regenerative air cooler

1. REGENERATIVE AIR COOLER
(UNDER THE GUIDENCE OF-PROFF. H.P
GUPTA)
SUBMITTED BY:
YAMIKA PATEL
SUMIT SINGH
PREM KUMAR
PARUL DIXIT

2. MONTH JUNE-
AUG.
2013
SEPT-NOV
2013
JANAURY
2014
FEB
2014
MARCH
2014
APRIL
2014
ACTIVITY
1st
Data search
and study
ACTIVITY
2nd
Selection of
materials &
design layout
ACTIVITY
3rd
Design &
Assembling
of parts
ACTIVITY
4th
Assembling
continued
…
ACTIVITY
5th
Testing of
system
ACTIVITY
6th
Preparation
of report
PROJECT SCHEDULE FORMAT

3. REGENERATIVE AIR COOLER

4. Main Components Of The
Regenerative Air Cooler Are:
1. Blower
2. Blower
3. Motor
4. Duct
5. Water Pad
6. Heat Exchanger
7. Water Pad
8. Water Tank
9. Water Pump

5. MATERIAL SELECTION:

6. HEAT EXCHANGER

7. FABRICATION OF THE HEAT
EXCHANGER
Obtain MS rods
for studs
Order GI sheets
for heat
exchanger
Obtain
insulation for
HE
Purchase
adhesive
Manufacture 5 studs
Cut the sheets
and drill holes
Cut the insulation
into proper size
Assemble the heat exchanger

8. TOP VIEW OF REGENERATIVE AIR
COOLER

9. RIGHT HAND SIDE OF
REGENERATIVE AIR COOLER

10. LEFT HAND SIDE VIEW OF
REGENERATIVE AIR COOLER

11. FRONT VIEW OF REGENERATIVE
AIR COOLER

12. BACK SIDE VIEW OF
REGENERATIVE AIR COOLER

13. MOTOR PERFORMANCE
CHARECTORISTICS:
POWER=95 W
BLADE VELOCITY= ПDN/60=П×.18×1300/60=12.25
m/sec
AC SUPPLY=230 V
CURRENT=.4 amp
FREQUENCY=50 Hz

14.
CALCULATION FOR BLOWER 2
Entering Air velocity, Vin= 6.62 m/sec
Inlet area, Ain=П/4(.16^2+.12^2)
Ain=.0314 m2
Q = Ain×Vin
By applying Bernoulli’s equation
Q=const
Q= .208 m3
/sec=12.4 m3
/min
Outlet area, out=.13×.13= .0169 m2
Exit Air velocity, Vout=Q/A=.208/.0169=12.3 m/sec

15. CALCULATION FOR DUCT
Inlet area, Ain=.13×.13= .0169 m2
Entering Air velocity, Vin=Q/A=.208/.0169=12.3 m/sec
By applying Bernoulli’s equation
Q=constant
Q= .208 m3
/sec=12.4 m3
/min
Outlet area, Aout=.205×.13=.02665 m2
Exit Air velocity, Vout =Q/A=.208/.02665 = 7.8 m/sec

16. CALCULATION FOR WATER
PAD
DBT1=38°C
WBT1=21°C
RH1=20%
H1=60.25 KJ/Kg of air
Spe. Vol, Vs1=.89 m3
/Kg
DBT2=36.5°C
WBT2=27°C
RH2=50%

17. H2=85 KJ/Kg of air
SPE. VOL, Vs2=.91 m3
/Kg
Mass of Air Supplied
ma=Q/Vs=12.8/.91=14.06Kg/min
Ma=.234 Kg/sec

18.
CALCULATION FOR HEAT EXCHANGER
TH1=38°C
TH2=?
TC1=36.5°C
TC2=?
Capacity rate of air, (C) min=ma
U=

19. H=40, l=.0005, k=52
U= =19.996
A=16(l × b) =16(.23×.23) =.8464
UA=19.996×.8464=16.92
N=70.58

20. ε=.92 TH1=38°C TH2=?
TC1=36.5°C TC2=?
ε =
ε =

21. (38-36.5) ε =38- TH2
(38-36.5) ε =38- TH2
1.5× .92= 38- TH2
1.38 = 38- TH2
TH2 =36.62 °C
1.38 = TC2-36.5
TC2=37.88°C
Temp. Of atm air, TH1=38°C
Temp. Of treated atm air, TH2=36.62°C
Temp. Of cooled air from water pad, TC1=36.5°C
Temp. Of exit air, TC2=37.88°C

22. 1.
Refrigeration and air conditioning………………..Khurmi and Gupta
2. Refrigeration and air conditioning………………..Arora and Domkundwar
3. Refrigeration and air conditioning………………..C.P. Arora
4. Refrigeration and air conditioning………………..P.L. Balany