Instruction Manual Manual de Instrucción Manual de Instruções
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1
THERMAL CONDUCTIVITY BY GUARDED PLATE METHOD
AIM:
To find the thermal conductivity of the specimen by two slab guarded hot
plate method.
APPARATUS REQUIRED:
1. Ammeter
2. Voltmeter
3. Thermocouple
4. Temperature indicator
SPECIFICATIONS:
Thickness of the specimen plate = 0.005 m
Specimen diameter = 0.140 m
Area A = 0.0153 m2
FORMULA USED:
Heat transferred through the specimen
dX
dTKAQ
Where,
Q – Heat transfer rate, w
K – Thermal conductivity of the specimen plate, W/mK
A – Surface area of the test plate, m2
dT – Temperature drop across the specimen, K
dX – Thickness of the specimen = 0.005m
005.0
34 TT
dX
dT
dT
dx
A
QK , W/mK
2
PROCEDURE:
1. Connect the three pin plug to the 230 v, 50 Hz, 15 amps main supply and
switch on the unit.
2. Turn the regulator knob clockwise, set the heat input by fixing the voltmeter
and ammeter readings and note down the heat input Q in the table.
3. Adjust the regulator for guard heater so that the main heater temperature is
less than that of the guard heater temperature.
4. Allow water through the cold plate at a steady rate
5. Allow the unit to attain the steady state condition.
6. When the steady state condition is reached note down the temperature
indicated in the temperature indicators.
7. In the temperature indicator, the temperatures T1, T6 represents the cold plate
temperature, T2, T5 represents the main heater temperature T3, T4 represents
the guard heater temperature T7, T8 represents the water temperature. These
values are noted in the table.
8. Calculate the thermal conductivity of the given specimen by using the given
formula and note the value in the table.
9. Repeat the experiment from step 2 to step 8 by varying the heat input to the
system.
3
The
rma
l
Con
du
ctivity
of
spe
cim
en
K
W/m
K
dT
/dX
Wa
ter
Tem
p
˚C T
8
T7
Gu
ard
he
ate
r
Tem
p
˚C
T4
T3
Ma
in h
ea
ter
Tem
p
˚C
T5
T2
Cold
pla
te
Tem
p
˚C T
6
T1
Q=
VI
Wa
tts
Am
me
ter
Rea
din
g
(am
ps)
I
Vo
ltm
ete
r
Rea
din
g
(va
olts)
V
S
No
TA
BL
E:
4
Figure 1. Two slab guarded hot plate
5
RESULT:
Thus the thermal conductivity of the given specimen was calculated.
6
THERMAL CONDUCTIVITY OF PIPE INSULATION USING LAGGED PIPE
APPARATUS
AIM:
To determine the thermal conductivity of the given insulating material by
using lagged pipe apparatus.
APPARATUS REQUIRED:
1. Ammeter
2. Voltmeter
3. Thermocouple
4. Temperature indicator
SPECIFICATIONS:
1. Heater diameter, d1 = 0.02m
2. Heater with asbestos diameter, d2 = 0.04m
3. Heater with asbestos + sawdust diameter, d3 = 0.08m
4. Length, L = 0.50m
FORMULA USED:
Heat transfer rate,
)ln(
)(2
)ln(
)(2
2
3
2
1
2
1
r
r
TLK
r
r
TLK
Q
Where, Q – Heat transfer rate, watts
K1 – Thermal conductivity of asbestos in W/mK
K2 – Thermal conductivity of sawdust in W/mK
L – Length of the pipe, 0. 5 m
ΔT– Temperature difference in K
r1 – Heater radius, 0.01m
r2 – Heater with asbestos, 0.02m
r3 – Radius with asbestos and sawdust, 0.04m
7
Thermal conductivity of asbestos
)(2
ln1
2
1TL
r
rQ
K
Where,
ΔT = Tavg (Heater) – Tavg (Asbestos)
Thermal conductivity of sawdust
)(2
ln2
3
1TL
r
rQ
K
Where,
ΔT = Tavg (Asbestos) – Tavg (Sawdust)
8
PROCEDURE:
1. Connect the three pin plug to the 230 v, 50 Hz, 15 amps main supply and
switch on the unit.
2. Turn the regulator knob clockwise, set the heat input by fixing the voltmeter
and ammeter readings and note down the heat input Q in the table.
3. Allow the unit to attain the steady state condition.
4. When the steady state condition is reached note down the temperature
indicated by the temperature indicators.
5. In the temperature indicator, the temperatures T1, T2, T3 represents the
temperature of the heater, T4, T5, T6 represents the temperature of the
asbestos and T7, T8 represents the temperature of the sawdust lagging by
using the multipoint digital temperature indicator. These values are noted in
the table.
6. Calculate K1 (Thermal conductivity of asbestos) and K2 (Thermal conductivity
of asbestos), by using the given formula and note the value in the table.
7. Repeat the experiment from step 2 to step 6 by varying the heat input to the
system.
9
Sa
w d
ust
K2
W/m
k
Asb
esto
s
K1
W/m
k
Sa
wdu
st
Tem
p
˚C
Tavg
T8
T7
Asb
esto
s T
em
p
˚C
Tavg
T6
T5
T4
Hea
ter
Tem
p.
˚C
Tavg
T3
T2
T1
Q
Wa
tts
Am
me
ter
Rea
din
gs
Am
ps
A
Vo
ltm
ete
r
Rea
din
g
s
Vo
lts
V
S
No
TA
BL
E:
10
Figure 2. Lagged pipe apparatus
11
RESULT:
Thus the thermal conductivity of the given insulating material (Asbestos
and Saw dust) has been calculated.
12
NATURAL CONVECTION HEAT TRANSFER FROM A VERTICAL CYLINDER
AIM:
To determine the actual heat transfer co-efficient and theoretical heat
transfer coefficient by natural convection.
APPARATUS REQUIRED:
1. Voltmeter
2. Ammeter
3. Thermocouple
4. Heater
5. Temperature indicator
SPECIFICATION:
Length of the rod, L = 0.50m
Diameter of the rod, D = 0.02m
Thermal conductivity of air at mean film temperature, (Tf), K
Area of the rod, A =DL = 0.0314 m2
FORMULA USED:
Theoretical heat transfer co-efficient (htheoretical)
For laminar flow
Nu = hL /k = 0.59(GrPr) 0.25 for 10 4<GrPr<10 9
For turbulence flow
Nu = hL /k = 0.10(GrPr) 0.33 for 10 9<GrPr<10 12
Where,
Nu - Nusselt Number
h - Heat transfer coefficient, W/m2 K
k – Thermal conductivity of airm W/mK
13
Grashoff number, 2
3
TlgGr
Where,
g – Acceleration due to gravity, 9.81 m/s2
β – Co-efficient of expansion, 273
1
fT
2
TTT s
f
Ts - Surface temperature in ˚C
Tα - Air temperature in ˚C
l – Length = 0.5m
ΔT – Ts - Tα, K
γ – Kinematic Viscosity at mean film temperature (Tf) from HMT data
book
Pr – Prandtl number
Actual heat transfer co-efficient (hact)
TAhQ act
Where,
Q – Heat transfer rate = VI, watts
` hact – Actual heat transfer co-efficient, W/m2K
A = Surface area of the heater = DL = 0.0314 m2
T = Ts - Tα
4
5432 TTTTT
2
61 TTT
Where,
Tω = Surface temperature in ˚C
Tα = Air temperature in ˚C
14
PROCEDURE:
1. Connect the three pin plug to the 230 v, 50 Hz, 15 amps main supply and
switch on the unit.
2. Turn the regulator knob clockwise, set the heat input by fixing the voltmeter
and ammeter readings and note down the heat input Q in the table.
3. Keep on the temperature indicator switch in the first position
4. Allow the unit to attain the steady state condition.
5. When the steady state condition is reached note down the temperature
indicated by the temperature indicators
6. In the temperature indicator, T2, T3, T4 & T5 represents the temperature of the
heater at different points. T1 represent the inlet temperature of the air and T6
represents the outlet temperature of the air. These values are noted in the
table.
7. Calculate the theoretical heat transfer coefficient (h theoretical) and actual heat
transfer coefficient (h actual) by using the given formulas.
8. Repeat the experiment from step2 to step 7 by varying the heat input to the
system.
15
hA
ctu
al
(W/m
2K
)
hT
heore
tica
l
(W/m
2K
)
Ou
tle
t
Tem
p o
f
Air
˚C
T6
Inle
t
Tem
p o
f
Air
˚C
T1
Hea
ter
tem
pe
ratu
re
˚C
T5
T4
T3
T2
Q
Wa
tts
Am
me
ter
Rea
din
g
Am
ps
I
Vo
ltm
ete
r
Rea
din
g
Vo
lts
V
S.
No
TA
BL
E:
16
Figure 3: Natural convection apparatus
17
RESULT: The theoretical and actual heat transfer coefficient has been calculated by
using natural convection apparatus
18
FORCED CONVECTION INSIDE TUBE
AIM:
To determine the actual heat transfer and theoretical heat transfer
coefficient using forced convection.
APPARATUS REQUIRED:
1. Voltmeter
2. Ammeter
3. Thermocouple
4. Temperature indicator
5. Blower
6. Manometer
SPECIFICATION:
Diameter of the pipe, d1 – 0.04m
Diameter of the orifice, d2 – 0.02m
Length of the pipe, L – 0.5m
FORMULA USED:
Actual heat transfer co-efficient,
hactual =TA
Q
, w/m2k
Where,
Q - Heat input rate= V x I, Watts
A – Surface area of the pipe = πDL = 0.62 m2
TTT s ˚ C
Ts – Wall temperature, ˚ C
19
4
4321 TTTTTs
Tα – Air temperature, ˚ C
2
65 TTT
Theoretical heat transfer co-efficient, htheoretical
K
hDNu , W/m2k
Where,
Nu – Nusselt number
h – Theoretical heat transfer co-efficient, w/m2k
d – Diameter of pipe, m
k –Thermal conductivity of air at Tf , w/mK (From HMT data book)
Air flow head,
1210
a
whhh
, m
Where,
h1, h2 = Manometer readings, m
ρw = Density of water, 1000 kg/m3
ρa = Density of air, 1.1465 kg/m3
Volume flow of air, 2
2
2
1
21 2...
aa
ghaaCdQv
o
, m3/sec
Where,
Cd = Co-efficient of discharge
a2 = Area of orifice, m2 = 2
24
d
= 3.14 x 10-4 m2
a1 = Area of pipe, m2 = 2
14
d
= 1.25 x 10-3 m2
20
Velocity of air, V = A
QV, m/ sec
Where,
A – Area of pipe, = 2
14
d
= 1.25 m2
Reynolds Number (Re) =
1Vd
Where,
d1 – Diameter of pipe, m
V – Velocity of air, m/sec
- Kinematics viscosity at Tf, m2/sec (From HMT data book)
Nusselt number (NU) = 0.023 (Re) 0.8 (Pr) 0.4
Where,
Pr – prandtl number for air at Tf, m2/sec (From HMT data
book)
Re – Reynolds number
D
KNuhthe
21
PROCEDURE:
1. Connect the three pin plug to the 230 v, 50 Hz, 15 amps main supply and
switch on the unit.
2. Turn the regulator knob clockwise, set the heat input by fixing the voltmeter
and ammeter readings and note down the heat input Q in the table.
3. Keep on the temperature indicator switch in the first position
4. Allow the unit to attain the steady state condition.
5. When the steady state condition is reached note down the temperature
indicated by the temperature indicators
6. In the temperature indicator, T1, T2, T3 & T4 represents the temperature of the
heater at different points. T5 represent the inlet in let temperature of the air
and T6 represents the outlet temperature of the air h1&h2 are the manometer
reading. These values are noted in the table.
7. Calculate the theoretical heat transfer coefficient (h theoretical) and actual heat
transfer coefficient (h actual) for forced convection by using the given formulas.
8. Repeat the experiment from step3 to step7 by varying the heat input to the
system.
22
23
h
W/m
2K
hac
hth
e
Ma
no
me
ter
Rea
din
g
m
h2
h1
Ou
tle
t
Tem
p o
f
Air
˚C
T6
Inle
t
Tem
p o
f
Air
˚C
T5
Hea
ter
tem
pe
ratu
re
(˚C
)
T3
T3
T2
T1
Q
Am
me
ter
Rea
din
g
Am
ps
A
Vo
ltm
ete
r
Rea
din
g
Vo
lts
V
S.N
o
TA
BL
E:
24
Figure 4. Forced convection apparatus
25
RESULT:
The theoretical and actual heat transfer coefficient has been calculated
using forced convection apparatus
26
HEAT TRANSFER FROM PIN-FIN APPARATUS
AIM:
To determine the temperature distribution of a PIN-FIN for forced
convection and to find the FIN efficiency.
APPARATUS REQUIRED:
1. Ammeter
2. Voltmeter
3. Heater
4. Blower
5. Fin specimen
6. Thermocouple
7. Temperature indicator
SPECIFICATION:
Duct width, B = 0.150m
Duct height , W = 0.100m
Orifice diameter, do = 0.020m
Orifice coefficient, Cd = 0.6
Fin length , L = 0.145m
Fin diameter, Df = 0.012m
FORMULA USED:
1. Surface temperature CTTTTTTT
Ts o
7
7654321
2. Ambient temperature, Tα = T8 , ˚C
3. Mean film temperature 2
TTsT f , ˚C
27
4. Volume flow rate, ghaACQ sd 2. , sec
3m
Where,
Cd = co-efficient of discharge, 0.6
As = Orifice area = 2
4d
= 1.25 x 10
-3, m
2
ha = Drop in manometric head, m
hha
wa
ρw – Density of water, 1000 kg/m3
ρw – Density of air, 1.14 kg/m3
h – Manometer differential head = h1 - h2, m
5. Velocity of air , BW
QV
. , m/sec
Where,
W = Width, m
B = Breadth, m
6. Reynolds number,
fVdRe
Where,
V = Velocity, m/sec
df = Diameter of fin, m
= Kinematic viscosity at Tf, m2/sec (From HMT data book)
8. Nusselt number, 333.0333.0PrRe989.0 Nu , for 1< Re < 4
333.0385.0PrRe911.0 Nu , for 4< Re < 40
333.0466.0PrRe683.0 Nu , for 40< Re < 4000
333.0618.0PrRe913.0 Nu , for 4000< Re < 40000
333.0805.0PrRe0266.0 Nu , for Re > 40000
Where,
Pr = Prandtl number at Tf (From HMT data book)
28
8. Heat transfer coefficient, fD
kNuh
. , w/m2K
Where,
K =Thermal conductivity at Tf, w/mK
Df = Diameter of the fin, m
9. Fin efficiency, %100)tanh(
mL
mLfin
Where,
kAhPm
p = Perimeter = πDf = 0.0376 m
A = Surface area of the pin fin = πDfL = 5.27 x 10-3 m2
L = Length of the pin fin, m
29
PROCEDURE:
1. Connect the three pin plug to the 230v, 50 Hz, 15 Amps main supply and
switch on the unit.
2. Turn the regulator knob clockwise, set the heat input by fixing the voltmeter
and ammeter readings and note down the heat input Q in the table.
3. Keep the thermocouple selectors switch in first position.
4. Allow the unit to attain the steady state condition.
5. Now switch ON the blower.
6. Set the air flow rate to the system by keeping the valve in 1/4th position.
7. The difference in U tube manometer limb levels h1, h2 is noted in the table.
8. Note down the temperatures by temperature indicator.
9. In the temperature indicator, T1, T2, T3, T4, T5, T6 and T7 represent the
temperature of the fin surface. These values are noted in the table and Tavg is
calculated.
10. Also note down the atmospheric temperature T8 in the table by using
temperature indicator.
11. Thus the fin efficiency is calculated using the given formula.
12. Repeat the experiment from step 2 to step 11 by varying the air flow rate to
1/2, 3/4, and fully opened position.
13. Tabulate the readings and calculate for different conditions.
.
30
TABLE:
S. No Valve
position
Manometer readings
Fin surface Temperature
C
Ambient temperature
C
Efficiency
%
h1 h2 T1 T2 T3 T4 T5 T6 T7 Tavg T8
1 1/4
2 2/4
3 3/4
4 Full open
31
Figure 5. PIN – FIN appararus
32
RESULT:
Thus the temperature distribution is determined and the fin efficiency is
tabulated.
33
DETERMINATION OF STEFAN-BOLTZMAN CONSTANT
AIM
To find out the Stefan-Boltzman constant using concentric hemisphere.
APPARATUS REQUIRED
1. Voltmeter
2. Ammeter
3. Thermocouple
4. Heater
5. Temperature indicator
SPECIFICATION
Mass of the disc, m = 0.005 kg
Diameter of the disc, d = 0.025m
Material of the disc = copper
Disc weight = 0.008 kg
Specific heat, Cp = 0.381 Kj/kgk
FORMULA USED
Radiation heat transfer
44
dh TTAQ
44
dh TTA
Q
Where,
- Stefean boltzman constant W/m2K4
- Emissivity of the black body = 1
TmCQ P
m - Mass of the disc, kg
34
Cp – Specific heat of copper = 0.381Kj/KgK
T – dT/dt
dT – Change in temperature, ˚C
dt – Change in time, sec
A - Area of disc, = 2
4d
=4.9 x 10-6 m2
Th - Average temperature of hemisphere, K
5
54321 TTTTTTh
Td - Temperature of disc, K
PROCEDURE
1. Allow the water to flow through the heater unit and through the hemisphere
2. Remove the disc from the bottom of hemisphere.
3. Switch on the heater and allow the hemisphere to reach steady state
temperature.
4. Note down the temperatures T1, T2, T3, T4, and T5 from the temperature
indicator and also note the steady state temperature of the disc T6 (Td).
These values are noted in the table.
5. The average of T1, T2, T3, T4, and T5 is hemisphere temperature.
6. Close the disc from the bottom of the hemisphere.
7. Allow the unit to attain steady state.
8. When the steady state is reached note down the temperature in the table.
9. Calculate the Stefan - boltzman constant by using the given formula.
10. Repeat the experiment from step 3 to step 9 by changing the heat input to the
system.
35
36
TABLE:
Sl.No.
Hemisphere Temperature oC
Average Temp of hemisphere
oC
Time
Steady state
Temp of disc (Td)
oC
Stefen boltzman constant,
σ W/m2K4
T6 T1 T2 T3 T4 T5 Th (secs)
37
Figure 6. Stefan-Boltzman apparatus
38
RESULT Thus the Stefan Boltzman constant of the given concentric hemisphere is calculated.
39
DETERMINATION OF EMMISIVITY OF A GREY SURFACE
AIM
To measure the emissivity of the given test plate surface.
APPARATUS REQUIRED
1. Ammeter
2. Voltmeter
3. Heater
4. Test plate
5. Black body
6. Thermocouple
7. Temperature indicator
SPECIFICATION
Diameter of the test plate = 0.150 m
Diameter of the black plate = 0.150 m
FORMULA USED
Emissivity of the test plate,
44
44
CP
CBBB
TT
TTEE
Where,
Emmisivity of black body, EB = 1
Average temperature of block body, Tab = Tb(Avg) + 273, K
Average temperature of polished body, Tpa = Tp(Avg) + 273, K
Temperature of the chamber, TC = T7 + 273, K
40
PROCEDURE
1. Connect the three pin plug to the 230V, 50Hz, 15 amps main supply and
switch on the Unit
2. Turn the regulator knob clockwise, set the heat input by fixing the voltmeter
and ammeter readings and note down the heat input Q in the table.
3. Keep the thermocouple selectors switch in first position.
4. Keep the toggle switch in position 1. By operating the energy regulators
power will be fed back to black plate.
5. Now keep the toggle switch in position 2 and operate the regulator 2 and feed
power to the test surface.
6. Allow the unit to stabilize.
7. Make sure that the power inputs to the black and test surface are set at equal
values.
8. Turn the thermocouple selector switch clockwise step by step and note down
the temperatures indicated by the temperature indicator.
9. In the temperature indicator the temperatures T1, T2, T3 represents the
polished body temperature, T4, T5, T6 represents the black body temperature
and T7 represents the chamber temperature. These values are noted in the
table.
10. Calculate the emmisivity by using the given formula.
11. Repeat the experiment from step 2 to step 10 by changing the heat input to
the system.
41
Em
issiv
ity
E
Ch
am
ber
Tem
p
˚C
T7
A
ve
rag
e
Tem
p
˚C
Tb(A
vg)
Bla
ck B
ody
Tem
p
˚C
T6
T5
T4
Ave
rag
e
Tem
p
˚C
Tp(A
vg)
Po
lish
ed
Bo
dy
Tem
p
˚C
T3
T2
T1
Q=
VI
Wa
tts
Am
mete
r
Re
ad
ing
Am
ps
I
Vo
ltm
ete
r
Re
ad
ing
Vo
lts
V
S.
No
TA
BL
E:
42
Figure 7. Emmisivity apparatus
43
RESULT The emissivity of the given polished plate was found out and it is tabulated.
44
EFFECTIVENESS OF PARALLEL AND COUNTER FLOW HEAT
EXCHANGER
AIM:
To find the overall heat transfer co-efficient and the effectiveness in
parallel flow and counter flow heat exchanger.
APPARATUS REQUIRED:
1. Heat Exchanger Apparatus
2. Temperature indicator
3. Thermocouple
4. Stopwatch
5. Water heater
SPECIFICATION:
Inner copper tube
Inner diameter, d1 = 0.012m
Outer diameter, d2 = 0.015m
Outer GI tube
Inner diameter, d3 = 0.04m
FORMULA REQUIRED:
1. Parallel flow
Heat ttransfer rate, LMTDUAQ
Overall heat transfer co-efficient, ).(LMTDA
QU , W/m2K
Where,
oiphh ThThCmQ . , W
mh – Mass of hot water, kg
45
Cph – Specific heat of hot water = 4.186 Kj/kgK
A – Outer area of inner copper tube = Ld2 = 0.025 m2
LMTD – Logarithmic Mean Temperature difference
0
0
ln
)(
T
T
TTLMTD
i
i
P
ΔTi = Thi - Tci
ΔTo = Tho - Tco
Thi – Hot water inlet temperature, K
Tci – Cold water inlet temperature, K
Tho – Hot water outlet temperature, K
Tco – Cold water outlet temperature, K
2. Counter flow
0
0
ln
)(
T
T
TTLMTD
i
i
C
ΔTi = Thi – Tco
ΔTo = Tho – Tci
Thi – Hot water inlet temperature, K
Tci – Cold water inlet temperature, K
Tho – Hot water outlet temperature, K
Tco – Cold water outlet temperature, K
3. Effectiveness of heat transfer,
For parallel flow,
ii
i
TcTh
ThTh
0
For counter flow,
ii
io
TcTh
TcTc
46
PROCEDURE:
1. Connect water supply at the back of the unit. The inlet water flows through the
geyser and inner pipe of the heat exchanger and flows out.
2. Also the inlet water flows through the annulus gap of the heat exchanger and
flows out.
3. For parallel flow open valve V2, V4 and V5.
4. Control the hot water flow approximately 2lit./min and cold water flow
approximately 5 lit./min.
5. Switch ON the geyser. Allow the temperature to reach steady state.
6. Note temperature T1 and T2 (hot water inlet and outlet temperature
respectively) in the table.
7. Under parallel flow condition T3 is the cold water inlet temperature and T4 is
the cold water outlet temperature. Note the temperature T3 and T4 in the
table.
8. Under counter flow condition T4 is the cold water inlet temperature T3 is the
cold water outlet temperature. Note the temperatures T3 andT4 in the table.
9. Note the time for 1 litre flow of hot and cold water and calculate the mass flow
rate by using the given formula.
10. For counter flow open valve V3, V1 and V5 and repeat the experiment from
step 5 to step 9 and calculate the mass flow rate by using t he given formula.
47
TABLE:
For Parallel Flow:
S No
Hot water, oC Cold water, oC Time taken
for 1 lit. of hot
water flow
(sec)
Time taken for
1 lit. of cold
Water flow
(sec)
Inlet,
Thi
Outlet,
Tho
Inlet,
Tci
Outlet,
Tco
T1 T2 T3 T4
For Counter Flow:
S
No
Hot water, oC Cold water, oC Time taken
for 1 lit. of
hot water
flow
(sec)
Time taken for
1 lit. of cold
Water flow
(sec)
Inlet,
Thi
Outlet,
Tho
Inlet,
Tci
Outlet,
Tco
T1 T2 T3 T4
48
Figure 8. Parallel flow and Counter flow heat exchanger
49
RESULT:
Thus the test on parallel and counter flow heat exchanger is performed
and the overall heat transfer co-efficient and the effectiveness of the heat
exchanger are determined.
50
DETERMINATION OF COP OF A REFRIGERATION SYSTEM
AIM:
To conduct the test on refrigeration test rig with isobutene and propane as
a refrigerant to determine the Coefficient of Performance (COP).
APPARATUS REQUIRED:
1. Refrigeration test rig
2. Thermometer
3. Stopwatch
FORMULA:
1. Co-efficient of performance (Actual)
if
fiw
actualEE
TTQ
inputEnergy
effectnfrigeratioPOC
860
Re)..(
55
if
fiw
EE
TTQ
860
3600
186.455
Where,
C.O.P – Co-efficient of performance
QW – Weight of water in evaporator, kg
T5i – Initial temperature of water, ˚C
T5f – Final temperature of water, ˚C
Ei – Initial energymeter reading, Kwhr
Ef – Final energymeter reading, Kwhr
2. Co-efficient of performance (Theoretical)
From p-h diagram of
Propane – R290 – Chart a
Iso butane – R600a – Chart b
51
(i). Stage1 – Compressor Inlet / Evaporator outlet
2
111
hbhah
Where,
h1 – Enthalpy at temperature T1, Kj/kg
ha1- From propane pressure – enthalpy chart for temperature T1, &
pressure P1
hb1- From iso-butane pressure – enthalpy chart for temperature T1,
& pressure P1
(ii). Stage2 – Compressor outlet
2
222
hbhah
Where,
h2 – Enthalpy at temperature T2, Kj/kg
ha2- From propane pressure – enthalpy chart for temperature T2, &
pressure P2
hb2- From iso-butane pressure – enthalpy chart for temperature T2,
& pressure P2
(iii). Stage3 – Evaporator inlet (Before throttling)
2
333
hbhah
Where,
h3 – Enthalpy at temperature T3, Kj/kg
ha3- From propane pressure – enthalpy chart for temperature T3, &
pressure P3
hb3- From iso-butane pressure – enthalpy chart for temperature T3,
& pressure P3
52
(iv). Stage4 – Evaporator inlet (After throttling)
h3 = h4
Where,
h4 – Enthalpy at temperature T4, Kj/kg
12
41..hh
hhPOC lTheoretica
3. Co-efficient of performance (Relative)
ltheoretica
actualrelative
POC
POCPOC
..
....
PROCEDURE:
1. Fill up the evaporator with known quantity of water.
2. After 5 min, when the system attains steady state the initial energy meter
reading is noted and also the water temperature in the evaporator is
noted.
3. After a known period of time say 30 min, note down the energymeter
reading and water temperature. Before noting the water temperature,
physically stir the water to ensure that uniform temperature is attained in
the evaporator.
4. Note down the pressure (P1, P2, P3 and P4), temperature (T1, T2, T3 and
T4) also temperature of water T5 and energymeter reading in the table.
5. Calculate actual COP, theoretical COP and relative COP by using the
given formula.
6. Repeat the experiment from step 3 to step 5 and the readings are noted in
the table.
53
En
erg
ym
ete
r
rea
din
g
Kw
hr
E
Tem
pe
ratu
re
of
wate
r
˚C
T5
Tem
pe
ratu
re
˚C
T4
T3
T2
T1
Pre
ssu
re,
Psig
P4
P3
P2
P1
Tim
e t
ake
n
(min
)
t
S N
o
T
AB
LE
:
1
ba
r =
0.0
694
76
Psig
54
P1 – Compressor inlet/Evaporator outlet
P2 – Compressor outlet/Condenser inlet
P3 – Evaporator inlet (Before throttling)
P4 – Evaporator inlet (After throttling)
V – Compressor changing valve
Figure 9. Refrigeration test rig
Rota
met
er
Fil
ter
Liquid received
stage cylinder
Condenser
Compressor Evaporator
P4
P3
P2
P1
55
RESULT:
Thus, the performance on a refrigeration test rig with isobutene and
propane refrigerant is conducted and the relative Coefficient of Performance was
found out.
56
PERFORMANCE TEST ON TWO STAGE RECIPROCATING AIR
COMPRESSOR
AIM:
To determine the volumetric efficiency of the cylinder at normal
temperature conditions and to draw various performance characteristics curves.
APPARATUS REQUIRED:
Two stage reciprocating air compressor.
SPECIFICATIONS:
Type = Two stage, single acting
Speed , N = 700 rpm
Type of cylinder cooling system = Air cooled
Low pressure cylinder (LP) bore dia, d1 = 89.9 mm
High pressure cylinder (HP) bore dia, d2 = 63 mm
Stroke length, L = 88.9 mm
Max. Pressure = 300 kg/cm2
Orifice diameter, do = 0.01m
Energymeter constant = 200 rev/Kwhr
FORMULA USED:
1. Density of air, aa
aa
TR
P
. , kg/m3
Where,
Pa – Atmospheric pressure = 1.013 x 105 N/m2
Ra – Universal gas constant = 287 J/kgk
` Ta – Room temperature, K
57
2. Pressure head in terms of air column, ha
T
wwa
hh
, m
Where,
w - Density of water = 1000 kg/m3
hw – Head of water column, m
a - Density of air = 1.145 kg/m3
3. Velocity of air through orifice, Va
aa ghV 2 , m/sec
4. Area of orifice, Ao
2
4oo dA
, m2
Where,
do – Orifice diameter, m
58
PROCEDURE:
1. Connect the three pin plug to the 230V, 50Hz, 15 amps main supply and
switch on the Unit
2. The valve is provided at the top of LP and HP cylinders, water drain cock and
the air outlet valves are closed after the motor has gained its speed. The
increase in pressure of air in the receiver tank is indicated by pressure
gauge.
3. The pressure of air is maintained constant to the desired valve say (2kgf/cm2)
by adjusting at the opening of the compressed air outlet valve in the reservoir
manually.
4. The following observations are to be made by keeping reservoir pressure
constant (2 kgf/cm2)
a. Delivery pressure
b. Manometer reading (hw)(pressure difference across orifice)
c. Temperature T1, T2, T3, T4 after attaining the steady state
d. Time taken for 5 revolution of energy meter disc
5. The same procedure is repeated for the observations of other reservoir
pressure (4, 6, 8, 10, 12 kgf/cm2).
6. Then the motor is switched off after releasing the valve provided at the top of
LP and HP cylinders.
7. The Volumetric efficiency, Input power and Cooling factor of the cylinder has
been calculated from the given formula and the performance characteristic
curves are drawn.
GRAPH:
Following graphs are plotted
1. Delivery pressure on X – axis Vs Volumetric efficiency on Y – axis
2. Delivery pressure on X – axis Vs Input power on Y – axis
59
Coo
ling
facto
r
Inp
ut
po
wer
Kw
Vo
l.
Eff
%
Tim
e f
acto
r
for
5 r
ev o
f
en
erg
ym
ete
r
dis
c
se
cs
Tem
pe
ratu
re, T
4
˚C
T3
˚C
T2
˚C
T1
˚C
Ma
no
me
teric
Rea
din
g h
1- h
2
cm
h2
cm
h1
cm
Deliv
ery
pre
ssu
re
Kg
/cm
2
S N
o
T
AB
LE
:
60
RESULT:
The performance test on air compressor was conducted, the results were
tabulated and graphs are drawn for above parameters.