CPD Programme
CPD Programme
The CIBSE Journal CPD Programme programme is free and can be used by any
Members of the Chartered Institution of reader. It is organised jointly by CIBSE Journal
Building Services Engineers (CIBSE) and other and London South Bank University, and will
professional bodies are required to maintain help you to meet CIBSE’s requirement for
their professional competence throughout CPD. It will equally assist members of other
their careers. Continuing professional institutions, who should record CPD activities
development (CPD) means the systematic in accordance with their institution’s guidance.
maintenance, improvement and broadening Simply study the module and complete the
of your knowledge and skills, and is therefore questionnaire on the final page, following the
a long-term commitment to enhancing your instructions for its submission. Modules will
competence. CPD is a requirement of both remain available online at www.cibsejournal.
CIBSE and the Register of the Engineering com/cpd while the information they contain
Council (UK). remains current. You can also undertake the
CIBSE Journal is pleased to offer this questionnaire online, and receive your results
module in its CPD programme. The by return email.
The psychrometrics of
air conditioning systems
The CPD articles in the April, August, November and December
2009 editions of the CIBSE Journal have set out the principles of the
Psychrometric Chart and how to use the chart to plot individual
processes, determine the respective component loads and combine
these into commonly employed sub-systems
T
his article will further develop the previously described. It is the function of the Establishing the supply air mass
use of the chart to combine the individual processes that determine the plant flow rate
processes to develop a basic constant requirements although, of course, the final As a design decision (based on the strategy
volume, full fresh air system – this components are normally restricted to those being used to distribute air in the room) the
technique may then be used to determine the commercially available. To provide an example minimum supply air temperature will be
required components of an air conditioning a system will be developed to satisfy the room 16C (8K lower than the room temperature).
system and may also provide the basis for loads for a 8m wide by 5m deep by 3m high If, alternatively, a low level supply system
comparisons of the performance of different single room that have been determined as:- was to be used, the supply air temperature
systems or operating regimes. • summer (cooling) room load of 5kW may well have been limited to being no more
Symbols are defined in the box at the end (sensible gain) 1.2kW (latent gain) than 5K cooler than the room temperature.
of the article – you may also find it useful to • winter (heating) load of 1.9kW (sensible (See CIBSE Guide Section B2.4.2 [1] for more
be able to refer to those earlier articles (they loss) and 0.8kW (latent gain). details.) It is normally the requirements of
are all available online at www.cibsejournal. The design conditions for the room are room cooling as opposed to heating (in
com). To help understand the example in 20C to 24C operative temperature and 35 per air conditioning applications) that will
this article you are recommended to build cent to 50 per cent saturation (ie a range of determine the air supply rate in a constant
up the process on a psychrometric chart. acceptable conditions that can vary season volume system. Hence the mass flow rate, ,
to season). The room is of medium thermal of the supply air can be calculated from the
Combining the processes weight (medium to fast response) and has room sensible cooling load, Φ
SC
= C
p
(θ
RS
–
The psychrometry for the complete air a moderate amount of glazing, and so at θ
SC
) where C
p
is the air specific heat capacity
conditioning system can be developed this stage it would be reasonable to assume of air that can be taken as 1.012 kJ⋅kg
-1
⋅K
-1
. So
from the combination of the individual that values of air temperatures equivalent to = Φ
SC
/ [C
p
(θ
RS
– θ
SC
)] = 5.0 / [1.012 x (24 –
components and sub-systems that have been values of operative temperatures be used. 16)] = 0.62 kg⋅s
-1
. >
www.cibsejournal.com March 2010 CIBSE Journal 59
CIBSEmar10 pp59-62 cpd.indd 59 2/25/10 4:25:21 PM
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