Design and Education

1. a. All Outside Air supplied
b. Outside air supplied with External Bypass system

2. c. All Outside air supplied with Re – Heater
d. Outside and Recirculated air supplied

3. e. Outside air and Re-circulated air supplied with Bypass (Re-circulated air)
f. Outside and Re-circulated air supplied with Re-heater
By: Engr. Yuri G. Melliza

4. Example: Outside and Re-circulated air supplied with Re-heater
A commercial building to be air conditioned has a sensible heat load of 36 KW and a
latent heat load of 10.2 KW. The building is to be maintained at 26C and 50% RH.
Outside air is at 32C DB and 24C WB. Forty five percent of the supply air is fresh air and
the rest is re-circulated air. Conditioned air leaves the AC unit at 14C and 100% RH then
it is reheated to 19C and is supplied to building. Determine
a. The fan capacity in m 3/sec to the building
b. The tonnage capacity of the AC unit (Assume tw = 14C )
c. The heat required by the re-heater
Figure
Qs = 36 KW
Q L = 10.2 KW
t 4 = 26C ; RH4 = 50%
t 0 = 32C DB ; tw0 = 24C
t 2 = 14C ; RH2 = 100%
0.45m = m 0
0.55m = m R
t 2 = 14C ; t3 = 19C
Other Data
hw = 58.679 KJ/kg
h 0 = 71.944 KJ/kgda ; W0 = 0.015538 kg m/kgda
h 4 = 52.943 KJ/kgda ; W 4 = 0.010518 kgm /kgda
h 2 =39.293 KJ/kgda ; W2 = W 3 = 0.009981 kg m/kgda
h f g at 19C=2456.49 KJ/kg
Processes:
014 – Adiabatic Mixing
1 to 2 – Cooling & De-humidifying
2 to 3 – Sensible Heating
3 to 4 – Heating &Humidifying

5. Qs  m(1.02)(t
4
36  m(1.02)(26
 t3)
- 19)
m  5.042 kg/sec
Q L  m(W
4
- W 3 )hfg
where : h fg  2500 KJ/kg (For air conditioni ng applicatio n)
10.2  5.042(0.01
W 3  0.0097
0518 - W 3 )2456.49
kgm
kgda
36  5.042(52.9 43  h 3 )
KJ
h 3  45.803
kgda
From Psychromet ric chart and formulas
at t 3  19  C and h 3  45.803 KJ/kgda
m
υ 3  0.8421
3
kgda
A) Fan Capacity  5.042(0.84 21)  4.245
mo
m
 0.45 
h1  h4
ho  h 4

h 1  52.943
0.45 
71.944  52.943
KJ
h 1  61.49
kgda
W 1  0.0128
m
3
sec
W1  W4
Wo  W 4

W 1  0.010518
0.015538
 0.010518
kgm
kgda
m 0  0.45(5.042
)
m o  2.27 kg/sec
m R  m  m 0  5.042  2.27
m R  3.15 kg/sec
Q  m(h 1 - h 2 ) - m(W 1 - W 2 )h w
Q  5.042(61.4 9  39.293)  5.042(0.01
28  0.009981)5
Q  111.92  0.834
Q  111.086
Tonnage
211
KJ
min
KW  6665.16
KJ
min
Capacity
 1 Ton of Refrigerat ion
Q  31.58 TONS
Heat required by the re - heater
Q R  m(h
3
- h 2 )  5.042(45.8 03 - 39.293)
Q R  32.82 KW
8.679

6. Example: Dehumidifier – Heater
Moist air at 32C and 60% RH enters the refrigeration coil of a De-humidifier with a flow
rate of 1.5 kg/sec. The air leaves saturated at 10C and is re-heated to 16C before entering
to the conditioned space. Condensate leaves the de-humidifier at 10C . Determine
a. The condensate removed in kg/sec
b. The Capacity of the de-humidifier unit (AC – unit) in Tons of refrigeration
c. The heat required by the Re-heater
Figure:
Psychrometric
Chart
h1
60%
1
W1
To conditioned
space
h3
h2
2
10°C
S
Processes:
1 to 2 – Cooling & Dehumifying
2 to 3 – Sensible Heating
Using Fundamental Formulas or Psychrometric chart
kgm
W 1  0.01807
kgda
KJ
h 1  78 . 429
kgda
W 2  0 . 007632
h 2  29 . 278
kgm
kgda
KJ
kgda
W 3  W 2  0 . 007632
kgda
KJ
h 3  35 . 393
h w  41 . 92
kgm
kgda
KJ
kg
a. Condensate removed
m w  m(W 1  W 2 )
m w  1 . 5( 0 . 01807  0 . 007632 )
m w  0 . 016
kg
sec
b. Capacity of AC-unit in TR
Q  m ( h 1  h 2 )  mwhw
Q  m ( h 1  h 2 )  m ( W 1  W 2 )h w
Q  1 . 5( 78 . 429  29 . 278 )  0 . 016 ( 41 . 92 )
Q  73 . 15 KW  4389
KJ
min


KJ  1 TR 

  20 . 8 Tons of Refrigerat ion
Q  4389
min  211 KJ 


min 

n
atio
atur
3
16°C
W2 = W3
ve
Cur
32°C

7. Example: Outside air and re-circulated air supplied
An auditorium is to be maintained at 27C and 50% RH. The calculated sensible heat load in the space
Qs = 145 KW and latent heat load QL = 95 KW (hfg = 2437.4 KJ/kg). The air mixture at 29C DB and
22C WB is cooled to 17C DB and 15C WBin a Chilled-water AC unit and delivered as supply-air to the
space, calculate
a. The kg/sec of supply air
b. The kg/sec of re-circulated air
c. The kg/sec of outside air if t0 = 34C and 50% RH
d. The kg/sec of condensate from AC-unit (assume hw = 121.63 KJ/kg)
e. The refrigeration capacity of the AC-unit in TR
Re-circulated Air
Exhaust air
3
mR
h0
50% RH
Exhaust Fan
3
h1
Q
m
m0
m
Fan
2
1
0
1
22°C
Qs = 145 KW
QL = 95 KW
Supply
air
W0
h3
Auditorium
m
0
Psychrometric Chart
3
4
W3
3
h2
W1
AC Unit
15°C
Processes:
013 – Adiabatic mixing
1 to 2 – Cooling & dehumidifying
2 to 3 – Heating & humidifying
From Chart or Fundamenta
kgm
W 0  0.016813
h 0  77.281
kgda
KJ
m0
m
m0
14.2

kgm
KJ
m  14 . 2
kgda
From
mR
KJ
m
kgda
mR
kgm
14.2
kgda
KJ

sec
h0  h1
h0  h3

77 . 281  64 . 298
77 . 281  55 . 626
kg
sec
64 . 298  55 . 626
77 . 281  55 . 626
kg
sec
 mass flow rate of outside
air
m w  m ( W 1  W 2 )  14 . 2 ( 0 . 013759  0 . 009831 )
m w  0 . 056
kg
sec
 mass flow rate of condensate
leaving AC - unit
Q  m(h 1 - h 2 ) - m w h w  14 . 2 ( 64 . 298  41 . 984 )  0 . 056 ( 121 . 63 )
Q  316.8588  0.38143168
Q 
317 . 24 ( 60 )
211
 317 . 24 KW
 90 . 21 TR  Capacity of AC - unit
 mass flow rate of supply air
theory of mixing
m R  8 . 51
kgda
kg
h0  h3
m 0  5 .7
34°C
145  m ( 1 . 02 )( 27  17 )
h1  h3

22°C 27°C 29°C
Qs  m ( 1 . 02 )( t 3  t 2 )
kgda
W 3  0.01117
kgda
W 2  0 . 009831
h 2  41.984
15°C 17°C
kgm
W 1  0.013759
h 1  64.298
To conditioned space
l Formulas
h 3  55.626
kgda
W2
2
 mass flow rate of re - circulated air