lectures notes of: refrigeration and air conditioning
TRANSCRIPT
1
Modern University For Information and Technology
Mechanical Engineering Department
Lectures Notes Of: Refrigeration
and Air Conditioning
MENG 319
Prepared By Dr: Mona Mousa
(First Edition 2021)
Refrigeration
1. Introduction Refrigeration is the action of cooling, and in practice this requires removal of of heat energy that can be converted into work is limited. As the heat flows
from hot to cold a certain amount of energy may be converted into work and
extracted. It can be used to drive a generator, for example.
The minimum amount of work to drive a refrigerator can be defined in terms of the
absolute temperature scale. Figure 1.1 shows a reversible engine E driving a
reversible heat pump P; Q and W represent the flow of heat and work. They are
called reversible machines because they have the highest efficiency that can be
visualised, and because there are no losses, E and P are identical machines.
The arrangement shown results in zero external effect because the reser- voirs
experience no net gain or loss of heat. If the efficiency of P were to be higher, i.e.
if the work input required for P to lift an identical quantity of heat Q2 from the
cold reservoir were to be less than W, the remaining part of W could power
another heat pump. This could lift an additional amount of heat.
The result would be a net flow of heat from the low temperature to the high temperature
without any external work input, which is impossible. The rela- tionship between Q1, Q2
and W depends only on the temperatures of the hot and cold reservoirs. The French
physicist Sadi Carnot (1796–1832) was the The ideal ‘never-attainable-in-practice’ ratio
of work output to heat input (W/Q1) of the reversible engine E equals: Temperature
Difference (T1 T0) divided by the Hot Reservoir Temperature (T1).
In Figure 1.1 the device P can be any refrigeration device we care to invent,
and the work of Kelvin tells us that the minimum work, W necessary to lift a quantity of
heat Q2 from temperature T0 to temperature T1
The temperatures must be measured on an absolute scale i.e. one that starts at absolute zero.
The Kelvin scale has the same degree intervals as the Celsius scale, so that ice melts at
273.16 K, and water at atmospheric pressure boils at 373.15 K. On the Celsius scale,
absolute zero is –273.15°C. Refrigeration
‘efficiency’ is usually defined as the heat extracted divided by the work input. This is
called COP, coefficient of performance. The ideal or Carnot COP takes.
Figure 1.1 Ideal heat engine, E, driving an ideal refrigerator (heat pump), P
The Refrigeration Cycle
In this cycle a unit mass of fluid is subjected to four processes after which it returns to
its original state. The compression and expansion processes, shown as vertical lines,
take place at constant entropy. A constant entropy (isentropic) process is a reversible or
an ideal process. Ideal expansion and compression engines are defined in Section 1.2.
The criterion of perfection is that no entropy is generated during the process, i.e. the
quantity ‘s’ remains constant. The add- ition and rejection of heat takes place at constant
temperature and these pro- cesses are shown as horizontal lines. Work is transferred into
the system during compression and out of the system during expansion. Heat is
transferred across the boundaries of the system at constant temperatures during
evaporation and condensation. In this cycle the net quantities of work and heat are in
propor- tions which provide the maximum amount of cooling for the minimum amount
of work. The ratio is the Carnot coefficient of performance (COP).
This cycle is sometimes referred to as a reversed Carnot cycle because the original
concept was a heat engine and for power generation the cycle operates in a clockwise
direction, generating net work.
SIMPLE VAPOUR COMPRESSION CYCLE
The vapour compression cycle is used for refrigeration in preference to gas cycles;
making use of the latent heat enables a far larger quantity of heat to be extracted for a
given refrigerant mass flow rate. This makes the equipment as compact as possible.
A liquid boils and condenses – the change between the liquid and the gaseous states –
at a temperature which depends on its pressure, within the limits of its
Heat is put into the fluid at the lower temperature and pressure thus pro- viding the
latent heat to make it vaporize. The vapour is then mechanically compressed to a higher
pressure and a corresponding saturation temperature at which its latent heat can be
rejected so that it changes back to a liquid. The cycle is shown in Figure 1.2. The cooling
effect is the heat transferred to the working fluid in the evaporation process, i.e. the
change in enthalpy between the fluid entering and the vapour leaving the evaporator.
In order to study this process more closely, refrigeration engineers use a pressure–enthalpy
or P–h diagram. This diagram is a useful way of describing the liquid and gas phase of a
substance. On the vertical axis is pressure, P, and on the horizontal, h, enthalpy. The
saturation curve defines the boundary of pure liquid and pure gas, or vapour. In the
region marked vapour, the fluid is superheated vapour. In the region marked liquid, it is
subcooled liquid. At pres- sures above the top of the curve, there is no distinction between
liquid and vapour. Above this pressure the gas cannot be liquefied. This is called the
critical pres- sure. In the region beneath the curve, there is a mixture of liquid and vapour.
The simple vapour compression cycle is superimposed on the P–h dia- gram. The
evaporation process or vaporization of refrigerant is a constant pressure process and
therefore it is represented by a horizontal line. In the compression process the energy
used to compress the vapour turns into heat and increases its temperature and enthalpy,
so that at the end of compression the vapour state is in the superheated part of the
diagram and outside the sat- uration curve. A process in which the heat of compression
raises the enthalpy of the gas is termed adiabatic compression. Before condensation can
start, the vapour must be cooled. The final compression temperature is almost always
above the condensation temperature as shown, and so some heat is rejected at a
temperature above the condensation temperature. This represents a deviation from the
ideal cycle. The actual condensation process is represented by the part of the horizontal
line within the saturation curve.
When the simple vapour compression cycle is shown on the temperature– entropy
diagram, the deviations from the reversed Carnot cycle can be identified by shaded
areas. The adiabatic compression process continues beyond the point where the
condensing temperature is reached. Expansion is a constant enthalpy process. It is
drawn as a vertical line on the P–h diagram. No heat is absorbed or rejected during this
expansion, the liquid just passes through a valve. Since the reduction in pressure at this
valve must cause a corresponding drop in temperature, some of the fluid will flash off
into vapour to remove the energy for this cooling. The volume of the working fluid
therefore increases at the valve by this amount of flash gas, and gives riseto its name,
the expansion valve. No attempt is made to recover energy from the expansion process,
e.g. by use of a turbine. This is a second deviation from the ideal cycle. The work that
could potentially be recovered is represented by the shaded rectangle
Example1.1
A refrigeration circuit is to cool a room at 0°C using outside air at 30°C to reject the heat.
The refrigerant is R134a. The temperature difference at the evaporator and the condenser is
5 K. Find the Carnot COP for the process, the Carnot COP for the refrigeration cycle and the
ideal vapour compression cycle COP when using R134a.
Solution
Multi-Evaporator And Cascade Systems
The objectives of this lesson are to:
1. Discuss the advantages and applications of multi-evaporator systems
compared to single stage systems.
2. Describe multi-evaporator systems using single compressor and a pressure
reducing valve with:
a) Individual expansion valves.
b) Multiple expansion valves.
3. Describe multi-evaporator systems with multi-compression, intercooling and
flash gas removal.
4. Describe multi-evaporator systems with individual compressors and multiple
expansion valves.
5. Discuss limitations of multi-stage systems.
6. Describe briefly cascade systems.
7. Describe briefly the working principle of auto-cascade cycle.
At the end of the lecture, the student should be able to:
1. Explain the need for multi-evaporator systems
2. Evaluate the performance of:
a) Multi-evaporator systems with single compressor and individual
expansion valves
b) Multi-evaporator systems with single compressor and multiple
expansion valves
3. Evaluate the performance of multi-evaporator systems with multi-
compression, intercooling and flash gas removal 4. Evaluate the performance of multi-evaporator systems with individual
compressors and multiple or individual expansion valves 5. Evaluate the performance of cascade systems 6. Describe the working principle of auto-cascade systems
13.1. Introduction
As mentioned in Chapter 12, there are many applications where refrigeration is required at different temperatures. For example, in a typical food processing plant, cold air may be required at –30
oC for freezing and at +7
oC for cooling of food
products or space cooling. One simple alternative is to use different refrigeration systems to cater to these different loads. However, this may not be economically viable due to the high total initial cost. Another alternative is to use a single refrigeration system with one compressor and two evaporators both operating at
−30oC. The schematic of such a system and corresponding operating cycle on P-h
diagram are shown in Figs.1(a) and (b). As shown in the figure the system consists of a single compressor and a single condenser but two evaporators. Both evaporators-I and II operate at same evaporator temperature (-30
oC) one evaporator (say
Evaporator-I) caters to freezing while the other (Evaporator-II) caters to product cooling/space conditioning at 7
oC. It can be seen that operating the evaporator at –
30oC when refrigeration is required at +7
oC is thermodynamically inefficient as the
system irreversibilities increase with increasing temperature difference for heat transfer. The COP of this simple system is given by:
In addition to this there will also be other difficulties such as: evaporator catering to space cooling (7
oC) may collect frost leading to blockage of air-flow passages, if a liquid is to
chilled then it may freeze on the evaporator and the moisture content of air may become too low leading to water losses in the food products. In such cases multi-stage systems with multiple evaporators can be used. Several multi- evaporator combinations are possible in practice. Some of the most common ones are discussed below.
13.2. Individual evaporators and a single compressor with a pressure-reducing valve
13.2.1. Individual expansion valves:
Figures 2 (a) and (b) show system schematic and P-h diagram of a multi-
evaporator system that uses two evaporators at two different temperatures and a single compressor. This system also uses individual expansion valves and a pressure regulating valve (PRV) for reducing the pressure from that corresponding to the high temperature evaporator to the compressor suction pressure. The PRV also maintains the required pressure in high temperature evaporator (Evaporator-II). Compared to the earlier system, this system offers the advantage of higher refrigeration effect at the high temperature evaporator [(h6-h4) against (h7-h5)]. However, this advantage is counterbalanced by higher specific work input due to the operation of compressor in
Heat
j t
Condenser
3 2
4
Evaporator-I (-30
oC)
Refrigeration
at –30oC
4
Evaporator-II at –30
oC
Refrigeration at +7
oC
1 1
Compressor
1
P
3
2
4 1 -30oC
h
Fig.1(a) & (b): A single stage system with two evaporators
superheated region. Thus ultimately there may not be any improvement in system
COP due to this arrangement. It is easy to see that this modification does not result in
significant improvement in performance due to the fact that the refrigerant vapour at the
intermediate pressure is reduced first using the PRV and again increased using
compressor. Obviously this is inefficient. However, this system is still preferred to the
earlier system due to proper operation of high temperature evaporator.
- I
Heat rejection
Condenser
3 2
4
Evaporator - II 6
+7oC
Refrigeration at
+7oC
8 1
PRV
Compressor
5
Evaporator - I 7
-30oC
Refrigeration at
-30oC
P
3
2
4 +7
oC
5
-30oC
6
7 1 8
h
Fig.2(a) & (b): Multi-evaporator system with single compressor and individual
expansion valves
The COP of the above system is given by:
. .
where .
m I and .
m II are the refrigerant mass flow rates through evaporator I and II
respectively. They are given by:
Enthalpy at point 2 (inlet to compressor) is obtained by applying mass and
energy balance to the mixing of two refrigerant streams, i.e.,
. .
If the expansion across PRV is isenthalpic, then specific enthalpy h8 will be equal to h6.
13.2.2. Multiple expansion valves:
Figures 3 (a) and (b) show system schematic and P-h diagram of a multi-
evaporator with a single compressor and multiple expansion valves. It can be seen
from the P-h diagram that the advantage of this system compared to the system with
individual expansion valves is that the refrigeration effect of the low temperature
evaporator increases as saturated liquid enters the low stage expansion valve. Since
the flash gas is removed at state 4, the low temperature evaporator operates more
efficiently.
The COP of this system is given by:
.
where m I and m II are the refrigerant mass flow rates through evaporator I and II
respectively. They are given by: .
I
Condenser
3
2 4
Evaporator - II 7
5
9 1
PRV
Compressor - I
6
Evaporator - I
8
P
3
2
+7oC
5 4
6
-30oC
7
8 1 9
h
Fig.3(a) & (b): Multi-evaporator system with single compressor and multiple
expansion valves
Enthalpy at point 2 (inlet to compressor) is obtained by applying mass and energy
balance to the mixing of two refrigerant streams, i.e.,
. .
If the expansion across PRV is isenthalpic, then specific enthalpy h7 will be equal to h9.
COP obtained using the above multi-evaporator systems is not much higher compared to single stage system as refrigerant vapour at intermediate pressure is first
throttled then compressed, and compressor inlet is in superheated region. Performance
can be improved significantly if multiple compressors are used in place of a single
compressor.
3. Multi-evaporator system with multi-compression, intercooling
and flash gas removal
Figures 4(a) and (b) show the schematic and P-h diagram of a multi- evaporator system which employs multiple compressors, a flash tank for flash gas removal and intercooling. This system is good for low temperature lift applications with different refrigeration loads. For example one evaporator operating at say –40
oC for
quick freezing of food products and other evaporator operating at –25oC for
storage of frozen food. As shown in the system schematic, the pressure in the high temperature evaporator (Evaporator-II) is same as that of flash tank. Superheated vapour from the low-stage compressor is cooled to the saturation temperature in the flash tank. The low temperature evaporator operates efficiently as flash gas is removed in the flash tank. In addition the high-stage compressor (Compressor-II) operates efficiently as the suction vapour is saturated. Even though the high stage compressor has to handle higher mass flow rate due to de-superheating of refrigerant in the flash tank, still the total power input to the system can be reduced substantially, especially with refrigerants such as ammonia.
The COP of this system is given by:
. .
.
where m I
and .
m e,II are the refrigerant mass flow rates through evaporator I and II
respectively. They are given by: .
m II is the mass flow rate of refrigerant through the high-stage compressor which can
be obtained by taking a control volume which includes the flash tank and high temperature evaporator (as shown by dashed line in the schematic) and applying mass and energy balance:
mass balance:
. . . . . . . .
m5 + m 2 = m7 + m3 ; m5 = m II = m3 & m 2 = m I = m7
energy balance:
6
3
2
. .
m5 h 5 + m2 h 2 + Qe,II = m7 h 7 + m3h 3
from known operating temperatures and evaporator loads (Qe,I and Qe,II) one can get the mass flow rate through the high stage compressor and system COP from the above equations.
Condenser
5 6
Evaporator - II
4
3b 3
Compressor - II
Qe,II
3a
6
Flash chamber
Control volume for
finding mass flow rate
through Compressor-II
2
7 1 Compressor - I
8
Evaporator - I
Qe,I
P
5 4
7
8 1
h
Fig.4(a) & (b): Multi-evaporator system with multiple compressors and a flash tank for
flash gas removal and intercooling
4. Multi-evaporator system with individual compressors and
multiple expansion valves
Figures 5(a) and (b) show the schematic and P-h diagram of a multi-
evaporator system which employs individual compressors and multiple expansion
valves.
The COP of this combined system is given by:
.
where .
m I and .
m II are the refrigerant mass flow rates through evaporator I and II
respectively. They are given by:
.
The inlet to the condenser (state 5) is obtained by applying mass and energy
balance to the process of mixing of refrigerant vapours from Compressors I and II.
5. Limitations of multi-stage systems
Though multi-stage systems have been very successful, they have certain
limitations. These are:
a) Since only one refrigerant is used throughout the system, the refrigerant used
should have high critical temperature and low freezing point.
b) The operating pressures with a single refrigerant may become too high or too low. Generally only R12, R22 and NH3 systems have been used in multi-stage systems as other conventional working fluids may operate in vacuum at very low evaporator temperatures. Operation in vacuum leads to leakages into the system and large compressor displacement due to high specific volume.
c) Possibility of migration of lubricating oil from one compressor to other leading to
compressor break-down.
The above limitations can be overcome by using cascade systems.
7
1
5 Condenser
6
2 7 Wc,II
Evaporator - II
8
Qe,II
9
Evaporator - I
1
Compressor - II
3
4
Compressor - I
Wc,I
Qe,I
P
6 2 5 4
8
9 3
h
Fig.5(a) & (b): Multi-evaporator system with individual compressors and multiple
expansion valves
6. Cascade Systems
In a cascade system a series of refrigerants with progressively lower boiling
points are used in a series of single stage units. The condenser of lower stage system
is coupled to the evaporator of the next higher stage system and so on. The component
where heat of condensation of lower stage refrigerant is supplied for vaporization of
next level refrigerant is called as cascade condenser. Figures 16(a) and (b) show the
schematic and P-h diagrams of a two-stage cascade refrigeration system. As shown, this
system employs two different refrigerants operating in two individual cycles. They
are thermally coupled in the cascade condenser. The refrigerants selected should have
suitable pressure-temperature characteristics. An example of refrigerant combination is the use of carbon dioxide (NBP = -78.4
oC, Tcr = 31.06
oC) in low
temperature cascade and ammonia (NBP = -33.33oC, Tcr = 132.25
oC) in high
temperature cascade. It is possible to use more than two cascade stages, and it is also possible to combine multi-stage systems with cascade systems.
Applications of cascade systems:
i. Liquefaction of petroleum vapours
ii. Liquefaction of industrial gases
iii. Manufacturing of dry ice
iv. Deep freezing etc.
Advantages of cascade systems:
i. Since each cascade uses a different refrigerant, it is possible to select a
refrigerant that is best suited for that particular temperature range. Very high
or very low pressures can be avoided
ii. Migration of lubricating oil from one compressor to the other is prevented
In practice, matching of loads in the cascade condenser is difficult, especially during
the system pull-down. Hence the cascade condensers are normally oversized. In
addition, in actual systems a temperature difference between the condensing and
evaporating refrigerants has to be provided in the cascade condenser, which leads to loss
of efficiency. In addition, it is found that at low temperatures, superheating (useful
or useless) is detrimental from volumetric refrigeration effect point-of-view, hence in
cascade systems, the superheat should be just enough to prevent the entry of liquid into
compressor, and no more for all refrigerants.
Optimum cascade temperature:
For a two-stage cascade system working on Carnot cycle, the optimum
cascade temperature at which the COP will be maximum, Tcc,opt is given by:
Tcc,opt = Te .Tc (13.18)
where Te and Tc are the evaporator temperature of low temperature cascade and condenser temperature of high temperature cascade, respectively.
2
3’ 2’
2
4’ 1’
4 1
h
Fig.6(a) & (b): A two-stage cascade refrigeration system
For cascade systems employing vapour compression refrigeration cycle, the
optimum cascade temperature assuming equal pressure ratios between the stages is
given by:
⎛ ⎞ ⎜ ⎟
T = ⎜ b1
+ b 2 ⎟
⎠
Low temp. refrigerant Pdischarge
High temp. refrigerant
ΔT
Psuction
7. Auto-cascade systems
An auto-cascade system may be considered as a variation of cascade system,
in which a single compressor is used. The concept of auto-cascade system was first
proposed by Ruhemann in 1946. Figure 13.7(a) shows the schematic of a two-stage auto-
cascade cycle and Fig.137(b) shows the vapour pressure curves of the two
Qc,out
Partial condenser
Compressor
Condenser
Evaporator
Qe,in
Fig.7(a): Schematic of a two-stage auto-cascade system
refrigerants used in the cycle on D˘hring plot.
In a two-stage auto-cascade system two different working fluids; a low boiling point (low temperature) refrigerant and a high boiling point (high temperature) refrigerant are used. The vapour mixture consisting of both these refrigerants is compressed in the compressor to a discharge pressure (Pdischarge). When this high pressure mixture flows through the partial condenser, the high temperature refrigerant
P
Te Te,h Tc,l Tc
T
Fig.7(b): Schematic illustrating principle of two-stage auto-
cascade system on D˘hring plot
can condense by rejecting heat (Qc,out) to the external heat sink, if its partial
pressure in the mixture is such that the saturation temperature corresponding to the
partial pressure is higher than the external heat sink temperature. Since the saturation temperature of the low temperature refrigerant is much lower than the external heat sink
temperature at its partial pressure, it cannot condense in the partial condenser, hence, remains as vapour. Thus it is possible theoretically to separate the high temperature
refrigerant in liquid form from the partial condenser. Next this high temperature, high
pressure liquid is expanded through the expansion valve into the condenser operating at a pressure Psuction. Due to the expansion of the high temperature refrigerant
liquid from Pdischarge to Psuction, its temperature drops to a sufficiently low value
(Te,h) so that when the low temperature, high pressure refrigerant vapour comes
in contact with the high temperature, low pressure refrigerant in the condenser it can condense at a temperature Tc,l. This condensed, high pressure, low temperature
refrigerant is then throttled to the suction pressure and is then made to flow through the
evaporator, where it can provide the required refrigeration effect at a very low temperature Te. Both the high temperature refrigerant from condenser and low
temperature refrigerant vapour from evaporator can be mixed as they are at the same
pressure. This mixture is then compressed in the compressor to complete the cycle. Thus using a single compressor, it is possible to obtain refrigeration at very low
temperatures using the auto-cascade system. In practice, more than two stages with more than two refrigerants can be used to achieve very high temperature lifts.
However, in actual systems, it is not possible to separate pure refrigerants in the partial condenser as some amount of low temperature refrigerant condenses in the partial
condenser and some amount of high temperature refrigerant leaves the partial condenser in vapour form. Thus everywhere in the system, one encounters refrigerant mixtures of
varying composition. These systems are widely used in the liquefaction of natural gas.
Questions: 1. Multi-evaporator systems are:
a) Widely used when refrigeration is required at different temperatures
b) When humidity control in the refrigerated space is required
c) When the required temperature lift is small
d) All of the above
Ans.: a) and b)
2. Multi-evaporator systems with a single compressor and a pressure reducing valve:
a) Yield very high COPs compared to multi-evaporator, single stage systems
b) Yield lower compressor discharge temperature compared to single stage systems
c) Yield slightly higher refrigeration effect in the low temperature evaporator
compared to single stage systems
d) Yield slightly higher refrigeration effect in the high temperature evaporator
compared to single stage systems
Ans.: d)
3. Compared to individual expansion valves, multiple expansion valves:
a) Yield higher refrigeration effect in the low temperature evaporator
b) Yield higher refrigeration effect in the high temperature evaporator
c) Yield lower compressor discharge temperature
d) Decrease the quality of refrigerant at the inlet to low temperature evaporator
Ans.: a) and d)
4. Compared to multi-evaporator and single compressor systems, multi-evaporator
systems with multiple compressors:
a) Yield higher COP
b) Decrease maximum cycle temperature
c) Yield higher refrigeration effect
d) All of the above
Ans.: a) and b)
5. In multi-stage systems:
a) The refrigerant used should have high critical temperature and high freezing point
b) The refrigerant used should have high critical temperature and low freezing point
c) There is a possibility of migration of lubricating oil from one compressor to other
d) Operating pressures can be too high or too low
Ans.: b), c) and d)
6. In cascade systems:
a) Different refrigerants are used in individual cascade cycles
b) There is no mixing of refrigerants and no migration of lubricating oil
c) Higher COPs compared to multi-stage systems can be obtained
d) Operating pressures need not be too high or too low
Ans.: a), b) and d)
7. Cascade systems are widely used for:
a) Large refrigeration capacity systems
b) Applications requiring large temperature lifts
c) Applications requiring very high efficiencies
d) All of the above
Ans.: b)
8. For a two-stage cascade system working on Carnot cycle and between low and high temperatures of –90
oC and 50
oC, the optimum cascade temperature at which the COP
will be maximum is given by:
a) –20oC
b) –30oC
c) –67oC
d) 0oC
Ans.: b)
9. In a two stage, auto-cascade system:
a) Two compressors and two refrigerants are used
b) A single compressor and a single refrigerant are used
c) A single compressor and two refrigerants are used
d) Two compressors and a single refrigerant are used
Ans.: c)
10. In a two stage, auto-cascade system:
a) Compressor compresses refrigerant mixture
b) Refrigerants are separated in partial condenser
c) Condensing temperature of low temperature refrigerant at discharge pressure is higher
than the boiling temperature of high temperature refrigerant at suction pressure d)
Condensing temperature of low temperature refrigerant at discharge pressure is lower
than the boiling temperature of high temperature refrigerant at suction pressure
Ans.: a), b) and c)
11. The figure given below shows a multi-evaporator, vapour compression refrigeration system working with ammonia. The refrigeration capacity of the high temperature evaporator operating at –6.7
oC is 5 TR, while it is 10 TR for the low
temperature evaporator operating at –34.4oC. The condenser pressure is 10.8 bar.
Assuming saturated conditions at the exit of evaporators and condenser, ammonia vapour to behave as an ideal gas with a gas constant of 0.4882 kJ/kg.K and isentropic index (cp/cv) of 1.29, and isentropic compression:
a) Find the required power input to compressor in kW b) Find the required power input if instead of using a single compressor,
individual compressors are used for low and high temperature evaporators.
Use the data given in the table:
10.8 bar
-6.7oC
5 TR
-34.4oC
10 TR
T,oC
Psat
(kPa)
hf (kJ/kg)
(sat.liquid) hg( kJ/kg) sat. vapour
-34.4 95.98 44.0 1417
-6.7 331.8 169.1 1455
27.7 1080.0 330.4 1485
Data for Problem 11
Ans.:
a) Single compressor: The P-h diagram for the above system is shown below:
P
3
2
4 -6.7
oC
5
-34.4oC
6
7 1 8
h
The required mass flow rate through the low temperature evaporator (mr,l) is given by:
mr,l = Qe,l/(h7 − h5) = (10 X 3.517)/(1417 − 330.4) = 0.03237 kg/s
The required mass flow rate through the high temperature evaporator (mr,h) is given by:
mr,h = Qe,h/(h6 − h4) = (5 X 3.517)/(1455 − 330.4) = 0.01564 kg/s
Assuming the refrigerant vapour to behave as an ideal gas, and assuming the variation
in specific heat of the vapour to be negligible, the temperature of the refrigerant after
mixing, i.e., at point 1 is given by:
T1 = (mr,l.T7 + mr,h.T6)/(mr,l + mr,h) = 247.6 K
Assuming isentropic compression and ideal gas behaviour, the power input to the
compressor,Wc
where mr is the refrigerant flow rate through the compressor (mr = mr,l + mr,h), R is the
gas constant (0.4882 kJ/kg.K), Pc and Pe are the discharge and suction pressures and k
is the isentropic index of compression ( = 1.29).
Substituting these values, the power input to the compressor is found to be:
1
Wc = 18.67 kW (Ans.)
Since the refrigerant vapour is assumed to behave as an ideal gas with constant
specific heat, and the compression process is assumed to be isentropic, the discharge
temperature T2 can be obtained using the equation:
Wc = mr.Cp(T2 – T1) = 18.67 kW
Substituting the values of mr, Cp (=2.1716 kJ/kg.K) and T1, the discharge temperature is found to be:
T2 = 427.67 K = 153.5oC
b) Individual compressors:
The P-h diagram with individual compressors is shown below:
P
6 2 5 4
7
8 3
h
The mass flow rates through evaporators will be same as before.
The power input to low temperature compressor (process 3 to 4), Wc,l
Similarly, for the high temperature compressor (process 1-2), the power input Wc,h
Therefore total power input is given by:
Wc = Wc,l + Wc,h = 12.13 + 2.75 = 14.88 kW (Ans.)
The compressor discharge temperatures for the low temperature and high temperature
compressor are found to be:
T4 = 411.16 K = 138.0oC
T2 = 347.27 K = 74.10oC
Comments: 1. Using individual compressors in place of a single compressor, the power input to
the system could be reduced considerably (≈ 20.3%). 2. In addition, the maximum compressor discharge temperature also could be reduced
by about 15oC.
3. In addition to this, the high temperature compressor operates at much lower
compression ratio, leading to low discharge temperatures and high volumetric
efficiency.
These are the advantages one could get by using individual compressors, instead of a
pressure regulating valve and a single compressor. However, in actual systems these
benefits will be somewhat reduced since smaller individual compressors generally
have lower isentropic and volumetric efficiencies.
4. A cascade refrigeration system shown in the figure given below uses CO2 as refrigerant for the low-stage and NH3 as the refrigerant for the high-stage. The system has to provide a
refrigeration capacity of 10 TR and maintain the refrigerated space at –36oC, when the
ambient temperature (heat sink) is at 43oC. A temperature difference of 7 K is required for
heat transfer in the evaporator, condenser and the cascade condenser. Assume the temperature lift (Tcond-Tevap) to be same for both CO2 and NH3 cycles and find a) Total power input to the
system; b) Power input if the cascade system is replaced with a single stage NH3 system
operating between same refrigerated space and heat sink.
The actual COP of the vapour compression system (COPact) can be estimated using
NH3 condenser
CO2 evaporator
43oC
NH3 Wc2
Cascade condenser
CO2
Wc1
- 36o
⎦
the equation:
= ⎡ −
T − T ⎤COPact
where
0.85 COPCarnot ⎢1 ⎣
c e
265 ⎥
COPCarnot = Carnot COP Tc =Condensing Temp., Te= Evaporator Temp.
Ans.: Since a temperature difference of & K is required for heat transfer, the CO2
evaporator and NH3 condenser temperatures are given by:
Te,CO2 = −36 −7 = -43
oC = 230 K
Tc,NH3 = 43 + 7 = 50oC = 323 K
In the cascade condenser,
Tc,CO2 = Te,NH3 + 7
Since the temperature lifts of CO2 and NH3 cycles are same,
(Tc,CO2 − Te,CO2) = (Tc,NH3 − Te,NH3)
From the above 4 equations, we obtain:
Tc,CO2 = 280 K Te,NH3 = 273 K
Substituting the values of temperatures in the expression for actual COP, we obtain:
COPCO2 = 3.17, and
COPNH3 = 3.77
The power input to CO2 compressor is given by,
Wc,CO2 = Qe,CO2/COPCO2 = 10 X 3.517 /3.17 = 11.1 kW
Since the heat rejected by the condenser of CO2 system is the refrigeration load for
the evaporator of NH3 system, the required refrigeration capacity of NH3 system is
given by:
Qe,NH3 = Qc,CO2 = Qe,CO2 + Wc,CO2 = 46.27 kW
Hence power input to NH3 compressor is given by:
Wc,NH3 = Qe,NH3/COPNH3 = 46.27 /3.77 = 12.27 kW
Therefore, the total power input to the system is given by:
Wc.total = Wc,CO2 + Wc,NH3 = 23.37 kW (Ans.)
b) If instead of a cascade system, a single stage NH3 is used then, the actual COP of
the system is:
COPNH3,1st = 1.363
Power input to single stage ammonia system is given by:
Wc,NH3,1st = Qe/ COPNH3,1st = 35.17/1.363 = 25.8 kW (Ans.)
Comments:
1) Using a cascade system the power consumption could be reduced by about 9.5 %. 2) More importantly, in actual systems, the compared to the single stage system, the compressors of cascade systems will be operating at much smaller pressure ratios, yielding high volumetric and isentropic efficiencies and lower discharge temperatures. Thus cascade systems are obviously beneficial compared to single stage systems for large temperature lift applications. 3. The performance of the cascade system can be improved by reducing the temperature difference for heat transfer in the evaporator, condenser and cascade condenser, compared to larger compressors.
Refrigeration Cycle components
Evaporators
An evaporator, like condenser is also a heat exchanger. In an evaporator, the refrigerant boils or
evaporates and in doing so absorbs heat from the substance being refrigerated. The name evaporator refers
to the evaporation process occurring in the heat exchanger
Types of Evaporators 1.Shell and Tube Evaporator
Liquid cooling evaporators may be direct expansion or flooded type. Flooded evaporators have a body
of fluid boiling in a random manner, the vapour leaving at the top. In the case of ammonia, any oil
present will fall to the bottom and be drawn off from the drain pot or oil drain connection.
Shell and Tube Evaporator
2.Direct Expansion Type-Shell and Tube
Direct Expansion Type-Shell and Tube
3.Double Pipe Type Evaporator
Double Pipe Type Evaporator 4.Direct Expansion Fin and Tube
Direct Expansion Fin and Tube
5.Embede Tube, Plate Surface Evaporator
Embede Tube, Plate Surface Evaporator
Condenser Classification of condensers:
Based on the external fluid, condensers can be classified as:
a) Air cooled condensers
b) Water cooled condensers, and
c) Evaporative condensers
-------------------------------------------------------------------------------------------
a) Air cooled condensers
As the name implies, in air-cooled condensers air is the external fluid, i.e., the refrigerant rejects heat to air
flowing over the condenser. Air-cooled condensers can be further classified into natural convection type or
forced convection type.
1-Natural Convection And Forced Convection Type b- Water Cooled Condenser
In water cooled condensers water is the external fluid. Depending upon the construction, water cooled
condensers can be further classified into:
1. Double pipe or tube-in-tube type
2. Shell-and-coil type
3. Shell-and-tube type ---------------------------------------------------------------------------------------
1. Double pipe or tube-in-tube type
2. Shell-and-coil type
3. Shell-and-tube type
c- Evaporative Condenser.
Air cooled vs water cooled condensers:
The expansion device
The expansion devices used in refrigeration systems can be divided into fixed opening type or variable
opening type. As the name implies, in fixed opening type the flow area remains fixed, while in variable
opening type the flow area changes with changing mass flow rates. There are basically seven types of
refrigerant expansion devices. These are:
1. Hand (manual) expansion valves
2. Capillary Tubes
3. Orifice
4. Constant pressure or Automatic Expansion Valve (AEV)
5. Thermostatic Expansion Valve (TEV)
6. Float type Expansion Valve
a) High Side Float Valve
b) Low Side Float Valve
7. Electronic Expansion Valve
One of the above seven types, Capillary tube and orifice belong to the fixed opening type, while the rest
belong to the variable opening type. Of the seven types, the hand operated expansion valve is not used when
an automatic control is required. The orifice type expansion is used only in some special applications. Hence
these two are not discussed here.
2-A capillary tube is a long, narrow tube of constant diameter. The word “capillary” is a misnomer since
surface tension is not important in refrigeration application of capillary tubes. Typical tube diameters of
refrigerant capillary tubes range from 0.5 mm to 3 mm and the length ranges from 1.0 m to 6 m.
The pressure reduction in a capillary tube occurs due to the following two factors:
The refrigerant has to overcome the frictional resistance offered by tube walls. This leads to some pressure
drop, and the liquid refrigerant flashes (evaporates) into mixture of liquid and vapour as its pressure
reduces. The density of vapour is less than that of the liquid. Hence, the average density of refrigerant
decreases as it flows in the tube. The mass flow rate and tube diameter (hence area) being constant, the
velocity of refrigerant increases since m = ρVA. The increase in velocity or acceleration of the refrigerant
also requires pressure drop.
3. Orifice
In variable area type expansion devices, such as automatic and thermostatic expansion
valves, the pressure reduction takes place as the fluid flows through an orifice of varying
area. Let A1and A2 be the areas at the inlet and the outlet of the orifice where, A1> A2. Let
V1 and V2 be the velocities, P1 and P2 are the pressures and ρ1and ρ2 be the densities at
the inlet and outlet respectively of the orifice as shown
4. Constant pressure or Automatic Expansion Valve (AEV)
An Automatic Expansion Valve (AEV) also known as a constant pressure expansion valve acts in such a
manner so as to maintain a constant pressure and thereby a constant temperature in the evaporator. The
schematic diagram of the valve is shown in Figure. As shown in the figure, the valve consists of an
adjustment spring that can be adjusted to maintain the required temperature in the evaporator. This exerts
force Fs on the top of the diaphragm.
The atmospheric pressure, Po also acts on top of the diaphragm and exerts a force of Fo = P oAd,
Ad being the area of the diaphragm. The evaporator pressure Pe acts below the diaphragm. The force due
to evaporator pressure is Fe= Pe Ad. The net downward force Fo + Fs - Fe is fed to the needle by the
diaphragm. This net force along with the force due to follow-up spring Ffs controls the location of the needle
with respect to the orifice and thereby controls the orifice opening.
If Fe+ Ffs> Fs+ Fo the needle will be pushed against the orifice and the valve will be fully closed.
7
5. Thermostatic Expansion Valve (TEV)
6. Float type Expansion Valve
Float type expansion valves are normally used with flooded evaporators in large capacity
refrigeration systems. A float type valve opens or closes depending upon the liquid level as sensed
by a buoyant member, called as float. The float could take the form of a hollow metal or plastic ball,
a hollow cylinder or a pan. Thus the float valve always maintains a constant liquid level in a chamber
called as float chamber. Depending upon the location of the float chamber, a float type expansion
valve can be either a low-side float valve or a high-side float valve Version
a) High Side Float Valve
b) Low Side Float Valve
7. Electronic Expansion Valve
An electronic expansion valve consists of an orifice and a needle in front it. The needle moves up and down
in response to magnitude of current in the heating element. A small resistance allows more current to flow
through the heater of the expansion valve, as a result the valve opens wider. A small negative coefficient
thermistor is used if superheat control is desired. The thermistor is placed in series with the heater of the
expansion valve. The heater current depends upon the thermistor resistance that depends upon the
refrigerant condition. Exposure of thermistor to superheated vapour permits thermistor to selfheat thereby
lowering its resistance and increasing the heater current.
This opens the valve wider and increases the mass flow rate of refrigerant. This process continues until the
vapour becomes saturated and some liquid refrigerant droplets appear. The liquid refrigerant will cool the
thermistor and increase its resistance. Hence in presence of liquid droplets the thermistor offers a large
resistance, which allows a small current to flow through the heater making the valve opening narrower. The
control of this valve is independent of refrigerant and refrigerant pressure; hence it works in reverse flow
direction also. It is convenient to use it in year-round-air-conditioning systems, which serve as heat pumps
in winter with reverse flow. In another version of it the heater is replaced by stepper motor, which opens
and closes the valve with a great precision giving a proportional control in response to temperature sensed
by an element.
Compressors The purpose of the compressor in the vapour compression cycle is to compress
THE PISTON COMPRESSION PROCESS
The piston type is very widely used, being adaptable in size, number of cylinders, speed and method of
drive. It works on the two-stroke cycle (see Figure 4.4). Automatic pressure-actuated suction and
discharge valves are used as shown in Figure 4.4. As the piston descends on the suction stroke, the
suction valve opens to admit gas from the evaporator. At the bottom of the stroke, this valve will close
again as the compression stroke begins. When the cylinder pressure becomes higher than that in the
discharge pipe, the discharge valve opens and the compressed gas passes to the condenser.
Schematic of Linde’s horizontal, double acting compressor
Double acting cylinder
Air Conditioning
Introduction to Air Conditioning Air conditioning may be required in buildings which have a high heat gain and as a result a high
internal temperature. The heat gain may be from solar radiation and/or internal gains such as people,
lights and business machines. The diagram below shows some typical heat gains in a room.
If the inside temperature of a space rises to about 25
oC then air-conditioning will probably be
necessary to maintain comfort levels. This internal temperature (around 25oC) may change depending
on some variables such as: Window blinds or shading methods
Heat absorbing glass
Heat reflecting glass
Operable windows
Higher ceilings
Smaller windows on south facing facades
Alternative lighting schemes.
If air conditioning is the only answer to adequate comfort in a building then the main choice of system can be considered.
Full comfort air conditioning can be used in summer to provide cool air (approx. 13oC to
18oC) in summer and warm air (approx. 28
oC to 36
oC) in winter.
Also the air is cleaned by filters, dehumidified to remove moisture or humidified to add
moisture.
Air conditioning systems fall into three main categories, and are detailed in the following pages;
Plant Central systems.
Room Air Conditioning Units.
Fan Coil Units.
1. Central plant systems have one central source of conditioned air which is distributed in a
network of ductwork.
Room air conditioning units are self-contained package units which can be positioned in
each room to provide cool air in summer or warm air in winter.
Fan coil units are room units and incorporate heat exchangers piped with chilled water and a fan
to provide cool air.
There are other forms of air conditioning such as:
Chilled Beams.
Induction Units.
Variable Air Volume Units.
Dual Duct Systems.
Chilled Ceiling.
But we will consider the more commonly used methods first. Typical central plant air
conditioning system.
The system shown above resembles a balanced ventilation system with plenum heating but with
the addition of a cooling coil.
In winter the heater battery will be on and the cooling coil will probably be switched off for the
majority of buildings. In summer the heater battery will not need to have the same output and the
cooling coil will be switched on.
A humidifier may be required to add moisture to the air when it is 'dry’. This is when outdoor air
has a low humidity of around 20% to 30%.
In the U.K. low humidity are rare and therefore humidification is sometimes not used. In
dryer regions humidification is required through most of the year whereas in tropical air
conditioning one of the main features of the system is the ability to remove moisture from warm
moist air.
Dampers are used in air conditioning central plant systems to control the amount of air in each
duct. It is common to have 20% fresh air and 80% recirculated air to buildings. In buildings with
high occupancy the fresh air quantity should be calculated based on C.I.B.S.E. data, this may
require a higher percentage of fresh air (i.e. more than 20%).
Filters are required to remove particles of dust and general outdoor pollution. This filter is
sometimes called a coarse filter or pre-filter. A removable fibreglass dust filter is positioned in
the fresh air intake duct or in larger installation oil filled viscous filter may be used.
The secondary filter, after the mix point, is used to remove fine dust particles or other
contaminant picked up in the rooms and recirculated back into the plant. A removable bag filter is
generally used for this where a series of woven fibre bags are secured to a framework which can
be slid out of the ductwork or air handling unit (A.H.U.) for replacement.
1-Air Handling Units
Air handling units (A.H.U.) are widely used as a package unit which incorporates all the main
plant items as shown below. Pipe work, ductwork and electrical connections are made after
the unit is set in place on site.
Since air conditioning plant rooms tend to be at roof level, the larger A.H.U.'s are lifted into
place by crane before the roof is fixed.
In some cases it is usual to place the fan in front of (that is upstream of) the heater battery and
cooling coil. This is because fans operate best if the system resistance is at the outlet rather than
the inlet of the impeller. This is shown on the schematic diagrams above.
The photograph below shows a typical air handling unit with handles on the doors for access
to equipment.
2-Room Air Conditioning Units
These units use refrigerant to transfer cooling effect into rooms.
Room air conditioning units fall into two main categories:
Split type
Window/wall units.
Split Air Conditioners
Split air conditioners have two main parts, the outdoor unit is the section which generates the
cold refrigerant gas and the indoor unit uses this cold refrigerant to cool the air in a space. The
outdoor unit uses a compressor and air cooled condenser to provide cold refrigerant to a
cooling coil in the indoor unit. A fan then blows air across the cooling coil and into the room.
The indoor unit can either be ceiling mounted (cassette unit), floor mounted or duct type. The
drawing below shows a ceiling mounted (cassette unit).
The photograph above shows a ceiling mounted cassette and an outdoor unit. Window / Wall Units
Window or wall units are more compact than split units since all the plant items are contained
in one box.
Window units are installed into an appropriate hole in the window and supported from a metal
frame.
Wall units like the one shown below are built into an external wall and contain all the necessary
items of equipment to provide cool air in summer and some may even provide heating in winter.
A small
Hermetically sealed compressor is used to provide refrigerant gas at the pressure required to
operate the refrigeration cycle. The condenser is used to condense the refrigerant to a liquid
which is then reduced in pressure and piped to the cooling coil.
Generally central plant systems are used in large prestigious buildings where a high quality
environment is to be achieved. Cassette units and other split systems can be used together
with central plant systems to provide a more flexible design.
Each system has its own advantages and the following is a summary of some of the main
advantages and disadvantages.
Central Plant Systems - Advantages:
1. Noise in rooms is usually reduced if plant room is away from occupied spaces.
2. The whole building can be controlled from a central control station. This means that
optimum start and stop can be used and a weather compensator can be utilised.
3. Also time clocks can bring air conditioning on and off at appropriate times.
4. Maintenance is centralised in the plant room. Plant is easier to access
Central Plant Systems - Disadvantages:
1. Expensive to install a complete full comfort air-conditioning system throughout a
building.
2. Space is required for plant and to run ductwork both vertically in shafts and horizontally
in ceiling spaces.
3. Individual room control is difficult with central plant.
4. Many systems have been tried such as Variable Air Volume (VAV), dual duct systems
and zone re-heaters. Zone re-heaters are probably more successful than the rest.
Room Air Conditioning Units - Advantages:
1. Cheaper to install.
2. Individual room control.
3. Works well where rooms have individual requirements.
4. No long runs of ductwork.
5. Can be used to heat as well as cool if a reversing valve is fitted.
Room Air Conditioning Units - Disadvantages:
1. Sometimes the indoor unit fan becomes noisy.
2. Noisy compressor in outdoor unit.
3. Each unit or group of units has a filter, compressor and refrigeration pipe work that needs
periodic maintenance and possible re-charging. Units have course filters therefore
filtration is not as good as with AHU’s.
4. The installation may require long runs of refrigerant pipe work which, if it leaks into the
building, can be difficult to remedy.
5. Not at robust as central plant.
6. The majority of room air conditioners just recalculate air in a room. With no fresh air
supply although most manufacturers make units with fresh air capability.
7. Cooling output is limited to about 9 kW maximum per unit, Therefore many units would
be required to cool rooms with high heat gains.
Air Conditioning process The aim of this section is to allow students to size air conditioning plant such as;
cooling coil, heater battery and
humidifier.
The notes are divided into several sections as
follows:
1. Psychometric for air conditioning.
2. The Psychometric chart.
3. Examples of psychometric properties.
4. Air conditioning plant for summer & winter.
5. Basic processes.
6. Typical air conditioning processes.
7. Annotation and room ratio.
8. Summer and winter cycles.
9. Examples.
The first section deals with Psychometric for air conditioning and discusses some properties of
moist air. A simplified psychometric chart is shown for familiarization, and some examples of
how to find air properties are provided.
A diagram of an air conditioning system is shown in schematic form in the section entitled AIR
Conditioning plant for summer & winter. Before sizing takes place the student should also
understand the processes that take place in air conditioning systems. There are four basic
processes for summer and winter air conditioning systems.
The following basic processes are
explained:
1. Mixing
2. Sensible Cooling and Heating
3. Cooling with Dehumidification
4. Humidification
The section on Typical Air Conditioning Processes shows winter and summer schematic
diagrams and psychometric charts. There are some more details that may be useful to
the designer of air conditioning systems. Further information is as follows: Annotation, Room
ratio when the processes have been superimposed onto a psychometric chart then calculations
may commence. These are as detailed in the following sections of the notes.
Summer and winter Cycles
1. Summer cycle psychometrics
2. Summer cycle calculations
3. Winter cycle psychometrics
4. Winter cycle calculations
5. Duct and Fan gains.
The final section is seven examples of plant sizing using psychometric
charts.
Psychometric for Air Conditioning
Psychometric is the study of air and water vapour mixtures.
Air is made up of five main gases i.e.
Nitrogen 78.03%, Oxygen 20.99%, Argon 0.94%, Carbon Dioxide 0.03%, and Hydrogen 0.01%
by volume.
The Ideal Gas Laws are used to determine psychometric data for air so that the engineer can
carry out calculations.
To make life easier a chart has been compiled with all the relevant psychometric data indicated.
This is called the Psychometric Chart. A typical chart is shown below.
Air at any state point can be plotted on the psychometric chart.
The information that can be obtained from a Psychometric Chart is as follows:
1. Dry bulb temperature
2. Wet bulb temperature
3. Moisture content
4. Percentage saturation
5. Specific enthalpy
6. Specific volume.
The following is a brief description of each of the properties of air.
1. Dry bulb temperature This is the air temperature measured by a mercury-in-glass thermometer.
2. Wet bulb temperature This is the air temperature measured by a mercury-in-glass thermometer which has the mercury bulb wetted by gauze that is kept moist by a reservoir of water. When exposed to
the environment the moisture evaporates from the wetted gauze, which gives a lower
reading on the thermometer. This gives an indication of how ‘dry’ or how ‘moist’ the air
is, since in ‘dry’ air the water will evaporate quickly from the gauze, which depresses the
thermometer reading.
3. Moisture content or specific humidity
This is the amount of moisture in air given in kg of moisture per kg of dry air e.g. for
room air at 21oC dry bulb and 15
oC wet bulb, the moisture content is about 0.008 kg/kg
d. a. This is a small mass of moisture (0.008 kg = 8 grams) per kg of dry air or 9.5 grams
per cubic metre of air.
4. Moist air It is a mixture of air and water vapour. The amount of water vapour present in the air
depends upon the absolute pressure and temperature of the mixture.
5. Standard air
It is moist air when the air has diffused the maximum amount of water vapour into it.
6. Dew point temperature It is the temperature of air recorded by a thermometer when the moisture present in it
begins to condense at constant pressure, thus the dew point temperature is the saturation
temperature corresponding to partial pressure of water vapour.
7. Relative humidity It is the ratio of the actual water vapour pressure of the air to the saturated water vapour
pressure of the air at the same temperature (PV, actual / PV, saturated).
8. Percentage saturation The Percentage saturation is another indication of the amount of moisture in air. This is
the ratio of the moisture content of moist air to the moisture content of saturated air at the
same temperature. When air is saturated it is at 100% saturation and cannot hold any
more moisture.
9. Specific enthalpy This is the amount of heat energy (kJ) in air per kg. If heat is added to the air at a heater
battery for example, then the amount to be added can be determined from Specific
enthalpy change.
10.Specific volume
This is the volume of moist air (dry air + water vapour) per unit mass. The units of
measurement are m3
per kg. Also specific volume = 1 / density.
11.Latent heat It is the heat which causes a change in phase with no change in the temperature.
12.Sensible heat It is the increase in heat content of air when the temperature rises as heat is added, or the
heat which causes a change in temperature.
The Psychometric Chart
The six properties of air previously discussed can be shown on one chart called a Psychometric
Chart. One of the purposes of the Psychometric Chart is to size heater batteries, cooling coils and
Psychometric Chart is to size heater batteries, cooling coils and humidifiers. A simplified
Psychometric Chart is shown below.
This chart is only for demonstration purposes. If accurate assessments are to be carried out use a
C.I.B.S.E. chart.
Using the Psychometric Chart
If any two properties of air are known then the other four can be found from the
psychometric chart.
Examples of Psychometric Properties
EXAMPLE 1
Find the moisture content of air at 25oC dry-bulb temperature and 25
oC wet-bulb
temperature.
Referring to the chart below, a vertical line is drawn upwards from 25oC dry-bulb
temperature until it intersects at 25oC wet-bulb temperature. This intersection point
happens to be on the 100% saturation line. The intersection point is highlighted and a
horizontal line is drawn to the right to find the corresponding moisture content. The
moisture content is therefore 0.020 kg/kg dry air.
EXAMPLE 2 Find the specific volume and wet-bulb temperature of air at 20
oC dry-bulb temperature
and 50% saturation.
Referring to the chart below, a vertical line is drawn upwards from 20oC dry-bulb
temperature until it intersects with the 50% saturation curve.
The intersection point is sometimes referred to as the state point.
The specific volume is found to be 0.84 m3/kg and the wet-bulb temperature is 14
oC
EXAMPLE 3 Find the specific volume, percentage saturation and moisture content of air at 15
oC
dry-bulb temperature and 10oC wet-bulb temperature.
Referring to the chart below, a vertical line is drawn upwards from 15oC dry-bulb
temperature until it intersects with the 10oC wet-bulb temperature line. This
intersection is the state point. The specific volumes found to be 0.823 m3/kg, the
percentage saturation 52% and the moisture content 0.0054 kg/kg d. a.
EXAMPLE 4 Find the specific volume, wet-bulb temperature, moisture content and specific enthalpy
of air at 35oC dry-bulb temperature and 30% saturation.
Referring to the chart below, a vertical line is drawn upwards from 35oC dry-bulb
temperature until it intersects with the 30% saturation curve.
This intersection is the state point.
The specific volume is found to be 0.883 m3/kg, the wet-bulb temperature is 22
oC,
the moisture content 0.011kg/kg d. a. and the specific enthalpy 65 kJ/kg.
Air conditioning plant for summer and winter
In the summer time when cooling is required by the air conditioning plant it will be necessary to
operate the cooling coil, re-heater and possibly other plant as well. In winter time the pre-heater
and re-heater battery will probably be on to provide warm air to overcome heat losses. Other
plant may be switched on as well. These plant items are shown in the diagram below.
The photographs below show some plant items.
Basic Air Conditioning Processes 1. Mixing
Where two air streams are mixed the psychometric process is shown as a straight line between two
air conditions on the psychometric chart, thus points 1 and 2 are joined and the mix point 3 will lie
on this line. Two air streams are mixed in air conditioning when fresh air (m1) is brought in from
outside and mixed with recirculated air (m2). The resulting air mixture is shown below as (m3).
The mixing ratio is fixed by dampers. Sometimes, in more sophisticated plant, modulating dampers
are used which are driven by electric motors to control the mixture of air entering the system. The
diagrams below show mixing of two air streams.
By the conservation of mass formula: m1 + m2 = m3
By the conservation of energy formula: m1 h1 + m2 h2 = m3 h3
Where: m = mass flow rate of air (kg/s)
h = Specific enthalpy of air (kJ/kg) found from psychometric chart.
2. Sensible Cooling and Heating
When air is heated or cooled sensibly, that is, when no moisture is added or removed, this
process is represented by a horizontal line on a psychometric chart.
For sensible heating:
The amount of heating input to the air approximates to H1-2 = m * Cp * (t2 - t1) Or more
accurately from psychometric chart: H1-2 = m * (h2 - h1) Where: H = Heat or cooling energy (kW)
m Cp
t
= Mass flow rate of air (kg/s)
= Specific heat capacity of air, may be taken as 1.01 kJ/kg oC.
= Dry bulb temperature of air (oC)
h = Specific enthalpy of air (kJ/kg) found from psychometric chart.
3. Cooling with Dehumidification
The most commonly used method of removing water vapour from air (dehumidification) is to cool
the air below its dew point.
The dew point of air is when it is fully saturated i.e. at 100% saturation.
When air is fully saturated it cannot hold any more moisture in the form of water vapour.
If the air is cooled to the dew point air and is still further cooled then moisture will drop out of the air
in the form of condensate.
This can be shown on a psychometric chart as air sensibly cooled until it becomes fully saturated
(the dew point is reached) and then the air is cooled latently to a lower temperature.
This is apparent on the psychometric chart as a horizontal line for sensible cooling to the 100%
saturation curve and then the process follows the 100% saturation curve down to another point at a lower
temperature.
This lower temperature is sometimes called the Apparatus dew Point (ADP) of the cooling coil. In
reality the ADP of the cooling coil is close to the cooling liquid temperature inside the coil. Chilled
water or refrigerant may be the cooling liquid.
The psychometric process from state point 1 to 2 to 3 may be shown as a straight line for
simplicity as shown above with a yellow line.
Typical Air Conditioning Processes
The schematic diagram below shows a typical plant system for summer air conditioning.
The psychrometric diagram below shows a typical summer cycle.
Schematic diagram below shows a typical plant system for winter air conditioning.
The psychrometric diagram below shows a typical winter cycle.
Where Q2
U
= =
Amount of heat produced due to heat transmission through wall Transmission coefficient U of Wall (carrier H.B; page 66)
ΔTequivalent = =
Equivalent temp Equivalent temp difference (carrier H.B; page 62) +Correction factor
(carrier H.B; page, 63
Thermal loads sources include a) Solar heat gain. b) Transmission heat gain.
c) Internal heat gain.
d) Ventilation ,and infiltration.
Thermal load types divided into two types a) Sensible Heat. b) Latent Heat.
How to calculate sensible and latent heat effect on the building causes increasing internal
temperature of building?
a) Solar Heat Gain
By using the equation, Q=A*sc*q
Where Q = Aamount of heat produced due to solar radiation BTU/HR
A
SC
q
=
=
=
Area of window (FT2)
Overall factor for solar heat gain through glass (carrier H.B ; page 52)
Solar heat gain through ordinary glass (carrier H.B ; page 44)
b) Transmission Heat Gain include 1-Windows
Can be calculated by using equation Q1 = A * U * ΔT Where Q1 = Amount of heat produced due to heat transmission through window.
A
U
To
Tr
=
=
=
=
Area of window (FT2)
Transmission coefficient U of windows (carrier H.B ; page 76)
Out side temp F
Room temp, F
2-WALL
Can be calculated by using equation Q2 = A * U * ΔTequivalent
CFM = (space volume (FT3)*n of air change)/60
OR = Amount of air required for each person *number of person Qs
Ql
Gr/Ib
= =
=
CFM * 1.08 * (To-Ti) 0.68 ( (Gr/Ib)o – (Gr/Ib)i ) Moisture content
BTU/HR BTU/HR
3, 4, and 5 Floor, Paptions, Ceiling
Can be calculated by using equation Q3,4,5 = A * U * ΔT
Where Q3,4,5 = Amount of heat produced due to heat transmission through floor, part ion, and ceiling (BTU/HR)
A = Area of through floor, part ion, and ceiling.
ΔT = 0 or 5 when your neighbour hood space was conditioned . (To-
10) – Ti when your neighbour hood space was not conditioned .
(To- – Ti) when your neighbour hood space was not conditioned , and very hot .
6-ROOF
Can be calculated by using equation Q6 = A * U * ΔT
Q6
A
= =
Amount of heat produced due to heat transfer through roof Roof area
U = Transmission coefficient U –flat roof ((carrier H.B; page 71)
ΔTequivalent = =
Equivalent temp Equivalent temp difference (carrier H.B; page 63) +Correction factor
(carrier H.B; page, 63)
1- Internal heat gain 1. lighting
Q = Power (watt) * 3.4 BTU/HR for normal lighting
= Power (watt) * 4.2 BTU/HR for folarseat lighting
2. Electrical equipment
Q = Power (watt) * 3.4 BTU/HR
3. Electrical motors
Q = Power (HP) * 2545 BTU/HR
4. People
Human body dissipate two types of j heat Sensible heat gain can be calculated as follow
Qs = number of persons * Ssensible heat gain for each person.
Latent heat gain can be calculated as follow
QL= number of persons * Latent heat gain for each person.
D) Ventilation and infiltration
Thermal load due to infiltration can be neglected.
Thermal load due to ventilation can be calculated as follow
Amount of air required for ventilation can be calculated by using 2 ways
From points A, B, C, and D
A/C Capacity = Ggrand sensible heat gain + Grand latent heat gain BTU/HR
Air transmission
Flow of air through ducts To overcome the fluid friction and the resulting head, a fan is required in air conditioning systems. When a
fan is introduced into the duct through which air is flowing, then the static and total pressures at the section
where the fan is located rise. This rise is called as Fan Total Pressure (FTP). Then the required power
input to the fan is given by:
It can be easily shown that when applied between any two sections 1 and 2 of the duct, in which the fan is
located, the FTP is given by:
Estimation of pressure loss in ducts
As air flows through a duct its total pressure drops in the direction of flow.
The pressure drop is due to:
1. Fluid friction
2. Momentum change due to change of direction and/or velocity
The pressure drop due to friction is known as frictional pressure drop or friction loss, Δpf. The pressure
drop due to momentum change is known as momentum pressure drop or dynamic loss, Δpd. Thus the
total pressure drop Δpt is given by:
Dynamic losses in ducts
where K is the dynamic loss coefficient, which is normally obtained from experiments
Sources of Dynamic Losses
Sudden Enlargement
Finding Pressure Loss By The equal friction factor method The equal method of sizing ducts is often preferred because it is quite easy to use. The method
can be summarized to
1- Compute the necessary air flow volume in every room and branch of the system.
2- Se to compute the total air volume in the main system
3- Determine the maximum acceptable air flow velocity in the main duct.
4- Determine the major pressure drop in the main duct.
5- Use the major pressure drop for the main duct as a constant to determine the duct sizes
throughout the distribution system.
6- Determine the total resistance in the duct system by multiplying the static resistance with the
equivalent length of the longest run.
7- Compute balancing dampers.
Example:
References : 1-Refrigeration and air conditioning Fourth Edition, G.F.Hundy, A.R. Trott, T.C.Welch 2- adermacher, R., & Hwang, Y. (2005). Vapor compression heat pumps with refrigerant mixes. Boca Raton, FL: Taylor & Francis. 3- Haile, J. M. (2002). Lectures in Thermodynamics: Macatea Productions. 4-Flow Rates in Heating System. (n.d.). Retrieved 27th May 2015 from http://www.engineeringtoolbox.com/water-flow-rates-heating-systems-d_659.html. 5- Refrigeration and air conditioning, S.K. Mondoal. 6- Industrial Refrigeration Hand Book,Wilbert F. Stoecker.