1. Define psychrometry.
Most commonly, psychrometry is defined as the science of measuring the water-vapour content of air. This definition is unnecessarily restrictive - a better definition is the study of gas mixtures containing a vapour near condensation.
2. Explain the difference between relative and specific humidity?
Specific or absolute humidity is the ratio of mass of water vapour in a dry gas mixture to the mass of dry gas. Unless water vapour is introduced or removed from the system, there is no change in state of the gas mixture that can alter the specific humidity.
The relative humidity is a measure of the mass of water vapour in a dry gas mixture relative to the maximum mass of water vapour the gas mixture is able to hold without condensation occurring. For an ideal gas, this mass ratio is equivalent to the ratio of vapour pressure (partial pressure of the water vapour) to the saturation pressure. Even without water vapour introduction or removal from the system, a change in state of the gas mixture can result in a change of relative humidity.
3. How will the relative and specific humidity change with temperature?
As discussed in part 2, the specific humidity will not change with temperature, however the relative humidity will change. As the temperature increases, the saturation pressure will also increase (i.e. the denominator in the relative humidity). If the total system pressure is unchanged, then the partial pressure of water vapour is also unchanged (i.e. the numerator in the relative humidity). The relative humidity therefore decreases with increasing temperature.
The reverse effect (i.e. decreasing temperature) will increase the relative humidity. Logically, there is a limit to this increase of 100%RH. The temperature reached at this state is the dew point temperature and any further cooling will result in condensation. The relative humidity below the dew point temperature will then remain 100%.
4. Can water vapour in air be treated as an ideal gas? Why?
Intermolecular interaction in a gas or gas mixture is reduced at low pressures. When a real gas becomes part of a mixture with ideal gases, the mixture can be treated as ideal if the partial pressure of the real gas is sufficiently low enough for molecular interactions to become negligible. A classic example of this is the treatment of water vapour (strongly non-ideal - as we know through the use of steam tables for solving steam turbine problems) and dry air mixtures as an ideal gas mixture. This approach is only accurate when the partial pressure of the water vapour in the mixture is less than about 10 kPa.
5. What is the dew point temperature?
As covered in part 3, when you take a gas and cool it at constant pressure, the dew point temperature is the temperature at which the relative humidity reaches 100%. The dew point is the highest temperature at which you will start to see condensation from the mixture at a particular system pressure.
6. What is the wet bulb temperature and how does it relate to the adiabatic saturation temperature?
A wet bulb temperature measurement is made with a thermometer that is kept in direct contact with liquid water while air-flow of at least 3m/s passes. This temperature measurement achieves a balance between sensible heat flow from the surroundings and the latent heat losses due to the evaporation of the water. It is therefore an estimate of the adiabatic saturation temperature.
Note that if the airflow is less than 3m/s, the equilibrium state reached is air velocity dependent and will need slight correction. At 3m/s and higher, the equilibrium state reached is essentially no longer velocity dependent.
7. On a psychrometric chart, sketch a dry heating, cooling, humidification and de-humidification process.
8. Describe how the dry bulb temperature, relative humidity, specific humidity, enthalpy and wet bulb temperature change during the processes of Dry heating, Dry cooling, Humidification, De-humidification?
9. On a psychrometric chart, sketch the function of an evaporative air-conditioning system. What is the limit of the cooling effectiveness of an evaporative cooler?
The limit to evaporative cooling is reached when the wet bulb temperature is the same as the dry bulb
temperature (i.e. at 100% RH). No further liquid water provided will evaporate at the air is already saturated
and no further cooling is possible.
10. How do we deal with the mixing of two air-streams in air conditioning applications? When are we likely to encounter the mixing of two air-streams?
When two airstreams at states 1 and 2 are mixed adiabatically, the state of the mixture lies on the straight
line connecting the two states. Division of the line is inversely proportional to the respective mass flow
rates. Recirculating already conditioned air is generally economical however AC design codes require
minimum ventilation with fresh air. To combine the recirculated air with the fresh air required, a mixing
chamber is used.
11. Why do we study the reversed Carnot cycle even though it is not a realistic model for refrigeration cycles?
The reversed Carnot cycle is totally reversible, with all heat exchange occurring during
isothermal processes. It therefore has the highest possible Coefficient of Performance
possible and is therefore useful as the benchmark for all other (usually more practical)
cycles.
12. Why is the reversed Carnot cycle executed within the saturation dome not a realistic model for refrigeration cycles?
Because the compression process involves the compression of a liquid-vapour mixture
which requires a compressor that will handle two phases. Also, the expansion process
involves the expansion of high-moisture content refrigerant.
13. Why do we use a throttling valve in the ideal vapour-compression refrigeration system?
In practice, it is less complex and more economical to use a throttling valve instead. The
ideal vapour-compression refrigeration cycle therefore does not include a turbine to make
the cycle more closely approximate the actual cycle.
14. The COP of vapour-compression refrigeration cycles improves when the refrigerant is subcooled before it enters the throttling valve. Can the refrigerant be subcooled indefinitely to maximise this effect, or is there a lower limit? Explain.
The minimum temperature that the refrigerant can be cooled to before throttling is the
temperature of the sink (the cooling medium) since heat is transferred from the refrigerant
to the cooling medium.
Also Read
15. Heat engines and heat pumps (refrigerators) are energy conversion devices altering amounts of energy transfer between Q and W. Which conversion direction (Q to W or W to Q) is limited and which is unlimited according to the second law.
Since work energy has the highest quality or availability, W to Q conversion can
theoretically be 100% efficient. The reverse process is subject to the second law limitations
since the quality or availability of heat energy is dependent on the temperature of the
source. Finite temperatures imply that heat energy is always less available than work
energy and the conversion from Q to W is limited.
16. An engineer suggests changing the electrical heating in a house to a heat pump system. Is that a good idea?
Yes (installation cost not withstanding). The COP of a heat pump is always greater than or
equal to 1. An electric heater can at best achieve an efficiency of 100%. Therefore, even
when operating at its worst, the heat pump is at least as efficient as the electric heater.
This should lead to the thought of how is it possible to provide more heat energy via a
heat pump than the work energy input. The answer is of course that the work input is not
being converted directly to heat output. The work input in a heat pump simply drives the
compressor. It is the evaporation and condensation action of the refrigerant that causes
heat to be absorbed (mostly latent heat of vapourisation) from the cool space and released
to the warm space.
17. It is proposed to use water instead of refrigerant-134a as the working fluid in air-conditioning applications where the minimum temperature never falls below the freezing point. Would you support this proposal? Explain.
No. Assuming the water is maintained at 10C in the evaporator, the evaporator pressure
will be the saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not
practical to design refrigeration or air-conditioning devices that involve such extremely
low pressures.
18. In gas refrigeration cycles, can the turbine be replaced by an expansion valve as is done with vapour-compression refrigeration cycles? Why?
No. For ideal gases, the enthalpy of the gas depends only on temperature (i.e. h = h(T)).
Since the throttling process is an isenthalpic process (i.e. h1 = h2), the temperature of gas
will not drop during a throttling process. The high pressure part of the cycle (i.e.
condenser) will then be at the same temperature as the low pressure part of the cycle (i.e.
evaporator) hence negating any possible refrigeration effect.
19. How can very low temperatures be achieved using gas refrigeration cycles?
Using reversed Brayton cycle gas refrigeration, regeneration is required to achieve low gas
refrigeration temperatures. Without regeneration, the lowest turbine inlet temperature is
the temperature of the surroundings or any other cooling medium. With regeneration, the
high-pressure gas is further cooled before expanding in the turbine. Lowering the turbine
2
inlet temperature automatically lowers the turbine exit temperature, which is the
minimum temperature in the cycle.
20. Why is the Carnot cycle not a realistic model for steam power plants?
The Carnot cycle is not a realistic model for steam power plants because (1) limiting the
heat transfer processes to two-phase systems to maintain isothermal conditions severely
limits the maximum temperature that can be used in the cycle, (2) the turbine will have to
handle steam with a high moisture content which causes erosion, and (3) it is not practical
to design a compressor that will handle two phases.
21. How do actual vapour power cycles differ from idealized ones?
The actual vapour power cycles differ from the idealized ones in that the actual cycles
involve friction and pressure drops in various components and the piping, and heat loss to
the surrounding medium from these components and piping.
22. Is it possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water entering at 20°C?
Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much
higher than the temperature of the cooling water.
23. During a regeneration process, some steam is extracted from the turbine and is used to heat the liquid water leaving the pump. This does not seem like a smart thing to do since the extracted steam could produce some more work in the turbine. How do you justify this action?
This is a smart idea because we waste little work potential but we save a lot from the heat
input. The extracted steam has little work potential left, and most of its energy would be
part of the heat rejected anyway. Therefore, by regeneration, we utilize a considerable
amount of heat by sacrificing little work output.
24. How do open feed-water heaters differ from closed feed-water heaters?
In open feed-water heaters, the two fluids actually mix, but in closed feed-water heaters
there is no mixing.
25. How do the following quantities change when the simple ideal Rankine cycle is modified with regeneration?
Assume the mass flow rate through the boiler is the same. a) turbine work
output, b) heat supplied, c) heat rejected and d) moisture content at turbine exit.
Moisture content remains the same, everything else decreases
26. How is the utilization factor for co-generation plants defined? Could the utilization factor be unity for a co-generation plant that does not produce any power?
The utilization factor of a co-generation plant is the ratio of the energy utilized for a useful
purpose to the total energy supplied. It could be unity for a plant that does not produce
any power.
27. What is the difference between the clearance volume and the displacement volume of reciprocating engines?
The clearance volume is the minimum volume formed in the cylinder whereas the
displacement volume is the volume displaced by the piston as the piston moves between
the top dead center and the bottom dead center.
28. An ideal Otto cycle with a specified compression ratio is executed using (a) air, (b) argon, and (c) ethane as the working fluid. For which case will the thermal efficiency be the highest? Why?
The thermal efficiency will be the highest for argon because it has the highest specific heat
ratio, k = 1.667.
29. For fixed maximum and minimum temperatures, what is the effect of the pressure ratio on (a) the thermal efficiency and (b) the net work output of a simple ideal Bray-ton cycle?
For fixed maximum and minimum temperatures, (a) the thermal efficiency increases with
pressure ratio, (b) the net work first increases with pressure ratio, reaches a maximum,
and then decreases.
30. How do the inefficiencies of the turbine and the compressor in a simple Bray-ton cycle affect (a) the back work ratio and (b) the thermal efficiency of a gas-turbine engine?
Back work ratio is the ratio of the compressor (or pump) work input to the turbine work
output. It is usually between 0.40 and 0.6 for gas turbine engines. As a result of turbine
and compressor inefficiencies, (a) the back work ratio increases, and (b) the thermal
efficiency decreases.
31. How does regeneration affect the efficiency of a Bray-ton cycle, and how does it accomplish it?
Regeneration increases the thermal efficiency of a Bray-ton cycle by capturing some of the
waste heat from the exhaust gases and preheating the air before it enters the combustion
chamber.
32. Somebody claims that at very high pressure ratios, the use of regeneration actually decreases the thermal efficiency of a gas-turbine engine. Is there any truth in this claim? Explain.
Yes. At very high compression ratios, the gas temperature at the turbine exit may be
lower than the temperature at the compressor exit. Therefore, if these two streams are
brought into thermal contact in a re-generator, heat will flow to the exhaust gases instead
of from the exhaust gases. As a result, the thermal efficiency will decrease.
33. In combined gas–steam cycles, what is the energy source for the steam?
The energy source of the steam is the waste energy of the exhausted combustion gases.
34. Why is the combined gas–steam cycle more efficient than either of the cycles operated alone?
Because the combined gas-steam cycle takes advantage of the desirable characteristics of
the gas cycle at high temperature, and those of steam cycle at low temperature, and
combines them. The result is a cycle that is more efficient than either cycle operated alone.
35. What is the difference between the binary vapour power cycle and the combined gas–steam power cycle?
In binary vapour power cycles, both cycles are vapour cycles. In the combined gas-steam
power cycle, one of the cycles is a gas cycle.
36. Why is mercury a suitable working fluid for the topping portion of a binary vapour cycle but not for the bottoming cycle?
Because mercury has a high critical temperature, relatively low critical pressure, but a
very low condenser pressure. On the downside, it is also toxic, expensive, and has a low
enthalpy of vaporization.
37.Write the general energy balance equation for simple compressible systems.
(𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡) − (𝑊𝑜𝑢𝑡 − 𝑊𝑖𝑛) + (𝐸𝑚𝑎𝑠𝑠,𝑖𝑛 − 𝐸𝑚𝑎𝑠𝑠,𝑜𝑢𝑡) = 𝛥𝐸𝑠𝑦𝑠𝑡𝑒𝑚 = 𝐸𝑖𝑛 − 𝐸𝑜𝑢t
38. What are the different mechanisms for transferring energy to or from a control volume? Consider
simple compressible systems only.
Heat transfer, work transfer and mass flow.
39. What are the two energy components accounted for by enthalpy?
Internal energy and flow energy i.e. ℎ = 𝑢 + 𝑃𝑣
40. What is an adiabatic process?
A process during which there is no heat transfer across the system boundaries.
41.On a P-V diagram, what does the area under the process curve represent?
The boundary work for a quasi-equilibrium process.
42. An ideal gas at a given state expands to a fixed final volume first at constant pressure and then at constant temperature. For which case is the work done greater?
The work done is greater for the constant pressure case. Compare how the two different cases
appear on the P-V process diagram and the difference in the size of the area (i.e. work) under the
curves.
43. Somebody claims that the mass and mole fractions for a mixture of CO2 and N2O are identical. Is this true? Why?
The mass of each component gas in a mixture is related to the number of moles of this gas via the molecular
weight. Generally then, for any mixture of gases, the mass and mole fractions of each constituent are not the
same as their molecular weights are likely to be different.
In this particular case however, the molecular weights are identical (i.e. 44 kg/kmol) and therefore the mass
of each component relative to the other is in the same ratio as the number of moles of each component. The
mass and mole fractions are therefore identical.
44. What is the apparent molar mass of a gas mixture?
The average molar mass (or molecular weight) for a mixture may be estimated from the mole fraction
weighted addition of the molar masses for each of the component gases. This may be thought of as the
summation of all the component gas masses divided by the total number of moles of gas present (i.e. the
summation of all the component gas moles). Since this average molar mass is simply the molar mass that the
mixture appears to have (i.e. the mixture mass divided by the mixture moles) it is referred to as the apparent
molar mass.
The apparent molar mass may be used to determine the apparent (or average) specific gas constant for the
mixture via the universal gas constant.
45. Is a mixture of ideal gases an ideal gas?
It is highly likely that if each component gas in a non-reacting mixture behaves with negligible
intermolecular interaction over the pressure and temperature range of interest (i.e. they act as ideal gases),
then the mixture will also. Separation of like molecules or a decrease in partial pressure is generally useful
in improving ideal gas behaviour by decreasing the likelihood of any interactions.
46. Under what circumstances can a mixture of ideal and real gases be treated as an ideal gas?
All gases are real. Over a range of conditions many gases may be treated as ideal. Some gases that have
small, symmetrical molecules (e.g. He gas) behave as an ideal gas over a very large range, while others with
large, non-symmetrical or polar molecules may only be modelled as an ideal gas in exceptional
circumstances. Generally, intermolecular interaction is reduced (or its impact becomes less significant) at
low pressures and high temperatures.
When a real gas becomes part of a mixture with ideal gases, the mixture can be treated as ideal if the partial
pressure of the real gas is sufficiently low enough for molecular interactions to become negligible. A classic
example of this, which we will re-visit in the psychrometry and air-conditioning topic next week, is the
treatment of water vapour (strongly non-ideal - as we know through the use of steam tables for solving
steam turbine problems) and dry air mixtures as an ideal gas mixture. This approach is only accurate when
the partial pressure of the water vapour in the mixture is less than about 10 kPa.
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