Example 1
Example 1
Example 1
SUMMER CYCLE
A room is to be maintained at 22oC dry-bulb temperature, 50% saturation, when
the sensible heat gain is 10.8 kW in summer. The latent heat gain is 7.2 kW.
Determine the cooling coil and reheater outputs required by using a psychometric
chart if the plant schematic is as shown below.
DATA: Outdoor condition is 28oC, 80% saturation. The outdoor air and
recirculated air ratio is 20%/80%. The Apparatus Dew Point ADP is 8oC Neglect
the cooling coil contact factor.
solution
1
5. finding point m :
VoTo+VrTr
Tm ¿ Vs
Tm=0.2*28+0.8*22
Tm=23.2 C
6. find RRL
2
10.
find Vs
Qs
Vs= 1.22 ( Te−Ts )
=10.8/(1.22*(22-15)=1.264 m3/s
Qcc=1.2 Vs(hm-hADP)
=1.2*1.64*(50-25)
= 37.93 KW
Qrh=1.2 Vs(hs-hADP)
=1.2*1.64*(31.5-25)
= 9.932 KW
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Example 2. SUMMER CYCLE (Cooling Coil contact factor)
An office is to be maintained at 22oC dry-bulb temperature, 50% saturation in
summer.
DATA:
Outdoor condition is 28oC, 80% saturation.
The outdoor air and recirculated air ratio is; 20% / 80%.
The Apparatus Dew Point ADP is 8oC
4
solution
5. find point m :
VoT 0+VrTr
Tm ¿ Vs
Tm=0.2*28+0.8*22
Tm=23.2 C
6. find RRL
0.8 =(hm-hw)/(50.5-25)
5
hw=30.1 kJ/kg
11. find Vs
Qs
Vs= 1.22 ( Te−Ts )
=10.8/(1.22*(22-17)=1.311 m3/s
Qcc=1.2 Vs(hm-hadp)
=1.2*1.311*(50-25)
= 39.34 KW
Qrh=1.2 Vs(hs-hw)
=1.2*11.311*(36.5-30.1)
= 10 KW
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Example 3 SUMMER CYCLE (air flows to be
calculated)
A Lecture Theatre measures 15 m x 10 m x 6 m high.
DATA:
Outdoor condition is 28oC, 80% saturation.
The Apparatus Dew Point ADP is 7.5oC.
The latent heat gain is 10.0 kW.
The sensible heat gain is 12.0 kW.
Maximum occupancy is 200 people.
The recommended total air supply rate is 6 – 10 air changes per hour
The cooling coil contact factor is unknown at present and should be
calculated.
Determine air flow rates and calculate the mass flow rate of fresh air and
supply air to the room.
The room is a non-smoking area.
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Solution
Fresh air rate = 8 l/s/p x 200 people = 1600 l/s = 1.6 m3/s
The specific volume at the outside condition may be determined from a
psychrometric chart. It is 0.88 m3/kg.
Mass flow rate = Volume flow rate / specific volume
Mass flow rate (Fresh Air) = 1.6 / 0.88 = 1.82 kg/s .
If the maximum ventilation supply air rate is taken from table B2.3 to be
10.0 air changes per hour, then the mass flow rate can be calculated.
Volume flow rate (m3/h) = Volume of room (m3) x air change rate
(ac/h)
Volume of room (m3) = 15 x 10 x 6 = 900 m3
Volume flow rate (m3/h) = 900 (m3) x 10 (ac/h)
Volume flow rate (m3/h) = 9000 m3/h
Volume flow rate (m3/s) = 9000 / 3600 = 2.5 m3/s.
Mass flow rate = Volume flow rate / specific volume
The specific volume at the supply condition may be approximated at this
stage from a psychrometric chart. It is 0.834 m 3/kg.
Mass flow rate (Supply Air) = 2.5 / 0.834 = 3.0 kg/s.
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The ratio by mass is therefore;
1.82 / 3.00 = 0.6, i.e. 60% fresh air and therefore 40% recirculated air.
Qs=1.22Vs(Tr-Ts)
Ts=22-(12/1.22)
=18 C
2. Point ADP can be indicated and lines drawn between these points as
shown.
4. Point S is then shown on the chart, on the room ratio line at 18 oC.
5. A horizontal line is then drawn from point S towards the line O – ADP
6. Point W can then be found where the horizontal line W - S intersects the
line O - ADP.
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The heat ratio is 12 kW sensible / 22 kW total = 0.545.
7.
Qcc=1.2 Vs(hm-hadp)
=1.2*2.5*(64-24)
= 120 KW
Qrh=1.2 Vs(hs-hw)
=1.2*2.5*(36-26.5)
= 28.5 KW
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