of independent variables.
C= the number of components.
Φ = the number of phases present in equilibrium.
For the single component (C=1) Two phase system (Φ=2) system, the number of
independent variables needed is :
N=1-2+2=1
For example, water (C=1) at the triple point has three phases (Φ=3 ) and thus n=0. That is
none of the properties of a pure substance at the triple point can be varied. From this rule
we can know that a pure substance which exists in a single phase (Φ=1) will have two
independent variables (n= 2 ). That means there are two independent intensive properties
required to be specified to fix up the state of the system at equilibrium.
Problem
A reheat cycle operating between 30 and 0.04 bar has a superheat and a reheat
temperature of 450 deg C. The first expansion takes place till the stem is dry saturated and
then reheat is given. Neglecting feed pump work determine the reheat cycle efficiency.
9
Solution :- given data ;- P1=30 BAR; P2=0.04 BAR, T1=450 deg C and T3=450degC,and
x 2 = 1.0 .
1
450 deg C
3
T
30 bar
2
6
5
0.04 bar
4
S
From steam tables : at 30 bar at 450 deg C : h1= 3344 kJ/kg ; S1= 7.08 Kj/kg K
h1 = 3344kJ / kg
s1 = 7.08kJ / kgK
At 0.04 bar;
h f 4 = 121kJ / kg, s f 4 = 0.423kJ / kgK
h fg 4 = 2433kJ / kg , s fg 4 = 8.05kJ / kgK
1-2 ISENTROPIC PROCESS
S1=S2=7.08 Kj/kg K
S2=Sg .. dry saturated steam
P2= Psat at Sg
From steam table :
P2 =2.3 bar.
At 2.3 bar :
10
h2 = 2712.6kJ / kg ,
h3 = 3381.0kJ / kg , s3 = 8.306kJ / kgK
3-4
Isentropic process
S3=S4=8.306 Kj/kg/K
S 4 = S f 4 + x 4 × S fg = 0.423 + x 4 × 8.053
X4 =0.98
h4 = h f 4 + x4 × h fg 4 = 121.4 + 0.98 × 2433
=2505 Kj/kg.
The cycle efficiency :
η
=
=
 (h1 − h2 ) + (h3 − h4 ) 


(
h
−
h
)
+
(
h
−
h
)
 1 5
3
2 
 (3344 − 2712.6) + (3381 − 2505) 

 x100
(
3344
−
121
.
4
)
+
(
3381
−
2712
.
6
)


Efficiency =0.3873 =38.73% Ans/
11
AERO -THERMODYNAMICS PROBLEMS.
Problem –1.
Three identical bodies of A,B and C constant heat capacity are at temperatures of 300, 100 and
300K respectively. A heat engine is operated between A and B and a heat pump working as a
refrigerator between B and C. The heat pump is operated by the output of heat engine. If no
work or heat is supplied from outside, find the highest temperature to which any one of the body
can be raised by the operation of heat engine or a refrigerator.
Tf
T f/
C 300K
QS1
W = QS1 − QR1
HE
HP
QR1
Tf
B 100K
/
Let T f be the final temperature of bodies A and B. T f be the final temperature of body C
and C be the final heat capacity of these three identical bodies.
(∆ S)Universe > 0
………………………(1)
(∆ S)Univ = (∆ S)A + (∆ S)B + (∆ S)C + (∆ S)HE + (∆ S)HP.
But (∆ S)HE & (∆ S)HP =0
Therefore
C ln
(∆ S)Univ = (∆ S)A + (∆ S)B + (∆ S)C > 0
T f/
Tf
Tf
+ C ln
+ C ln
≥0
300
100
300
∴ C ln
T f2 × T f/
(300 × 300 × 100)
[ ∆s =CP ln
T2
]
T1
≥0
For minimum value of Tf (∆ S)Univ =0
12
∴ C ln
T f2 × T f/
(3 × 10 6 )
=0
T f2 × T f/
= ln 1
(3 × 10 6 )
Since ln 1=0 the above equation becomes
∴ C ln
T f2 × T f/ = 9 × 10 6.............(2)
and also QS 1 = QR1 + QR 2
C(300-Tf ) =C(Tf -100) + C (Tf/ -300)
[ because Q=m C (T2 – T1 )]
300-Tf = Tf –100 + Tf/ -300
Tf'=700-2 x Tf …………………(3)
From 2 &3
T f2 (700 − 2 × T f ) = 9 × 10 6
T f = 300 K
Tf' =700-2 x Tf
=700-2 x 300
=100K
The maximum temperature can be raised for 100K body as 300 K of B
PROBLEM 2.
Two reversible engines A & B are arranged in series. A rejecting heat directly to B engine ,
receives 200 kJ at a temperature of 421deg C from a hot source, while engine B is in
communication with a cold sink at a temperature of 4.4 degC. If the work output of A is twice
that of B, find :
i)
The intermediate temperature between A and B
ii)
The efficiency of each engine, and
iii)
The heat rejected to the cold sink.
Solution:
Work output from engine A,
WA= QS1 - QR1 =200- QR1
13
For reversible heat engine,
TH
TH Q S 1
=
..........(1)
T
QR1
QS1
WA =2 WB
A
694 200
=
T
QR1
QR1
Therefore QR1=0.288t….. (2)
QS2
B
WB
QR2
TL
So WA =200-0.288 T
But WB =100-0.144 T……..(3)
And also WB= QS2 - QR2
= 0.288T - QR2
….
(4)
(Because QR1 = QS2 )
Eliminating (3) and (4)
100-0.144T =0.288T- QR2
QR2 =0.432T –100 …….(5)
Similarly, for reversible engine B,
T
Q
= S2
TL Q R 2
T
Q
So
= R1 ......as QS2 = QR1
277.4 QR 2
0.288T
0.288T
=
=
. Therefore T=416.42 K or 143.42 deg C…..Ans
QR 2
0.432T − 100
So QR1 = 0.288 x 416.42=119.93 KJ.
And QR2 =0.432 x416.42 – 100 =79.89 KJ. Ans
Q
119.93
Efficiency of engine A, = 1- R1 = 1 =40.04 %
QS1
200
Efficiency of engine B, =
1-
QR2
79.89
= 1=33.39%. …Ans.
119.93
QS 2
(because QS2 = QR1) )
14
Problem 4: Throttling process:
Steam at 1Mpa and 0.9 dry is throttled to a pressure of 200 kPa. Using steam table, find the
quality of steam and change of entropy. Check your answer using Mollier chart. State whether
the process is reversible or reversible.
SOLUTION:
Given data :- P1=1 Mpa=10 bar.
X=0.9. Throttling process P2= 200 kPa=2 bar.
To find : quality of steam and change of entropy.
During the throttling process enthalpy remains constant, i.e. h2=h1.
From steam tables at 1 Mpa or 10 bar:
h fg1 = 2013.6kJ / kg
h f 1 = 726.6kJ / kg
s fg1 = 4.445kJ / kgK
s f 1 = 2.138kJ / kgK
h1 = 762.2 + 0.9 × 2013.6 : h1 = 2574.8kJ / kg
s1 = s f 1 + xs fg1 : s1 = 2.138 + 0.9 × 4.445
s1 = 6.1385kJ / kg
h f 2 = 504.7 kJ / kg : h fg 2 = 2201.6kJ / kg
s f 2 = 1.53kJ / kgK
s fg 2 = 5.597kJ / kgK
h2 = h f 2 + xhfg 2
h2 = 504.7 + x2 × 2201.6
Since : h1 = h2 : 2574.84 = 504.7 + x2 × 2201.6
Ans : x2 = 0.94
Quality of steam is wet
15
s 2 = 1 . 53 + 0 . 94 ( 5 . 59 )
s 2 = 6 . 79 kJ / kgK
change .. in .. entropy
= ∆ s = s 2 − s1 = 6 . 79 − 6 . 138
∆ s = 0 . 652 kJ / kgK
Generally throttling is an irreversible process.
From Mollier chart,
For x1=0.9 at 10 bar line, note the enthalpy of steam,
S1=6.1 kJ/kgK.
Since the throttling process is a constant enthalpy process, draw horizantal line in the
Mollier chart upto 2 bar as shown in figure below. Now read theentropy of final state of
steam,
S1= 6.76 kJ/kgK.
Change in entropy = (s2-s1)= 0.66kJ/kgK.
Since ∆s is positive, the process is irreversible……Answer.
10
ba
r
h
ar
2b
X=0.96
h2=h1
X=0.94
1
2
X=
0.9
0
Sp. Entropy (s)
Problem 5
Steam turbine receives steam at a pressure of 20 bar superheated at 300 degC. The exhaust
pressure is 0.07 bar and expansion takes place isentropically. Using steam steam tables calculate
the following ;
a)Heat supplied assuming that the feed pump supplies water to the boiler at 20 bar.
16
b) Heat rejected.
c)work done.
d) Thermal efficiency.
e) Theoretical steam consumption.
300 C
T
1
p1=20 bar
4
P2=0.07 bar
2
3
ENTROPY (S)
SOLUTION:
Given data :- p1=20 bar : T1=Tsup=300deg C ; P2=0.07 bar
To find Qs, Qr, W, Efficiency (Rankine), and SSC.
From super heated steam table at 20 bar and 300 degC,
h1 = 302 . 5 kJ / kg
s1 = 6 . 77 kJ / kgK
from .. steam .. tables ..at .. 0 . 07 .bar
h f 2 = 163 . 42 kJ / kg
h fg 2 = 2409 . 2 kJ / kg
s f 2 = 0 . 559 kJ / kgK
s fg 2 = 7 . 718 kJ / kgK
v f 2 = 0 . 001007 m 3 / kg
we .. know .. that
s1 = s 2 = s f 1 + x 2 × s fg 2
h2 = h f 2 + x 2 × h fg 2
6 . 77 = 0 . 558 + x 2 × 7 . 718
h2 = 163 . 4 + 0 . 8 × 2409 . 2 = 2090 . 76 kJ / kg .
x = 0 .8
pump .work : W P = h4 − h3 = v 3 × ( p 4 − p 3 )
= v f 2 ( p1 − p 2 ) = 0 . 001007 × ( 2000 − 7 )
W P = 2 . 0068 kJ / kg
heat sup plied : Q S = h1 − ( h f 2 + w p ) = 3025 − (163 . 4 + 2 . 0069 )
Q S = 2859 . 79 kJ / kg .
heat .. rejected : Q R = h2 − h3 = h2 − h f 2 = 2090 . 76 − 163 . 4
Q R = 1927 . 36 kJ / kg .... Ans
work ..done ..W = Q S − Q R ..or ( h1 − h2 ) − W P = 2859 . 7 − 1927 . 37
= 932 . 34 kJ / kg ....... Ans
Thermal ..efficiency =
W
932 . 34
=
= 0 . 326 = 32 . 6 %
Q S 2859 . 79
17
Theoretical steam consumption :
=
3600 3600
=
= 3.86kg / kW − hr
W
932.34
IMPORTANT FORMULAE IN STEAM OR PURE SUBSTANCE
PROPERTIES
ENTHALPY ( h )
KJ/kg
WET STEAM
DRY STEAM
SUPER HEATED STEAM
hwet = h f + x.h fg
hg = h f + h fg
hsup = hg + C P (TSUP − TS )
SPECIFIC
VOLUME, (v) in
m3/kg
Density (ρ ) in
kg/m3
v wet = x.v g
v dry = v g
ρ wet =
ρ wet =
Work done (W)
in KJ/kg
WWET
Wsup
Internal energy
in KJ/kg
U WET = hwet − WWET U G = hG − WG
U SUP = hSUP − W SUP
Specific Entropy
(s) in KJ/kg.K
S wet = S f + x.S fg
T
S SUP = S G + C PS ln SUP
 TS
v sup =
1
1
ρg =
v wet
vg
= 100 × p.v g W g = 100 × p.v g
S DRY = S f + S fg
v g × TSUP
TS
1
v sup
= 100 × p.v sup



FOR NON-FLOIW PROCESS.
WORK DONE
HEAT TRANSFER
CONSTANT VOLUME
W=0
Q = (h2 − h1 ) − ( p2 v sup − p1 x1v g1 )
CONSTANT
PRESSURE
CONSTANT
TEMPERATURE
HYPERBOLIC
W=p(v2-v1)
Q =h2 – h1
W=p(v2-v1)
Q= h2 - h1
v 
W = p1v1 ln 2 
 v1 
v
W = p1 v 1 ln  2
 v1
ISENTROPIC
POLYROPIC
W=U1- U2
Q= 0
PROCESSES
W=
p1v1 − p2 v 2
n −1
Q=

 + ( h2 − h1 )

n( p1v1 − p2 v 2 )
+ (h2 − h1 )
n −1
18
FOR FLOW PROCESS.
SYSTEMS
WORK DONE
HEAT TRANSFER
BOILER
TURBINE
CONDENSOR
NOZZLE
W=0
W=h2-h1
W=0
W=0
Q=h2-h1
Q=0
Q=h1 –h2
Q=0
APPLICATION OF STEADY FLOW ENERGY EQUATIONS TO VARIOUS
ENGINEERING SYSTEMS.
1.0
BOILER OR STEAM GENERATOR.
Q = h2 − h1 …………….in KJ.
2.O
CONDENSOR.
h1 + Q = h2
2.0
NOZZLE.
V2 =
(2 × (h
2
− h1 ) + V12 )
The expansion of the fluid in a nozzle is treated as reversible adiabatic or isentropic.
V2 =
(2 × C
P
(T1 − T2 ) + V12 )
γ −1


 P2  γ

2 
V2 =  2 × C P (T1 − T1   + V1 


 P1 


P 
as…….. T2 =  2 
 P1 
γ −1
γ
γ −1


 P2  γ 

V2 = 2 × C P × T1 1 −  
+ V12 ……………m/sec
  P1  


19
TURBINE.
Turbine is a device which converts potential energy of working fluid into mechanical work.
The turbine is fully insulated. Therefore there is no heat transfer (Q=0).
h1 = h2 + W
AIR COMPRESSOR.
h1 = h2 − W
h1 − Q = h2 − W
……..IF ‘Q’ IS
…..FOR NO HEAT LOSS
REJECTED TO SURROUNDINGS.
20
THROTTLING PROCESS.
When a liquid expands through a minute aperture orifice or a slightly opened valve, the
process is called throttling process.
P2 ,V2 , U1
P1 ,V1 , U1
Q=0,
W=0,
AND
Enthalpy is constant i.e. h1=h2
Relation between specific heats at constant volume and pressure.
The ratio between specific heat
CP
IS DENOTED BY GAMMA, i.e. ‘ γ ’
CV
CP
=γ
CV
C P − CV = γ
PROBLEM 6.
A refrigerator works on an ideal vapor compression refrigeration cycle, uses R-12 as the
working fluid. The minimum and maximum pressure of the cycle is 0.15 Mpa and 0.9Mpa
respectively. If the mass flow rate of the refrigerant is 0.045 kg/sec; determine (a) the rate of heat
removal from the refrigerated space and the power input to the condenser. (b)the rate of heat
rejection to the environment and (c) COP of the refrigerant.
SOLUTION:
21
In an ideal vapour compression refrigeration cycle the compression process is isentropic and the
refrigerant nters the compressor as a saturated vapour at the evaporator pressure and leaves the
condenser as saturated liquid at the condenser pressure.
IRREVERSIBLE PROCESS
BECAYSE OF
T
THROTTLING
h3=h4
2
p1=0.9 MPa
Q1
3
Win
4
1
P2=0.15 MPa
Q2
ENTROPY (S)
The enthalpy of the refrigerant at all four states is obtained from the refrigerant R-12
TABLES.
At
P1 =0.15Mpa,
h1 = hg = 178.64kJ / kg : S1 = S g = 0.7089kJ / kgK
At
P1 =0.9 Mpa,
h2 = 210.43kJ / kg : and ..S 2 = S1 = 0.7089 kJ / kgK
At
P3 =0.9 Mpa,
h3 = h f = 71.93kJ / kg : and ..h4 = h3 : throttling = 71.93kJ / kg
The rate of heat removal and the power input to the compressor respectively is given by :
22
.
.
Q L = m (h1 − h4 ).....and ............W IN = m (h2 − h1 )
.
Q L = 0.045 × (178.641 − 71.93) = 4.8kW
.
W IN = 0.045 (210.43 − 178.64) = 1.43kW
(b) The rate of heat rejected from the refrigerant to the environment is :
•
m×(h2 − h ) = 0.045 × (210.43 − 71.93) = 6.33kW
3
Also QH is given by :
.
Q H = Q L +Win = 4.80+1.43=6.23kW
( c ) COP=
Q L 4.80
=
= 3.36... Ans
W IN 1.43
PROBLEM :7
In an aircraft engine compressed air at 3 bar and 450K enters a combustion chamber in which heat is added at
constant pressure to the air by the combustion of the fuel.Air is thus heated to a temperature of 1250K.It then enters
a turbine with a negligible velocity . It expands in the turbine until its temperature falls to 100K.The velocity of air
leaving the turbine is 50m/sec.Air then enters a convergent divergent nozzle wherein it expands until its temperature
drops to 800K.Flow through both turbine and nozzle may be considered as reversible adiabatic. Determine :
(a)
(b)
(c)
(d)
1
Heat added in the co0mbustion chamber/kg of air.
Work done in the turbine per kg of air.
Velocity of air leaving the nozzle.
Pressure of air leaving the turbine and at the exit from the nozzle.
Compressed
air at 3 bar ,
450K
2
Combustion
chamber
3 bar,
1250K
w
1000K, 50 m/sec
Air
turbine
3
v4
nozzle
23
(a) combustion chamber.
Heat added to air Q = ( h2 − h1 ) = C PO (T2 − T1 ) = 1.0035(1250 − 450) = 804kJ / kg
(b) Turbine :
work done in the turbine
V22 − V32
W = (h2 − h3 ) + (
) + Q as Q=0
2
V32
50 2
W = C PO (T2 − T3 ) −
= 1.0035(1259 − 1000) −
= 251kJ / kg
2
2000
note that the kinetic terms are negligible.
( c ) Velocity of air leaving nozzle.
V4 = 2(h3 − h4 ) + V32 = 2C PO (T3 − T4 ) + V32
V4 = 2 × 1003.5(1000 − 800) + 50 2 = 636m / sec
(d) Pressure of air leaving the turbine is given by:
k
1.4
 T  k −1
 1000  0.4
P3 = P2  3  = 3.0 × 
 = 1.374bar
T
1250


 2
(e) pressure of air at exit of the nozzle is given similarly by:
T 
P4 = P3  4 
 T3 
k
k −1
1.4
 800  0.4
= 1.374 × 
 = 0.63bar
 1000 
PROBLEM 8
An aircraft is flying at a speed of 1000km.p.h. The ambient air pressure and temperature is 0.35
bar and –15 deg C. respectively. What is the pressure of air entering the compressor after inlet
24
diffuser section, and what is the pressure ratio required for compressor for the pressurization of
the aircraft?
Solution:
Speed of aircraft
V=
1000 × 1000
= 277.8m / sec
3600
from the SSSF conditions for the diffuser.
h2 = h1 +
V12
2
Turbo jet engine
Fuel flow
Exhaust gases from exit
Ambient air
Diffuser intake
Combustion chamber
3
COMB CHAMBER
COMPRESSOR
Turbine
4
TURBINE
DIFFUSER
1
NOZZLE
5
2
6
4
T
3
5
2
6
1
S
25
V12
C PO T2 = C PO T1 +
2
Whence the temperature of air leaving inlet diffuser,
V12
277.8 2
T2 = T1 +
= 258 +
= 258 + 38.5 = 296.5 K
2C PO
2 × 1003.5
k
1.4
 T2  k −1
 296.5  0.4
P2 = P1   = 0.35 × 
 = 0.57bar
 258 
 T1 
Pressure ratio of compression required for cabin pressurization:
=
1.01325
= 1.78
0.57
…Answer
PROBLEM : 9
An engine operates on the air –standard cycle. The compression ratio is 18. The pressure and
temperature at the start of compression process are 100kPa and300K.The heat added is 1800
KJ/kg of air. Determine the maximum pressure and temperature in the cycle, the thermal
efficiency and the mean effective pressure.
QH
T
2
3 ρMAX
S=CONSTANT
4
QL
1
v
State 2:
26
v 
T2 = T1.  1 
 v2 
γ −1
= 300(18) 0.4 = 953.3 K
γ
v 
P2 = P1.  1  = 100(18)1.4 = 5720kPa.
 v2 
state :3
Q H 2 = 2 Q3 = h3 − h2 = C PO (T3 − T2 )
1800 = 1.0035(T3 − 953.3);∴T3 = 2744.3 K
Maximum temperature and pressure of the cycle.
Tmax=T3=2744.3K
Pmax=P2 =p3=5720 kPa.
State 4 :
v3 =
RT3 0.287(2744.3)
=
= 0.1377m 3 / kg
P3
5720
v 4 = v1 =
RT1 0.287(300)
=
= 0.861m 3 / kg
P1
100
v 
T4 = T3.  3 
 v4 
γ −1
 0.1377 
= 2744.3

 .861 
γ
0.4
= 1318.3 K
v 
 0.1377 
P4 = P3.  3  = 5720
 = 439.4kPa
 .861 
 v4 
1.4
Heat rejected:
Q L = 4 Q1 = U 1 − U 4 = CVO (T1 − T4 ) = 0.7165(300 − 1318.3) = −731kJ / kg
Net work :
W NET = Q H − Q L = 1800 − 731 = 1069kJ / kg
Thermal efficiency:
η theor
W NET 1069
=
= 0.594(59.4%)
QH
1800 1
Mean effective pressure:
27
pmean =
1.0
W NET
1069
=
= 1419kPa.
(v1 − v 2 ) 0.861 − (0.86 / 18)
PROBLEM 10.
Consider the entropy change of an ideal gas following reversible path :
1-a (isothermal) and a-2 (constant volume )
& 1-b (adiabatic & b-2 (constant volume )
as shown below. Show that the path combination produces the same entropy.
1
p
T=C
a
V=C
Q=0
2
b
v
For path 1-a-2:
S A − S1 = R ln
va
v
T
= R ln 2 : and .S 2 − S a = CVO ln 2
v1
v1
T1
So that by adding we get;
S 2 − S1 = R ln
v2
T
+ CVO ln 2 …………………….(A)
v1
T1
For path 1-b-2:
S B − S 1 = 0..as..Q = 0(adiabatic ) ∴ S 2 − S B = C VO ln
T2
TB
γ −1
For adiabatic process
= CVO ln
T1  vb 
T
v
=   . ∴ S 2 − S B = CVO ln[ 2 × ( 2 ) λ −1 ]
TB  v1 
T1 v1
T2
v
+ CVO ( K − 1) ln 2
T1
v1
28
T2
v
+ R ln 2 ………(B) EXPRESSIO SAME AS (A)
T1
v1
C
R
.as.....C P −C V = R ∴ P − 1 =
ie..R = CVO ( K − 1)
CV
CVO
= C VO ln
2.0
REVERSIBLE POLYTROPIC PROCESS FOR AN IDEAL GAS.
P2
T
T2=T1
2S
P1
2N
REVERSIBLE
ADIABATIC
N=K
2T
ISOTHERMAL N=1
REVERSIBLE
POLYTROPIC
WORK
Q
we ..have ... pv n = C .
dp
dv
dp
p
+n
= 0..... ∴
= −n
p
v
dv
v
S
also ...... p1v n1 = p2 v n 2 = C = pv n
log p1 + n log v1 = log p1 + n log v 2
∴n =
ln( p1 / p2 )
ln( v 2 / v1 )
c
p 1 v 1n
p 2 v 2n
also ... p =
=
=
vn
vn
vn
2
Hence ..1 W 2 =
∫
1
2
pdv = C
dv
∫v
n
1
p 2 v 2n v 11 − n − p 1 v 1n v 11 − n
 v 12 − n − v 11 − n 
= C
 = 1W 2 =
1− n
 1− n 
p v − p1v 1
= 2 2
1− n
p 2 v 2 − p1v 1
.......... .......... .....( 2 )
1W 2 =
1− n
29
1
v 2  p1  n
p1v1  p2 nn−1 
Substituting
=   ....∴1 W 2 =
( ) − 1
v1  p2 
1 − n  p1

R
[T 2 − T1 ]
1− n
now ..... q = ( u 2 − u 1 ) + W
∴1 W 2 =
R
) × (T 2 − T1 )
1− N
k − n
∴ q = C VO (
) × (T 2 − T1 )
1− n
k − n
∴q = (
)× 1W 2
1− n
= ( C VO +
3.0
PROBLEM 11.
Air expands irreversibly from 3 bars, 200 deg C, to 1.5 bars and 105deg C.Compute the entropy
change/kg of air.
Solution.
Although the process is irreversible, entropy being a property change can be computed by
integrating along a reversible path between the two specified states.
S 2 − S1 = C PO ln
= 1.0035 ln
4.0
T2
P
− R ln 2
T1
P1
378
1.5
− 0.278 ln
= −0.0323kJ / kgK
473
3
PROBLEM 12.
An insulated tank is divided into equal parts by volume. One part contains Argon gas at 30 degC and 5 bar. The
other part is vacuum. The partition between these two parts is broken and the gas fills the whole tank. Determine
final pressure and entropy change per kg of Argon gas.
Solution.
Considering it as an ideal gas, we have :
30
T2 = T1 ,..as...u2 = u1.. for .. free.. exp ansion.. process .
v  1
∴ p2 = p1  1  = p1 = 2.5bar
 v2  2
Entropy change ( choosing the reversible path)
= R ln
v 2 8.344
=
ln 2 = 0.1441kJ / kgK as molecular weight of Argon gas =40
v1
40
5.0
PROBLEM 13
A Single stage reciprocating air compressor has a swept volume of 2000cm3 and runs at 800rpm.
It operates on a pressure ratio of 8, with a clearance of 5 % of the swept volume. Assume NTP
room conditions and at inlet (p=101.3 kpa, t=15 deg C), and polytropic compression and
expansion with n=1.25; calculate (a)indicated power (b) volumetric efficiency c) mass flow rate
d) FAD (free air delivery ) e) isothermal efficiency f)the actual power needed to drive the
compressor, if mechanical efficiency is 0.85.
3
P
2 ρMAX
P2
PVN =C
P1
4
VC
1
VS
V
SUCTION AT CONSTANT TEMPERATURE T1.
P1=101.3kpa;P2=8P1=810.4kpa; T1=288K, VS=2000cm3
V3==VC=0.05*VS=100 cm3.
V1=VC+VS=2100 cm3.
31
1
1
 P n
n
P3V3 = P4V4n .∴V4 =  3  × V3 = (8 )1.25 × 100 = 528cm 3
 P4 
V1-V4=2100-528=1572 cm3
n −1


n


P
n
2
W =
P1 (V1 − V4 )   − 1

 P1 
n −1


OR
n −1


n


P
n
2
W =
(m1 − m 4)   − 1

 P1 
n −1


W=
n −1
1.25
101.3 × 10 3 × 1572 × 10 6 × (8) n − 1 = 411J

0.251

NOTE : If the number of stages is ‘N’ then work done =Nx W
Indicated power=
speed ..in..rpm × work..done
watts
60
411 × 800 × 10 −3
Indicated power
kW
60
(b) volumetric efficiency =
5.46 Kw
1572 × 100
= 78.6%
2000
pV 101.3 × 10 3 × 1572 × 10 −6
c) mass of compressed air per cycle : m =
=
= 1.93 × 10 −3
RT
287 × 288
−3
THEREFORE MASS FLOW RATE= 1.93 × 10 × 800 = 1.54kg / min
−6
3
d) FAD (free air delivery ) 1572 × 10 × 800 = 1.26m / min
e)
W1 = P1 (V1 − V4 ) × ln
p2
= 101.3 × 10 3 × 1572 × 10 −6 ln 8 = 331J
p1
0.331 × 800
= 80.6%
5.47 × 60
5.47
Input .. Power ..
= 6.44 kW
0.85
η ISOTHERMAL =
32
6.0
EFFICIENCY OF COMPRESSOR
The efficiency of a compressor is defined the ratio of isothermal work done to the actual work
done by the compressor.
P1V1 ln
Efficiency of a compressor = η =
P2
P1
P1V1 ln
WT
=
WC
W COMP
N × nRT1
The work done by the compressor W C =
n −1
=
V1
V2
W COMP

 PX
 P
 1



n −1
n


− 1


Where N=number of stages of compression and Px is the intermediate pressure.
The optimum pressure Px= P1 × P3 for two stages. For number of stages the optimum pressure is
given by
1
 P N
PX
=  D  Where Pd=delivery pressure and Ps=suction pressure and Px=intermediate
P1
 PS 
pressure.
7.0
PROBLEM 14
A two stage air compressor with perfect inter-cooling takes in air at 1 bar pressure at 27deg C .
The law of compression in both the stages is PV 1.3 = cons tan t .The compressed air is delivered
at 9 bar from the HP cylinder to an air reservoir. Calculate per kg of air a) the minimum work
done and b) the heat rejected to the intercooler.
SOLUTION.
The minimum work required is a two stage compressor is given by :-
N × nRT1
WC =
n −1

 PX
 P
 1



n −1
n


− 1


33
=
2 × 1.3 × 0.287 × 300 1  0.3 
(3)1.3 − 1 ……..as Px=3

0.3

=26x0.287x100x0.287=214.16kJ/kg
T2  P2 
= 
T1  P 1 
n −1
n
=3
0.3
1.3
= 1.2886
therefore T2=386.56 K
Heat rejected to the intercooler =1.005(386.56-300)=86.99kJ/kg…Ans
8.0
PROBLEM 15.
GAS POWER CYCLE OTTO CYCLE.
An engine working in the Otto cycle is supplied with air at 0.1mPa, 35 deg C. The compression
ratio is 8. Heat supplied is 2100 KJ/kg.Calculate the max pressure and temperature of the cycle,
the cycle efficiency and the mean effective pressure.(for air Cp=1.005, Cv==0.718, and
R=0.287kJ/kgK).
P
3
OTTO
CYCLE
T2  V21 
= 
T1  V2 
γ −1
= (8)
0.4
3
T
0.884
V2 =
= 0.11m 3γ /S=kgC
Pv = C
8
Q1
2
WE
WC
4
= 2.3 : .......T2 = 2.3 × 308 = 708.4 K
1
1
Q1 = CV (T3 −T 2) = 2100 kJ / kg
: .........T3 − 708.4 =
V
T1 = 273 + 35 = 308 K P  V
T3 = TMAX = 3633 K .......... ......... 2 2=  1
P1 = 0.1MPa = 100kN /Pm
1
 V2
4
Q2
2100
= 2925 K
0.718
S
γ

 = (8 )1.4 = 18.37........ P2 = 1.873 MPa

P3V3 Q
P1V= 2100kJ / kg ,..rk = 8...γ = 1.4363
= 2 2 .......... ..P3 = PMAX = 1.837 ×
= 9.426 MPa
T3
T2
1
1
1 708
η cyc = 1 −
= 1−
= 1−
= 0.565..or ..56.5%
W NET = Q1 × η CYCLE
2.3 .5kJ / kg
rkγ −=1 2100 ×800..4565 = 1186
1186 .5
V
RT
0.287 × 308
= PM × (V11 −=V82,......
)=P
=3
= 1.533 MPa ......... Ans
VM1 (=0.8841 == 0.11)... ∴ PM = MEP
0.844m
0.774
V2
P1
100
34
9.0
PROBLEM 16
A diesel engine has a compression ratio of 14 and cut off takes place at 6% of the stroke. Find
the air standard efficiency.
QH
P
3 ρMAX
2
S=CONSTANT
PV γ = C
4
QL
1
v
rk =
V1
= 14 : .........V3 − V2 = 0.06 (V1 − V2 ) = 0.06 (14 × V2 − V2 ) = 0.78V2
V2
V3 = 1.78V2 : CUT ..OFF .. RATIO ...
η diesel
V3
= 1.78 = rc
V2
rcγ − 1
= 1 − × γ −1 ×
γ rk
rc − 1
1
1
1
1
1.781.4 − 1
× 0 .4 ×
1.4 14
1.78 − 1
1.24
1 − 0.248 ×
= 0.605 = 60 .5%
0.78
= 1−
10.0
PROBLEM 16.
35
A turbojet aircraft flies with a velocity of 300 m/sec at an altitude where the air is at 0.35 bar and
–40 degC. The compressor has a pressure ratio of 10, and the temperature of the gases at the
turbine inlet is 1100deg C. Air enters the compressor at a rate of 50 kg/sec. Estimate a) the
temperature and pressure of the gases at the turbine exit. b) the velocity of gases at the nozzle
exit and c) the propulsive efficiency of the cycle.
NOZZLE
DIFFUSER
COMB
CHAMBER
PRODUCT
GAS
COMPRESSOR
TURBINE
6
1
2
3
5
4
T
4
Q1
3
P=C
5
2
6
1
Q2
S
For isentropic flow of air in the diffuser,
Q12 − W1− 2 = h2 − h1 +
V 22 − V12
2
V12
0 = C P × (T2 − T1 ) −
2
2
V1
300 2 × 10 − 3
: ... ∴ T2 = 277.8 K
T2 = T1 +
= 233 +
2×CP
2 × 1.005
1 .4
γ
T
 277.8  0.4
P2 = P1 × ( 2 ) γ −1 = 35kN / m 2 × 
 9 = 64.76kPa
T1
 233 
P3 = r p × P2 = 10 × 64.76 = 647.6kPa.
P
T3 =  3
 P2



γ −1
γ
× T2 = 277.8 × (10)
0 .4
1.4
: ......T3 = 536.6 K M
36
W C = WT
h3 − h2 = h4 − h5 : ...T3 − T2 = T5 − T5
T5 = T4 − T3 + T2 = 1373 − 536 .66 + 277 .78 = 1114 .2 K
γ
γ −1
T 
 1114 .12 
P5 =  5  × P4 = 647 .6

 1373 
 T4 
P5 = 311 .69 kPa .
3.5
b) for isentropic expansion of gases in the nozzle:
P 
T6 = T5 ×  6 
 P5 
γ −1
γ
 35 
= 1114 .2 × 

 311.69 
0.286
= 596.12
Neglecting the KE of gas at nozzle inlet,
1
V6 = [2 × C P × [T5 − T6 ] × 1000]2
V = [2 × (h − h )]12 m / sec  = 1029.4m / sec
 6

5
6


c) The propulsive efficiency of a turbo-jet engine is the ratio of the propulsive power developed
Wp to the total heat transfer to the fluid.
W P = W [V EXIT − V INT ]× V AIRCRAFT
= 50 × [1020.4 − 300]× 300 = 10.80 MW
Q1 = w × (h4 − h3 ) = 50 × 1.005 × (1373 − 536.66) = 42.026 MW
η=
10.806
= 0.257 = 25.7%........ Ans
42.026
11.0
JET PROPULSION
T = ( m& air + m& fuel ) × V E + m& AIR × V∞ + ( PE − P∞ ) × AE
Where T =thrust in Newtons, m& air and m& fuel are mass flow of air and fuel respectively.
37
V∞ ….velocity of free stream air or aircraft velocity
VE …velocity of jet exhaust
PE …..pressure at nozzle exit
P∞ ….free stream pressure or static pressure at the altitude where aircraft is flying.
12
ROCKET MOTOR.
The specific Impulse of the rocket motor is given by :
 2 × γ × R × To 
PE
I SP = 
1 − ( )
γ −1
PO


γ −1
γ
 

 
1
2
PE
is the ratio of the nozzle exit
PO
pressure to the rocket chamber pressure, R is the gas constant, and γ = ratio of the specific
heat for the combustion gases taken as 1.4 for adiabatic flow.
where Isp=specific impulse; To =chamber temperature
The mass flow through the nozzle is given by :-
PO × A
m& =
TO
∗

 γ
 R



2


+
1
γ


γ +1
γ −1




Where A∗ is the area of throat cross section.
13.0
STEADY STATE STEADY FLOW ENERGY EQUATIONS.……SSSFE.
w1 (u1 + p1 v1 +
(u1 + p1 v1 +
v12
v2
dQ
dW
)+
= w 2 (u2 + p2 v 2 + 2 ) −
….is energy flow per unit time
2
dτ
2
dτ
v12
v2
dQ
dW
)+
= (u 2 + p 2 v 2 + 2 ) −
………for energy flow per unit mass
2
dm
2
dm
In differential form :-
38
dQ = dh + VdV + gdz + dW
Where dQ = du + pdv
dh = du + pv
dh = du + pdv + vdp
dh = dq + vdp
dq = dh − vdp
Tds = dh − vdp
For closed system the work done by the gas is pdv while for flow systems the work done is
given by
14.0
vdp .
PROBLEM 17.
NOZZLE FLOW FOR A GAS FOR A POLYTROPIC PROCESS.
Since this is a flow system i.e. not a closed system the work done should be a
nozzle flow analyses is taken as an example to make things more clearer.
vdp
term. A
2
1
V1=70 m/S
V1=? m/S
.v1=0.1m 3 /kg
.v1=0.6m 3 /kg
P1=15 bar
P1=1.5 bar
A system above is steady flow nozzle ideally insulated device so that no heat transfer takes
place with the surroundings, while the flow takes place through it. A gas expands through the
nozzle following a reversible polytropic law pv γ = C . There is no change in PE but the
pressure drop from 15 bar to 1.5 bar and the specific volume increases from 0.1 m3 to 0.6m3.
39
If the entrance velocity to the nozzle is 70 m/sec determine the exit velocity.
We have;
dQ = dh + VDV + gdz + dW
here Q=0 ; W=0 as no external work is added and the potential energy is =0 as z is in the same
datum.
Therefore,
dq = du + pdv + vdp + VdV
dq = dq + vdp + VdV
− vdp = VdV
2
∫ − vdp = work..done.in.a. polytropic.. process
1
=
n
( p1v1 − p2 v 2 )
n −1
2
V 2 − V12 
VdV = ∫ VdV =  2

2 

1
V22 − V12  1.3
5
5
5
5
 2  = 0.3 (15 × 10 × 0.1 − 1.5 × 10 × 0.6 ) = 4.33 × 0.6 × 10 = 6 × 10 J


2
2
V2 − 70
= 2.6 × 10 5 J
2
V2 = 2 × 2.6 × 10 5 + 4900 = 724 .5m / s
15.0
Reversible adiabatic flow of a fluid with area change.
Consider reversible adiabatic flow of a fluid in a passage of varying cross sectional area.
The following equations can be written for the flow :Continuity equation : ρ × A × V = m& = cons tan t
Differentiating we have
dρ
dA dV
+
=0
ρ
A V
SSSF equations (neglecting PE changes )
+
40
dh = VdV = 0
Tds = dh −
dp
ρ
Therefore combining SSSF equations with the property relation we get;
dV =
1 dp
V ρ
substituting for , dV ,from the continuity equations
dA
dρ dV
dρ
dp
=−
−
=−
+
ρ
ρ ρV 2
A
V
dA
dp
=
A ρV 2


dρ
 − V 2
+ 1
dp 

Since the flow is reversible adiabatic ,
dp
V2
2
=C = 2
dρ
M
Substituting the same in the above we obtain
dA
dp
=
(1 − M 2 ) ………………………………….(1)
2
A ρV
For nozzles always dp <0
While for diffusers it is dp >0.
Hence on closer analyses of the above equation we can see the following :FOR NOZZLES :-
For subsonic portion of the nozzle the nozzle cross section converges i.e. when M<1;
And the nozzle diverges for supersonic speeds i.e. M>1.0.
FOR DIFFUSERS:
It is the reverse of nozzles i.e. at the convergent portion diverges or M>1.0 and converges
for supersonic flow.
41
EXAMPLE E7-3 A 4-cylinder SI engine with a cylinder displacement of 0.5 L and a
clearance volume of 62.5 mL is running at 3000 RPM. At the beginning of the compression
process, air is at 100 kPa and 20oC. The maximum temperature during the cycle is 1800
K. Employing the cold-air standard Otto cycle, determine (a) the power developed by the
engine, (b) the thermal efficiency, and (c) the MEP. (d) What-if-Scenario: How would
the answer in part (b) change if the maximum temperature were raised to 2200 K? (e)
How would the answer in part (b) change if the ideal gas model were used? [Manual Solution]
[TEST Solution]
Answers: (a) 30.7 kW, (b) 58.6%, (c) 614 kPa, (d) 58.5%, (e) 53.5%
42
43
EXAMPLE E7-5 An air standard Diesel cycle has a compression
ratio of 18. The heat transferred to the working fluid per cycle is
2000 kJ/kg. At the beginning of the compression process the
pressure is 100 kPa and the temperature is 25oC. Employing the
perfect gas (PG) model for the working fluid, determine (a) the
pressure at each point in the cycle, (b) the cut-off ratio, (c) the
thermal efficiency, (d) the net work per unit mass, and (e) the
MEP. (f) What-if-Scenario: How would the thermal efficiency
change if the heat transfer were 1500 kJ/kg? [Manual Solution] [TEST
Solution]
Answers: (a) 5728 kPa, 487 kPa, (b) 3.1, (c) 57.8%, (d) 1153 kJ/kg, (e) 1430 kPa, (f)
60.6%
EXAMPLE E7-6 The Diesel cycle described in Example
E7-5 is
modified into a Dual cycle by breaking the heat addition process
into two equal halves so that 1000 kJ/kg of heat is added at
constant volume followed by 1000 kJ/kg of heat addition at
constant pressure. Determine (a) the temperature at each point in
the cycle, (b) the thermal efficiency, (c) the net work per unit
mass, and (d) the MEP. (e) What-if-Scenario: How would the
thermal efficiency change if 75% of the total heat transfer took
place at constant volume? [Manual Solution] [TEST Solution]
Answers: (a) T2 = 947 K, T3 = 2340 k, T4 = 3335 K, T5 = 1209 K, (b) 67.3%, (c) 1346
kJ/kg, (d) 1666 kPa, (e) 61.1%
44
45
46
EXAMPLE E8-2 Air enters the compressor of a 50 MW simple
gas turbine at 100 kPa, 298 K. The pressure ratio of the
compressor is 11 and the turbine inlet temperature is 1500 K.
Using the cold-air standard Brayton cycle to model the power
plant, determine (a) the mass flow rate, (b) the thermal
efficiency, and (c) the back work ratio. (d) What-if-Scenario:
How would the efficiency and mass flow rate change if the
compression ratio is varied within the range 4-50? [Manual
Solution] [TEST Solution]
Answers: (a) 11.61 kg/s, (b)
47
48
49
50
EXAMPLE E9-2 A steam power plant operates on an ideal Rankine
cycle. Superheated steam flows into the turbine at 3 MPa and 400oC
with a flow rate of 100 kg/s and exits the condenser at 50oC as
saturated water. Determine (a) the net power output, (b) the thermal
efficiency, and (c) the quality of steam at the turbine exit. (d) Whatif-scenario: How would the efficiency change if the condenser
temperature can be lowered to 30oC? [Manual Solution] [TEST
Solution]
Answers: (a) 108,179 kW, (b) 33.2%, (c) 91.3%, (d) 36.5%
51
52
10-1-7 A refrigerator uses R-12 as the working fluid and operates
on an ideal vapor compression refrigeration cycle between 0.15 MPa
and 1 MPa. If the mass flow rate is 0.04 kg/s, determine (a) the
tonnage of the system, (b) compressor power, and (c) the COP. (d)
What-if-scenario: How would the answer in part (c) change if R12 was replaced with R-134a, a more Environmentally benign
refrigerant? [Manual Solution] [TEST Solution]
Answers: (a) 1.165 tons, (b) 1.36 kW, (c) 3.01, (d) 3.33
53
54
Notes prepared by M.Sundaresan.
EXAMPLE E10-5 A two-stage cascade refrigeration plant uses R-22
as the working fluid in both the stages. The lower cycle operates
between the pressure limits of 120 kPa and 380 kPa and the topping
cycle has a condenser pressure of 1200 kPa. The heat exchanger
that couples the two cycles requires a minimum temperature
difference of 5oC between the heating and the heated streams. If the
mass flow rate in the lower cycle is 0.08 kg/s, determine (a) the
mass flow rate in the upper cycle, (b) the cooling capacity, and (c)
the COP. (d) What-if-Scenario: How would the COP change if a
single cycle operated between 1200 kPa and 120 kPa? [Manual Solution]
Answers: (a) 0.11 kg/s, (b) 4.53 tons, (c) 2.7, (d) 2.58
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