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4. METHODOLOGY

Our aim is to develop a non-electric solar refrigerator-freezer. The refrigerator uses an aqua-ammonia absorption system similar to that used in propane refrigerators. The refrigerator consists of two separate units; the solar collector-generator and the refrigerator box.     

The collector-generator consists of a thermal solar collector and most of the refrigeration works. It needs to be mounted in a sunny location, the same as any solar collector.

The refrigerator box is the refrigerator per se and can be placed wherever is convenient, presumably the kitchen.When the sun shines, the collector-generator produces ammonia refrigerant, which is stored until night when the actual cooling takes place. To keep the refrigerator cold through the day and during cloudy weather, there is built-in storage. There are no moving parts.

In use, our solar refrigerator is little different from any other refrigerator. Currently, the drawbacks are a greater temperature variation since there is only one cycle per day.

We are also studying a very small unit where the refrigerator box is under the solar collector so it is all one unit that is left outside. This is meant to be used for vaccines and to be as low cost and simple as possible. We intend to include a freezer for a limited ice-making capacity, as this simplifies the design.

We have been looking seriously at auxiliary cooling for prolonged cloudy spells. We are trying an external heat pipe for colder locations and thermoelectric cooling for warmer locations. These have the advantage of being relatively inexpensive and can be offered as add-ons instead of requiring extensive redesigning.

The other technology, which can be implemented, is keeping the cold storage outside at the solar collector and using a pump to circulate the cold into the refrigerator. This would add significantly to the expense and require electricity, but would simplify installation and keep ammonia out of the living space.

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5.Absorption refrigeration using solar energy

Refrigeration is gaining very much importance in today’s world. But the refrigeration that we see us uses electricity-powered generation.

In some places electricity is not readily available and also it is becoming costlier, so we have to go for non-conventional type of refrigeration. Solar powered refrigeration is useful where sunshine is in abundance in countries like India.

5.1 Components:

The main components of solar refrigerator using intermittent absorption cycle are as follows. (Ref Fig 5.1.1)

1. Flat Plate Collector:

The flat plate collector acts as a generator during daytime and as absorber during night. It consists of an upper header and a lower header of GI material and the two headers connected by means of a number of collector tubes in between.

Ammonia – water solution is placed inside this system. During the daytime, ammonia gets evaporated from the solution by absorbing solar energy. During nighttime, ammonia vapor is absorbed back in the weak solution.

2. Condenser

During the day time ammonia vapor gets collected and condensed in the condenser evaporator. During nighttime the condensed ammonia absorbs heat from the system to be cooled, vaporizes and returns to the collector.

3. Valves

There are totally three valves in the system one valve is in the charging line, second valve line is between upper header and the condenser. The third valve is in the vapor return line.

5.2 Working of the system

The generator is charged with pre-generated amount of water and ammonia to give the required concentration of ammonia in the solution. At the commencement of the refrigeration period the line is opened between the upper header and the condenser, at the end of regeneration period, the condenser is isolated from the rest of the system and the generator absorber is allowed to cool. To carry out refrigeration the valve between the evaporator and the collector is opened.

So the ammonia liquid in the evaporator absorbs the heat from the water to be cooled, gets evaporated and this vapor returns to the collector where it is absorbed again. This cycle is repeated again the next day.

     As this process of evaporation continues, the temperature of water gets reduced and thus refrigeration is achieved.

5.3  WORKING FLUID (REFRIGERANT)

5.3.1   Definition:

      Any substance that absorbs heat through expansion or vaporization may be called a refrigerant. A broader definition may include such secondary cooling media as brine solution and cold water.

5.3.2   Requirements for a refrigerant:

       These are certain desirable characteristics which a fluid used as a refrigerant should possess.

1)                  It should be non-poisonous

2)                  It should be non-explosive

3)                  Condensing pressure should not be excessive

4)                  Low boiling temperature at atmospheric pressure

5)                  High critical temperature

6)                  High latent heat of vaporization

7)                  Low specific heat of liquid

8)                  Low specific volume of vapor

9)                  It should be non-corrosive

10)              Chemically stable

11)              Ease of locating leaks

12)              Availability, low cost and ease of handling

13)              Satisfactory heat transfer and viscosity coefficients

14)              Freezing temperature of the liquid should be below any temperature at which the evaporator might operate

15)              Low compressor discharge temperatures are desirable

5.3.3        Classification of refrigerants:

       The National Refrigeration Society Code, U.S.A., catalogues all the refrigerants into three groups.

These are:

         Group-One  :( safest of refrigerants).

R-113, R-611, R-11, R-21, R-114, R-12,

R-30, R-22, R-744, R-502, R-13, R-14, R-500.

         Group-Two : ( toxic and somewhat flammable refrigerants).

R-1130, R-611, R-160, R-764, R-40, R-717.

         Group-Three :( flammable refrigerants)

R-600, R-601, R-290, R-170, R-1150, R-50.

We used ammonia as the refrigerant. Ammonia comes under group two refrigerants and is denoted by R-717. It was one of the first refrigerants used.

It is used in large industrial installations. It is colorless. It’s boiling temperature at atmospheric pressure is –33.8C and the melting point from the solid is –77.8C.It is somewhat flammable and forms an explosive mixture with air. It attacks bronze in the presence of a little moisture but does not corrode iron and steel. This refrigerant is extremely soluble in water.

 Ammonia leaks can be easily detected by smell or by burning Sulphur candles or wicks, which generate a dense cloud of white smoke in the presence of ammonia vapor.

6. THERMODYNAMIC ANALYSIS

1)            Amount of ice to be produced is                     5 kg

2)            Refrigerant used is ammonia                                     (NH₃)

3)            Absorbent used is water                                 (H₂O)

4)            Average ambient temperature at  Davangere.

                  During day time    =                               32⁰C

                  During night time  =                               27⁰C

5)            Amount of heat to be removed from water to convert it into ice

         Q     =        mS ( ta – tb) + mL

         Where

         M   =        Mass of ice to be produced                              =   5 kg

         S_W  =      Secific heat of water                                        =   4.187 kJ/ kg k

         T_A  =       Ambient temperature           =    +32⁰C             =   305⁰k

         T_B  =       Temperature of ice               =    –10⁰C              =   263⁰K

         L    =        Latent heat of water at 0^oC                                =  335 kJ/kg

         Q   =        5 [ 4.187(305 – 273) + 2.0935 (273 – 263) + 335 ]

         Q   =        12449.535 

                =       2450 kJ/kg

This is the heat load or refrigerator effect. Considering heat flow through insulation let the heat load be equal to 2450 kJ.

(6)     Condenser temperature,   t_c   =         40⁰C

(7)     Evaporater temperature,  t_{e     =              –10}⁰_C

_{(8)        Enthalpy of ammonia fluid at condenser temperature}

                                                    h    =         390.587 kJ/kg

(9)     Enthalpy of vapor ammonia at evaporator temperature

                                             =  hg    =          1450.22 kJ/kg

(10)     Refrigerating effect,

                                             Qe   =      hg  –   hf 

                                                =  145.22  –  390.587

                                                =  1059.63 kJ/kg

         This refrigerating effect is to be obtained in 12 hours. Therefore, heat removed,

         Qe             =       2450 / ( 12  x  3600)

                           =       0.0567 kJ/S

 (11)      Let  m_a  be the mass of ammonia required

                  Qe    =   m_a   –   q_e

Therefore,           

                  m_a                 =            Qe/q_e

                                       =            0.0567/1059.53

                                       =            5.35  x 10^–5  x  12 x  3600

                                       =            2.312  kg

         This is the mass of ammonia required to obtain the refrigerating effect for 12  hours, let it be  2.5  kg.

 (12)     Calculation of coefficient of performance of refrigeration (COP)

                                              T2 ( T3  –  T1)

                  [COP]     =     ---------------------------

                                              T3 (T1  – T2)    

                  Where

                  T1           =    Condenser temperature               =  40⁰C    =  313⁰k

                  T2           =    Evaporater temperature               =  –10⁰C  =  263⁰C

                  T3           =    Generator temperature                 =    70⁰C   =   343⁰k

                                                                                                                                                                                 263 ( 343  –  313 )

                  [COP]      =   ------------------------------  =  0.46

                                             343( 313 – 263 )

                  Hence co-efficient of performance pf the system

                  [COP]          =    0.46

6.1    Design of Flat Plate Collector

Volume and area calculations:

i)    area of the corrugated sheet

                                 =          length x breath

                                 =          ( 95 x 94 ) cm²

                                            =      8930 cm²   =  0.8930 m²

ii)     volume of the upper header

                        length, l             =    1.1 m  =  110 cm

                        diameter,d         =     4”      =  10.16 cm

        volume  = pd2/4 x l  =  p x (10.16)2/4  x 110

                    = 8918.05 cm³

                    = 8.91805 liters

iii)    volume of the lower header

                                    length ,l                        = 1.1 m            =  110 cm

                                    diameter.d                   = 2”                  =  5.08 cm

            volume                 =         pd²/4 x l        

                    =  p(5.08)²/(4) x 110

                   =  2229.5 cm3        

                   =  2.23 liters

iv)     volume of the connecting pipes

                        length of each pipe        =  112 cm

                diameter of each pipe     =  1.27 cm

                number of pipes             =  12 nos

        volume  =  pd²/4 x l x12

                    =  p (1.27)²/4 x 112 x 12             

                    =  1702.5 cm³ =  1.702 liters

v)     volume of the collector

                             =    volume of upper header + volume of lower

                       header +  volume of connecting pipes

                   =  8.918 + 2.23 + 1.702

           =    12.85 liters

CALCULATIONS F PLAE EFFICIENCY FACTOR AND OVERALL COLLECTOR EFFICIENCY:

CALCULATIONS:

(1)  to find the overall heat loss coefficient, U_L

        U       =   Ut   +   Ub

        Ut      =   top loss coeffient

        Ub     =   back loss coeffient

Where

        Ut      =   [ (l/h_pc  +  h_rpc) + (l/h_w +  h_rcs)]^–1

Where

        h_pc     =   Heat tranfer coefficient from plate to cover

(glass)

        h_rpc    =   radiation heat tranfer coefficient from plate to

                       cover plate

hw      =   heat tranfer coefficient of wind blowing over collector

h_rcs    =   radiation heat transfer coeffient from cover (glass) to sky

h_pc     =   [(1– 0.0018)  (T-10)] x 1.14   dt^0.31] ^0.070

T        =   average temperature between plate and cover

dt       =   difference in temperature between plate and cover

l         =   distance between plate and cover

hrpc    =   g ( T²p + T²c) (Tp + Tc)

              –––––––––––––––––––––––

                (l/Ep)  +  (l/Ec)  –1

Tp       =  plate yemperature in ⁰K

Tc       =  cover glass temperature in K

Ep       =  plate emittance

Ec       =  cover emittance

g         =  Stefan boltzman constant

h         =  Ec g (T²c + T²s)² (Tc + Ts) 

Tc       =   cover glass temperature

Ts       =   sky glass temperature + ambient temperature

hw      =   5.7  +  3.8 (V)

V        =   velocity of wind blowing over collector

TO FIND THE HEAT LOSS CO–EFFICIENT, U_L

Plate-to-cover spacing                                    =   8 cm                       =  l

Plate emittance                                               =   0.821                      = Ep

Ambient air + sky temperature                       =   32⁰C                       =  Ts

Wind speed                                                     =   5m/sec                   = v

(from wind mill data)                                      

back insulation thickness                                =   10cm         

insulation conductivity                                     =   0.065 w/m-⁰C       

mean plate temperature                                 =   90⁰C

(assumed)

Ub                    = K/L

                         = 0.065/8 x 10^–2

                        =   0.8125 W/m^2-0C

Ut                    =   [(l/h_pc + h_rps)  +  (l/h_w + h_rcs)]^–1

Hpc                 =   [l–0.0018 (T – 10)] x 1.14  dt0.310

                                                            –––––––

                                                             l0.070

T                      =   45 + 90/2                            =   67.5 ⁰C

dt                     =   90 – 45                               =   45 ⁰C

L                      =   8 cms

hpc                  =   [( l – 0.0018 x (67.5 – 10)] x 1.14 x   (45)^0.31/(8)^0.07

                        =   [ l – 0.0018 (57.5)] x 1.14 x 3.25/1.15

                        =   2.88 W/m^2–0C

hrpc                 =       (T²p + T²c)  (T_p + T_c)

                             –––––––––––––––––––––

                                    (1/Ep) + (1/Ec) – 1

=       4.87 x 10^–8 [(90+ 73)²+(45+273)²] [(90+273)+(45+273)]

    –––––––––––––––––––––––––––––––––––––––––––––––––

                   1/0.821 + 1/0.837 – 1

                        =   6.33 W/m^2–0C

hrcs                 =   Ec g (T²c +  T²s) (tc + ts)

                        =   0.837 x 4.87 x 10^–8 (318² + 305² ) (318 + 305)

                        =   5.73 W/m^2–0C

hw                   =   5.7  +  3.8 (V)

                        =   5.7  +  3.8  (5)

                        =   24.7

U_{t                            =   [1/(hpc +hrps) + 1/(hs + hrcs)]}

_{hpc                          =   2.88 W/m}²

_{hrpc                         =   6.33 W/m}²

_{hw                            =   24.7 W/m}²

_{hrcs                          =   5.73  W/m}²

Ut                     =   [1/2.88 + 6.33) + 1/(24.7 + 5.73)]– 1

                        =   7.074 W/m^2–0C

Bottom layer coefficient, Ub:

Ub                    =    K/L

K                      =    thermal conductivity of insulation

                              (Glass wool)

                        =    0.056    ~Kcal/m²hr⁰C

                        =    0.056/0.86

                        =    0.065W/m⁰C

L                      =   insulation thickness    =   8 cm

U_b                    =      0.035/8 x 10–2

                        =   0.8125

U_L                    =   U_t + U_b   =   7.07  +  0.8125

                        =  7.9

II      PLATE EFFICIENCY FACTOR, Fp:

            I                      WUL        WUL        WUL         W

      ––––             =   –––––– + ––––––– + –––––– + –––––––

       Fp                       pdh       pdk’/m        Cb        b + 2bF   

            Kp                   =    thermal conductivity of plate material

                                     =   59.86 W/m⁰C 

            W                    =   pitch  =  0.1 m

PLATE EFFICIENCY FACTOR. Fp = 69.53 %

QUANTITY OF AQUA – AMMONIA USED

Since the volume of the collector is 12.85 liter, the quantity of ammonia solution to be used is about 12 liters.

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