lec 5 Vapour Absorption Refregiration Cycle.ppt

Vapour Absorption Refrigeration System

• Peculiar property of some substances to have more
affinity for another substances at some temperature and
pressure conditions while less affinity at other
conditions
• Ferdinand Carre, a Frenchman, invented the absorption
system in 1860
• The absorption cycle is similar in certain respects to the
vapour- compression cycle
• In the absorption system also, refrigeration is produced
by the evaporation of a liquid refrigerant in the
evaporator
• The difference between the two systems lies in the
principle of converting the refrigerant vapour back to
the liquid
Absorption Refrigeration System

•Economically attractive - source of inexpensive heat energy
at a temperature of 100 to 200 oC. Also used where there are
severe limits to the electrical power available.
• The overall energy used is greater than with the
compression cycle, so the COP is lower.
• Heat energy to the generator may be any form of low-grade
energy such as hot oil, natural gas, hot water, steam, Solar
radiation etc.
• NH3–H2O and H2O-LiBr absorption systems are common.
• Major advantage is that liquid is compressed instead of
vapour.
• The COP of actual systems is usually much less than 1.

• A refrigeration cycle will operate with the condenser,
expansion valve and evaporator as shown
• From the evaporator, the low pressure vapours are
transformed into high pressure vapour and delivered to
the condenser
• The absorption system first absorbs the low pressure
vapour in an appropriate absorbing liquid
• Since this process is similar to condensation, so heat
must be rejected during the process
• In the next step, the pressure of the liquid is elevated
with a pump
• In the final step, the vapours are again released from the
absorbing liquid by addition of heat

• Vapour compression ---- work operated cycle ---
compression requires work
• Absorption cycle ----- heat operated cycle --- operating
cost associated with provision of heat,
• Some work in absorption system also, to drive pump,
but amount of work for a given quantity of refrigeration
is minor compared with that needed in vapour-
compression cycle

The Basic Absorption Cycle - The condenser, expansion valve and the
evaporator are similar as in a standard vapour-compression cycle

• The compression operation is now provided by the
assembly in the left-half of the diagram
• Low pressure vapour from the evaporator is absorbed
by the liquid solution in the absorber
• As the refrigerant vapour enters into the solution in the
absorber, the temperature of the solution tend to rise
• To resist the tendency, a cooling coil removes this heat
of solution
• The solution in the absorber is called a strong solution
because it is rich in refrigerant

• The pump draws the strong solution from the absorber,
elevates the pressure of the solution, and forces the
strong solution into the generator
• In the generator the addition of heat raises the
temperature, which drives off some of the refrigerant as
a vapour at high pressure and temperature
• Solar energy, waste heat from the process industry,
exhaust gases from automobile, power plants, steel
plants, gas power plants, etc.
• As the refrigerant vapour leaves the solution in the
generator, the solution becomes weak or have a low
concentration of refrigerant
• The weak solution flows back to the absorber through a
throttling valve whose purpose is to ???
• From the generator the refrigerant proceeds through the
condenser, expansion valve and evaporator

• The pattern for the flow of heat to and from the four-heat
exchange components in the absorption cycle is as
follows:-
o High temperature heat enters the generator while
low temperature heat from the refrigerated space
enters the evaporator
o The heat rejection from the cycle occurs at the
absorber and condenser at temperatures such that
the heat can be rejected to atmosphere

S.# Refrigerant Absorber Absorber State
1. Ammonia Water Liquid
2. Ammonia Sodiumthiocynate Solid
3. Ammonia Lithiumnitrate Solid
4. Ammonia Calcium chloride Solid
5. Water Lithium bromide Solid
6. Water Lithium chloride Solid
7. Methylene
chloride
Dimethyl ether or tetra
ethylene glycol
Liquid
Refrigerant-absorber pairs

Lithium Bromide (LiBr) Water Absorption Cycle
• LiBr is a solid salt crystal, in the presence of water vapour it
will absorb the vapour and become a liquid solution
• If two vessels were connected as shown in the figure, one vessel
containing LiBr-water solution and the other pure water, each liquid would
exert a water-vapour pressure that is a function of the solution
temperature and the concentration of the solution.
• At equilibrium the water-vapour pressure exerted by the two
liquids would be equal

The refrigerant is absorbed by a transport medium and compressed in liquid form. The
most widely used absorption refrigeration system is the ammonia-water (aqua-
ammonia) system, where ammonia serves as the refrigerant and water as the transport
medium. Dissolution of NH3 into water is exothermic and inversely proportional to
temperature. The work input to the pump is usually very small, and the COP of
absorption refrigeration systems is defined as
gen
L
in
p
gen
L
abs
Q
Q
W
Q
Q
input
Work
effect
Cooling
COP




,

Temperature-pressure concentration diagram for LiBr-water
solutions
• Concentration is the abscissa of the graph and water-vapour
pressure could be considered as the ordinate on the vertical scale
on the right
• The saturation temperature of pure water corresponding to
these vapour pressures is shown as the ordinate on the left
• The chart applies to saturated conditions where the solution
is in equilibrium with water vapour
• If the temperature of pure water is 40 °C, the vapour pressure
the liquid exerts is 7.38 kPa, which can be determined from the
opposite side of vertical scale
o A LiBr solution with a concentration x of 59 % &
temperature of 80°C also develops a water-vapour pressure of 7.38
kPa
• o If the solution had a concentration x of 54% & temperature
of 70°C the water-vapour pressure would again be 7.38 kPa

Compute the rate flow of refrigerant (water) through the condenser
and evaporator in the cycle shown in Figure below if the pump
delivers 0.6 kg/s and the following temperatures prevail: generator,
100C; condenser, 40C; evaporator. 10C; and absorber, 30C

?
4
3 



m
m
s
kg
m /
6
.
0
1 











6
.
0
1
3
2 m
m
m







2
2
1
1 x
m
x
m
)
664
.
0
(
2

m
s
kg
m /
452
.
0
2 



3
m
The basic LiBr water vapour cycle is shown in the figure
Where
The two mass flow balances can be written around the generator
Total Mass flow balance:-
LiBr balance:-
From chart of LiBr-water solution
Since the pressure in condenser & generator must be same, the pressure in
condenser is equal to the 7.38 kPa corresponding to a saturation temperature of
40°C
Similarly, the pressure in evaporator & absorber must be same, the pressure in
absorber is equal to 1.23 kPa corresponding to a saturation temperature of 10 °C
So from charts:- x1 = 50% & x2 = 66.4%
Therefore (2) → 0.6 (0.5) =
And (1) ----- 0.148 kg/s
(1)
(2)

Enthalpy of LiBr Solutions
• For thermal calculations on the absorption cycle, enthalpy
data must be available for the working substances at all crucial
positions in the cycle
• Water in liquid and vapour forms flows in and out of the
condenser & evaporator, so enthalpies at these points can be
determined from a table of properties of water
• In the generator and absorber, LiBr-water solutions exist for
which enthalpy is a function of both - solution temperature and
concentration
• Figure represents the enthalpy data for LiBr-water solution

1
1
2
2
3
3 h
m
h
m
h
m





4
4
3
3 h
m
h
m



1
1
5
5
2
2 h
m
h
m
h
m





4
4
5
5 h
m
h
m



For the absorption system of previous example, compute qg, qc, qa, qe
and the COP
h1 = h at 30 °C and x of 50% = -168 kJ/kg
h2 = h at 100 °C and x of 66.4% = -52 kJ/kg
The enthalpies of water liquid & vapour are found from steam tables:-
h3 = h for saturated vapour at 100 °C = 2676 kJ/kg
h4 = h for saturated liquid at 40 °C = 167.4 kJ/kg
h5 = h for saturated vapour at 10 °C = 2519.9 kJ/kg
qg =
= 473.3 kW
=371.2 kW
= 450.3 kW
= 348.2 kW
COP = qe/qg = 348.2/473.3 = 0.736
qc =
qa =
qe =

Absorption cycle with heat exchanger
•The heat-exchanger transfers heat between the two streams of solutions
•It heats the cool solution from the absorber on its way to the generator
and cools the solution returning from the generator to the absorber
•By addition of heat exchanger, the COP increases as well

Aqua-Ammonia System
In aqua-ammonia absorption system, water is used as an absorbent while
ammonia is used as a refrigerant
The system consists of all the components i.e., generator, absorber,
condenser, evaporator, and heat exchanger---- plus a rectifier & analyzer

• Additional components as refrigerant vapours released at
generator contains water vapour as well
• Normally aqua-ammonia system operate at evaporating
temperature below 0 °C
• If large amounts of water vapours are present in the
evaporator, chance that they may get converted to ice & block the
lines
• So to remove as much water vapour as possible, the vapours
driven off from the generator first flows through the rectifier, which
is a direct-cooled heat exchanger
• In the rectifier, the vapours from the generator first flow
counter-current to the incommoding strong solution from the
absorber
• Next the solution passes through the analyzer which is a
water-cooled heat exchanger, condensing some water rich liquids
which drains back to the rectifier

Problem: A water-LiBr absorption refrigeration system is shown (see figure). The
temperature at point 2 is 52 °C. The mass flow rate delivered by the solution pump is
0.6 kg/s. What are the rates of energy transfer at each of the components and the
COP of this cycle? Also, what is the temperature at state 4?

lec 5 Vapour Absorption Refregiration Cycle.ppt

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