increasing cop of ac.docx

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.




P = 5 bar , ta = 18.164 ⁰C, ha = 474.50 Kj/Kg


P = 15 bar, tb =? (to find it temperature from sb = sa


P = 15 bar, tc = 58.48 ⁰C, hc = 298.50 Kj/Kg


P = 5 bar, hd = hc = 298.50 Kj/Kg


P = 15 bar, te = 58.59 ⁰C, he = 494.75 Kj/Kg

Entropy at a:


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




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:


= 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:


z = (hc - hf)/(ha - hf) ………………………………………………..(1)

z = (ξ1L- ξ1d)/ ξ1

L- ξ1V)………………………………………………(2)




E = (ha – hd) / (hb – ha)

= (474.5 – 298.5)/ (506.4875 – 474.5)

= 5.50215



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


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%




P = 5 bar , ta = 17.858 ⁰C, ha = 463.75 Kj/Kg


P = 15 bar, tb =? (to find it temperature from sb = sa)


P = 15 bar, tc = 58.03 ⁰C, hc = 295.0 Kj/Kg


P = 5 bar, hd = hc = 295.0 Kj/Kg


P = 15 bar, te = 58.164 ⁰C, he = 483.25 Kj/Kg

Entropy at a:


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




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:


= 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:


z = (hc - hf)/(ha - hf) ………………………………………………..(1)

z = (ξ1L- ξ1d)/ ξ1

L- ξ1V)………………………………………………(2)




E = (ha – hd) / (hb – ha)

= (463.75 – 295.0)/ (493.83259 – 463.75)

= 5.60956



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


E = (h1 - h4)/(h2 - h1)

= (408.59045 – 284.4252)/(431.2381151 – 408.59045)

= 5.482474.



=(1/ER134a-1/Emix)/(1/ER134a)

= 2.2655%




P = 5 bar , ta = 17.52 ⁰C, ha = 452.5 Kj/Kg




P = 15 bar, tc = 57.55 ⁰C, hc = 292.25 Kj/Kg


P = 5 bar, hd = hc = 292.25 Kj/Kg


P = 15 bar, te = 57.7 ⁰C, he = 472 Kj/Kg

Entropy at a:


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




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:


= 472 + 0.4×1.5908×(64.225-57.7) + 0.6×1.34955×(64.225-57.7)

= 481.4354763 Kj/Kg

Temperature at d:


z = (hc - hf)/(ha - hf) ………………………………………………..(1)

z = (ξ1L- ξ1d)/ ξ1

L- ξ1V)………………………………………………(2)




E = (ha – hd) / (hb – ha)

= (452.5 – 292.25)/ (481.4354763 – 452.5)

= 5.538184



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


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


E = (h1 - h4)/(h2 - h1)

= (408.41 – 283.66)/(431.0476 – 408.41)

= 5.51.



=(1/ER134a-1/Emix)/(1/ER134a)

= 0.5089%




P = 5 bar , ta = 17.13 ⁰C, ha = 441 Kj/Kg




P = 15 bar, tc = 57.04 ⁰C, hc = 289 Kj/Kg


P = 5 bar, hd = hc = 289 Kj/Kg


P = 15 bar, te = 57.18 ⁰C, he = 460.25 Kj/Kg

Entropy at a:


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




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:


= 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:


z = (hc - hf)/(ha - hf) ………………………………………………..(1)

z = (ξ1L- ξ1d)/ ξ1

L- ξ1V)………………………………………………(2)




E = (ha – hd) / (hb – ha)

= (441 – 289)/ (468.55 – 441)

= 5.51724



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


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


E = (h1 - h4)/(h2 - h1)

= (408.205 – 282.836)/(430.85 – 408.205)

= 5.536.

% increment in COP = no increment in COP is obtained


=(1/ER134a-1/Emix)/(1/ER134a)

= no power saving i.e., power consumption is increasing.




P = 5 bar , ta = 16.72 ⁰C, ha = 430.25 Kj/Kg




P = 15 bar, tc = 56.5 ⁰C, hc = 286.25 Kj/Kg


P = 5 bar, hd = hc = 286.25 Kj/Kg


P = 15 bar, te = 56.596 ⁰C, he = 448.75 Kj/Kg

Entropy at a:


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




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:


= 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:


z = (hc - hf)/(ha - hf) ………………………………………………..(1)

z = (ξ1L- ξ1d)/ ξ1

L- ξ1V)………………………………………………(2)




E = (ha – hd) / (hb – ha)

= (430.25 – 286.25)/ (455.9124 – 430.25)

= 5.61132



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


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


E = (h1 - h4)/(h2 - h1)

= (407.9345 – 281.9368)/(430.632 – 407.9345)

= 5.55



=(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.


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.


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/

http://webbook.nist.gov/

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