Example
2. Winter Cycle
A room has a 18.0 kW
sensible heat loss in winter and a 4.5 kW latent heat gain from the occupants.
Determine the supply air
temperature and heater battery load using the following information.
DATA:
Indoor condition: 21oC dry-bulb temperature, 50% saturation.
Outdoor condition: -2oC d.b., 80% saturation.
The outdoor air and recirculated air ratio is 20%/80%.
No preheating or humidification takes place in this simplified example.
Procedure
(Winter Cycle)
1. Draw schematic diagram
of air-conditioning plant (see above)
2. Plot room condition R on psychrometric chart.
3. Plot outside condition O on psychrometric chart.
4. No Preheater condition P
5. Join points O and R
6. Find the mix point M by
measuring the length of the line O-R
and multiply this by the mixing ratio.
On a full size CIBSE psychrometric
chart this measures 110mm.
The ratio of recirculated air is 0. therefore; 110mm x 0.8 = 88mm
Measure up the O-R line from point O by 88mm.
This determines point M .
If there is more recirculated air than outside air at the
mix point, then point M will be closer to point R than point O.
7. Find
the room ratio.
This is the sensible to total heat ratio.
Neglect signs ie. the total heat for the room will be
Sensible loss plus Latent gain.
Total heat = 18 kW sensible + 4.5 kW
latent = 22.5 kW total.
Heat ratio = 18 / 22.5 = 0.8
Plot this ratio on the
protractor, top segment, on the psychrometric chart and transfer this line onto
the chart so that it passes through point R.
8. Find the supply air dry bulb
temperature by calculation.
9. Plot the supply air condition S
on the room ratio line.
This is on a
horizontal line from point M to the right hand side of the chart, and
intersects with the RRL.
The supply air Temperature is found to
be 32.5oC.
Supply
Air Flow Rate
When the sensible heat loss
and supply air temperature in winter are known then the mass flow rate of air
is calculated from the following formula:
Hs = ma x Cp ( ts - tr )
where:
Hs = Sensible heat loss (kW)
ma = mass
flow rate of air (kg/s)
Cp = Specific
heat capacity of humid air (approx.1.01 kJ/kg degC)
tr = room temperature (oC)
ts = supply air temperature (oC)
..............therefore:
ma = Hs / Cp ( ts - tr )
ma = 18 / 1.01 (32.5 - 21)
ma = 18 / 11.615
ma = 1.55 kg/s
Heater
The heater battery output
is as follows:
H reheater battery = ma ( hS
- hM)
where:
H reheater battery = Reheater battery output (kW)
ma = mass flow rate of air (kg/s)
hS = specific enthalpy at condition S (kJ/kg)
hM = specific enthalpy at condition M (kJ/kg)
The specific enthalpies at points S and M are shown on the psychrometric chart below.
Hheater battery = ma ( hS - hM)
Hheater battery = 1.55 ( 50 - 34)
Hheater battery = 24.8 kW
Therefore the heater
battery load is 24.8
kW.