REGENERATIVE AIR COOLER(UNDER THE GUIDENCE OF-PROFF. H.P

GUPTA)

SUBMITTED BY:YAMIKA PATELSUMIT SINGHPREM KUMARPARUL DIXIT

MONTH JUNE-AUG.2013

SEPT-NOV 2013

JANAURY 2014

FEB 2014

MARCH 2014

APRIL2014

ACTIVITY

1st

Data search and study

ACTIVITY2nd

Selection of materials & design layout

ACTIVITY3rd

Design & Assembling of parts

ACTIVITY4th

Assembling continued…

ACTIVITY5th

Testing of system

ACTIVITY6th

Preparation of report

PROJECT SCHEDULE FORMAT

REGENERATIVE AIR COOLER

Main Components Of The Regenerative Air Cooler Are:1. Blower2. Blower3. Motor4. Duct5. Water Pad 6. Heat Exchanger7. Water Pad 8. Water Tank9. Water Pump

MATERIAL SELECTION:

HEAT EXCHANGER

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

TOP VIEW OF REGENERATIVE AIR COOLER

RIGHT HAND SIDE OF REGENERATIVE AIR COOLER

LEFT HAND SIDE VIEW OF REGENERATIVE AIR COOLER

FRONT VIEW OF REGENERATIVE AIR COOLER

BACK SIDE VIEW OF REGENERATIVE AIR COOLER

MOTOR PERFORMANCE CHARECTORISTICS:POWER=95 WBLADE VELOCITY=

ПDN/60=П×.18×1300/60=12.25 m/secAC SUPPLY=230 VCURRENT=.4 ampFREQUENCY=50 Hz

  

 

 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

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

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%

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 

 

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

U=

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

ε=.92 TH1=38°C TH2=? TC1=36.5°C TC2=?

ε =

ε =

(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

1.Refrigeration and air conditioning………………..Khurmi and Gupta2. Refrigeration and air conditioning………………..Arora and Domkundwar 3. Refrigeration and air conditioning………………..C.P. Arora 4. Refrigeration and air conditioning………………..P.L. Balany