Impact of Fouling on VCRS and its
experimental study
Guided By: Submitted by:
Mr. Ankit Kulshreshtha
(Project Guide) Piyush Kumar(1101440033)
Rishabh Pandey(1101440040)
Shantanu Kaushik(1101440051)
Shashank Pandey(1101440052)
4/12/2015 1
CONTENTS
Topics
Test rig
Experimental Setup Description
Properties of Refrigerant
Actual Cycle
Losses in VCRS
Mathematical Calculations
Variation of COP with different
parameters
Comparison of Results
Fouling Mitigation
Conclusion
References
Page No.
3
4-9
10-12
13
14
15-20
21-32
33
34-35
36-37
38
4/12/2015 2
Fig No. 1 Refrigeration Test Rig [6]
4/12/2015 3
Experimental setup description[6]
1-Thermocouples: Six thermocouples are used in our test rig.
The readings of the thermocouples are taken by knob by
rotating it .first thermocouple fit at outlet of the condenser.
Second and third thermocouples are fitted in the condenser
tank and evaporator tank respectively. And remaining
thermocouples are fitted in inlet water tank, inlet to
condenser and outlet to evaporator.
2-Condenser and evaporator: - as shown in the figurethe tub
and tube type condenser and evaporator are used in the kit
and the evaporator having insulator on it. These tanks contain
the two mechanicalagitators as shown in figure, for the
continuous mixture the water,the agitator can be controlled
by switch . These tanks have two valves at the lower side of
tanks for the water outlet. A copper tube is used for the flow
of the refrigerant in spiral form in the tub.
4/12/2015 4
Fig No.2 Condenser[6]
4/12/2015 5
Experimental setup description:-
3-Display devices: our refrigeration test-rig contains LED type
display device. All the readings are shown through these
display devices.
4-Stirer: A stirer is used for the proper mixing of water and to
maintain the temperature uniformity throughout the tank and
it is of 1 to 70 hp and 45 watt, 50 Hz and 230 Volt.
5-Rotameter and water tank: rota meter is fitted in
between water tank and the condenser and evaporator
tank and it contains a motor that is used to raise the
pressure of the water and it contains a switch by which
the flow of water can be controlled.
4/12/2015 6
Fig No.3 Evaporator[6]
4/12/2015 7
Experimental setup description:-
6-Expansion valve,compressor and drier: in our refrigeration
test-rig a reciprocating compressor is used to compress the
refrigerant before condenser and it is placed between the
evaporator and condenser and it is of 230 volt,50 Hz and
single phase. The drier is used to absorb the moisture content
and it is placed between the inlet to evaporator and outlet to
the condenser. A copper capillary tube type expansion device
is being used for the expansion of the refrigerant and it is
placed between drier and evaporator.
4/12/2015 8
Experimental setup description:-
7-Pressure gauges: – The two pressure gauges are used , one at
high pressure side and other one is placed at the lower
pressure side to measure the condenser pressure and
evaporator pressure respectively. Means one is suction
pressure gauge and other one is discharge pressure gauge,
suction pressure gauge having pressure range from 0 to 150
psi and discharge pressure gauge having pressure range from
0 to 300 psi.
4/12/2015 9
Properties of the refrigerant-
Molecular weight: 102.03 g/mol.
Critical point
• Critical temperature: 100.95 °C
• Critical pressure: 40.6 bar
• Critical density: 512 kg/m3
4/12/2015 10
Properties of the refrigerant-
Liquid phase
• Liquid density (1.013 bar and 25 °C (77 °F)) : 1206
kg/m3
• Boiling point (1.013 bar) : -26.55 °C
• Latent heat of vaporization (1.013 bar at boiling
point) : 215.9 kJ/kg
• Vapor pressure (at 20 °C or 68 °F) : 5.7 bar
• Vapor pressure (at 5 °C or 41 °F) : 3.5 bar
• Vapor pressure (at 15 °C or 59 °F) : 4.9 bar
• Vapor pressure (at 50 °C or 122 °F) : 13.2 bar
4/12/2015 11
Properties of the refrigerant-
Gaseous phase
• Gas density (1.013 bar at boiling point) : 5.28
kg/m3
• Gas density (1.013 bar and 15 °C (59 °F)) : 4.25
kg/m3
• Compressibility Factor (Z) (1.013 bar and 15 °C
(59 °F)) : 1
• Specific gravity : 3.25
• Specific volume (1.013 bar and 15 °C (59 °F)) :
0.235 m3/kg
4/12/2015 12
Fig No.4 ACTUAL VAPOR COMPRESSION CYCLE [5]
4/12/2015 13
Fig No.5 Schematic diagram of a simple refrigeration cycle and various
losses in VCR system. [5]
4/12/2015 14
MATHEMATICAL CALCULATION [6]
1.Expansion valve:
h6 = h7 , at P1 and T1, h1 = 297.1 kJ/kg from table,
at T6 = 40.8 and P6 = 1040 kPa , h6 = 109.7 kJ/kg from
table,
pressure loss in condenser = P1 – P6
= 1385.4 – 1040
= 345.4 kPa.
Ploss
cond = 25% of Pc
h6 =h7 = 109.7 kj/kg
4/12/2015 15
MATHEMATICAL CALCULATION[6]
2.Evaporator :-
T7 = -6.90C, h6 =h7 = 109.7 kj/kg, P7 = 225.6 kPa
h7 = hf + x hfg at 225.6 kPa
109.7 = 42.51 + x(246.32 – 42.51)
x = 0.329
T4 = 10.30C, Pe = P4 = 170 kPa , h4 = 262.0 kJ/kg
Pressure loss in evaporator = P7 – P4 =55.6 kPa
Ploss
evap = 25% of Pevap
(Qevap) R = mR ( h4 – h7) = 0.008 (262 – 109.7) = 1.218 kW
(Qevap) W = mW.CPW . ( Two – Twin) = mw.Cpw (T5 –T3)
= 0.046 Х 4.186 (18.3 – 12.4) = 1.134 kW
4/12/2015 16
Qloss
s= 1.218 – 1.134 = 0.08 approx 5% of Qevap
(Qmax)evap = mW.CPW ( Twine – Trine )
= mW.CPW ( T5 – T7 ) = 0.046 Х 4.186 (18.3 + 6.9) = 4.84
kW
Ɛ = (Qevap)w / Qmax = 1.134 / 4.84 = 0.23 = 23%.
Ɛ = (Qevap)R / Qmax ) = 1.218 / 4.84 = 0.25 = 25%.
Qloss
s = mR ( h4 – h7) - mW.CPW . ( Two – Twin)
Refrigeration effect = mR ( h4 – h7) - 0.05 mR ( h4 – h7) = 1.134 kW
Now, Ɛ = 1 – exp(-NTU)
NTU = UA / Cmin
UAe = Cmin. ln(1/ 1 - Ɛ evap)
U = (mW.CPW /Ae) ln(1/1-0.23)
U = ( 0.046 Х 4.186)/0.6619 ln(1/0.77)
U = 75.69 watt/m2 0C.4/12/2015 17
MATHEMATICAL CALCULATION:
3.Compressor:
h4 = 262 kJ/kg and S4 = 1.010 kJ/kgK
at 1385.4 = Pb = Pc , Sg = 0.9108 kJ/kgK
since, Sg< S4 hence point lies in the superheated region.
At1385.4 kPa, Sb = 1.006 kJ/kgK at 800C, hb = 308.4 kJ/kg
Ƞ = 0.72 , ƞisentropic = hb – h4 / ha – h4
ha = 262 + (308.4 – 262) / 0.72
= 326.44 kJ/kg
At 326.44 kJ/kg , P = 1385 kPa and T = 920C
4/12/2015 18
MATHEMATICAL CALCULATION:
Heat loss in compressor :
mR( h4 - ha) - mR( h4 – hb) = 0.008 (326.44 – 262) – 0.008(308.4
– 262)
= 0.144 kW
Qloss
discharge + Qloss
comp = mR( h1 - ha) = 0.008(326.44 – 297.1) =
0.234 kW
Qloss
discharge = 0.234 - .144 = 0.09 kW
(Wcomp)actual = mR (ha - h4) = 0.008 (326.44 – 262) = 0.515 kW
(Wcomp)isentropic = mR (hb - h4) = 0.008(308.4 – 262) = 0.371 kW
COP = RE / Wcomp = 1.134 / 0.515 = 2.2
4/12/2015 19
MATHEMATICAL CALCULATION:
4.Condenser:
(Qcond)w =mW.CPW . ( Twoc – Twinc) = 0.046 Х 4.186 (26 – 18.3)
= 1.48 kW
(Qcond)R = mR (h1 – h6) = 0.008(297.4 – 109.7) = 1.49 kW
Qmax = mW.CPW (69.1 – 18.3) = 9.76 kW
Ɛcond = (Qcond)w / Qmax = 1.48 / 9.76 = 0.1516i.e. 15.16%
Now, Ɛ = 1 – exp(-NTU)
NTU = UA / Cmin
UAc = Cmin. ln(1/ 1 - Ɛcond)
U = (0.046 Х 4.186) / 0.3301.ln(1/1-0.15)
U = 94.66 watt/m2 0C
4/12/2015 20
VARIATION OF COP, REFRIGERATION EFFECT AND
COMPRESSOR WORK WITH MASS FLOW RATE:-
Input
variables
RE Wcomp COP UCond Uevap
mw1
0.046(kg/s)
1.134 kW 0.515 kW 2.04 94.66
(w/m2 0C)
75.69
(w/m2 0C)
mw2
0.07(kg/s)
1.002 kW 0.552 kW 1.80 144
(w/m2 0C)
115.7
(w/m2 0C)
mw3
0.09(kg/s)
0.990 kW 0.672 kW 1.47 185
(w/m2 0C)
148.7
(w/m2 0C)
4/12/2015 21
Fig No.6 Graph (Mass flow rate vs COP)
C
O
P
Mass flow rate of water (kg/s)
0
0.5
1
1.5
2
2.5
0.046 0.07 0.09
COP
COP
4/12/2015 22
Fig No.7 Graph (COP vs Ucond)
COP
0
20
40
60
80
100
120
140
160
180
200
2.04 1.8 1.47
Ucond (w/m2 0C)
Ucond (w/m2 0C)
4/12/2015 23
Fig No.8 Graph (COP vs Uevap)
COP
0
20
40
60
80
100
120
140
160
2.04 1.8 1.47
Uevap (w/m2 0C)
Uevap (w/m2 0C)
4/12/2015 24
Fig No.9 Graph (Mass flow rate of water vs Refrigeration Effect)
Mass flow rate of water (kg/sec)
0.9
0.95
1
1.05
1.1
1.15
0.046 0.07 0.09
REFRIGERATION EFFECT (kW)
REFRIGERATION EFFECT (kW)
4/12/2015 25
Fig No.10 Graph (Compressor Work vs COP)
Compressor work (kW)
0
0.5
1
1.5
2
2.5
0.515 1.002 0.99
COP
COP
4/12/2015 26
Fig No.11 Graph (COP vs Wcomp)
COP
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
2.04 1.8 1.47
Wcomp (kW)
Wcomp (kW)
4/12/2015 27
Fig No.12 Graph (Mass flow rate of water in condenser(kg/sec)
vs COP)
C
O
P
Mass flow rate of water in condenser(kg/sec)
0
0.5
1
1.5
2
2.5
3
3.5
4
0.028 0.046 0.052 0.066
4/12/2015 28
Fig No.13 Graph (Compressor work(watt) vs COP)
C
O
P
Compressor work(watt)
0
0.5
1
1.5
2
2.5
3
3.5
4
208 245 266 308
4/12/2015 29
Fig No.15 Graph (Condenser pressure(bar) vs COP)
C
O
P
Condenser pressure(bar)
0
0.5
1
1.5
2
2.5
3
9.3 10.3 10.8 11.4
4/12/2015 30
VARIATION OF COP WITH TEMPERATURE:
Input temperature (TWinc) (0C) COP
21.2 2.7
25.2 2.47
28 2.04
28.7 1.88
4/12/2015 31
Fig No.16 Graph (Input Temperature vs COP)
COP
4/12/2015 32
COMPARISON BETWEEN MATHEMATICAL AND EXPERIMENTAL RESULTS:
INPUTS MATHEMATICAL
RESULT
EXPERIMENTAL
RESULT
ERROR
( %age)
T1 (0C) 70 69.1 1
T6 (0C) 41 40.8 0.5
T7 (0C) -7 -6.9 1
T4 (0C) 10 10.3 2
COP 2.2 2.04 7
RE (kW) 1.134 1.134 0
Wcomp (kW) 0.515 0.555 7
Ucond (watt/m2 0C) 94.66 94.66 0
Uevap (watt/m2 0C) 75.69 75.66 0
Twoe (0C) 12 12.4 3
Twoc (0C) 26.14 26 0.5
4/12/2015 33
Fouling Mitigation
Offline techniques clean heat exchangers by mechanical and/or chemical
means while the system is down, which results in lost production time.
They are labor-intensive and expensive; as much as 8% of the
maintenance costs in a typical industrial plant are due to fouling mitigation
in heat exchangers. When offline mitigation requires the use of aggressive
chemicals, the company is saddled with additional operating costs, and
new problems relating to the increased safety hazards for company
personnel and disposal of toxic waste. Moreover, the fouling process starts
again immediately after cleaning, and the gradual accumulation of
deposits reduces performance until the next cleaning treatment is
initiated. These drawbacks make heat exchanger operation with regular
offline cleaning expensive, cumbersome and inefficient for the entire
industrial process.
4/12/2015 34
Fouling Mitigation
The online method is an ongoing process which uses mechanical
means to keep the heat exchanger clean while it operates.
Some cleaning systems also use chemicals (which must be
carefully adapted to each process), but online mechanical-
only cleaning is both environmentally responsible and highly
cost-effective.
Online mechanical cleaning boosts performance in two ways: it
does not require system shutdown with temporary loss of
operation; and it keeps performance optimal and energy-
efficient through continuous cleaning, which does not allow
any fouling to occur. In addition, online mechanical cleaning
does not only eliminate staff time for cleaning services, but
also purchase of chemicals and waste disposal.
4/12/2015 35
CONCLUSION:-
We have successfully validated the performance of VCR system with
the help of our mathematical model and we have also studied the
effect of various parameters like mass flow rate of water, condenser
pressure, evaporator pressure, input temperatures in condenser
and evaporator,etc on the COP and vapor compression refrigeration
system.
We have also study the fouling phenomenon in the condenser and
evaporator of VCR system.After drawing the curves among various
parameters we conclude that:
1) As mass flow rate of water increases the COP of VCR system
decreases.
2) As overall heat transfer coefficient of evaporator increases the COP
of the VCR system decreases.
3) As overall heat transfer coefficient of condenser increases the COP
of the VCR system decreases.
4/12/2015 36
CONCLUSION:-
4) As mass flow rate of water increases the refrigeration effect
decreases.
5) As compressor work increases the COP of the VCR system
decreases.
6) As the condenser pressure increases the COP of the VCR system
decreases.
7) As the evaporator pressure decreases the COP of the VCR system
decreases.
8) As the input temperature increases the COP of the VCR system
decreases.
4/12/2015 37
References
1.Cabello, R., Navarro, J., & Torrella, E. (2005). Simplified steady-state
modelling of a single stage vapour compression plant. Model development
and validation.Applied thermal engineering, 25(11), 1740-1752.
2.Qureshi, B. A., & Zubair, S. M. (2012). The impact of fouling on performance
of a vapor compression refrigeration system with integrated mechanical sub-
cooling system. Applied Energy, 92, 750-762.
3.Qureshi, B. A., & Zubair, S. M. (2014). The impact of fouling on the
condenser of a vapor compression refrigeration system: An experimental
observation.International Journal of Refrigeration, 38, 260-266.
4.Cabello, R., Torrella, E., & Navarro-Esbrı́, J. (2004). Experimental evaluation
of a vapour compression plant performance using R134a, R407C and R22 as
working fluids. Applied thermal engineering, 24(13), 1905-1917.
5.Arora,C.P. (2014) , Actual Vapour Compression Refrigeration Cycle.
Refrigeration and Air Conditioning, 114-115
6. Katiyar,C.,Yadav,A.,Singh,R.P.,Soni,R.K. & Awasthi,Hrydesh.(2014), Impact of
Fouling on VCR System.
4/12/2015 38

Impact of Fouling on VCR System

  • 1.
    Impact of Foulingon VCRS and its experimental study Guided By: Submitted by: Mr. Ankit Kulshreshtha (Project Guide) Piyush Kumar(1101440033) Rishabh Pandey(1101440040) Shantanu Kaushik(1101440051) Shashank Pandey(1101440052) 4/12/2015 1
  • 2.
    CONTENTS Topics Test rig Experimental SetupDescription Properties of Refrigerant Actual Cycle Losses in VCRS Mathematical Calculations Variation of COP with different parameters Comparison of Results Fouling Mitigation Conclusion References Page No. 3 4-9 10-12 13 14 15-20 21-32 33 34-35 36-37 38 4/12/2015 2
  • 3.
    Fig No. 1Refrigeration Test Rig [6] 4/12/2015 3
  • 4.
    Experimental setup description[6] 1-Thermocouples:Six thermocouples are used in our test rig. The readings of the thermocouples are taken by knob by rotating it .first thermocouple fit at outlet of the condenser. Second and third thermocouples are fitted in the condenser tank and evaporator tank respectively. And remaining thermocouples are fitted in inlet water tank, inlet to condenser and outlet to evaporator. 2-Condenser and evaporator: - as shown in the figurethe tub and tube type condenser and evaporator are used in the kit and the evaporator having insulator on it. These tanks contain the two mechanicalagitators as shown in figure, for the continuous mixture the water,the agitator can be controlled by switch . These tanks have two valves at the lower side of tanks for the water outlet. A copper tube is used for the flow of the refrigerant in spiral form in the tub. 4/12/2015 4
  • 5.
  • 6.
    Experimental setup description:- 3-Displaydevices: our refrigeration test-rig contains LED type display device. All the readings are shown through these display devices. 4-Stirer: A stirer is used for the proper mixing of water and to maintain the temperature uniformity throughout the tank and it is of 1 to 70 hp and 45 watt, 50 Hz and 230 Volt. 5-Rotameter and water tank: rota meter is fitted in between water tank and the condenser and evaporator tank and it contains a motor that is used to raise the pressure of the water and it contains a switch by which the flow of water can be controlled. 4/12/2015 6
  • 7.
  • 8.
    Experimental setup description:- 6-Expansionvalve,compressor and drier: in our refrigeration test-rig a reciprocating compressor is used to compress the refrigerant before condenser and it is placed between the evaporator and condenser and it is of 230 volt,50 Hz and single phase. The drier is used to absorb the moisture content and it is placed between the inlet to evaporator and outlet to the condenser. A copper capillary tube type expansion device is being used for the expansion of the refrigerant and it is placed between drier and evaporator. 4/12/2015 8
  • 9.
    Experimental setup description:- 7-Pressuregauges: – The two pressure gauges are used , one at high pressure side and other one is placed at the lower pressure side to measure the condenser pressure and evaporator pressure respectively. Means one is suction pressure gauge and other one is discharge pressure gauge, suction pressure gauge having pressure range from 0 to 150 psi and discharge pressure gauge having pressure range from 0 to 300 psi. 4/12/2015 9
  • 10.
    Properties of therefrigerant- Molecular weight: 102.03 g/mol. Critical point • Critical temperature: 100.95 °C • Critical pressure: 40.6 bar • Critical density: 512 kg/m3 4/12/2015 10
  • 11.
    Properties of therefrigerant- Liquid phase • Liquid density (1.013 bar and 25 °C (77 °F)) : 1206 kg/m3 • Boiling point (1.013 bar) : -26.55 °C • Latent heat of vaporization (1.013 bar at boiling point) : 215.9 kJ/kg • Vapor pressure (at 20 °C or 68 °F) : 5.7 bar • Vapor pressure (at 5 °C or 41 °F) : 3.5 bar • Vapor pressure (at 15 °C or 59 °F) : 4.9 bar • Vapor pressure (at 50 °C or 122 °F) : 13.2 bar 4/12/2015 11
  • 12.
    Properties of therefrigerant- Gaseous phase • Gas density (1.013 bar at boiling point) : 5.28 kg/m3 • Gas density (1.013 bar and 15 °C (59 °F)) : 4.25 kg/m3 • Compressibility Factor (Z) (1.013 bar and 15 °C (59 °F)) : 1 • Specific gravity : 3.25 • Specific volume (1.013 bar and 15 °C (59 °F)) : 0.235 m3/kg 4/12/2015 12
  • 13.
    Fig No.4 ACTUALVAPOR COMPRESSION CYCLE [5] 4/12/2015 13
  • 14.
    Fig No.5 Schematicdiagram of a simple refrigeration cycle and various losses in VCR system. [5] 4/12/2015 14
  • 15.
    MATHEMATICAL CALCULATION [6] 1.Expansionvalve: h6 = h7 , at P1 and T1, h1 = 297.1 kJ/kg from table, at T6 = 40.8 and P6 = 1040 kPa , h6 = 109.7 kJ/kg from table, pressure loss in condenser = P1 – P6 = 1385.4 – 1040 = 345.4 kPa. Ploss cond = 25% of Pc h6 =h7 = 109.7 kj/kg 4/12/2015 15
  • 16.
    MATHEMATICAL CALCULATION[6] 2.Evaporator :- T7= -6.90C, h6 =h7 = 109.7 kj/kg, P7 = 225.6 kPa h7 = hf + x hfg at 225.6 kPa 109.7 = 42.51 + x(246.32 – 42.51) x = 0.329 T4 = 10.30C, Pe = P4 = 170 kPa , h4 = 262.0 kJ/kg Pressure loss in evaporator = P7 – P4 =55.6 kPa Ploss evap = 25% of Pevap (Qevap) R = mR ( h4 – h7) = 0.008 (262 – 109.7) = 1.218 kW (Qevap) W = mW.CPW . ( Two – Twin) = mw.Cpw (T5 –T3) = 0.046 Х 4.186 (18.3 – 12.4) = 1.134 kW 4/12/2015 16
  • 17.
    Qloss s= 1.218 –1.134 = 0.08 approx 5% of Qevap (Qmax)evap = mW.CPW ( Twine – Trine ) = mW.CPW ( T5 – T7 ) = 0.046 Х 4.186 (18.3 + 6.9) = 4.84 kW Ɛ = (Qevap)w / Qmax = 1.134 / 4.84 = 0.23 = 23%. Ɛ = (Qevap)R / Qmax ) = 1.218 / 4.84 = 0.25 = 25%. Qloss s = mR ( h4 – h7) - mW.CPW . ( Two – Twin) Refrigeration effect = mR ( h4 – h7) - 0.05 mR ( h4 – h7) = 1.134 kW Now, Ɛ = 1 – exp(-NTU) NTU = UA / Cmin UAe = Cmin. ln(1/ 1 - Ɛ evap) U = (mW.CPW /Ae) ln(1/1-0.23) U = ( 0.046 Х 4.186)/0.6619 ln(1/0.77) U = 75.69 watt/m2 0C.4/12/2015 17
  • 18.
    MATHEMATICAL CALCULATION: 3.Compressor: h4 =262 kJ/kg and S4 = 1.010 kJ/kgK at 1385.4 = Pb = Pc , Sg = 0.9108 kJ/kgK since, Sg< S4 hence point lies in the superheated region. At1385.4 kPa, Sb = 1.006 kJ/kgK at 800C, hb = 308.4 kJ/kg Ƞ = 0.72 , ƞisentropic = hb – h4 / ha – h4 ha = 262 + (308.4 – 262) / 0.72 = 326.44 kJ/kg At 326.44 kJ/kg , P = 1385 kPa and T = 920C 4/12/2015 18
  • 19.
    MATHEMATICAL CALCULATION: Heat lossin compressor : mR( h4 - ha) - mR( h4 – hb) = 0.008 (326.44 – 262) – 0.008(308.4 – 262) = 0.144 kW Qloss discharge + Qloss comp = mR( h1 - ha) = 0.008(326.44 – 297.1) = 0.234 kW Qloss discharge = 0.234 - .144 = 0.09 kW (Wcomp)actual = mR (ha - h4) = 0.008 (326.44 – 262) = 0.515 kW (Wcomp)isentropic = mR (hb - h4) = 0.008(308.4 – 262) = 0.371 kW COP = RE / Wcomp = 1.134 / 0.515 = 2.2 4/12/2015 19
  • 20.
    MATHEMATICAL CALCULATION: 4.Condenser: (Qcond)w =mW.CPW. ( Twoc – Twinc) = 0.046 Х 4.186 (26 – 18.3) = 1.48 kW (Qcond)R = mR (h1 – h6) = 0.008(297.4 – 109.7) = 1.49 kW Qmax = mW.CPW (69.1 – 18.3) = 9.76 kW Ɛcond = (Qcond)w / Qmax = 1.48 / 9.76 = 0.1516i.e. 15.16% Now, Ɛ = 1 – exp(-NTU) NTU = UA / Cmin UAc = Cmin. ln(1/ 1 - Ɛcond) U = (0.046 Х 4.186) / 0.3301.ln(1/1-0.15) U = 94.66 watt/m2 0C 4/12/2015 20
  • 21.
    VARIATION OF COP,REFRIGERATION EFFECT AND COMPRESSOR WORK WITH MASS FLOW RATE:- Input variables RE Wcomp COP UCond Uevap mw1 0.046(kg/s) 1.134 kW 0.515 kW 2.04 94.66 (w/m2 0C) 75.69 (w/m2 0C) mw2 0.07(kg/s) 1.002 kW 0.552 kW 1.80 144 (w/m2 0C) 115.7 (w/m2 0C) mw3 0.09(kg/s) 0.990 kW 0.672 kW 1.47 185 (w/m2 0C) 148.7 (w/m2 0C) 4/12/2015 21
  • 22.
    Fig No.6 Graph(Mass flow rate vs COP) C O P Mass flow rate of water (kg/s) 0 0.5 1 1.5 2 2.5 0.046 0.07 0.09 COP COP 4/12/2015 22
  • 23.
    Fig No.7 Graph(COP vs Ucond) COP 0 20 40 60 80 100 120 140 160 180 200 2.04 1.8 1.47 Ucond (w/m2 0C) Ucond (w/m2 0C) 4/12/2015 23
  • 24.
    Fig No.8 Graph(COP vs Uevap) COP 0 20 40 60 80 100 120 140 160 2.04 1.8 1.47 Uevap (w/m2 0C) Uevap (w/m2 0C) 4/12/2015 24
  • 25.
    Fig No.9 Graph(Mass flow rate of water vs Refrigeration Effect) Mass flow rate of water (kg/sec) 0.9 0.95 1 1.05 1.1 1.15 0.046 0.07 0.09 REFRIGERATION EFFECT (kW) REFRIGERATION EFFECT (kW) 4/12/2015 25
  • 26.
    Fig No.10 Graph(Compressor Work vs COP) Compressor work (kW) 0 0.5 1 1.5 2 2.5 0.515 1.002 0.99 COP COP 4/12/2015 26
  • 27.
    Fig No.11 Graph(COP vs Wcomp) COP 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 2.04 1.8 1.47 Wcomp (kW) Wcomp (kW) 4/12/2015 27
  • 28.
    Fig No.12 Graph(Mass flow rate of water in condenser(kg/sec) vs COP) C O P Mass flow rate of water in condenser(kg/sec) 0 0.5 1 1.5 2 2.5 3 3.5 4 0.028 0.046 0.052 0.066 4/12/2015 28
  • 29.
    Fig No.13 Graph(Compressor work(watt) vs COP) C O P Compressor work(watt) 0 0.5 1 1.5 2 2.5 3 3.5 4 208 245 266 308 4/12/2015 29
  • 30.
    Fig No.15 Graph(Condenser pressure(bar) vs COP) C O P Condenser pressure(bar) 0 0.5 1 1.5 2 2.5 3 9.3 10.3 10.8 11.4 4/12/2015 30
  • 31.
    VARIATION OF COPWITH TEMPERATURE: Input temperature (TWinc) (0C) COP 21.2 2.7 25.2 2.47 28 2.04 28.7 1.88 4/12/2015 31
  • 32.
    Fig No.16 Graph(Input Temperature vs COP) COP 4/12/2015 32
  • 33.
    COMPARISON BETWEEN MATHEMATICALAND EXPERIMENTAL RESULTS: INPUTS MATHEMATICAL RESULT EXPERIMENTAL RESULT ERROR ( %age) T1 (0C) 70 69.1 1 T6 (0C) 41 40.8 0.5 T7 (0C) -7 -6.9 1 T4 (0C) 10 10.3 2 COP 2.2 2.04 7 RE (kW) 1.134 1.134 0 Wcomp (kW) 0.515 0.555 7 Ucond (watt/m2 0C) 94.66 94.66 0 Uevap (watt/m2 0C) 75.69 75.66 0 Twoe (0C) 12 12.4 3 Twoc (0C) 26.14 26 0.5 4/12/2015 33
  • 34.
    Fouling Mitigation Offline techniquesclean heat exchangers by mechanical and/or chemical means while the system is down, which results in lost production time. They are labor-intensive and expensive; as much as 8% of the maintenance costs in a typical industrial plant are due to fouling mitigation in heat exchangers. When offline mitigation requires the use of aggressive chemicals, the company is saddled with additional operating costs, and new problems relating to the increased safety hazards for company personnel and disposal of toxic waste. Moreover, the fouling process starts again immediately after cleaning, and the gradual accumulation of deposits reduces performance until the next cleaning treatment is initiated. These drawbacks make heat exchanger operation with regular offline cleaning expensive, cumbersome and inefficient for the entire industrial process. 4/12/2015 34
  • 35.
    Fouling Mitigation The onlinemethod is an ongoing process which uses mechanical means to keep the heat exchanger clean while it operates. Some cleaning systems also use chemicals (which must be carefully adapted to each process), but online mechanical- only cleaning is both environmentally responsible and highly cost-effective. Online mechanical cleaning boosts performance in two ways: it does not require system shutdown with temporary loss of operation; and it keeps performance optimal and energy- efficient through continuous cleaning, which does not allow any fouling to occur. In addition, online mechanical cleaning does not only eliminate staff time for cleaning services, but also purchase of chemicals and waste disposal. 4/12/2015 35
  • 36.
    CONCLUSION:- We have successfullyvalidated the performance of VCR system with the help of our mathematical model and we have also studied the effect of various parameters like mass flow rate of water, condenser pressure, evaporator pressure, input temperatures in condenser and evaporator,etc on the COP and vapor compression refrigeration system. We have also study the fouling phenomenon in the condenser and evaporator of VCR system.After drawing the curves among various parameters we conclude that: 1) As mass flow rate of water increases the COP of VCR system decreases. 2) As overall heat transfer coefficient of evaporator increases the COP of the VCR system decreases. 3) As overall heat transfer coefficient of condenser increases the COP of the VCR system decreases. 4/12/2015 36
  • 37.
    CONCLUSION:- 4) As massflow rate of water increases the refrigeration effect decreases. 5) As compressor work increases the COP of the VCR system decreases. 6) As the condenser pressure increases the COP of the VCR system decreases. 7) As the evaporator pressure decreases the COP of the VCR system decreases. 8) As the input temperature increases the COP of the VCR system decreases. 4/12/2015 37
  • 38.
    References 1.Cabello, R., Navarro,J., & Torrella, E. (2005). Simplified steady-state modelling of a single stage vapour compression plant. Model development and validation.Applied thermal engineering, 25(11), 1740-1752. 2.Qureshi, B. A., & Zubair, S. M. (2012). The impact of fouling on performance of a vapor compression refrigeration system with integrated mechanical sub- cooling system. Applied Energy, 92, 750-762. 3.Qureshi, B. A., & Zubair, S. M. (2014). The impact of fouling on the condenser of a vapor compression refrigeration system: An experimental observation.International Journal of Refrigeration, 38, 260-266. 4.Cabello, R., Torrella, E., & Navarro-Esbrı́, J. (2004). Experimental evaluation of a vapour compression plant performance using R134a, R407C and R22 as working fluids. Applied thermal engineering, 24(13), 1905-1917. 5.Arora,C.P. (2014) , Actual Vapour Compression Refrigeration Cycle. Refrigeration and Air Conditioning, 114-115 6. Katiyar,C.,Yadav,A.,Singh,R.P.,Soni,R.K. & Awasthi,Hrydesh.(2014), Impact of Fouling on VCR System. 4/12/2015 38