-

Southern Luzon State University

College of Engineering

Department of Mechanical Engineering

PROPOSED DESIGN OF A COLD STORAGE

STRUCTURE FOR THE TOTAL PRODUCT LOAD OF

MUSHROOM

AGPI, AARON DALE T.

BSME IV- GO

Engr. Ronnel S. Nombrefia

May 19, 2017

Page 1 of 27
PROJECT DESCRIPTION:

Cold and cool storage room is a walk-in storage facility. Safe operation of cold and cool

rooms in restaurants, hotel, food shops and supermarkets is crucial for preserving the quality of

the stored foods. The application of cold rooms are numerous, including packaging of foods and

beverages, short and long term food storage as well as freezing. With the swelling demand for food

due to continuous population increase, supermarkets emerged as a primary source of almost any

commodities available where purchase of goods is a daily activity.

Vegetables are known as “healthy foods” which contain several of nutrients necessary for

our bodies to function actively. But as time passes by, the quality of the vegetables lessen and this

becomes a major problem for many producers that wants the vegetables quality to last much longer.

In order to solve this problem the designer found a way to make the vegetables life much longer

by designing a cold storage for the purpose of maintaining its quality and life. The designer

selected a proper materials for a cold storage in walls, ceiling, doors and floor.

SITE PLAN

Page 2 of 27
 The dimensions of the storage are determined by estimating 60 crates to be used. The

dimension of the plastic crate is 0.500m x 0.300m x 0.145m. The crate rack on the right has a

dimension of .508m x 0.3048m x 0. 8128m that has a capacity of 4 crates per rack. The pathways

are included in the space considerations.

The figure below is a top view which shows the number of racks to be used in the cold

storage. There are a total of 20 racks consisting of 5 crates per 1 rack which result to a total of

100 crates.

1 Mushroom = .39kg

Maximum no. of crates in cold storage with no space consideration between crates

= = (8m) (4m)/(.500m)(.300m) = 233.33≈234 crates

For space considerations between crates, use: 100 no. of crates

Page 3 of 27
Volume of crate = (0.500m) (0.300m) (0.145m) = 0.02175 �

Volume of Mushroom = 1398.2 cm3

Maximum mass of the product per crate =

. ℎ .
� � � � �
. ℎ

=6≈

Maximum mass of the product per crate =

�

Total mass of the product = 100 crates x = 600 kg

I. GIVEN

Ambient Temperature 35℃

Room Temperature -10℃

Refrigerant R-134a

Room Size 8m x 4m x 4m

(length x width x height)

II. Product: Mushroom

III. REFRIGERATION LOAD COMPUTATION

Heat Load in Cold Storage

1. Transmission Load

a. Heat load in wall

b. Heat load in ceiling

c. Heat load in floor

d. Heat load in door

2. Miscellaneous Load

a. Heat load in appliances

Page 4 of 27
 b. Heat load in occupants

3. Infiltration Load

4. Product Load

a. Heat load in product

b. Heat load in respiration

1.) TRANSMISSION HEAT LOAD

Heat Load by means of Conduction

For conduction, we have the conduction through plane wall, and composite wall given by

the equations:

∆� ∆�
Q= and Q=
�

Where: Q = heat transmitted, W

A = heat transfer area, �

∆� = change in temperature, ℃ �� °�

L = thickness of the wall, m

� �
k = thermal conductivity, or
−℃ −°

�� = overall resistance, � -
�

Characteristics of the Material

Materials in the
Type of � �
Wall ,� ∙ �, � ∙ Thickness ,t, m
Material � � �

Concrete Block,

Lightweight Exterior 0.29 0.15

Aggregate

Page 5 of 27
 Asbestos –
Sheathing 1.73 0.03
cement

Expanded

Polystyrene, Insulation 27.7 0.07

Extruded

Stainless Steel Interior 0.0625 0.03

A.) HEAT LOAD COMPUTATION OF THE WALL

a.) Total Resistance of the Wall

Outside Air Film (moving air, cooling season)

= 0.044 −
�

Concrete Block, Lightweight Aggregate

R2 = 0.29 ∙
�

Asbestos – cement

R3 = (t) (

R3 = (0.03m) ( 1.73 � ∙ )
�

Page 6 of 27
 R3 = 0.0519 ∙�

Expanded Polystyrene, Extruded

R4 = (t) (

R4 = (0.07m) ( 27.7 � ∙ )
�

R4 = 1.939 ∙
�

Stainless Steel

R5 = (t) (

R5 = (0.0015m) (0.0625 � ∙ )
�

R5 = 0.001875 ∙
�

Inside air film (vertical, heat flow horizontal)

R6= 0.12 −
�

TOTAL RESISTANCE: RT= R1+R2+R3+R4+R5+R6

RT = 0.044 � − + 0.29 � ∙ + 0.0519 � ∙ + = 1.939 � ∙ + 0.001875 � ∙ + 0.12

� � � � �

� −
�

RT = 2.45 −
�

b.) Conduction through Walls

North Wall

Heat load of the wall by means of conduction

Page 7 of 27
 ∆�
� =
�

A = Area of the wall

= (8m)(4m)

A = 32 �

( )[ ℃− − ℃ ]
� =
. −
�

= 597.76 W

West Wall

Heat load of the wall by means of conduction

∆�
� =
�

A = Area of the wall – Area of the door

= (4m)(4m)

A = 16 �

( )[ ℃− − ℃ ]
� =
. −
�

=293.88 W

South Wall

Heat load of the wall by means of conduction

∆�
� =
�

A = Area of the wall

= (8m)(4m)-[2(2mx1m)]

A = 28 �

Page 8 of 27
 ( )[ ℃− − ℃ ]
� =
. −
�

= 514.29W

East Wall

Heat load of the wall by means of conduction

∆�
� =
�

A = Area of the wall

= (4m)(4m)

A = 16 �

( )[ ℃− − ℃ ]
� =
. −
�

=293.88 W

Total heat load in Walls

�� =� + � + � +�

= 597.76 W + 293.88 W+ 514.29 W+ 293.88 W

�� = 1699.81W

Composition of the Floor

Page 9 of 27
 Materials in the Characteristics of the Material

Floor Type of � R, � - Thickness, t, m

�, �
�−�
Material

Concrete,
Exterior 0.1 0.2
Lightweight

Polyethylene

sheet Vapor Retarder 0.33 0.02

(low-density)

Expanded

Polystyrene Insulation 0.03 0.05

Stainless Steel Interior 16 0.03

a.) Total Resistance of the Floor

Concrete, lightweight

� = (t) (

= (0.2m) ( �−
. �

=2 −
�

Polyethylene sheet (low-density)

� = (t) (

= (0.02m) (
.
�−
�

= 0.0606 −
�

Expanded Polystyrene

� = (t) (

Page 10 of 27
 = (0.05m) ( .
�−�

= 1.67 −
�

Stainless Steel

� = (t) (

= (0.03m) ( �−
�

= 0.001875 −
�

Inside air film (vertical, heat flow horizontal)

= 0.12 −
�

Total Resistance, �� = � + � + � + � + �

=2 − + 0.0606 − + 1.67 − + 0.001875 − + 0.12

� � � �

−
�

� = 3.85 −
�

b.) Conduction through Floor

Heat load of the floor by means of conduction

∆�
� =
�

A = Area of the floor

= (8m) (4m)

A = 32 �

( )[ ℃− − ℃ ]
� =
. −
�

= 1436.15W

Page 11 of 27
Total Heat load in Floor

� = 1436.15W

C. HEAT LOAD COMPUTATION OF THE CEILING

a.)Total Resistance of the Ceiling

Characteristics of the Material

Materials in the
Type of �
�
Ceiling ,� ∙ �, � ∙ Thickness ,t, m
Material � � �

Asphalt Shingles Roofing 0.08 0.15

Expanded

Polystyrene, Insulation 27.7 0.2

Extruded

Stainless Steel Interior 0.0625 0.03

Total Resistance of the Ceiling

Outside Air Film (moving air, cooling season)

R1 = 0.044 m2 ∙
�

Asphalt Shingles

R2 = 0.08 ∙
�

Expanded Polystyrene, Extruded

R3 = (t)(

R3 = (0.2m)( 27.7 � ∙ )
�

R3 = 5.54 ∙
�

Stainless Steel

R4 = (t)(

R4 = (0.0015m)(0.0625 � ∙ )
�

Page 12 of 27
 R4 = 0.001875 ∙�

Inside Air Film (Vertical, heat flow horizontal)

`R5 = 0.12 m2 ∙
�

Total Resistance, RT = R1 +R 2 +R3 + R 4 +R5

RT = 0.044 m2 ∙ + 0.08 ∙ + 5.54 ∙ + 0.001875 ∙ + 0.12 m2 ∙

� � � � �

RT = 5.79m2 ∙
�

b.) Conduction through Ceiling

Heat load of the ceiling by means of conduction

�=� ��

�= � �

�=

∆
�=

Materials in the Characteristics of the Material

Door Type of Material � R, � - Thickness, t, m

�, �
�−�

Stainless Steel Exterior 16 0.0015

0.077

Polyurethane Insulation 0.03

Foam

Stainless Steel Interior 16 0.0015

Page 13 of 27
 Composition of the door

D. HEAT LOAD COMPUTATION OF THE DOOR

a.) Total Resistance of the Door

Outside Air Film (moving air, cooling season)

� = 0.044 � −
�

Stainless Steel

� = (t) (

= (0.0015m) ( �−
�

� = 0.00009375 � −
�

Polyethylene sheet (low-density)

� = (t) (

= (0.077m) ( �−
. �

Page 14 of 27
 � =2.5667 � − �

Stainless Steel

� = (t) (

= (0.0015m) ( �−
�

� = 0.00009375 � −
�

Inside air film (vertical, heat flow horizontal)

� = 0.12 � −
�

Total Resistance, �� = � + � + � + � + �

= (0.044 + 0.00009375 + 2.5667 + 0.00009375 + 0.12) � −

�

� = 2.73089 −
�

b.) Conduction through Door

Heat load of the door by means of conduction

∆�
� =
�

A = Area of the door

= 2(2m x 1.25m)

A =5�

( )[ ℃− − ℃ ]
� =
. −
�

= 82.391 W

Total Heat load in Door

Page 15 of 27
 � = 82.391 W

TRANSMISSION LOAD, �

Section Computed Values

Wa��� 1699.81W

F���� 1436.15W

Ce���n� . �

D��� 82.391 W

Total ( � 1082.9154 W

2.) MISCELLANEOUS LOAD

a.) Heat generated by from mechanical and electrical equipment

No. of lights = 6

Wattage = 10 watts

Qapp ia s = �a��a�e × CLF × n�. �f ������

Qapp ia s = �a��� × . ×

����������� = . �

b.) Heat from Occupants

Rate of heat Emission for mixtures of males and females

Degree of Activity Sensible Latent Total

Seated, quiet .60 40 100 W

Seated, very light .70 45 115 W

work

Page 16 of 27
 Moderate Dancing .78 160 238 W

Athletics .80 215 295 W

Heavy machine work; .82 285 367 W

lifting

Standing, light work; .75 55 130 W

walking

Floor area for Cold Room Storage: 32 �

Activity: Standing, light work; walking

Occupancy Space Per Person =

For the number of occupants = approx. 6 persons (3 light work) and (3 Heavy Work)

Heat Gain Per Person = (heat gain) (1-sensible heat gain)

Therefore, � = [(130 W) (1-0.75)(3)] + [(367)(1-.82)(3)]

Total occupants’ heat load

� = 295.68 Watts

MISCELLANEOUS LOAD,

Computed Values

Appliances 74.4 W

Occupants 295.68 W

Total ( 370.08 W

Page 17 of 27
3.) INFILTRATION LOAD

V = volume of air per air change, � ℎ

Page 18 of 27
 = 8m x 4m x 4m

= 128� x
.

V = 4520.277 ��

N = number of air changes per 24 hours, in air changes/24hrs

N = 7.679 air changes per 24 hours (obtained from table 5 by interpolating 4,520.277 �� )

� = heat extracted in cooling outside air to temperature of the refrigerator, in ���/��

� = 3.096 (obtained from table 6 by interpolating ℉ or - ℃. Thus, to cool air from 95℉

or ℃ and 60% humidity to ℉ or - ℃)

�′ = infiltration load/service load

= NV�

= (7.679)(4520.277)(3.096)

�′ = 107,465.8971 or 4477.745 or 1313.47 W

ℎ ℎ

INFILTRATION HEAT LOAD

′ = 1313.47 W

4.) PRODUCT LOAD

When a product enters storage at a temperature above the temperature of the space, the

product will give off heat to the space until it cools to the space temperature. The heat gain from

the product is computed by the following equations:

� =� +� +�

Where � = product load, KJ

Page 19 of 27
 � = ℎ��� �� ���� ���� �������� ����������� ��

�������� �����������, ��

� = ℎ��� �� ������, ��

� = ℎ��� �� ���� ���� �������� ����������� ��

����� ������� �����������, ��

� = � � (� − � )

� =� ℎ

� = � � (� − � )

where: m = mass of the product, kg

� = �������� ℎ��� ����� ��������, ∙

� = �������� ℎ��� ����� ��������, ∙

� = �������� �����������, ˚C

� = �������� �����������, ˚C

� = ������� �����������, ˚C

ℎ = ������� ������ ℎ���,

Page 20 of 27
 Total Mass Specific Specific
Latent
of Stocked Heat Heat
Entering Freezing Heat of
Food Product Above Below
temperature temperature
Fusion
Items Freezing Freezing
kg (˚C) (˚C)
��
�� ��
��
�� ∙ � �� ∙ �

Mushroom
600 35 -0.9 3.89 1.84 307

a.) Product load

� = � � (� − � )

= (600)(3.89)[35-(-0.9)]

� = 83,790.6 KJ

� =� ℎ

= (600 kg)(307 )

� = 184,200KJ

� = � � (� − � )

= (600kg)(1.84)[-0.9-(-10)]

� = 10,046 KJ

� = 83,790.6 KJ + 184,200KJ + 10,046 KJ

Page 21 of 27
� = 278036.6 KJ

T in OC q in mW/kg

Product α b

Mushroom 211 0.15

� = heat of respiration, in mW/kg

�
=� �

. − ˚C
= �

= 0.0470805W/kg (600kg) (3600s/1hr) (24hrs/1day) (1KJ/1000J)

QR =2440.65 KJ

TOTAL PRODUCT LOAD,

� =� +�

= (278036.6 + 2440.65) KJ

= 280477.2531 KJ

, .
= ℎ 6
ay
�� ℎ

QP = 3246.2645 W

Page 22 of 27
TOTAL REFRIGERATION LOAD, �

Computed Values

Transmission Load 1082.9154 W

Miscellaneous Load 370.08 W

Infiltration Load 1313.47 W

Product Load 3246.2645 W

6012.13 W

Total Or

TR

COMPRESSOR CAPACITY

Compressor running time:

16 hours if the room is above 0℃

18 hours if the room is below 0℃

��
� =

. � ℎ � �
= = (8016.17 W)
ℎ � . �

= 2.28 TR

NET HEAT LOAD, � �

� � = 1.2 �� = 1.2 (2.28 TR) = 2.73 TR

IV. EQUIPMENT SIZING

Design of Component Of Cold Storage

Page 23 of 27
For ammonia used as refrigerant

Ideal vapour compression cycle

Saturated properties of R134a

Specific Specific Enthalpy Specific Entropy

Temperature Pressure volume Saturated Saturated Saturated Saturated

(oC) (kPa) (m3/kg) Liquid Vapor Liquid Vapor

(kJ/kg) (kJ/kg) (kJ/kg-K) (kJ/kg-K)

-10* 200.7 0.09960 38.53 244.55 0.1550 0.9378

45* 1161 0.01734 115.82 273.40 0.4184 0.9137

The refrigerant is superheated by 5oC; therefore temperature T1’= -10oC + 5oC= -5OC (268K).

Since the entropy at point 2 is equal to entropy at point 1, we can equate these two equations-

S2’=S2 + Cpln(T2’/T2) ------------- equation (2)

S1’=S1 + Cpln(T1’/T1) ------------- equation (3)

Where: Cp for R134a= 0.854kJ/kg·K

Equation (2) = Equation (3)

S2= 0.9137 + 0.854ln (T2’/318)

S1= 0.9378 + 0.854ln (268/263)

0.9137 + 0.854ln (T2’/318) = 0.9378 + 0.854ln (268/263)

 T2’=333.32K and S2=S1=0.946kJ/kg·K

Page 24 of 27
Calculation of COP

� � ℎ −ℎ
COP= =ℎ −ℎ

T3 is subcooled by 3oC, T3’=314K

h3’=h3 – Cp (T3-T3’) = 112.22kJ/kg -0.854kJ/kg·K (317-314) oK =109.658kj/kg=h4’

h1’=h1 – Cp (T1-T1’) = 237.74kJ/kg -0.854kJ/kg·K (262-257) oK =233.47kj/kg

h2’=h2 – Cp (T2-T2’) = 270.01kJ/kg -0.854kJ/kg·K (333.47-317)oK=256.005kj/kg

ℎ′ −ℎ′ . − .
COPactual = ℎ′ −ℎ′ = = 5.49
. − .

� � � .
Horse power required = = = . ℎ�
.

Taking 20% more of the calculated horse power = . ℎ� + 0.2* . ℎ� = 6.36hp ≈7 ℎ�.

Refrigerating effect = (h1 − h4) = (233.47− 109.658) = 123.812 kJ/kg

� . . .
ṁ= = =0.1738kg/s
� � .

Design of cooling tower

Cooling towers are used to cool down the temperature of water coming out of the condenser.

Assuming a 5m height for the cooling tower. For cooling, pump will be required for circulating

water,
∗ . ∗ . ∗
�� �� ���� = ���� = = . �/ = . ℎ�

Design of condenser

Condenser design depends on the amount of heat removed by the condenser. The capacity of

condenser is given by,

Page 25 of 27
 Qr = Qe + Wc

Where Qr=heat rejected by the system

Qe= heat load of the system

Wc=work done by the compressor to remove Qe

Qr= 6.114TR*3.52 + 6.36hp*0.746 = 26.27kW

Assuming that water is entering at temperature 30C and leaving at 44C.

Thus, for 14C change of temperature amount of heat gain will be,

�⁄
�= cpwΔT = 4.186kJ/kg-K * (14 K) =58.604 kJ/kg.
o

�
Therefore, the amount of water required = Qr/ ⁄� =26.27kW/(58.604 kJ/kg. )= 0.448kg/s

Design of throttling device

Throttling device must be capable of expanding 625kg or 78.12m3 of refrigerant per hour and must

operate in the pressure range of 1.57 bars to 11.29bars.

Design of evaporator

The capacity of evaporator will be equal to the amount of refrigerating effect(123.812 kJ/kg) or

(6.114TR).But since I only use one compressor, it is ideal to only have one evaporator.

Page 26 of 27
REFERENCES

https://energy.gov/sites/prod/files/2013/12/f5/webinar_hvac_calculatingloads_20110428.pdf

http://www.engineeringtoolbox.com/specific-heat-capacity-food-d_295.html

https://www.researchgate.net/publication/266672637_Physiological_Characteristics_of_Mushro

oms_Strawberries_Broccoli_and_Tomatoes_Respiration_in_Air_and_Modified_Atmosphere_an

d_Transpiration

https://www.ajol.info/index.php/aga/article/viewFile/1662/560

http://kolmetz.com/pdf/EGD2/ENGINEERING_DESIGN_GUIDELINES_refrigeration_systems

_rev_web.pdf

Page 27 of 27