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SOFTWARE FOR DESIGN OF COLD STORAGE FOR FRUITS AND VEGETABLES Paper No. CSBE 08-114

RAJARAMMANNA NITHYA Dr. ALAGUSUNDARAM

July14, 2008

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Outline…• Introduction

• Objectives

• Background

• Methodology

• Result & Conclusion

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Introduction• India is an “Agricultural Country”

• Second largest producer of Fruits and Vegetables

• Only 2%is processed commercially

• 20-50% Post harvest losses

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IntroductionWhat should be done? Proper preservation methods (a) Reducing the rate of respiration (b) Preservatives (c) Control/Modified atmosphere storage (d) Cold Storage

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Cold storage in India

• 3443 cold storages in India

• 10.35 MT capacity • India produces 32 MT fruits & 71 MT

vegetables

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Objectives

• To develop a software for the technical design of a cold storage for fruits and vegetables

• To calculate total cooling load and cost economics of cold storage

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Background• Working of Cold Storage

• Water requirement

• Electricity

• Labor

• Transportation

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Source: http://www.google.ca/search?hl=en&q=refrigeration+effect&meta

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Apple

Apple

Apple

Apple

Apple

Working of Cold Storage

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Methodology

• Development of Software Visual basic 6.0• User friendly

• No chance of mystification

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Cooling Load

• Striking a balance between heat gain and heat remove

• Four major source of heat gain (a) Building (b) Field load (c) Respiration (d) Service load

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(a) Heat transfer through wall, ceilings and floor

kJ/s

)( 0 iTTAUQ

Where:A = Area of the wall, ceiling, and floor (m2)

U = Over all heat transfer coefficient kJ /s m2 K

Ti = Internal temperature (C)

To = External temperature (C)

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kJ /s

Where:Q = Rate of heat transfer (kJ /d)

Cp = Specific heat of the product (J /kg K) t = Chilling time (d)m = Mass of the product (kg)

Tz = Desired final product temperature (oC)

Tj = Entering product temperature (oC)

(b) Heat gain from product

t

TzTjmCQ p )(

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(c) Heat given out by bulb

Q = Number of bulbs * W * (UF) * (AF) kJ/s

Where:UF = Use factor is 0.5 for industries

AF = Allowance factor is 1.25 for fluorescent tubes

W = Wattage of bulb used, i.e., 65 W

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(d) Heat generated by labour working

kJ/snSHGQ Where:

n = Number of people working in cold storage

SHG = Sensible heat gain per person, 1500 kJ

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(e) Heat given out by the power equipments

141.3746 skJefficiencyMotor

HpinmotorofrangePowerQ

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Total refrigeration required

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5.3

kJs

kJsinremovedheatTotalrequiredionrefrigeratTotal

1 Ton of refrigeration = 3.5 kJ /s

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Working of Software

Working of Software

Working of Software

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Constants used

• 1 ton of refrigeration removes 3.5 kJ of heat per second

• Ammonia (NH3) is the only refrigerant taken

• Ammonia R40 is selected

• 1 ton of refrigeration = 0.97 Hp (R40 ammonia)

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Result• Total cooling load

• Heat emitted by the products

• Heat transfer through walls, roofs and floors

• Heat emitted by light, labour, and lift

• Total electric charge per day

• Number of labours needed

• Amount of ammonia utilized

• Final cost of civil and electrical works

• Total income and expense per day

• Pay back period in years

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Conclusion• Can get idea about design, construction & cost

economics

• Accurate and reliable cooling load and cost calculation

• Need not entirely depend on construction and design engineers

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Limitations

• Can hold only 50 t

• Fruits (apple, orange, banana, pineapple, and grape)

• Vegetables (tomato, potato, cabbage, carrot, and onion)

• Ammonia R40 is the only coolant used in this design.

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INDIAN COUNCIL OF AGRICULTURAL RESEARCH

SHANMUGANATHAN COLD STORAGE, TRICHY, TAMIL NADU, INDIA

DR. DIGVIR S JAYAS, UNIVERSITY OF MANITOBA

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Acknowledgements

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THANK YOU…

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