1
ST.ALOYSIUS INSTITUTE OF TECHNOLOGY, JABALPUR,
DEPT.OF MECHANICAL ENGINEERING
ENGINEERING THERMODYNAMICS (ME

304
)
EN
GINEERING
THERMODYNAMICS
2
INDEX
S.
No.
Name of Experiment
Signature
Date
Remark
1
To find mechanical equivalent of heat
using
Joules apparatus.
2
To study working of impulse and reaction
steam turbine by models.
3
To calculate COP of
vapor
compression
refrigeration system and to plot on T

s, p

H
diagrams.
4
To study working of Gas turbines by models and
to
identify various processes of Brayton Cycle.
5
To plot specific fuel
consumption
versus rpm
diagrams for diesel and petrol engines
.
3
Experiment No. 1
Object:
To find mechanical equivalent of heat using Joules apparatus.
Joules
equivalent definition:

It is the ratio of work done to the heat transferred from a system during a cycle and denoted by J
c
which
is known as Joules constant. It is only a unit conservation factor.
J
c
= work/Heat
In SI system of units the value J
c
is unit i.e., 1Nm = 1J.
Joules apparatus:

The main component of joules apparatus is following there,
1.
Water bath
2.
Thermometer
3.
Weight
4.
Paddle wheel
5.
Pulley
6.
Rope
Fig:

4
Theory
The ratio of the mechanical (or electrical) unit for energy to the heat unit for energy is called the
mechanical equivalent of heat or Joule's equivalent. It is named after the British physicist Sir James
Joule (1818

1889) who was the first to measure the
ratio in 1850. He used a mechanical device in
which mechanical energy of rotating paddle wheels (turned by falling weights) heated water in a
calorimeter. In this experiment, we shall determine the ratio of the energy supplied by an electric
current to the
heat that it produces.
When a current flows in a conductor, the amount of work done to cause the current to flow is given by
W = VIt
(1)
where
W is the work done in J,
I is the current flowing in A,
V is the potential difference across the
ends of the conductor in V, and
t is the time that the current flows in seconds.
If the heat energy that this current produces raises the temperature of a certain amount of water
(contained in a calorimeter) from temperature T1 to T2o C, the heat energy
produced is given by
H = (mwcw + mccc + h)(T2

T1)
(2)
where
H is the heat energy produced in calories,
m w is the mass of the water in g,
m c is the mass of the calorimeter in g,
c w is the specific heat of water (1 cal/goC),
c c is the specif
ic heat of the calorimeter in cal/goC)
h is the heat capacity of the immersion heater, thermometer, and stirrer in cal/oC.
The value of the mechanical equivalent of heat may then be calculated from the ratio
W/H = IVt/{(mwcw + mccc + h)(T2

T1)} J/c
al
(3)
The accepted value for this ratio is 4.186 J/cal.
5
The Circuit
Electrical Calorimeter
Procedure
1.
Connect up the circuit as shown above. Do not switch on the power supply yet. The heating coil
will quickly burn out if the current is switched on and it is not immersed in water.
2.
Weigh the inner calorimeter and record its mass in grams in the table be
low.
3.
Weigh the stirrer and thermometer separately and record their values.
4.
Fill the calorimeter with sufficient water from the cold water tap to just cover the heating coils
when immersed. Reweigh the calorimeter with the water and replace it in it
s insulating container. Place
the heating coil in the calorimeter.
5.
Set the variable resistor for maximum resistance. Stir the water in the calorimeter and measure
its initial temperature, T1 to the nearest 0.1 oC.
6.
Switch on the power supply and qui
ckly adjust the variable resistor so that the ammeter reads
2.0 A. Start the stopwatch. Take the voltmeter reading and record it in the table. Adjust the variable
resistor periodically as needed to maintain the current at 2.0 A.
7.
Read the voltmeter at t
he beginning and end of the heating period and record the values in the
table. Stir the water periodically until its temperature has risen between 6 and 8 oC.
8.
Turn off the power supply and stop the stopwatch. Record the heating time in seconds in the
t
able. Continue stirring the water gently for a few minutes until it reaches its maximum temperature.
Read this value to the nearest 0.1oC and record the temperature, T2, in the table.
9.
Repeat procedures (4) through (8) for a current of 2.5 A.
Mass o
f calorimeter empty, mc...................................................... ______ g
Specific heat of calorimeter, cc...................................................... ______ cal/goC
Mass of stirrer .................................................
.............................. ______ g
Specific heat of stirrer ................................................................... ______ cal/goC
Mass of thermometer .................................................................... ______ g
Specifi
c heat of glass .................................................................... ______ cal/goC
6
Specific heat of mercury (or alcohol) ........................................... ______ cal/goC
Heat capacity of stirrer, thermometer and heater, h .....
................ ______ cal/oC
CALCULATIONS
Current I = 2.0 A
Current I = 2.5 A
Mass of calorimeter + water (g)
Mass of water (g)
Initial Temperature of water, T1 (oC)
Final Temperature of water, T2 (oC)
Difference in Temperature (T2

T1) (oC)
Initial Voltage (V)
Final Voltage (V)
Average value of the mechanical equivalent of heat = ________ J/cal.
Percent error ________
Calculations:
1.Using Equation (1), calculate the electrical energy used in J.
2.Calculate the heat
energy produced in calories using Equation (2). The heat capacity of the stirrer,
thermometer, and immersion heater may be determined from the relation:
h = mass of stirrer x specific heat of stirrer + mass of the thermometer x specific heat of glass + mas
s of
the thermometer x specific heat of mercury (or alcohol) + the heat capacity of the immersion heater
(about 1 cal/oC).
3.Calculate the value of the mechanical equivalent of heat using Equation (3) and determine its average
value from the two sets of d
ata.
Result:
mechanical equivalent of heat (J). It is the ratio of work done to the transferred from a system during a
cycle and denoted by J
c
which is known as Joule’s Constant. It is only a unit conservation factor.
7
Experiment No.
2
Object:
To study
working of impulse and reaction steam turbine by models.
Apparatus:
Model of Impulse and Reaction steam turbines.
Theory:
Steam turbines:
The steam turbine is a prime mover in which the potential energy of steam is
transformed into kinetic energy and lat
er in its turn is transformed into the mechanical energy of the
rotation of the turbine shaft.
Classification of steam turbine:
With respect to the action of steam, turbine are clas
s
ified as:

Impulse turbine
Reaction turbine
1.
Impulse turbine:
It
is a turbine, which runs by the impulse of steam jet. In this turbine, the steam
is first made to flow through a nozzle. Then the steam jet impinges on the turbine blades with are
curved like bucket and are mounted on the circumference of the wheel. The st
eam jet after impinges
glide over the concave surface of blades and final
l
y leave the turbine.
The top portion of Impulse turbine exhibits a longitudinal section through the upper half, the middle
portion shows o
ne set of nozzle which is f
lowed by a ring
of moving blades, while lower part indicate
changes in pres and velocity during the flow of steam through the turbine. The principle equation of
this turbine is the wel
l known “De level” turbine.
8
Figure of Impulse Turbine
2.
Reaction turbine:
In a Reaction turbine, the steam enters the wheel under pressure and flow over
the blades. The steam while gliding proper the blades and then makes them to move. The turbine runner
is rotated by the reactive forces of steam jets.
In this, there is a grad
ual pres
s
ure drop takes place continuously over the fixed and moving blades. The
fuel of fixed blades is that they after al
l
ow it expand to a larger velocity as the steam pas
s
es over the
moving blades. Its K.E. is absorbed by them a three stage Reaction tu
rbine.
Compounding:
If the steam is expended from the boiler pres
s
ure in one stage the speed of rotor
becomes tremendously high which drop up practical complicacies. The
y
are several methods of
reducing this speed to lower value, al
l
these methods utiliz
ed a multiple system of rotor in series.
Keyed on a common shaft and the steam pressure or jet velocity is absorbed in stage as the steam flows
over the blades. This is known as
compounding:

1. Velocity compounding:
Steam is expanded through a stationary
nozzle from the boiler or inlet
pressure to condenser pressure. So the pressure in the nozzle drops, the K. E. of steam increase due to
9
increase in velocity. A portion of this available energy is absorbed by a row of moving blades. The
steam then flow thr
ough the second row of the blades which are fixed. They redirect the steam flow
without altering its
velocity to the f
lowing nearest row moving blades. Where again work is done on
them and steam with a low velocity from the turbine.
2. Pressure compoundi
ng:
In this rings of fixed nozzle incorporated between ring of moving blades.
The steam of boiler pressure enters the first set of nozzle and expands partial
l
y. The K.E. of steam thus
obtained in absorbed by the moving blades. The steam then expands partia
l
l
y in the second set of
nozzles whose its pressure again fal
l
s and the velocity increases. The K.E. thus obtained is observed by
the second ring of moving blades. This is repeated in stage 3 and steam final
l
y leaves the turbine at low
velocity and pressur
e.
3.
Pressure

Velocity compounding:
This method is the combination of velocity and pres
s
ure
compounding. The total drop in steam pres
s
ure is divided into stages and velocity obtained in each
stage is also compounded. The ring of nozzle, are fired at beg
inning of each stage and pressure remains
constant during each stage.
10
Figure of Reaction Turbine
VIVA

VOCE
1. State principle of working of Reaction steam turbine?
2. Define the term stage e
f
ficiency of a turbine?
3. Define the term vacuum e
f
ficiency of a turbine?
4. What is prime mover?
11
5. What is compounding?
Experiment No.
3
Object:
To calculate COP of
vapor
compression refrigeration system and to plot on T

s, p

H diagrams.
Apparatus:
REFRIGERATION
CYCLE TEST RIG.
Theory:
Introduction:
Now a daysRefrigeration & air conditioning system have become important due to the wide applications. The
Refrigeration effect is produced either by vapor absorption cycle or Vapor Compression Cycle. In these system
the refrigerants used ma
y be Ammonia, CO
2
, Freon gases.
Equipments
:
1.
COMPRESSOR
: Hermetically sealed type KirloskarMake 1/3 T. Capacity.
2.
AIR COOLED CONDENSOR
: Copper Coils with fins & cooling fan.
3.
EXPANSION DEVICE
:
i) Thermostatic expansion valve.
ii) Capillary Tube
4.
EVAPORATOR
: Copper Coil immersed in water. The evaporator is installed from
outside to prevent heat loss.
5.
ROTAMETER
: For measuring flow of Refrigerant Eureka Make.
Calibrated for Freon : 134a Gas
6.
ENERGY METER
: One each
for power supply to the Measurement of Compressor & Evaporator Heater.
7.
DIMMERSTAT
: To control power supply to heater.
8.
PRESSURE GAUGE
: One each for the measurement of high &low pressure.
9.
ELECTRIC HEATER : Immersion type 1.5 Kw
10.
DIGITAL TEMPERATURE INDIC
ATOR
: To Measure the temperatures
at various Points i.
e.
Evaporator Inlet & OutletCondenser Inlet & Outlet temperatures of Refrigerant & Evaporator bath
Temperature.
11.
SOLINOID VALVE
: Castle, Italy, Mate.
12
12.
HP &
LP CUTOUT
: Safety device suitable for the low &
high pressure of compressor.
13.
SERVICE
VALVE
: Needle type for changing gas.
14.
AMMETER
: for compressor
15.
VOLTMETER
: 230 V, A.C. for compressor
16.
FILTER DRYER
: 1 No.
17.
THERMOSTAT
: Safety Device.
18.
SWITCHES
: For various controls
The refrigerant circuit is mounted on a
board. The unit is supported on frame. Do not operate the changing
valve. See the sufficient water is filled in the evaporator container.
Operation Instruction:
Connect the two plugs to main. Before ON the supply, confIrm that all the switches on pan
el are off
position. See the dimmerstat is at zero position. Then put ON the heater switch & give power to
heater. This will heat the water in evaporator & this can be seen dial thermometer. Adjust the heater
voltage such that the Temperature dial ther
mometer reading reaches 25

30
0
C.
Now ON the D.P. switches. Put ON the condenser fan switch & wait for 2

3 minutes. Now switch
ON the solenoid valve switch & the compressor switch. The refrigeration flow will start. This can be
confirmed on the
Rotameter. Now the ammeter, voltmeter will show the current & voltage for
compressor. Note down the time for 10 revolutions of energy for compression.
After some time we will see that the Temperature of water in the evaporator slowly goes down &
reaches
steady state.( Adjust this temp. at 28 to 30
0
C.
After the steady state note down the readings as
follows:
1.
HP Condenser pressure in Kg/cm
2
.
=
Kg/cm
2
2.
LP Evaporator Pressure in Kg/Cm
2
=
Kg/cm
2
3.
Rotameter in Reading LPH
=
LPH
4.
Condenser Inlet Tem
perature in
0
C
= Tci
13
5.
Condenser Outlet Temperature
0
C
= Tco
6.
Evaporator Inlet Temperature in
0
C
= Tci
7.
Evaporator Outlet Temperature
0
C
= Teo
8.
Time for 10 Pulses of heater energymeter = in sec
9.
Time for 10 Pulses of comp energymeter = in sec.
10.
Evaporator Bath Temp in
0
C =
0
C.
Conclusion:
From the high pressure, & low pressure & the temperatu
re draw the refrigeration cycle
on the pressure enthalpy
chart. The cycle will be some what as shown :
3 2
1.
Compressor Power ( Wact ) =
Nc x 3600 P 4 1
tc x Emc x ( Comp ) Pressure
in
Where,
kg/cm
2
Nc
= 10
H
Tc
= Time for 10 Pulses in sec
Enthalpy in K J / kg
Emc( Comp ) = Energy Meter constant.
2.
Heater Power ( Nact) :
Nh x 3600
Th x Emc x ( Heater )
Where,
Nh
= 10
Th
= Time for 10
Pulses in sec
Emc( Heater ) = Energy Meter constant.
3.
C.O.P ( Act ) :
Power CosumedBy Heater
=
Power Consumed By Compressor
Compressor Power Consumed
=
Heater Power Consumed
14
Nact
=
Wact.
Heo
–
Hei
4.
Theoretical C.O.P =
Hci
–
Heo
Where,
Heo= Enthalpy of gas at evaporator outlet
Hei = Enthalpy of gas at evaporator inlet
Hco = Enthalpy of gas at condenser outlet
Above values are obtained for P.H. chart
5.
Carnot C.O.P
T
L
T
H

T
L
.
Where,
T
H
= Saturated Temperature of Condenser Pressure in ° K.
T
L
= Saturated Temperature of Evaporator Pressure in ° K.
6.
COP Relative
Actual C.O.P
Theoretical C.O.P
Observations:
1.
Compressor Energymeter
Constant ( E M C)com = 3200 Pulses/kwatthr
2.
Heater Energymeter Constant ( E M C) h = 3200 Pulses/kwatthr
15
Observation Table:
S r.
No
H P
PSI
L P
PSI
Q
LPH
Tci
oc
Tco
oc
Tei
oc
Teo
oc
Tbath
oc
Tc
Sec
Th
Sec
Expansion Valve
1.
140
32
22
49
39
9
26
27
27.6
10.3
Capillary Tube
2
142
40
25
43
39
11
25
27
26.0
10.1
HP : Condenser Pressure in PSI
L.P. : Evaporator Pressure in PSI
Q :
Refrigerant flow Rate in LPH
Tci : Refrigerant inlet Temp. To Condenser in
0
C
Tco : Refrigerant Outlet Temp. from condenser in
0
C
Tei : Refrigerant Inlet Temp to Evaporator in
0
C
Teo : Refrigerant Outlet Temp from Evaporator in
0
C
Tbath
: Evaporator Container Temp. in
0
C
tc : Time for 10 Pulses of Compressor energy meter in sec.
Th : Time for 10 Pulses of Heater Energy meter in sec.
Sample Calculation:
1.
Compressor Power ( Wact )
Nc x 3600
=
tc x Emc comp
10 x 3600
=
27.6 x 3200
= 0.407Kwatt.
16
2.
Heater Power ( Nact )
N h x 3600
=
10.3 x 3200
10 x 3600
=
10.3 x 3200
= 1.092
3.
Actual C. O. P. of the
System
Refrigeration Effect Produced
=
Actual Work Done
Power Consumed by Heater
=
Power Consumed by Compressor
Nact
1.092
=
=
Wact
0.407
= 2.68
4.
Carnot C.O.P
T
L
=
T
H

T
L
Where
T
L
= Saturated Temp. of Evaporator Pressure in
0
K
T
H
= Saturated Temp. ofCondensor Pressure In
0
K
Where
LP
= 32 PSI
= ( 32 X 0.0731 ) + 1.0133
= 3.263
Kg/cm
2
T
L
= ( 3
+ 273 )
From properties of R

134a
17
= 276
0
K
HP = 140 PSI
= ( 140 x 0.07031 ) + 1.0133
= 10.856
Kg/cm
2
T
H
= ( 43 +273 ) From Properties of R

134a
= 316
0
K
276
Cornot C. O. P.=
( 316
–
276)
=
6.90
5.
Efficiency of the Plant (
)
Actual C.O.P.
=
Carnot C.O.P
2.68
=
X 100
6.90
=
38.84 %
6.
Theoretical C
. O. P.
Heo

Hei
=
Hci

Heo
413.1

212.2
=
423.5

413.1
= 19.31
7.
Relative C.O.P.
Actual C.O.P.
=
Theoretical c.o.p.
18
2.68
=
19.31
= 0.138
8.
Ton of Refrigeration ( T )
Refrigeration Effect Produced in K J/Min
=
210
=
N act X 60
210
= 1.092
x 60
210
=
0.312
ton of refrigerant
Conclusion:
The actual C.O.P. is less than the theoretical due to the losses at different points also the temperature are
measured at
somewhat
different positions.
Precautions:
1.
Do not operate the charging valve
2.
Before switching ON supply see that all the switches o
n
the panel board are off & the
dimmer stat is at
‘Zero’ position
3.
The
evaporator contains
adequate water.
4.
Proper earthing is provided.
5.
In case supply fails while the trial is going on. Off the switches & also dimmer stat restore before
switching ON confirm
that the H.P. & L.P. pressure gauges show equal readings. If there is difference of
pressure wait till the gauge pressure are equal.
6.
Stabilised supply for the equipment is necessary for better performance & the life of the compressor
19
Experiment No. 4
Object:
To study working of Gas turbines by models and to identify various processes of Brayton Cycle.
Theory:
A simple gas turbine power plant is shown in Fig.
5.1.
Air is first compressed adiabatically in process
a

b, it then
enters the combustion chamber where fuel is injected and burned essentially at constant pressure in process b

c,
and then the products of combustion expand in the turbine to the ambient pressure in process c

d and are thrown
out to the
surrou
ndings. The cycle is open. The state diagram on the p

v coordinates is shown in fig.
5.2.
Open
cycles are used in aircraft, automotive (buses and trucks) and industrial gas turbine installation.
20
The Brayton cycle is the air standard cycle for the gas
turbine power plant. Here air is first compressed
reversibly and adiabatically, heat is added to it
reversibly at constant pressure, air expands in the
turbine reversibly and adiabatically, and heat
is then rejected from the air reversibly at constant
pres
sure to bring it to the initial state. The Brayton cycle, therefore, consists of:
Two reversibly isobars and two reversible adiabatics.
21
The flow, p

v, and T

s diagrams are shown in fig.
5.3.
For m kg of air
(
)
(
)
Cycle efficiency,
Now
(
)
(
)
⁄
(
)
22
o
r
(
)
(
)
⁄
(
)
(
)
If
= compression ration =
the efficiency becomes
(
)
(
)
o
r
If
= pressure ration =
the efficiency may be expressed in the following form also
(
)
(
)
⁄
or
(
)
(
)
⁄
The
efficiency of the Brayton cycle, therefore, depends upon either the compression ratio or the
pressure ration. For the same compression ratio, the Brayton cycle efficiency is equal to the Otto cycle
efficiency.
A closed cycle gas turbine plant
(fig
. 5.3
)
is used in a gas

cooled nuclear reactor plant, where the source
is a high temperature gas

cooled reactor (HTGR) supplying heat from nuclear fission directly to the
working fluid (a gas).
23
Experiment No.5
AIM

To plot specific fuel consumption
versus rpm diagrams for diesel and petrol
engines
.
Apparatus

Four

Stroke Single

Cylinder Diesel Engine & Petrol Engine, Stop Watch, and Digital
Tachometer,
THEORY:

Under some circumstances (i.e Electric Generator) C. I. Engines are required to run at
constant
speed. For this purpose the test is to be performed at constant speed and the load is varied from zero to
maximum. When load on the engine increases its speed decreases. Accordingly the fuel supply is
adjusted to keep the engine speed constant. C
orresponding to each load setting, dynamometer readings
and fuel consumption rate are measured. The BP, BSFC, BMEP, A/F, and Mechanical Efficiency are
calculated from measured data and plotted against the load.
FORMULE USED:

(i) Torque,
T = 9.81
x W x R Effective
N

m.
Where R Effective = (D + d)/ 2 or (D + tBelt)/ 2 m,
W (Load) = ( S1

S2)Kg,
(ii) Brake Power,
B P = ( 2πN T ) / 60, 000
KW
Where N = rpm,
T = Torque N

m,
(iii) Fuel Consumption,
m f = ( 50 ml x 10

6 x ρ Fuel ) / (
t )
Kg/Sec.
24
Here; 1 ml = 10

3 liters, and 1000 liters = 1 m3
So, 1 ml = 10

6 m3
(iv) Brake Mean Effective Pressure,
BMEP =
(BP x 60,000)/ ( L Stroke x A x
N’)
N/ m2
Where L Stroke = Stroke m, A (Cross Section of the Cylinder)
= (π D2Bore)/
4
m2,
N’ (Number of Power Strokes/ min.) = N/ 2 per min. ; For Four

Stroke Engine.= N per min;
For Two

Stroke Engine., and N = rpm.
(v) Brake Specific Fuel Consumption,
BSFC = ( m f x 3600 ) / B P
Kg/ KW . hr (vi) Mass of
the Air,
m Air = Cd Ao √2 g Δh
ρ Air ρ Water
Kg/ Sec
Where Cd ( Co

efficient of Discharge ) = 0.6,
ρ Air
=
(
Pa x 102 ) / ( R x Ta ) Kg/ m3 A2
o ( Area of Orifice ) = (π do )/ 4 m2, Pa = 1.01325 Bar,
R =
0.287 KJ/Kg . K, Ta = ( ta + 273
) K, ta = Ambient Temperature
C
(vii) Air Fuel Ratio,
A/F = ( m Air / m f )
Kg/ Kg of Fuel
(viii) Mechanical
Efficiency,
η mechanical = BP / IP
PROCEDURE:

1. Before starting the engine check the fuel supply, lubri
cation oil, and availability of
cooling water.
2. Set the dynamometer to zero load.
3. Run the engine till it attains the working temperature and steady state condition.
4. Adjust the dynamometer load to obtain the desired engine speed. Note down the fuel consumption
rate.
5. Change the dynamo
meter load so that the engine speed Change, to maintain the engine speed
constant fuel consumption increases.
25
6. Note down the fuel consumption rate, speed, air inlet temperature, at this load setting.
7. Repeat steps 5 and 6 for various loads.
8.
Disengage the dynamometer and stop the engine.
9. Do the necessary calculation.
OBSERVATIONS:

Engine Speed, N
= 1500 rpm
No. of Cylinders, n
= Single
Bore Diameter, Dbore
= m
Stroke Length, Lstroke
= m
Calorific Value of Fuel, C.V
. = 38,000 KJ/Kg
Gas Constant, R = 0.287 KJ/Kg . K
Ambient Temperature, ta
= oC
Atmospheric Pressure, Pa
= 1.01325 Bar
Orifice Diameter, do
= 25 x 10

3 m
Co

efficient of Discharge, Cd
= 0.6
Specific Gravity of fuel, ρfuel
= 810 to 910
Kg/m3
Density of Water, ρwater
= 1,000 Kg/m3
Brake Drum Diameter, D
= 181.5 x 10

3 m
Rope Diameter, d
=
m
or
Belt thickness, t Belt
= 5.5 x 10

3 m
OBSRVATIONS:

S.No.
Engine Speed N
(rpm)
Dynamometer spring balance
Time taken
t(sec)
Manometer
fuel H (m)
S1 (kg)
S2 (kg)
1
2
3
4
26
CALCULATIONS:

S.No.
N
(rpm)
Torque
(N

m)
BP
(kW)
Mair
(kg/hr)
Mf
(kg/hr)
BSFC
(kg/kW.hr)
BMEP
(N/m2)
A/f
Ratio
Mech.
Effe. %
1
2
3
4
RESULT :

Performance curves are plotted & they are similar to the standard performance curves.
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