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Abstract
This paper presents the analytical method for increasing the performance of air conditioner using a binary mixture of two refrigerant R152a and R134a. The two refrigerant are non toxic and environmental friendly (zero ODP and low GWP) and also suitable for safety purpose.
Temperature composition diagram and enthalpy composition diagram of the above binary mixture of R152a and R134a are calculated at 5 bar and 15 bar which is our operating pressure of evaporator and condenser respectively. Then calculation of coefficient of performance (COP) of air-conditioner is done by varying the compositions of our binary mixture. The COP calculation is done for same AMTD in order to maintain condenser and evaporator size same as the size of air-conditioning machine is not changed.
The most suitable composition of this binary mixture is hence calculated for which the COP and power saving will be maximum in respect to R134a for same AMTD design.
Introduction
With the atmospheric threat such as ozone layer depletion and green house effect to the earth environment hydro chlorofluorocarbons (most notably HCFC-22) gained in popularity in early 1974. While HCFCs were still damaging to the ozone layer, but much less so than CFCs. HCFCs and HCFC mixtures were developed that could serve as drop-in replacements for most of the CFCs in use. However, the excitement over HCFCs was short-lived, as the Vienna Convention of 1995 not only accelerated the HCFC-reduction timetable, but also required that their production effectively cease by 2020, with a complete cessation by 2030. Japan and some European countries have established cut-off dates that begin much earlier. In Switzerland, for example, these are banned by 2005.
Once again, replacement refrigerants need to be found, but this time there are no obvious solutions. While some single component refrigerants present no obvious solutions. While some single component refrigerants present reduced-performance possibilities, the solution appears to lie with synthetic mixtures. These mixtures may be azeotropic, near aezeotropic, or zeotropic.
Selecting A Suitable Refrigerants
As most of the refrigerants, solvents, and other industrial chemicals that contain Chlorine (Cl), including CFCs, and HCFCs are being phased out. When these substances are released into the atmosphere, the chlorine they contain interacts with the earths ozone layer, and causes its depletion. This depletion reduces the protection from the sun’s ultra violet radiations that the ozone layer provides, and can cause crop failures, skin cancers and other hazards.
Also of concern is a global warming effect caused by the release of these substances. There are indications that global warming has caused stratospheric cooling, which makes chlorine even more effective and depleting the ozone layer.
Selecting a Suitable Refrigerants
As refrigerants with chlorine atoms have undesirable impacts on the environment and there are also many undesirable impacts on environment and there are also many undesirable factors related to them. So CFCs and HCFCs must be phased out. So the substitute or alternative for these must be developed and used in refrigeration. Some refrigerants manufacturer have developed interim refrigerant blend that can be used without making any changes to existing machinery.
To select a replacement refrigerant, there are many variables to understand:
(a) The environmental impact including Ozone Depletion Potential (ODP), Global Warming Potential (GWP), and Total Equivalent Warming Impact (TEWI) ratings.
(b) Safety and health issues regarding toxicity and exposure limits and flammability.(c) Chemical behavior.(d) Refrigeration capacity and relative efficiency.
Ideally an alternative refrigerant would have zero or low ODP, low direct GWP.
When selecting a refrigerant it is important to consider how well suited the refrigerant is to its application including operating temperatures, where and how it will be used.
Refrigerant safety is an issue in terms of toxicity and flammability. For any application, new refrigerant must be used safely.
Safety Classifications: Refrigerants are classified into safety groups by using a combination of letters A and B (for toxicity) and the numbers 1, 2, 3 (for flammability).
Class A: signifies refrigerants which are non toxic at concentration less than equal to 400particle per mole (ppm).
Class B: refrigerants are toxic at levels below 400 ppm.
Class 1: indicates no flame propagation.
Class 2: has a low flammability limit.
Class 3: are highly flammable.
Examples of each safety classifications
A1 Nearly all commercial refrigerants incl. R-12, R-502, R-22, R-134aA2 R-152a, R-142bA3 R-290 (propane)B1 R-123B2 R-717 (ammonia)B3 Vinyl chloride
So by considering all of the above discussed factors for selecting a suitable refrigerant we have a choice of refrigerant mixture of R-134a and R-152a of class A1 and class A2.
Favorable factors related to R-134a and R-152a:
R-134a Non-flammable. Evaporator pressure positive. Discharge and winding temperature lower than R-12.
R-152a Zero GWP. Energy consumption lower than R-134a. Lower starting torque motor required in comparison to R-134a.
Properties related to R-134a and R-152a:
Refrigerant R-134a R-152aChemical formula CF3CH2F CH3CHF2
Category HFC HFCNBP (⁰C) -26.15 -24.15Flammability Non flammable Slightly flammableP0 at 5⁰C (bar) 3.5 3.149Pk at 40⁰C (bar) 10.167 9.092
Temperature composition diagram for R-152a/R-134a mixture
Molecular masses (M)R-152a = 66.05 and (M)R-134a = 102.03
Let the subscripts 1 and 2 denote R-152a and R-134a respectively.
At 15 bar, t1sat = 60⁰C, and t2
sat = 55.2⁰C.
So the bubble temperature will vary between 55.2⁰C and 60⁰C.
We have done calculations for 55.6⁰C, 56⁰C, 56.4⁰C, 56.8⁰C, 57.2⁰C, 57.6⁰C, 58⁰C, 58.4⁰C,
58.8, 59.2, and 59.6 ⁰C bubble temperature.
At 5 bar, t1sat = 19.2⁰C and t2
sat = 16⁰C.
So the bubble temperature will vary between 16⁰C and 19.2⁰C.
We have done calculation for 16.4 ⁰C, 16.8 ⁰C, 17.2 ⁰C, 17.6 ⁰C, 18 ⁰C, 18.4 ⁰C, 18.8 ⁰C,
19 ⁰C bubble temperature.
All the above calculation has done by using liquid and vapor equilibrium composition formulae for R-152a:
For equilibrium vapor composition:
y1=( p1
sat−p1sat p2
sat )/ p
p1sat−p2
sat
For equilibrium liquid composition:
x1=p1
p1sat =
y1 p
p1sat
Equivalent mass fraction:
Mass fractions in liquid phase:
ξ1L=
x1 M 1
x1 M 1+ x2 M 2
Mass fraction in vapor phase:
ξ1V=
y1 M 1
y1 M1+ y2 M 2
By using the above formulae a computer program is made for calculating the mass fractions in liquid phase and vapor phase of R-152a with their respective bubble temperature.
Values obtained from above are tabulated for 5 bar and 15 bar as follows:
Vapor liquid equilibrium of R-152a and R-134a mixture at 15 bar pressure:
T (⁰C) P1sat P2
sat ξ1L ξ2
L
55.2 15 0 055.6 13.512 15.134 0.055086 0.04947456.0 13.643 15.282 0.118576 0.10722356.4 13.775 15.431 0.185511 0.16896756.8 13.908 15.580 0.255861 0.23485157.2 14.042 15.731 0.330641 0.30600457.6 14.117 15.883 0.409874 0.38269858.0 14.313 16.036 0.493983 0.46562058.4 14.450 16.190 0.583446 0.55557858.8 14.558 16.345 0.678802 0.65352059.2 14.727 16.502 0.780782 0.76068359.6 14.866 16.659 0.889070 0.87733260.0 15 1.00 1.00
Vapor liquid equilibrium of R-152a and R-134a mixture at 5 bar pressure
T (⁰C) P1sat P2
sat ξ1L ξ2
L
16.0 5.00 0 016.4 4.5851 5.1071 0.143180 0.13045416.8 4.6433 5.7123 0.238211 0.21918817.2 4.7020 5.2382 0.341001 0.31716617.6 4.7613 5.3046 0.452379 0.42577318.0 4.8211 5.3718 0.573630 0.54699018.4 4.8816 5.4395 0.706141 0.68319618.8 4.9426 5.5079 0.851369 0.83713919.0 4.9733 5.5424 0.929333 0.92187819.2 5.0000 1.000000 1.000000
Enthalpy Composition diagram for R152a and R134a mixture
The calculations for liquid and vapor enthalpies at 15 bar and 5 bar for bubble temperature range and their corresponding values of ξ1
L and ξ1V will be done by the help of following
formulae:
At given temperature t:
hL = ξ1Lh1
L + ξ1Lh2
L
hV = ξ1V[hg1-Cp1(t1
sat-t)] + ξ1V[hg2+Cp2(t-t2
sat)]
A program had been made for calculating the enthalpies in liquid phase and vapor phase of R-152a and R-134a mixture with their respective bubble temperature and composition by using the above formulae.
The values obtained from above are tabulated as
Saturated liquid and vapor enthalpies of R-152a and R134a mixture at 15 bar
T (⁰C) ξ1L ξ1
V h1L(KJ/Kg) h2
L(KJ/Kg) hL(KJ/Kg) hV(KJ/Kg)55.2 0 0 300.74 279.79 279.8 425.255.6 0.055086 0.049474 301.53 280.42 281.6 431.156.0 0.118576 0.107223 302.33 281.06 283.6 437.856.4 0.185511 0.168967 303.13 281.70 285.7 445.056.8 0.255861 0.234851 303.93 282.34 287.9 452.757.2 0.330641 0.306004 304.73 282.98 290.2 460.957.6 0.409872 0.383698 305.53 283.62 292.6 469.758 0.493983 0.465620 306.34 284.27 295.2 479.258.4 0.583446 0.555578 307.14 284.91 297.9 489.558.8 0.678802 0.653520 307.95 285.56 300.8 500.659.2 0.780782 0.760683 308.76 286.21 303.8 512.759.6 0.889070 0.877332 309.57 286.85 307.1 525.960.0 1.00 1.00 310.38 287.50 310.4 539.7
Saturated liquid and vapor enthalpies of R-152a and R134a mixture at 5 bar
T (⁰C) ξ1L ξ1
V h1L(KJ/Kg) h2
L(KJ/Kg) hL(KJ/Kg) hV(KJ/Kg)16.0 0 0 227.69 221.87 221.9 407.616.4 0.143180 0.130454 228.39 222.43 223.3 422.116.8 0.238211 0.219188 229.10 222.99 224.4 432.217.2 0.341001 0.317166 229.81 223.54 225.7 443.217.6 0.452379 0.425773 230.51 224.10 227.0 455.518.0 0.573630 0.546990 231.22 224.66 228.4 469.118.4 0.706141 0.683196 231.93 225.22 230.0 484.3
18.8 0.851369 0.837139 232.64 225.78 231.6 501.519.0 0.929333 0.921878 233.0 226.06 232.5 511.019.2 1.000000 1.000000 233.35 226.34 233.4 519.6
From the tables of properties above the graph is plotted between enthalpy and composition separately for 15 bar and 5 bar pressure for the binary mixture R-152a and R-134a.
CALCULATION TO FIND OUT COP
We are going to substitute R-134a by this binary mixture R-152a/R-134a in an air conditioner operating on simple saturation cycle. The condenser and evaporator pressures are 15 bar and 5 bar.
Now calculation work will be done separately for various compositions of our binary mixture R-152a/R-134a. The most suitable composition of this binary mixture will be that for which the COP will be maximum in respect to R-134a for same AMTD design.
We are calculating the COP of the binary mixture for same AMTD in order to maintain condenser and evaporator size same, because we are not going to change the size of air conditioning machine.
Before going any further calculation to find out COP for separate composition of the binary mixture R-152a/R-134a, we have found out the values of temperature and enthalpies for various composition from the graph which is shown in the tabulated form below:
Composition of binary mixture
Pressure 5 bar
Pressure 5 bar
Pressure 15 bar
Pressure 15 bar
Pressure 5 bar
Pressure 5 bar
Pressure 15 bar
Pressure 15 bar
%R-134a(by mass)
tliquid
tf(⁰C)tvapor
ta(⁰C)tliquid
tc(⁰C)tvapor
te(⁰C)hliquid
(Kj/Kg)hvapor
(Kj/Kg)hliquid
(Kj/Kg)hvapor
(Kj/Kg)25 18.524 18.58 59.008 59.164 230.5 491.25 303.0 511.7540 18.08 18.164 58.48 58.59 228.5 474.50 298.50 494.7550 17.756 17.858 58.03 58.164 227.5 463.75 295.0 483.2560 17.42 17.52 57.55 57.7 226.5 452.5 292.25 472.070 17.05 17.13 57.04 57.18 225.5 441.0 289.0 460.2580 16.648 16.72 56.5 56.596 223.75 430.25 286.25 448.75
Calculation for binary mixture of 25% R-134a and 75% R-152a by mass:
The cycle abcd with the mixture is shown on p-h diagram
State a: saturated vapor
P = 5 bar , ta = 18.58 ⁰C, ha = 491.25 Kj/Kg
State b: superheated vapor
P = 15 bar, tb =? (to find it temperature from sb = sa
State c: saturated liquid
P = 15 bar, tc = 59.088 ⁰C, hc = 303.0 Kj/Kg
State d: liquid vapor mixture after throttling
P = 5 bar, hd = hc = 303 Kj/Kg
State e: saturated vapor
P = 15 bar, te = 59.164 ⁰C, he = 511.75 Kj/Kg
Entropy at a:
sa = (ξs)R-152a + (ξs)R-134a + Δsm (Δsm is the entropy of the mixing)
at ta = 18.58 ⁰C
sR-152a = 2.1015 Kj/KgK
sR-134a = 1.742468 Kj/KgK
sa = (0.75×2.1015) + (0.25×1.742468) + Δsm
sa = 2.01174 + Δsm
Entropy at e:
At te = 59.164 ⁰C
sR-152a = 2.0671 Kj/KgK
sR-134a = 1.7268 Kj/KgK
se = (ξs)R-152a + (ξs)R-134a + Δsm
= (0.75×2.0671) + (0.25×1.7268) + Δsm
= 1.9820 + Δsm
To solve the discharge temperature at b
At te = 59.164 ⁰C Cp1 = 1.13134 Kj/KgK Cp2 = 1.372824 Kj/KgK
sb = se + (ξ1Cp1lnTb/Te) + (ξ2 Cp2lnTb/Te) = sa
2.01174 + Δsm = 1.9820 + Δsm + (0.75×1.13134×lnTb/332.164) + (0.25×1.372824×lnTb/332.16)
Solving this we get
Tb = 340.5576 K (tb = 67.55 ⁰C)
Enthalpy at b:
hb = he + ξ1Cp1(tb-te) + (ξ2Cp2(tb-te)
= 511.75 + 75×1.13134×(67.55-59.164) + 0.25×1.372824×(67.55-59.164)
= 521.744 Kj/Kg
Temperature at d:
Overall composition and enthalpy during throttling process remain same. Hence
z = (hc - hf)/(ha - hf) ………………………………………………..(1)
z = (ξ1L- ξ1d)/ ξ1
L- ξ1V)………………………………………………(2)
The two equation can be solved for td by iteration td lies between tf and ta.
By iteration, td = 18.55 ⁰C
Theoretical COP of the cycle
E = (ha – hd) / (hb – ha)
= (491.25 – 303)/ (521.744 – 491.25)
= 6.1733
Estimating power saving and increment in COP of air-conditioner with R-152a/R-134a mixture:
For equivalent R-134a cycle
By AMTD
Tk = (te + tc)/2
= (59.164 + 59.088)/2
= 59.126 ⁰C
t0 = (ta + td)/2
= (18.58 + 18.55)/2
= 18.565 ⁰C
At t0 = 18.565 ⁰C
h1 = 408.9693 Kj/Kg, s1 = 1.7186 Kj/KgK
At tk = 59.126 ⁰C
h3 = 286.0897 Kj/Kg, h2’ = 426.3915 Kj/Kg, s2’ = 1.702937 Kj/KgK, Cp = 1.372216 Kj/KgK
s1 = s2 = s2’ + Cp lnT2/T2’
1.7186 = 1.702937 + 1.372216 lnT2/332.126
T2 = 335.9387 K (t2 = 62.9387 ⁰C)
h2 = h2’ + Cp (t2 - t2’)
= 426.3915 + 1.372216 (62.9387 – 59.126)
= 431.6233 Kj/Kg
COP of equivalent R-134a cycle:
E = (h1 - h4)/(h2 - h1)
= (408.9693 – 286.0879)/(431.6233 – 408.9693)
= 5.4243.
Result:
No increment in COP and power consumption increasing.
Calculation for binary mixture of 40% R-134a and 60% R-152a by mass:
The cycle abcd with the mixture is shown on p-h diagram
State a: saturated vapor
P = 5 bar , ta = 18.164 ⁰C, ha = 474.50 Kj/Kg
State b: superheated vapor
P = 15 bar, tb =? (to find it temperature from sb = sa
State c: saturated liquid
P = 15 bar, tc = 58.48 ⁰C, hc = 298.50 Kj/Kg
State d: liquid vapor mixture after throttling
P = 5 bar, hd = hc = 298.50 Kj/Kg
State e: saturated vapor
P = 15 bar, te = 58.59 ⁰C, he = 494.75 Kj/Kg
Entropy at a:
sa = (ξs)R-152a + (ξs)R-134a + Δsm (Δsm is the entropy of the mixing)
at ta = 18.164 ⁰C
sR-152a = 2.097554 Kj/KgK
sR-134a = 1.718718 Kj/KgK
sa = (0.6×2..097554) + (0.4×1.718718) + Δsm
sa = 1.9460196 + Δsm
Entropy at e:
At te = 58.59 ⁰C
sR-152a = 2.049715 Kj/KgK
sR-134a = 1.703205 Kj/KgK
se = (ξs)R-152a + (ξs)R-134a + Δsm
= (0.6×2.049715) + (0.25×1.703205) + Δsm
= 1.911031 + Δsm
To solve the discharge temperature at b
At te = 58.59 ⁰C Cp1 = 1.60405 Kj/KgK Cp2 = 1.36354 Kj/KgK
sb = se + (ξ1Cp1lnTb/Te) + (ξ2 Cp2lnTb/Te) = sa
1.9460196 + Δsm = 1.911031 + Δsm + (0.6×1.60405×lnTb/331.59) + (0.4×1.36354×lnTb/331.59)
Solving this we get
Tb = 339.3443 K (tb = 66.3443 ⁰C)
Enthalpy at b:
hb = he + ξ1Cp1(tb-te) + (ξ2Cp2(tb-te)
= 494.75 + 0.6×1.60405×(66.3743-58.59) + 0.4×1.36354×(66.3743-58.59)
= 506.4875 Kj/Kg
Temperature at d:
Overall composition and enthalpy during throttling process remain same. Hence
z = (hc - hf)/(ha - hf) ………………………………………………..(1)
z = (ξ1L- ξ1d)/ ξ1
L- ξ1V)………………………………………………(2)
The two equation can be solved for td by iteration td lies between tf and ta.
By iteration, td = 18.12 ⁰C
Theoretical COP of the cycle
E = (ha – hd) / (hb – ha)
= (474.5 – 298.5)/ (506.4875 – 474.5)
= 5.50215
Estimating power saving and increment in COP of air-conditioner with R-152a/R-134a mixture:
For equivalent R-134a cycle
By AMTD
Tk = (te + tc)/2
= (58.59 + 58.48)/2
= 58.535 ⁰C
t0 = (ta + td)/2
= (18.164 + 18.12)/2
= 18.142 ⁰C
At t0 = 18.142 ⁰C
h1 = 408.768 Kj/Kg, s1 = 1.718729 Kj/KgK
At tk = 58.535 ⁰C
h3 = 284.486 Kj/Kg, h2’ = 426.22375 Kj/Kg, s2’ = 1.7032325 Kj/KgK, Cp = 1.36266 Kj/KgK
s1 = s2 = s2’ + Cp lnT2/T2’
1.718729 = 1.7032325 + 1.36266 lnT2/331.535
T2 = 335.3268 K (t2 = 62.3268 ⁰C)
h2 = h2’ + Cp (t2 - t2’)
= 426.22375 + 1.36266 (62.3268 – 58.535)
= 431.39 Kj/Kg
COP of equivalent R-134a cycle:
E = (h1 - h4)/(h2 - h1)
= (408.7681 – 284.486)/(431.39 – 408.7681)
= 5.4938.
% increment in COP = 0.152%
Power saving = (WR-134a –Wmix)/WR-134a
=(1/ER134a-1/Emix)/(1/ER134a)
= 0.1517%
Calculation for binary mixture of 50% R-134a and 50% R-152a by mass:
The cycle abcd with the mixture is shown on p-h diagram
State a: saturated vapor
P = 5 bar , ta = 17.858 ⁰C, ha = 463.75 Kj/Kg
State b: superheated vapor
P = 15 bar, tb =? (to find it temperature from sb = sa)
State c: saturated liquid
P = 15 bar, tc = 58.03 ⁰C, hc = 295.0 Kj/Kg
State d: liquid vapor mixture after throttling
P = 5 bar, hd = hc = 295.0 Kj/Kg
State e: saturated vapor
P = 15 bar, te = 58.164 ⁰C, he = 483.25 Kj/Kg
Entropy at a:
sa = (ξs)R-152a + (ξs)R-134a + Δsm (Δsm is the entropy of the mixing)
at ta = 17.858 ⁰C
sR-152a = 2.098013 Kj/KgK
sR-134a = 1.718871 Kj/KgK
sa = (0.5×2..098013) + (0.5×1.718871) + Δsm
sa = 1.908442 + Δsm
Entropy at e:
At te = 58.164 ⁰C
sR-152a = 2.050236 Kj/KgK
sR-134a = 1.703418 Kj/KgK
se = (ξs)R-152a + (ξs)R-134a + Δsm
= (0.5×2.050236) + (0.5×1.703418) + Δsm
= 1.876827 + Δsm
To solve the discharge temperature at b
At te = 58.164 ⁰C Cp1 = 1.59766 Kj/KgK Cp2 = 1.356742 Kj/KgK
sb = se + (ξ1Cp1lnTb/Te) + (ξ2 Cp2lnTb/Te) = sa
1.908442 + Δsm = 1.876827 + Δsm + (0.5×1.59766×lnTb/331.164) + (0.5×1.35674×lnTb/331.16)
Solving this we get
Tb = 338.32795 K (tb = 65.32795 ⁰C)
Enthalpy at b:
hb = he + ξ1Cp1(tb-te) + (ξ2Cp2(tb-te)
= 483.25 + 0.5×1.59766×(65.32795-58.164) + 0.5×1.35674×(65.32795-58.164)
= 493.83259 Kj/Kg
Temperature at d:
Overall composition and enthalpy during throttling process remain same. Hence
z = (hc - hf)/(ha - hf) ………………………………………………..(1)
z = (ξ1L- ξ1d)/ ξ1
L- ξ1V)………………………………………………(2)
The two equation can be solved for td by iteration td lies between tf and ta.
By iteration, td = 17.78 ⁰C
Theoretical COP of the cycle
E = (ha – hd) / (hb – ha)
= (463.75 – 295.0)/ (493.83259 – 463.75)
= 5.60956
Estimating power saving and increment in COP of air-conditioner with R-152a/R-134a mixture:
For equivalent R-134a cycle
By AMTD
Tk = (te + tc)/2
= (58.164 + 58.03)/2
= 58.097 ⁰C
t0 = (ta + td)/2
= (17.858 + 17.78)/2
= 17.819 ⁰C
At t0 = 17.819 ⁰C
h1 = 408.59045 Kj/Kg, s1 = 1.7188905 Kj/KgK
At tk = 58.097 ⁰C
h3 = 284.4252 Kj/Kg, h2’ = 426.0991 Kj/Kg, s2’ = 1.7034515 Kj/KgK,
Cp = 1.3557035 Kj/KgK
s1 = s2 = s2’ + Cp lnT2/T2’
1.7188905 = 1.7034515 + 1.3557035 lnT2/331.097
T2 = 334.887663 K (t2 = 61.887663 ⁰C)
h2 = h2’ + Cp (t2 - t2’)
= 426.0991 + 1.3557035 (61.887663 – 58.097)
= 431.2381151 Kj/Kg
COP of equivalent R-134a cycle:
E = (h1 - h4)/(h2 - h1)
= (408.59045 – 284.4252)/(431.2381151 – 408.59045)
= 5.482474.
% increment in COP = 2.318%
Power saving = (WR-134a –Wmix)/WR-134a
=(1/ER134a-1/Emix)/(1/ER134a)
= 2.2655%
Calculation for binary mixture of 60% R-134a and 40% R-152a by mass:
The cycle abcd with the mixture is shown on p-h diagram
State a: saturated vapor
P = 5 bar , ta = 17.52 ⁰C, ha = 452.5 Kj/Kg
State b: superheated vapor
P = 15 bar, tb =? (to find it temperature from sb = sa)
State c: saturated liquid
P = 15 bar, tc = 57.55 ⁰C, hc = 292.25 Kj/Kg
State d: liquid vapor mixture after throttling
P = 5 bar, hd = hc = 292.25 Kj/Kg
State e: saturated vapor
P = 15 bar, te = 57.7 ⁰C, he = 472 Kj/Kg
Entropy at a:
sa = (ξs)R-152a + (ξs)R-134a + Δsm (Δsm is the entropy of the mixing)
at ta = 17.52 ⁰C
sR-152a = 2.09842 Kj/KgK
sR-134a = 1.7190 Kj/KgK
sa = (0.4×2..09842) + (0.6×1.7190) + Δsm
sa = 1.870768 + Δsm
Entropy at e:
At te = 57.7 ⁰C
sR-152a = 2.0508 Kj/KgK
sR-134a = 1.70365 Kj/KgK
se = (ξs)R-152a + (ξs)R-134a + Δsm
= (0.4×2.0508) + (0.6×1.70365) + Δsm
= 1.84251 + Δsm
To solve the discharge temperature at b
At te = 57.7 ⁰C Cp1 = 1.5908 Kj/KgK Cp2 = 1.34955 Kj/KgK
sb = se + (ξ1Cp1lnTb/Te) + (ξ2 Cp2lnTb/Te) = sa
1.870768 + Δsm = 1.84251 + Δsm + (0.4×1.5908×lnTb/330.7) + (0.4×1.34955×lnTb/330.7)
Solving this we get
Tb = 337.2259 K (tb = 64.225 ⁰C)
Enthalpy at b:
hb = he + ξ1Cp1(tb-te) + (ξ2Cp2(tb-te)
= 472 + 0.4×1.5908×(64.225-57.7) + 0.6×1.34955×(64.225-57.7)
= 481.4354763 Kj/Kg
Temperature at d:
Overall composition and enthalpy during throttling process remain same. Hence
z = (hc - hf)/(ha - hf) ………………………………………………..(1)
z = (ξ1L- ξ1d)/ ξ1
L- ξ1V)………………………………………………(2)
The two equation can be solved for td by iteration td lies between tf and ta.
By iteration, td = 17.44 ⁰C
Theoretical COP of the cycle
E = (ha – hd) / (hb – ha)
= (452.5 – 292.25)/ (481.4354763 – 452.5)
= 5.538184
Estimating power saving and increment in COP of air-conditioner with R-152a/R-134a mixture:
For equivalent R-134a cycle
By AMTD
Tk = (te + tc)/2
= (57.7 + 57.55)/2
= 57.625 ⁰C
t0 = (ta + td)/2
= (17.52 + 17.44)/2
= 17.48 ⁰C
At t0 = 17.48 ⁰C
h1 = 408.41 Kj/Kg, s1 = 1.7190 Kj/KgK
At tk = 57.625 ⁰C
h3 = 283.66 Kj/Kg, h2’ = 425.9575 Kj/Kg, s2’ = 1.7036875 Kj/KgK,
Cp = 1.3483875 Kj/KgK
s1 = s2 = s2’ + Cp lnT2/T2’
1.7190 = 1.7036875 + 1.3483875 lnT2/330.625
T2 = 334.40 K (t2 = 61.40 ⁰C)
h2 = h2’ + Cp (t2 - t2’)
= 425.9575 + 1.3483875 (61.40 – 57.625)
= 431.0476 Kj/Kg
COP of equivalent R-134a cycle:
E = (h1 - h4)/(h2 - h1)
= (408.41 – 283.66)/(431.0476 – 408.41)
= 5.51.
% increment in COP = 0.51%
Power saving = (WR-134a –Wmix)/WR-134a
=(1/ER134a-1/Emix)/(1/ER134a)
= 0.5089%
Calculation for binary mixture of 70% R-134a and 30% R-152a by mass:
The cycle abcd with the mixture is shown on p-h diagram
State a: saturated vapor
P = 5 bar , ta = 17.13 ⁰C, ha = 441 Kj/Kg
State b: superheated vapor
P = 15 bar, tb =? (to find it temperature from sb = sa)
State c: saturated liquid
P = 15 bar, tc = 57.04 ⁰C, hc = 289 Kj/Kg
State d: liquid vapor mixture after throttling
P = 5 bar, hd = hc = 289 Kj/Kg
State e: saturated vapor
P = 15 bar, te = 57.18 ⁰C, he = 460.25 Kj/Kg
Entropy at a:
sa = (ξs)R-152a + (ξs)R-134a + Δsm (Δsm is the entropy of the mixing)
at ta = 17.13 ⁰C
sR-152a = 2.09897 Kj/KgK
sR-134a = 1.719135 Kj/KgK
sa = (0.3×2..09897) + (0.7×1.719135) + Δsm
sa = 1.8330855 + Δsm
Entropy at e:
At te = 57.18 ⁰C
sR-152a = 2.05142 Kj/KgK
sR-134a = 1.70391 Kj/KgK
se = (ξs)R-152a + (ξs)R-134a + Δsm
= (0.3×2.05142) + (0.7×1.70391) + Δsm
= 1.808163 + Δsm
To solve the discharge temperature at b
At te = 57.18 ⁰C Cp1 = 1.58321 Kj/KgK Cp2 = 1.3416 Kj/KgK
sb = se + (ξ1Cp1lnTb/Te) + (ξ2 Cp2lnTb/Te) = sa
1.8330855 + Δsm = 1.808163 + Δsm + (0.3×1.58321×lnTb/330.18) + (0.7×1.3416×lnTb/330.18)
Solving this we get
Tb = 336.05084 K (tb = 63.05084 ⁰C)
Enthalpy at b:
hb = he + ξ1Cp1(tb-te) + (ξ2Cp2(tb-te)
= 460.25 + 0.3×1.58321×(63.05084-57.18) + 0.7×1.3416×(63.05084-57.18)
= 468.55 Kj/Kg
Temperature at d:
Overall composition and enthalpy during throttling process remain same. Hence
z = (hc - hf)/(ha - hf) ………………………………………………..(1)
z = (ξ1L- ξ1d)/ ξ1
L- ξ1V)………………………………………………(2)
The two equation can be solved for td by iteration td lies between tf and ta.
By iteration, td = 17.07 ⁰C
Theoretical COP of the cycle
E = (ha – hd) / (hb – ha)
= (441 – 289)/ (468.55 – 441)
= 5.51724
Estimating power saving and increment in COP of air-conditioner with R-152a/R-134a mixture:
For equivalent R-134a cycle
By AMTD
Tk = (te + tc)/2
= (57.18 + 57.04)/2
= 57.11 ⁰C
t0 = (ta + td)/2
= (17.13 + 17.07)/2
= 17.1 ⁰C
At t0 = 17.1 ⁰C
h1 = 408.205 Kj/Kg, s1 = 1.71915 Kj/KgK
At tk = 57.11 ⁰C
h3 = 282.836 Kj/Kg, h2’ = 425.803 Kj/Kg, s2’ = 1.703945 Kj/KgK,
Cp = 1.34055 Kj/KgK
s1 = s2 = s2’ + Cp lnT2/T2’
1.71915 = 1.703945 + 1.34055 lnT2/330.11
T2 = 333.875 K (t2 = 60.875 ⁰C)
h2 = h2’ + Cp (t2 - t2’)
= 425.803 + 1.34055 (60.875 – 57.11)
= 430.85 Kj/Kg
COP of equivalent R-134a cycle:
E = (h1 - h4)/(h2 - h1)
= (408.205 – 282.836)/(430.85 – 408.205)
= 5.536.
% increment in COP = no increment in COP is obtained
Power saving = (WR-134a –Wmix)/WR-134a
=(1/ER134a-1/Emix)/(1/ER134a)
= no power saving i.e., power consumption is increasing.
Calculation for binary mixture of 80% R-134a and 20% R-152a by mass:
The cycle abcd with the mixture is shown on p-h diagram
State a: saturated vapor
P = 5 bar , ta = 16.72 ⁰C, ha = 430.25 Kj/Kg
State b: superheated vapor
P = 15 bar, tb =? (to find it temperature from sb = sa)
State c: saturated liquid
P = 15 bar, tc = 56.5 ⁰C, hc = 286.25 Kj/Kg
State d: liquid vapor mixture after throttling
P = 5 bar, hd = hc = 286.25 Kj/Kg
State e: saturated vapor
P = 15 bar, te = 56.596 ⁰C, he = 448.75 Kj/Kg
Entropy at a:
sa = (ξs)R-152a + (ξs)R-134a + Δsm (Δsm is the entropy of the mixing)
at ta = 16.72 ⁰C
sR-152a = 2.09952 Kj/KgK
sR-134a = 1.7193 Kj/KgK
sa = (0.2×2..09952) + (0.8×1.7193) + Δsm
sa = 1.795344 + Δsm
Entropy at e:
At te = 56.596 ⁰C
sR-152a = 2.052104 Kj/KgK
sR-134a = 1.704202 Kj/KgK
se = (ξs)R-152a + (ξs)R-134a + Δsm
= (0.2×2.052104) + (0.8×1.704202) + Δsm
= 1.7737824 + Δsm
To solve the discharge temperature at b
At te = 56.596 ⁰C Cp1 = 1.574842 Kj/KgK Cp2 = 1.33284 Kj/KgK
sb = se + (ξ1Cp1lnTb/Te) + (ξ2 Cp2lnTb/Te) = sa
1.795344 + Δsm = 1.7737824 + Δsm + (0.2×1.57484×lnTb/329.596) + (0.8×1.3328×lnTb/329.59)
Solving this we get
Tb = 334.7815 K (tb = 61.7815 ⁰C)
Enthalpy at b:
hb = he + ξ1Cp1(tb-te) + (ξ2Cp2(tb-te)
= 448.75 + 0.2×1.57484×(61.7815-56.596) + 0.8×1.33284×(61.7815-56.596)
= 455.9124 Kj/Kg
Temperature at d:
Overall composition and enthalpy during throttling process remain same. Hence
z = (hc - hf)/(ha - hf) ………………………………………………..(1)
z = (ξ1L- ξ1d)/ ξ1
L- ξ1V)………………………………………………(2)
The two equation can be solved for td by iteration td lies between tf and ta.
By iteration, td = 16.66 ⁰C
Theoretical COP of the cycle
E = (ha – hd) / (hb – ha)
= (430.25 – 286.25)/ (455.9124 – 430.25)
= 5.61132
Estimating power saving and increment in COP of air-conditioner with R-152a/R-134a mixture:
For equivalent R-134a cycle
By AMTD
Tk = (te + tc)/2
= (56.596 + 56.5)/2
= 56.548 ⁰C
t0 = (ta + td)/2
= (16.72 + 16.66)/2
= 16.69 ⁰C
At t0 = 16.69 ⁰C
h1 = 407.9345 Kj/Kg, s1 = 1.719305 Kj/KgK
At tk = 56.548 ⁰C
h3 = 281.9368 Kj/Kg, h2’ = 425.6344 Kj/Kg, s2’ = 1.704226 Kj/KgK,
Cp = 1.33212 Kj/KgK
s1 = s2 = s2’ + Cp lnT2/T2’
1.719305 = 1.704226 + 1.33212 lnT2/329.548
T2 = 333.29953 K (t2 = 60.29953 ⁰C)
h2 = h2’ + Cp (t2 - t2’)
= 425.6344 + 1.33212 (60.29953 – 56.548)
= 430.632 Kj/Kg
COP of equivalent R-134a cycle:
E = (h1 - h4)/(h2 - h1)
= (407.9345 – 281.9368)/(430.632 – 407.9345)
= 5.55
% increment in COP = 1.10%
Power saving = (WR-134a –Wmix)/WR-134a
=(1/ER134a-1/Emix)/(1/ER134a)
= 1.09279%
COMMENTS
Comments about the calculations done for respective compositions of binary mixture R-152a/R-134a:
25%R-134a and 75%R-152a by mass: for this composition of binary mixture the COP is decreasing and power consumption is increasing which is against to our desired results. So this composition is not suitable to R-134a in air conditioning machine, which is working between 5 bar evaporator pressure and 15 bar condenser pressure.
40%R-134a and 60%R-152a by mass: for this composition of binary mixture the COP is increasing by 0.152% and power saving is 0.1517% which is according to our desired results. So this composition may be a substitute to R-134a in air conditioning machine, which is working between 5 bar evaporator pressure and 15 bar condenser pressure.
50%R-134a and 50%R152a by mass: for this composition of binary mixture the COP is increasing by 2.318% and power saving is 2.2655% which is according to our desired results. So this composition may be a substitute to R-134a in air conditioning machine, which is working between 5 bar evaporator pressure and 15 bar condenser pressure.
60%R-134a and 40%R-152a by mass: for this composition of binary mixture the COP is increasing by 0.51% and power saving is 0.5089% which is according to our desired results. So this composition may be a substitute to R-134a in air conditioning machine, which is working between 5 bar evaporator pressure and 15 bar condenser pressure.
70% R-134a and 30%R-152a by mass: for this composition of binary mixture the COP is decreasing and power consumption is increasing which is against our desired results. So this composition is not suitable substitute to R-134a in air conditioning machine, which is working between 5 bar evaporator pressure and 15 bar condenser pressure.
80%R-134a and 20%R-152a by mass: for this composition of binary mixture the COP is increasing by 1.10% and power saving is 1.09279% which is according to our desired results. So this composition may be a substitute to R-134a in air conditioning machine, which is working between 5 bar evaporator pressure and 15 bar condenser pressure.
CONCLUSIONAfter gone through the intricate steps of calculation and graphical analysis, we
finally become able to increase the performance of air conditioning machine by using a binary mixture of composition 50% R-152a/50%R-134a.
This improves the performance by increasing COP 2.318% and power saving by 2.2655% with respect to single refrigerant R-134a.
Along with the advantage mention above the ingredients are non toxic and environment friendly (zero ODP and low GWP) and also suitable for safety purposes.
REFERENCES
BOOKS: “Refrigeration and Air-conditioning” by C.P. Arora “A text book of Refrigeration and Air-conditioning” by R.S. Khurmi & J.K.
Gupta. “Refrigeration and Air-conditioning” by R.K. Rajput. “A course in Refrigeration and Air-conditioning” by S.C. Arora and S.
Domkundwar. “Thermal Engineering” by P.L. Ballany.
Web sites:
http://webbook.nist.gov/ http://www.aircondition.com/ http://refrigerants.com/ http://icorinternational.com/