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MTX311 2019 re-Exam
Thermodynamics,MTX311 2019 re-Exam worked out solutions. The memorandum has solutions are 100% correct including alternative working methods. The solutions are done in a step by step matter
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On a hot and sunny day in a building close to the sea (36°C, 1 bar)200 g/s of air, with a temperature of 36°C and a relative humidity of 62% is cooled and de-
humidified in an air conditioning system. The target temperature is 22°C, 40% relative humidity. An
image of a simple air conditioner system (from the website explainthatstuff) is given in fig. 1; the
text that goes with it is presented in the text box. As you can read in the text, the people who wrote
explainthatstuff didn’t know about reheaters in air conditioning systems. Note that you do not HAVE
to read the text box; I just want to give you all the information that they give the readers)
Q1A (4) how much heat has to be extracted by the cooling and condensation process? (step 2)
Q1B (4) how much condensate is formed?
Q1C (4) In this design, the re-heating element is electric. How much electricity is used in the re-heat
process? (step 3)
1. Warm air from the room is sucked in through
a grille at the base of the machine
2. The air flows over some chiller pipes through
which a coolant fluid is circulating. This part of the
machine works just like the chiller cabinet in a
refrigerator. It cools down the incoming air and a
dehumidifier removes any excess moisture.
3. The air then flows over a heating element
(similar to the one in a fan heater). On a cold day,
this part of the unit may be turned right up so the
HVAC works as a heater.
4. A fan at the top blasts the air back through
another grille into the room. If the heating element is turned down, the air re-entering the
room is much cooler, so the room gradually cools down.
5. Meanwhile, coolant (a volatile liquid that evaporates easily) flows through the chiller pipes.
As it does so, it picks up heat from the air blowing past the pipes and evaporates, turning
from a cool liquid into a hotter gas. It carries this heat from inside the room to the outside of
the building, where it gives up its heat to the outside air. How? Just like in a refrigerator, the
coolant flows through a compressor unit and some condensing pipes, which turn it back into
a cool liquid ready to cycle round the loop again.
6. What happens to the heat? In the unit outside the building, there are lots of metal plates
that dissipate the heat to the atmosphere. An electric fan blows air past them to accelerate
the process.
, If (and ONLY IF) you got stuck in Q1, assume that the cooler has to extract 20kW in heat, the
reheater has add 5kW (electricity) and the process generates about 1kg/min in condensate.
Q2A (6) make an estimate of the average temperature of the gas during cooling. How much available
work is transferred from the chiller pipes to the air in the cooling process? (note that the OUTSIDE
temperature is 36°C)
Q2B (6) Small refrigerators have typical second law efficiencies of 50%. How much energy is
consumed by the whole air conditioning process? (you can ignore fan work).
It would be much smarter to use an air-conditioner system with a heat exchanger as a reheater. For
your convenience, I have taken a picture from the lecture slides. As a result of this, the re-heater
does not need any electricity anymore and we also need less cooling power in the cooling coils (the
chiller pipes in explainthatstuff).
Q2C (6) calculate the temperature at the inlet and outlet of the cooling cools and the amount of heat
that is removed by the cooling coils.
Q2D (8) is there condensation in the reheater? If so, how much? Why is this relevant?
Q2E (6) Assume again that the refrigerator (which removes heat from the cooling coils) has a second
law efficiency of 50%. How much electricity is needed in this process?
Q2F (6) The change in specific exergy of the air in this refrigeration process is about 0.5kW. What is
the second law efficiency of the “explainthatstuff” process? (comment)
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