ΕΡ&Σ
Π Α Ν Ε Π Ι Σ Τ Η Μ Ι Ο Θ Ε Σ Σ Α Λ Ι Α Σ
ΠΟΛΥΤΕΧΝΙΚΗ ΣΧΟΛΗ
ΤΜΗΜΑ ΜΗΧΑΝΟΛΟΓΩΝ ΜΗΧΑΝΙΚΩΝ
LFM&T
ΕΡΓΑΣΤΗΡΙΟ ΡΕΥΣΤΟΜΗΧΑΝΙΚΗΣ & ΣΤΡΟΒΙΛΟΜΗΧΑΝΩΝ
Α.Δ. Παπαθανασίου, Αν.Καθ.
Λεωφ. Αθηνών-Πεδίον Άρεως-38334 Βόλος - Τηλ. 24210 74094/11- email: fluids@mie.uth.gr - web: http://www.mie.uth.gr/labs/fluids
1Ο ΣΕΤ ΑΣΚΗΣΕΩΝ ΕΙΣ ΤΟ ΜΑΘΗΜΑ ΘΕΡΜΟΔΥΝΑΜΙΚΗ ΙΙ
Διδάσκων: Αθανάσιος Παπαθανασίου, Αν.Καθ.
Προβλήματα 5.44, 5.184 και 5.199 που επισυνάπτονται
Assumptions 1 This is a steady-flow process
since there is no change with time. 2 Potential
energy change is negligible. 3 There are no
work interactions.
Analysis We take the steam as the system,
which is a control volume since mass crosses
the boundary. The energy balance for this
steady-flow system can be expressed in the
rate form as
Energy balance:
400C STEAM
800
kPa
Q
10 m/s
300C
200
kPa
E  E out
in



Rate of net energy transfer
by heat, work, and mass
E system0 (steady)

0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out


V2 
V2 
m  h1  1   m  h2  2   Q out


2 
2 


V2
V 2 Q
h1  1  h2  2  out
2
2
m
or
since W  pe  0)
The properties of steam at the inlet and exit are (Table A-6)
P1  800 kPa v1  0.38429 m3/kg

T1  400 C  h1  3267 .7 kJ/kg
P2  20 0 kPa v 2  1.31623 m3/kg

T1  300 C  h2  3072 .1 kJ/kg
The mass flow rate of the steam is
m 
1
v1
A1V1 
1
(0.08 m2 )(10 m/s)  2.082 kg/s
3
0.38429 m /s
Substituting,
3267.7 kJ/kg 
(10 m/s) 2  1 kJ/kg 
V22  1 kJ/kg 
25 kJ/s

3072
.
1
kJ/kg





2 2
2 2
2
2  1000 m /s  2.082 kg/s
 1000 m /s 

V2  606 m/s
The volume flow rate at the exit of the nozzle is
V2  m v 2  (2.082kg/s)(1.31623 m3/kg)  2.74 m3/s
(Answer: Wnet=20,448MW)
Assumptions 1 This is a steady-flow process since there is no change with time. 2
Kinetic and potential energy changes are negligible. 3 The devices are adiabatic and
thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats.
Properties From the steam tables (Tables A-4 through 6)
P3  12.5 MPa 
 h3  3343.6 kJ/kg

T3  500 C
and
P4  10 kPa 
 h4  h f  x4h fg  191.81  0.92 2392.1   2392.5 kJ/kg
x4  0.92

From the air table (Table A-17),
T1  295 K 
 h1  295.17 kJ/kg
2
T2  620 K 
 h2  628.07 kJ/kg
Analysis There is only one inlet and one exit
for either device, and thus m in  m out  m . We
take either the turbine or the compressor as
the system, which is a control volume since
mass crosses the boundary. The energy
balance for either steady-flow system can be
expressed in the rate form as
E  E out
in



Rate of net energy transfer
by heat, work, and mass
E system0 (steady)

0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
For the turbine and the compressor it becomes
1 MPa
620 K
Air
comp
1
98 kPa
295 K
12.5 MPa
500C
3
Steam
turbine
4
10 kPa
Compressor:
Turbine:
 air h1  m
 air h2
W comp, in  m

 steam h3  W turb, out  m
 steam h4
m
 air (h2  h1 )
W comp, in  m
 steam (h3  h4 )
 W turb, out  m
Substituting,
Wcomp, in  10 kg/s 628.07  295.17  kJ/kg  3329 kW
W turb,out  25 kg/s 3343.6  2392.5  kJ/kg  23,777 kW
Therefore,
W net,out  W turb,out  W comp, in  23,777  3329  20,448 kW
5.199
The turbocharger of an internal combustion engine consists of a turbine and a
compressor. Hot exhaust gases flow through the turbine to produce work and the work
output of from the turbine is used and input to the compressor. The pressure of
ambient air increases as it flows through the compressor, before it enters the engine
cylinders. Thus, the purpose of the turbocharger is to increase the air pressure so more
air gets into the cylinder. As a result, more fuel can be burned and more power can be
produced from the engine.
In a given turbocharger exhaust gases enter the turbine as shown at the rate of 0.02
Kg/s. Air enters the compressor at the rate of 0.018 Kg/s. To avoid what is knows as
engine “knock”, an aftercooler is used to cool the air coming from the compressor,
with cool ambient air, before it enters the engine at a temperature no larger than 80C.
Treating the exhaust gases as air, determine the temperature at the exit of the
compressor and the flowrate of ambient air needed in the aftercooler.
(Answer: Tair,2=108.6 oC, Q=44.9 L/s)
Assumptions 1 All processes are steady
since there is no change with time. 2
Kinetic and potential energy changes are
negligible. 3 Air properties are used for
exhaust gases. 4 Air is an ideal gas with
constant specific heats. 5 The mechanical
efficiency between the turbine and the
compressor is 100%. 6 All devices are
adiabatic. 7 The local atmospheric pressure
is 100 kPa.
Properties The constant pressure specific
heats of exhaust gases, warm air, and cold
ambient air are taken to be cp = 1.063,
1.008, and 1.005 kJ/kg·K, respectively
(Table A-2b).
Analysis (a) An energy balance on turbine gives
Air
Turbin
e
Exhau
st
gases
Compressor
Cold
air
Aftercool
er
W T  m exh c p ,exh Texh,1  Texh,2   (0.02 kg/s)(1.06 3 kJ/kg  K)(400  350)K  1.063 kW
This is also the power input to the compressor since the mechanical efficiency
between the turbine and the compressor is assumed to be 100%. An energy balance on
the compressor gives the air temperature at the compressor outlet
W C  m a c p ,a (Ta,2  Ta,1 )
1.063 kW  (0.018 kg/s)(1.00 8 kJ/kg  K)( Ta,2  50)K 
 Ta,2  108.6 C
(b) An energy balance on the aftercooler gives the mass flow rate of cold ambient air
m a c p,a (Ta,2  Ta,3 )  m ca c p,ca (Tca,2  Tca,1 )
(0.018 kg/s)(1.00 8 kJ/kg  C)(108.6  80) C  m ca (1.005 kJ/kg  C)(40  30) C
m ca  0.05161 kg/s
The volume flow rate may be determined if we first calculate specific volume of cold
ambient air at the inlet of aftercooler. That is,
v ca 
RT (0.287 kJ/kg  K)(30  273 K)

 0.8696 m3/kg
P
100 kPa
Vca  m v ca  (0.05161kg/s)(0.8696 m 3 /kg)  0.0449m3 /s  44.9L/s