COMPRESSOR LESS PORTABLE REFRIGERATOR

BLOCK DIAGRAM

PELTIER EFFECT

The thermoelectric refrigerator works on the PELTIER effect that The

Peltier–Seebeck effect, or thermoelectric effect, is the direct conversion of

thermal differentials to electric voltage and vice versa. Related effects are

the Thomson effect and Joule heating. The Peltier–Seebeck and Thomson

effects are reversible (in fact, the Peltier and Seebeck effects are reversals

of one another); Joule heating cannot be reversible under the laws of

thermodynamics.

Seebeck effect

The Seebeck effect is the conversion of temperature differences

directly into electricity. This effect was first discovered, accidentally,

by the German physicist Thomas Johann Seebeck in 1821, who

found that a voltage existed between two ends of a metal bar when a

temperature difference ΔT existed in the bar.

The effect is that a voltage, the thermoelectric EMF, is created in the

presence of a temperature difference between two different metals or

semiconductors. This causes a continuous current to flow in the

conductors if they form a complete loop. The voltage created is of

the order of several microvolts per degree difference.

Thermo-electric cooling

Thermoelectric coolers are solid state heat pumps used in

applications where temperature stabilization, temperature cycling, or

cooling below ambient are required.

There are many products using thermoelectric coolers, including

CCD cameras (charge coupled device), laser diodes,

microprocessors, blood analyzers and portable picnic coolers.

The typical thermoelectric module is manufactured using two thin

ceramic wafers with a series of P and N doped bismuth-telluride

semiconductor material sandwiched between them

The N type material has an excess of electrons, while the P type

material has a deficit of electrons. One P and one N make up a

couple, as shown in Figure 1. The thermoelectric couples are

electrically in series and thermally in parallel. A thermoelectric

module can contain one to several hundred couples.

Designing of thermal emf refrigeration

Th = Tamb + (O) *(Qh) where

TH=The temperature of hot side

Tamb=The ambient temperature

O=thermal resistance of heat exch

Qh=heat realeased heat released to the hot side of the thermoelectric

(watts).

Qh = Qc + Pin

Where

Qh = the heat released to the hot side of the thermoelectric (watts).

Qc = the heat absorbed from the cold side (watts).

Pin = the electrical input power to the thermoelectric (watts).

The temperature difference across the thermoelectric (T) relates to Th

and Tc according to Equation

T = Th – Tc

The thermoelectric performance curves in Figures 2 and 3 show the

relationship between T and the other parameters.

Estimating Qc, the heat load in watts absorbed from the cold side is

difficult, because all thermal loads in the design must be considered.

Among these thermal loads are:

Active: I2R heat load from the electronic devices

Any load generated by a chemical reaction

Passive:

1. Radiation (heat loss between two close objects with different temperatures).

2. Convection (heat loss through the air, where the air has a different temperature than the object)

3. Insulation Losses

4. Conduction Losses (heat loss through leads, screws, etc.)

5. Transient Load (time required to change the temperature of an object)

As the thermoelectric operates, the current flowing through it has

two effects:

(1) the Peltier Effect (cooling) and

(2) the Joulian Effect (heating).

We know that joulian effect is proportional to the squire of the

current so heating effect will dominates the cooling effect that why

we can not increase the current to a maximum value called Imax for

themo-electric.

The thermal resistance of the heat sink causes a temperature rise

above ambient. If the thermal resistance of the heat sink is unknown,

then estimates of acceptable temperature rise above ambient are:

Natural Convection20°C to 40°C

Forced Convection10°C to 15°C

Liquid Cooling2°C to 5°C (rise above the liquid coolant temperature)

As we have done our design on a Melcor thermoelectric . The

specifications for the are(these specifications are at Th = 25°C):

1. Qmax = 51.4 watts

2. Vmax = 15.4 volts

3. Imax = 6.0 amps

4. Tmax = 67°C

To determine if this thermoelectric is appropriate for this application,

it must be shown that the parameters T and Qc are within the

boundaries of the performance curves.

5. Our main aim to maintain the temperature of container 5°C which

contain 16 litres of air in 0.5minute.

we know 1000litres =1m3

16 litres=0.016 m3

density of air=1.293 kg/m3.

mass of air =0.016*1.293=0.020688kg

specific heat of air=1kJ/Kg°k

As Q=m*s*(th-tc)=0.020688*1000*(35-5)=620.64 J

this is maintain in 0.5minutes so

Required power=620.64/(0.5*60)=22 watts

(As we assume that the ambient temperature Tamb=25c the rise in the

temperature due to sink resistance is 10°C

So final temperature will be =25+10=35°C)

Performance Curve (T vs. Qc)

Performance Curve (T vs. Voltage )

So by the graph

maximum current = 3.6amp .

corresponding voltage by graph between temperature and voltage

voltage=10v

Now we will determine the corresponding value of temperature by these

values of current and voltage.

We know that the temperature at hot side

Th = Tamb + (O) *(Qh)

Value of heat released at hot side

Qh = Qc + Pin

Now Pin that is the input power to produce this effect is

Pin=V*I

V=10volt

I=3.6amp

Pin=10*3.6=36watt

And Qh=Qc+Pin

=22+36

=58 watts

And Th= Tc +Rcon*Qh

Now the temperature is also rise due to its convective resistance , we

assume that the convective resistance of the sink is 0.15°C

So Th=25+0.15*58=33.7°C

The calculated Th is close enough to the original estimate of Th, to

conclude that the CP1.4-127-06L thermoelectric will work in the given

application

Material used for insulation

The material used for the assembly components deserves careful

thought. The heat sink and cold side mounting surface should be made

out of materials that have a high thermal conductivity (i.e., copper or

aluminum) to promote heat transfer.

However, insulation and assembly hardware should be made of

materials that have low thermal conductivity (i.e., polyurethane foam

and stainless steel) to reduce heat loss.

Environmental concerns such as humidity and condensation on the

cold side can be alleviated by using proper sealing methods. A

perimeter seal (Figure 4) protects the couples from contact with water or

gases, eliminating corrosion and thermal and electrical shorts that can

damage the thermoelectric module.

Typical thermoelectric from Melcor with a perimeter seal

Single Stage vs. Multistage

Given the hot side temperature, the cold side temperature and the

heat load, a suitable thermoelectric can be chosen. If T across the

thermoelectric is less than 55°C, then a single stage thermoelectric is

sufficient. The theoretical maximum temperature difference for a

single stage thermoelectric is between 65°C and 70°C.

If T is greater than 55°C, then a multistage thermoelectric should be

considered. A multistage thermoelectric achieves a high T by

stacking as many as six or seven single stage thermoelectrics on top

of each other.

Materials used to built thermocouples.

Silicon, Bismuth, Nickel ,Cobalt ,Palladium, Platinum, Uranium,

Copper, Manganese, Titanium, Mercury, Lead , Tin, Chromium,

Molybdenum ,Rhodinium ,Iridium ,Gold ,Silver , Aluminium, Zinc,

Tungsten, CadmiumIron, Arsenic, Tellurium, Germanium