Cengel and Boles + NPTEL Condensed

1. Refrigeration – Cengel and Boles + NPTEL(Condensed)
Can an object never be cooled to absolute zero because the thermal energy has to go somewhere
colder?
From a purely classical standpoint this is absolutely right. The classical way to cool a gas is to
compress it, in which case, according to the Ideal Gas Law (PV=nRT), it will heat up (as Pressure
rises Temperature must rise). We then put this gas through some type of heat exchanger, and
since it has a higher temperature than the cooling medium, heat flows from it into the cooling
medium. With the gas cooler now, it is allowed to expand (and depressurize), which according to
the Ideal Gas Law will cause the temperature to lower.
I pretty much just described how a refrigerator works, except in a refrigerator the gas gets so cold
that it condenses into liquid. This method alone should suffice to get you down into the single
digits of the Kelvin range. This is basically how you make Liquid Hydrogen (20 K boiling point) and
Liquid Nitrogen (77 K) or even Liquid Helium (4 K).
To get down into the sub Kelvin region requires some different techniques, and this is where
classical thermodynamics starts to break down. Classical Thermodynamics is based in the Zeroth
Law, which is so named because, though it is more fundamental than the First Law (energy neither
created nor destroyed), it was conceived of later than the First Law. It pretty much states that if
you have something in thermal equilibrium (read, same temperature) as two other things, then
those two things are in thermal equilibrium with each other. In symbolic terms, if you have body
A at temperature T(A), body B at T(B), and body C at T(C), and T(C) = T(A) and T(C) = T(B), then T(A)
= T(B). It's basically the Transitive Property for thermodynamics.
Theoretically, of course, you can use the technique I described above (compression ==>heat
exchanger ==> expansion) to get as cold as you want, but never reach zero (just as the asker
originally stated). Each time you compress your gas (cooling it with Liquid Helium, presumably),
you then expand it further and further. If you cooled your gas to 4K and then expanded it to double
its original volume, then the temperature would be 2 K (as long as you kept the pressure the same).
Quadruple the volume and you lower the temperature to 1 K, and so on and so on. But the problem
begins would you start getting to this low density, you begin to run into issues. And this is where
classical thermo really starts to break down. At some point the gas stops behaving like a gas, and
starts behaving like a group of individual atoms. CT is not designed to handle such things, and it's
fortunate that Statistical Mechanics came along to help describe what is going on. It turns out that
a better way to cool something down into the sub-kelvin range is to slowly bleed of some of the
gas, in a technique that is known as evaporative cooling. Your skin is cooled by evaporative cooling

2. when your sweat evaporates and carries heat away from your skin on a hot day. The same thing
can be done with really cold vapors, by bleeding of some of the hottest (fastest moving) atoms of
the vapor and letting it carry away some of the heat. Beyond this you need lasers to actually slow
down the atoms even more, by the simple (hahaha) technique of using the coherent photons in
the laser beam to slam into the atoms and stop them dead in their tracks (or as close as we can
get). This is not that different from two billiard balls running into each other at high speed and
then slowly drifting away from the collision. You can even use magnets to hold the atoms in place,
by exploiting the different charge of the proton and electron.
None of this answers the original question, though. By using the
compression/cooling/expansion technique (modern refrigeration), you can theoretically get as
cold as you want. But of course, this is basically Zeno's Paradox in temperature. You can only lower
the temperature to some fraction of the original temperature, and as long as it is a fraction you'll
never get to zero. So the original question wasn't entirely right, but it had a great insight, and that
makes it worthy of consideration. So no, we'll never get to absolute zero, but we're getting closer
every day.
PROPERTIES OF PURE SUBSTANCES
A pure substance does not have to be of a single chemical element or compound, however. A
mixture of various chemical elements or compounds also qualifies as a pure substance as long as
the mixture is homogeneous. Air, for example, is a mixture of several gases, but it is often
considered to be a pure substance because it has a uniform chemical composition. However, a
mixture of oil and water is not a pure substance. Since oil is not soluble in water, it will collect on
top of the water, forming two chemically dissimilar regions.
A mixture of two or more phases of a pure substance is still a pure substance as long as the
chemical composition of all phases is the same (Fig. 3–2). A mixture of ice and liquid water, for
example, is a pure substance because both phases have the same chemical composition. A mixture
of liquid air and gaseous air, however, is not a pure substance since the composition of liquid air
is different from the composition of gaseous air, and thus the mixture is no longer chemically
homogeneous. This is due to different components in air condensing at different temperatures at
a specified pressure.

3. Compressed Liquid and Saturated Liquid
Consider a piston–cylinder device containing liquid water at 20°C and 1 atm pressure (state 1, Fig.
3–5). Under these conditions, water exists in the liquid phase, and it is called a compressed liquid,
or a subcooled liquid, meaning that it is not about to vaporize. Heat is now transferred to the water
until its temperature rises to, say, 40°C. As the temperature rises, the liquid water expands slightly,
and so its specific volume increases. To accommodate this expansion, the piston moves up slightly.
The pressure in the cylinder remains constant at 1 atm during this process since it depends on the
outside barometric pressure and the weight of the piston, both of which are constant. Water is
still a compressed liquid at this state since it has not started to vaporize. As more heat is
transferred, the temperature keeps rising until it reaches 100°C (state 2, Fig. 3–6). At this point
water is still a liquid, but any heat addition will cause some of the liquid to vaporize. That is, a
phase-change process from liquid to vapor is about to take place. A liquid that is about to vaporize
is called a saturated liquid. Therefore, state 2 is a saturated liquid state.
Saturated Vapor and Superheated Vapor
Once boiling starts, the temperature stops rising until the liquid is completely vaporized. That is,
the temperature will remain constant during the entire phase-change process if the pressure is
held constant. At sea level (P = 1 atm), the thermometer will always read 100°C if the pan is
uncovered or covered with a light lid. During a boiling process, the only change we will observe is
a large increase in the volume and a steady decline in the liquid level as a result of more liquid

4. turning to vapor. Midway about the vaporization line (state 3, Fig. 3–7), the cylinder contains equal
amounts of liquid and vapor. As we continue transferring heat, the vaporization process continues
until the last drop of liquid is vaporized (state 4, Fig. 3–8). At this point, the entire cylinder is filled
with vapor that is on the borderline of the liquid phase. Any heat loss from this vapor will cause
some of the vapor to condense (phase change from vapor to liquid). A vapor that is about to
condense is called a saturated vapor. Therefore, state 4 is a saturated vapor state. A substance at
states between 2 and 4 is referred to as a saturated liquid–vapor mixture since the liquid and vapor
phases coexist in equilibrium at these states.
One the phase-change process is completed, we are back to a single phase region again (this time
vapor), and further transfer of heat results in an increase in both the temperature and the specific
volume (Fig. 3–9). At state 5, the temperature of the vapor is, let us say, 300°C; and if we transfer
some heat from the vapor, the temperature may drop somewhat but no condensation will take
place as long as the temperature remains above 100°C (for P = 1 atm). A vapor that is not about
to condense (i.e., not a saturated vapor) is called a superheated vapor. Therefore, water at state 5
is a superheated vapor.

5. The only reason water started boiling at 100°C was because we held the pressure constant at 1
atm (101.325 kPa). If the pressure inside the cylinder were raised to 500 kPa by adding weights on
top of the piston, water would start boiling at 151.8°C. That is, the temperature at which water
starts boiling depends on the pressure; therefore, if the pressure is fixed, so is the boiling
temperature.
At a given pressure, the temperature at which a pure substance changes phase is called the
saturation temperature Tsat. Likewise, at a given temperature, the pressure at which a pure
substance changes phase is called the saturation pressure Psat
Consequences of Tsat and Psat Dependence
Consider a sealed can of liquid refrigerant-134a in a room at 25°C. If the can has been in the room
long enough, the temperature of the refrigerant in the can is also 25°C. Now, if the lid is opened
slowly and some refrigerant is allowed to escape, the pressure in the can will start dropping until
it reaches the atmospheric pressure. If you are holding the can, you will notice its temperature
dropping rapidly, and even ice forming outside the can if the air is humid. A thermometer inserted
in the can will register -26°C when the pressure drops to 1 atm, which is the saturation
temperature of refrigerant-134a at that pressure. The temperature of the liquid refrigerant will
remain at -26°C until the last drop of it vaporizes.

6. Diagram
Add weights on top of the piston until the pressure inside the cylinder reaches 1 MPa. At this
pressure, water has a somewhat smaller specific volume than it does at 1 atm pressure. As heat is
transferred to the water at this new pressure, the process follows a path that looks very much like
the process path at 1 atm pressure, as shown in Fig. 3–15, but there are some noticeable
differences. First, water starts boiling at a much higher temperature (179.98C) at this pressure.
Second, the specific volume of the saturated liquid is larger (liquid being incompressible has left
effect of pressure on it) and the specific volume of the saturated vapor is smaller than the
corresponding values at 1 atm pressure. That is, the horizontal line that connects the saturated
liquid and saturated vapor states is much shorter. As the pressure is increased further, this
saturation line continues to shrink, as shown in Fig. 3–15, and it becomes a point when the
pressure reaches 22.06 MPa for the case of water. This point is called the critical point, and it is
defined as the point at which the saturated liquid and saturated vapor states are identical.

7. A piston–cylinder device that contains liquid water at 1 MPa and 150°C. Water at this state exists
as a compressed liquid. Now the weights on top of the piston are removed one by one so that the
pressure inside the cylinder decreases gradually (Fig. 3–18).
The water is allowed to exchange heat with the
surroundings so its temperature remains constant. As the pressure decreases, the volume of the

8. water increases slightly. When the pressure reaches the saturation-pressure value at the specified
temperature (0.4762 MPa), the water starts to boil. During this vaporization process, both the
temperature and the pressure remain constant, but the specific volume increases. Once the last
drop of liquid is vaporized, further reduction in pressure results in a further increase in specific
volume. Notice that during the phase-change process, we did not remove any weights. Doing so
would cause the pressure and therefore the temperature to drop [since Tsat =f(Psat)], and the
process would no longer be isothermal.

9. On P-v or T-v diagrams, these triple-phase states form a line called the triple line. The states on
the triple line of a substance have the same pressure and temperature but different specific
volumes. The triple line appears as a point on the P-T diagrams and, therefore, is often called the
triple point. For water, the triple-point temperature and pressure are 0.01°C and 0.6117 kPa,
respectively. That is, all three phases of water coexist in equilibrium only if the temperature and
pressure have precisely these values.

10. Superheated Vapor

11. But unlike superheated vapor, the compressed liquid properties are not much different from the
corresponding saturated liquid values.
THE JOULE-THOMSON COEFFICIENT
The temperature behavior of a fluid during a throttling (h = constant) process is described by the
Joule-Thomson coefficient, defined as

12. A careful look at its defining equation reveals that the Joule-Thomson coefficient represents the
slope of h = constant lines on a T-P diagram.
Some constant-enthalpy lines on the T-P diagram pass through a point of zero slope or zero Joule-
Thomson coefficient. The line that passes through these points is called the inversion line, and the
temperature at a point where a constant-enthalpy line intersects the inversion line is called the
inversion temperature. The temperature at the intersection of the P = 0 line (ordinate) and the
upper part of the inversion line is called the maximum inversion temperature. Notice that the
slopes of the h = constant lines are negative (μJT < 0) at states to the right of the inversion line and
positive (μJT > 0) to the left of the inversion line.
A throttling process proceeds along a constant-enthalpy line in the direction of decreasing
pressure, that is, from right to left. Therefore, the temperature of a fluid increases during a
throttling process that takes place on the right-hand side of the inversion line. However, the fluid
temperature decreases during a throttling process that takes place on the left-hand side of the
inversion line. It is clear from this diagram that a cooling effect cannot be achieved by throttling
unless the fluid is below its maximum inversion temperature. This presents a problem for
substances whose maximum inversion temperature is well below room temperature. For

13. hydrogen, for example, the maximum inversion temperature is -68°C. Thus hydrogen must be
cooled below this temperature if any further cooling is to be achieved by throttling.
REFRIGERATORS AND HEAT PUMPS

14. Schematic of a Carnot refrigerator and T-s diagram of the reversed Carnot cycle.
THE IDEAL VAPOR-COMPRESSION REFRIGERATION CYCLE
1-2 Isentropic compression in a compressor
2-3 Constant-pressure heat rejection in a condenser
3-4 Throttling in an expansion device
4-1 Constant-pressure heat absorption in an evaporator

15. A rule of thumb is that the COP improves by 2 to 4 percent for each °C the evaporating temperature
is raised or the condensing temperature is lowered.
The condenser and the evaporator do not involve any work, and the compressor can be
approximated as adiabatic. Then the COPs of refrigerators and heat pumps operating on the vapor
compression refrigeration cycle can be expressed as

16. ACTUAL VAPOR-COMPRESSION REFRIGERATION CYCLE
Two common sources of irreversibilities are fluid friction (causes pressure drops) and heat transfer
to or from the surroundings.
In the ideal cycle, the refrigerant leaves the evaporator and enters the compressor as saturated
vapor. But it may not be possible to control the state of the refrigerant so precisely. Instead, it is
easier to design the system so that the refrigerant is slightly superheated at the compressor inlet.
This slight overdesign ensures that the refrigerant is completely vaporized when it enters the
compressor. Also, the line connecting the evaporator to the compressor is usually very long; thus
the pressure drop caused by fluid friction and heat transfer from the surroundings to the
refrigerant can be very significant. The result of superheating, heat gain in the connecting line, and
pressure drops in the evaporator and the connecting line is an increase in the specific volume,
thus an increase in the power input requirements to the compressor since steady-flow work is
proportional to the specific volume.
The compression process in the ideal cycle is internally reversible and adiabatic, and thus
isentropic. The actual compression process, however, involves frictional effects, which increase
the entropy, and heat transfer, which may increase or decrease the entropy.

17. In the ideal case, the refrigerant is assumed to leave the condenser as saturated liquid at the
compressor exit pressure. In reality, however, it is unavoidable to have some pressure drop in the
condenser as well as in the lines connecting the condenser to the compressor and to the throttling
valve. Also, it is not easy to execute the condensation process with such precision that the
refrigerant is a saturated liquid at the end, and it is undesirable to route the refrigerant to the
throttling valve before the refrigerant is completely condensed. Therefore, the refrigerant is sub-
cooled somewhat before it enters the throttling valve. We do not mind this at all, however, since
the refrigerant in this case enters the evaporator with a lower enthalpy and thus can absorb more
heat from the refrigerated space. The throttling valve and the evaporator are usually located very
close to each other, so the pressure drop in the connecting line is small.
GAS REFRIGERATION CYCLES
The reversed Brayton cycle, better known as the gas refrigeration cycle.
Despite their relatively low COPs, the gas refrigeration cycles have two desirable characteristics:
They involve simple, lighter components, which make them suitable for aircraft cooling, and they
can incorporate regeneration, which makes them suitable for liquefaction of gases and cryogenic
applications.

18. Vapour refrigeration system is based on phase change principle (cooling effect).
Gas refrigeration is mainly based on sensible cooling.
For pure substance boiling point coincides with dew point temperature, similarly the freezing point
and melting point also coincides. (why not for impure substance)
1. Constant volume process: heating & cooling of a gas stored in a rigid vessel (non-flow process)
2. Constant pressure process: Heating/cooling of a gas in a piston cylinder assembly (non-flow
process)
3. Constant temperature process: Compression/expansion of a gas with heat transfer (non flow
process)
4. Adiabatic process: compression and expansion of a gas under perfectly insulated conditions.
5. Polytropic process: compression/expansion with heat transfer.

19. METHODS OF PRODUCING LOW TEMPERATURES
Sensible cooling
Endothermic mixing
Phase change processes
Expansion of liquids
Expansion of gases
Thermoelectric methods
Magnetic methods

23. Temperature drop is same whether we use throttling process or use a turbine; because of the initial
state of the working fluid.
Expansion in the subcooled region. Throttling gives lower temperature drop for the same pressure drop.
Also both the process gives lower temperature drop compared to expansion from saturated initial state.

28. Cold air assumptions won’t work for throttling process because if it would have been ideal gas there is
no reduction in the temperature as enthalpy is only function of temperature for ideal gas. Taking gas as
real gas we can explain throttling well.

29. Isothermal processes of reverse Carnot’s cycle is replaced by two isobaric processes in Reverse Brayton
cycle. Also here shaft is joined between turbine and the compressor. Part of work output is being
utilized within the cycle.

32. Closed systems can use dense air that is good volumetric efficiency reducing the sizes of compressor and
turbine, heat exchangers etc. Also in closed system we can use alternative for air also.

34. PSYCHROMETRY
At 50°C, the saturation pressure of water is 12.3 kPa. At pressures below this value, water vapor
can be treated as an ideal gas, even when it is a saturated vapor. Therefore, water vapor in air
behaves as if it existed alone and obeys the ideal-gas relation Pv = RT. Then the atmospheric air
can be treated as an ideal-gas mixture whose pressure is the sum of the partial pressure of dry
air* Pa and that of water vapor Pv:
The partial pressure of water vapor is usually referred to as the vapor pressure. It is the pressure
water vapor would exert if it existed alone at the temperature and volume of atmospheric air.

35. Since water vapor is an ideal gas, the enthalpy of water vapor is a function of temperature only,
that is, h = h(T). This can also be observed from the T-s diagram of water where the constant
enthalpy lines coincide with constant-temperature lines at temperatures below 50°C. Therefore,
the enthalpy of water vapor in air can be taken to be equal to the enthalpy of saturated vapor at
the same temperature.
Relative Humidity

37. As the air cools at constant pressure, the vapor pressure Pv remains constant. Therefore, the vapor
in the air (state 1) undergoes a constant-pressure cooling process until it strikes the saturated
vapor line (state 2). The temperature at this point is Tdp, and if the temperature drops any further,
some vapor condenses out. As a result, the amount of vapor in the air decreases, which results in
a decrease in Pv. The air remains saturated during the condensation process and thus follows a
path of 100 percent relative humidity (the saturated vapor line). The ordinary temperature and
the dew-point temperature of saturated air are identical.
The dew-point temperature of room air can be determined easily by cooling some water in a metal
cup by adding small amounts of ice and stirring. The temperature of the outer surface of the cup
when dew starts to form on the surface is the dew-point temperature of the air.
Knowing the dew-point temperature, we can determine the vapor pressure Pv and thus the
relative humidity. Another way of determining the absolute or relative humidity is related to an
adiabatic saturation process. The system consists of a long insulated channel that contains a pool
of water. A steady stream of unsaturated air that has a specific humidity of ω1 (unknown) and a
temperature of T1 is passed through this channel. As the air flows over the water, some water
evaporates and mixes with the airstream. The moisture content of air increases during this
process, and its temperature decreases, since part of the latent heat of vaporization of the water
that evaporates comes from the air. If the channel is long enough, the airstream exits as saturated
air (Φ = 100 percent) at temperature T2, which is called the adiabatic saturation temperature.
If makeup water is supplied to the channel at the rate of evaporation at temperature T2, the
adiabatic saturation process described above can be analyzed as a steady-flow process. The
process involves no heat or work interactions, and the kinetic and potential energy changes can
be neglected. Then the conservation of mass and conservation of energy relations for this two
inlet, one-exit steady-flow system reduces to the following:

38. WHY IT IS NOT CONTANT PRESSURE PROCESS???

40. A more practical approach is to use a thermometer whose bulb is covered with a cotton wick
saturated with water and to blow air over the wick. The temperature measured in this manner is
called the wet-bulb temperature Twb.
The basic principle involved is similar to that in adiabatic saturation. When unsaturated air
passes over the wet wick, some of the water in the wick evaporates. As a result, the temperature
of the water drops, creating a temperature difference (which is the driving force for heat transfer)
between the air and the water. After a while, the heat loss from the water by evaporation equals
the heat gain from the air, and the water temperature stabilizes. The thermometer reading at this
point is the wet-bulb temperature.
In general, the adiabatic saturation temperature and the wet-bulb temperature are not
the same. However, for air–water vapor mixtures at atmospheric pressure, the wet-bulb
temperature happens to be approximately equal to the adiabatic saturation temperature.
Therefore, the wet-bulb temperature Twb can be used in Eq. 14–14 in place of T2 to determine the
specific humidity of air.
The state of the atmospheric air at a specified pressure is completely specified by two independent
intensive properties. The rest of the properties can be calculated easily from the previous
relations. The constant wet- bulb-temperature lines are used as constant-enthalpy lines in some
charts. For saturated air, the dry-bulb, wet-bulb, and dew-point temperatures are identical.
Therefore, the dew-point temperature of atmospheric air at any point on the chart can be

41. determined by drawing a horizontal line (a line of ω = constant or Pv = constant) from the point to
the saturated curve. The temperature value at the intersection point is the dew-point
temperature.
The comfort of the human body depends primarily on three factors: the (dry-bulb) temperature,
relative humidity, and air motion.
The relative humidity of air decreases during a heating process even if the specific humidity ω
remains constant. This is because the relative humidity is the ratio of the moisture content to the
moisture capacity of air at the same temperature, and moisture capacity increases with
temperature. Mass of air and water remains constant and work transfer is also zero we have
Heating with Humidification
If steam is introduced in the humidification section, this will result in humidification with
additional heating (T3 > T2). If humidification is accomplished by spraying water into the airstream

42. instead, part of the latent heat of vaporization comes from the air, which results in the cooling of
the heated airstream (T3 < T2). Air should be heated to a higher temperature in the heating section
in this case to make up for the cooling effect during the humidification process.
Cooling with Dehumidification
The specific humidity of air remains constant during a simple cooling process, but its relative
humidity increases. If the relative humidity reaches undesirably high levels, it may be necessary to
remove some moisture from the air, that is, to dehumidify it. This requires cooling the air below
its dew-point temperature.

43. Evaporative Cooling
Evaporative cooling is based on a simple principle: As water evaporates, the latent heat of
vaporization is absorbed from the water body and the surrounding air. As a result, both the water
and the air are cooled during the process.
A porous jug or pitcher filled with water is left in an open, shaded area. A small amount of
water leaks out through the porous holes, and the pitcher “sweats.” In a dry environment, this
water evaporates and cools the remaining water in the pitcher

44. The evaporative cooling process is essentially identical to the adiabatic saturation process since
the heat transfer between the airstream and the surroundings is usually negligible. Therefore, the
evaporative cooling process follows a line of constant wet-bulb temperature on the psychrometric
chart. (Note that this will not exactly be the case if the liquid water is supplied at a temperature
different from the exit temperature of the airstream.)
Adiabatic Mixing of Airstreams