Cold Room Refrigeration Project: Design & Analysis

​HIGHER SCHOOL OF TECHNOLOGY AND EXPERIMENTAL SCIENCE
Degree in Mechanical Engineering
Subject: EM 1044 - Air Conditioning and Refrigeration
Installations (2019-2020)
“PROJECT OF THE REFRIGERATION PART:
Cold room for a retail store”
Mª Amparo Almela Pitarch
Rubén Jaén Peláez
Aarón Sánchez Ferrando
Pau Alpuente Sebastiá
Castellón de la Plana, Diciembre 2019
Mechanical Engineering Area
INDEX
1. INTRODUCTION ……………….……………………………………………...….. 2
2. THERMAL LOAD CALCULATION ……………………………………………..... 3
3. SIZING AND SEALING OF THE ROOM’S ENCLOSURE ……………..……. 16
4. REFRIGERATION SYSTEM …………………………………...……………..... 17
5. BUDGET OF THE MAIN COMPONENTS ………………………………..…… 32
6. BIBLIOGRAPHY ……………………………………………………....….……… 37
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1. INTRODUCTION
_____________________________________________________________________________________________________
This project is based on the design of a cold store, installed in a small store in a
ground floor in Valencia. In our cold room we will store the perishable products
(strawberries, watermelons and plums) that are not in the cold exhibitors.
Firstly, we must calculate the thermal loads of our cold store: product cooling,
breathing, packaging and pallets, air renovation, staff, lighting, and enclosures heat
transfer.
After, we will have to choose the best refrigerant to our application, depend of their
properties, that will absorb the heat of our system. After this, we will select the
compressor according to the temperature difference we need it and the evaporator.
Based on these data, we will calculate the total refrigerant system and the economic
budget of all the components with a legally, technologically and economically viable
solution.
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2. THERMAL LOAD CALCULATION
_____________________________________________________________________________________________________
First, we will calculate the thermal loads manually and then we'll check the
calculations with the programme FRIO. Our cold storage is located in Valencia, with
an exterior temperature of 31.8 ºC and operating hours of 20 with a coefficient of
security of 15%.
Image 1. Project in software FRIO
In the software, we have chosen the properties that are determined by the UNE for
Valencia (Manises).
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2.1 - PRODUCT COOLING
To calculate the product cooling load, we have to obtain the mass of the new daily
product for every fruit.
First, we search the properties of our fruits on page 383 of ​Fundamentos de
Refrigeración.
M​T ​(tn)
Cp,Specific heat fresh
product (KJ/Kg K)
Freezing
temperature (ºC)
Storage density
(Kg/m²)
Water (%)
Strawberries
2
3,85
-0,8
140
90
Watermelons
5
3,95
-0,4
140
93
Plums
3
3,72
-0,8
140
86
Fruit
Table 1. Properties of fruits
Our daily input factor is x=0,5
x=
MD
MT
(1)
x = Daily input f actor (day)
M D = N ew daily product (tn/day)
M T = Amount of stored product (tn)
●
M D to plum is → 1,5 (tn/day)
●
M D to strawberries is → 1 (tn/day)
●
M D to watermelons is → 2,5 (tn/day)
Our cold store will keep the fruit at T C =1ºC to prevent freezing, as our goal is to
maintain, not freezing and an 85% of relative humidity. In addition, our products
enter with a temperature of T I =12ºC and with a regime time of n=8h. This time is
the hours that the product needs to obtain the temperature of the cold store. It
depends on the product, the velocity of the air and the form. We considered that the
product enters all in one time, and for this, the valours is between 6 and 12 hours.
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1000
QE = M D ·C p, f resh ·(T I − T C )· n·3600
(2)
QP = T hermal cooling load (KW )
T I = T emperature when enters (ºC)
M D = N ew daily product (tn/day)
T C = S torage temperature (ºC)
C p, f resh = S pecif ic heat f resh product (KJ/Kg ºC)
n = Regime time (hours)
●
1000
QE, P LU M = 1, 5·3, 72·(12 − 1)· 8·3600
= 2, 13 kW
●
1000
QE, ST RAW = 1·3, 85·(12 − 1)· 8·3600
= 1, 47 kW
●
1000
QE, W AT = 2.5·3, 95·(12 − 1)· 8·3600
= 3, 77 kW
2.2 - PACKAGING AND PALLETS
Our products are packed in wooden boxes under a pallet. We must obtain the
percentage that exists packaging and pallets in our products. This is in tables in
page 338/339 of ​Fundamentos de Refrigeración​ or in our notes of class.
Image 2. Tables of packaging
We have considered for all our fruits the percentage of packaging of 10% like
material generic and for the pallets to 4%. The specific heat of wood is 2,72 kJ/kgºC,
choosing the maximum value to calculate the maximum load.
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❖ Thermal load of packaging
1000
QP ACK = M D ·aE ·C p, pack ·(T I − T C )· n·3600
(3)
QP ACK = T hermal load packaging (KW )
M D = N ew daily product (tn/day)
C p, pack = S pecif ic heat of material (KJ/Kg ºC)
T I = T emperature when enters (ºC)
T C = S torage temperature (ºC)
aE = % of packaging mass
n = Regime time (hours)
●
10
1000
QP ACK, P LU M = 1, 5· 100
·2, 72·(12 − 1)· 8·3600
= 0, 155 kW
●
10
1000
QP ACK, ST RAW = 1· 100
·2, 72·(12 − 1)· 8·3600
= 0, 104 kW
●
10
1000
QP ACK, W AT = 2, 5· 100
·2, 72·(12 − 1)· 8·3600
= 0, 259 kW
❖ Thermal load of pallets
1000
QP ALLET = M D ·aE ·C p, pal ·(T I − T C )· n·3600
(4)
QP ALLET = T hermal load of pallet (KW )
aE = % of packaging mass
M D = N ew daily product (tn/day)
n = Regime time (hours)
C p, pal = S pecif ic heat of material (KJ/Kg ºC)
T C = S torage temperature (ºC)
T I = T emperature when enters (ºC)
●
4
1000
QP ALLET , P LU M = 1, 5· 100
·2, 72·(12 − 1)· 8·3600
= 0, 062 kW
●
4
1000
QP ALLET , ST RAW = 1· 100
·2, 72·(12 − 1)· 8·3600
= 0, 042 kW
●
4
1000
QP ALLET , W AT = 2, 5· 100
·2, 72·(12 − 1)· 8·3600
= 0, 104 kW
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2.3 - BREATHING
Concerning breathing, the vegetables and the fruits are living organisms (if not
cooked) and they keep breathing releasing heat to the ambient. This is only
considered if the temperature of the storage is above 0ºC and it not freezing. In ou
case, our products are at 1ºC and we must consider it.
1
QRP = [(1 − x)·M T ·q a (T i ) + x·M T ·q a (T m )]· 1000000
(5)
QRP = T hermal load breathing(KW )
M T = Amount of stored product (tn)
q a (T i ) = B reathing load a storage temperature (kJ/T n 24h)
q a (T m ) = B reathing load at average temperature between the entering and the storage (kJ/T n 24h)
x = Daily input f actor (day)
n = Regime time (hours)
In our case the value of new daily product is x=0,5 for all fruits.
For obtain the values the breathing load for every fruit, we have a table in page
396/397 of ​Fundamentos de Refrigeración​.
Fruit
0ºC
5ºC
10ºC
15ºC
20ºC
Strawberries
36,4/52,4
48,5/98,4
145,5/281,3
210,5/273,5
303,1/581,0
Watermelons
--
--
22,3
--
51,4/74,2
Plums
5,8/8,7
11,6/26,7
26,7/33,9
35,4/36,9
53,3/77,1
Table 2. Specific power due to breathing (mW/Kg)
For q a (T i ) we take the value of 0ºC, since it refers to the temperature of the
chamber and our it is at 1ºC. And for q a (T m ) is the value between the entering and
the storage; in our case it is 12ºC so we will take the average between the values ​of
15ºC and 10ºC.
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26,7+35,4
1
)]· 1000000
2
●
QRP , P LU M = [(1 − 0, 5)·3000·5, 8 + 0, 5·3000·(
●
QRP , ST RAW = [(1 − 0, 5)·2000·36, 4 + 0, 5·2000·(
●
QRP , W AT = 0 kW
= 0, 055 kW
145,5+210,5
1
)]· 1000000
2
= 0, 2144 kW
In the case of the watermelons, is a fruit without breathing in these temperatures.
Then, we have performed these same calculations with the software FRIO and have
obtained the following thermal loads shown in the following images.
Some results are different because are used values default in the program and we
have used other sources of values, although in conclusion the load is quite similar.
-
Summary:
Summary of the thermal loads of the product, packaging, pallets and respiration
obtained with FRIO:
Image 3. Thermal loads of Plum
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Image 4. Thermal loads of Strawberries
Image 5. Thermal loads of Watermelons
Fruit
Manually
Program
Strawberries
1,83 kW
1,87 kW
Watermelons
4,13 kW
4,33 kW
Plums
2,40 kW
2,36 kW
Total cold store
8,36 kW
8,32 kW
Table 3. Manual and program value comparison
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2.4 - TRANSMISSION THROUGH ENCLOSURES
In the calculation of the thermal loads of the enclosures, we have made a
comparison with different insulators. In addition, we have added a concrete layer,
which is not considered in the software.
The data used are as follows:
Exterior convection
Surface heat transfer coefficient
(m²)
(W/m² ºC)
Interior convection
heat transfer
coefficient (W/m² ºC)
Temperature difference
(ºC)
Roof
24
10
10
5
Floor
24
10
20
5
North wall
21
9
9
5
South wall
21
9
9
2
East wall
14
9
9
3
West wall
14
9
9
3
Table 4. Data of the enclosures.
For all cases, T​e​=31,8ºC and T​c​=1ºC. The insulation will have a thickness of 10,5 cm
and the concrete of 12 cm, as proposed on page 573 of ​Fundamentos de
Refrigeración.
Qc =
(T e − T c) + ΔT
1 + ea + ep + 1
he *A K a *A K p *A hi *A
​(6)
QC = T hermal load transmission through enclosures (KW )
T e = T emperature of the ambient (K)
T c = T emperature of the cold store (K)
h = C onvection heat transf er coef f icient (W /m² K)
K = C onduction heat transf er coef f icient (W /m K)
A = S urf ace (m²)
e = W all thickness (m)
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Material
Expanded
polyurethane
Kp
0,0231
East (kW) West (kW) North (kW) South (kW)
Floor (kW)
Roof (kW) Total (kW)
0,098
0,098
0,155
0,142
0,180
0,178
0,851
Polyurethane in
plates
0,026
0,109
0,109
0,173
0,159
0,202
0,199
0,951
Polystyrene
0,0288
0,120
0,120
0,191
0,175
0,222
0,219
1,046
Fiberglass with
polystyrene
0,0361
0,148
0,148
0,235
0,215
0,274
0,270
1,289
Table 5. Load of enclosure according to insulation.
As we can see the material that produces less load losses through the enclosures is
expanded polyurethane, so we have chosen this insulating material.
Image 6. Thermal loads of the enclosures according to the software
We note that the load calculated in the FRIO software is quite different from that
calculated by us, possibly due to the different differences in material values.
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2.5 - AIR RENOVATION
The air renewal load will take into account the volume of the chamber, the air
density, the number of daily renewals and the enthalpies of the air in indoor and
outdoor conditions.
First, we have that the volume of the camera is 6 x 4 x 3,5 = 84m³.
Secondly, we have considered an air density, ρ​e ​= 1,225 kg/m³.
On the other hand, we have a number of daily renewals. This value is provided by
the program, ​FRIO.​ Since it has been considered that the working conditions are
going to be labour intensive, the number of renewals will be equal to 20.
Finally, enthalpies have been calculated for outdoor conditions (Ts=31.8ºC and
relative humidity of 46%) and indoor conditions (Ts=1ºC and relative humidity of
85%). This calculation can be done by means of the psychometry diagram or, in our
case, by means of an online calculator, which is mentioned in slide 51 of topic ​1.2 Thermal loads in refrigeration.​
h​e​ = 66,7 kJ/kg; h​i​ = 9,66 kJ/kg
QRV =
N REV
24·3600 ·V
·ρe ·(he − hi )
​ (7)
QRV = T hermal load air renovation (KW )
he = E xterior convection heat transf er coef f icient (W /m² K)
V = C old storage volume (m³)
hi = I nterior convection heat transf er coef f icient (W /m² K)
N REV = Daily renewals
ρe = Outdoor air density (kg/m³)
●
QRV =
20
·84·1, 225·(66, 7
24·3600
− 9, 66) = 1, 357 kW
Image 7. Thermal loads of air renewal according to the software.
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2.6 - STAFF
For the calculation of the thermal power of the people, we have considered that in
the chamber there will be only one worker, who will be inside during 1h and 30 min a
day. In this time it is considered, the time of storage of the chamber and the
extraction of products outside the store.
The software calculates this power with a work of 24h. Therefore, a difference will be
seen when calculating it.
To obtain the value of q​pers​, we have interpolated the value of the operator's load
from table 12.21 on page 375 of the book ​Fundamentos de Refrigeración.​
T [ºC]
0
5
q​pers​ [W]
270
240
Table 6. Operator load values as a function of temperature
Our chamber will be at a temperature of 1ºC, therefore:
5−0
240 − 270
=
5−1
240 − q pers
→ q pers = 264 W
x
1
Qpers. = N pers ·q pers · 24
· 1000
(8)
Qpers. = T hermal load staf f (KW )
q pers = S urf ace of the storage room (m²)
N pers = N umber of personal
x = Operating hours (hours)
●
24 1
Qpers. = 1·264· 24
· 1000 = 0, 264 kW
●
Qpers. = 1·264· 1.5
· 1 = 0, 0165 kW
24 1000
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2.7 - LIGHTING
For the calculation of the thermal power of illumination, the software considers that
the lights are in operation during the 24h of the day. In our case, we have considered
that the lights will be on 1h and 30 min a day, the same time in which there is
someone inside the camera. With this difference of hours of operation of the light can
be observed the difference of thermal power of illumination provided by the software
and the one that we realize.
On the other hand, the book ​Fundamentos de Refrigeración on page 374, says that
the lighting in cameras is usually low, between 5 and 10 W/m². In this case, we have
considered a surface illumination power of 9W/m².
x
1
Qlight = P light ·S f loor · 24
· 1000
(9)
Qlight. = T hermal load light (KW )
S f loor = S urf ace of the storage room (m²)
P light = Lighting power (W /m²)
x = Operating hours (hours)
●
24 1
Qlight = 9·24· 24
· 1000 = 0, 216 kW
●
Qlight = 9·24· 1,5
· 1 = 0, 0135 kW
24 1000
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2.8 - FANS
For the load of the fans, 6% of the total load has been considered.
Qf ans =
6
100 ·(QE
Qf ans =
6
100 ·(7, 33
+ Qpack + Qpallet + QRP + Qc + QRV + Qpers. + Qlight. )
(10)
+ 0, 519 + 0, 208 + 0, 262 + 0, 686 + 1, 33 + 0, 266 + 0, 216) = 0, 649 kW
Summary of total thermal loads
Image 8. Total thermal load of the cold store
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3. SIZING AND SEALING OF THE ROOM’S ENCLOSURE
_____________________________________________________________________________________________________
Regarding the design of the distribution of our cold store, we take as reference the
specifications on page 348 of the book ​Fundamentos de Refrigeración ​which
determines the need to leave a series of spaces to meet the needs of provision,
maintenance and circulation of air and wheelbarrows.
So according to the book our products stored in wooden boxes must be separated
by 10 cm from each other, leaving a separation with the walls of 50 cm.
On the other hand, in our case, we access the product from the front, so we have not
created staff movement aisles, since all the space to move is in front.
The evaporator must be at least 1 meter away from our product, in our case when
leaving space of movement when entering it is 2 meters away.
Another rule to consider, is that the minimum height of our roof that we must have is
3m and in our case is 3,5m and our door will have a width of 150 cm for the
convenience of the staff to be able to remove the material from the chamber without
any difficulty.
Finally, the evaporators must be placed in the door and the compressor and
condenser will be located outside the chamber in a small separate room in a corner
of the chamber assuming there are no adjacent buildings.
Below we attach the plan of the distribution of our cold store.
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CREADO CON UNA VERSIÓN PARA ESTUDIANTES DE AUTODESK
CREADO CON UNA VERSIÓN PARA ESTUDIANTES DE AUTODESK
CREADO CON UNA VERSIÓN PARA ESTUDIANTES DE AUTODESK
Responsible department:
INGENIERÍA
MECÁNICA
Scale:
N/S
Created by:
Maria Amparo Almela Pitarch, Rubén Jaén Pelaez, Aarón Sánchez Ferrando
y Pau Alpuente Sebastiá
Document type:
Legal owner:
Plant layout flat
Document status:
Format:
Title, Supplementary Title
DEPARTAMENTO
DE TECNOLOGÍA
Cold Room for a Retail Store
Plant Layout
Finished
A3
Units:
Centimeters
Edition date:
2/12/2019
Language:
ANG
Page:
1/1
CREADO CON UNA VERSIÓN PARA ESTUDIANTES DE AUTODESK
Mechanical Engineering Area
4. REFRIGERATION SYSTEM
_____________________________________________________________________________________________________
❖ REFRIGERANT SELECTION
The R450A has been chosen as it is an alternative to the R134a that is currently
used so much. In comparison with the previous one, three improvements are
highlighted:
● The GWP is reduced more than half.
● Its use will persist for longer because it contaminates less.
● It has greater energy efficiency.
On the other hand we have the following factors:
● Lower volumetric efficiency.
● Less availability of specific compressors.
In this second counter it should be noted that the vast majority of compressors
designed for the R134a are compatible with the R450A, although in the compressor
data sheets we cannot know with certainty the refrigerant capacities of the
compressor for our refrigerant because they are only for R134a.
Another feature to highlight in the R450A is that it is an azeotropic refrigerant,
therefore it has a temperature slide although it is very low (0.6K), so it will not affect
our installation.
Image 9. Properties of R450A
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For our installation, we verify in FRIO what type of condenser provides greater
system efficiency, either by water or by air.
The air condensers are widely used in domestic, commercial and industrial
applications and the rise in the use, compared to the water ones, is due to that the
water sources not necessary, especially in urban spaces and is scarce and
expensive.
However, they present two drawbacks in comparison to the water ones:
● Higher temperature difference and hence higher condensation temperature
and pressure (specific heat of the dry air is relatively low and lower convection
coefficient)
● Noise
❏ Air condenser
Image 10. Refrigeration System by air
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❏ Water condenser
Image 11. Refrigeration System by water
As we observe, we obtain in our refrigeration system we obtain greater efficiency
with a water condenser.
Our refrigeration system it's a simple cycle with a mode of operation of a refrigeration
machine, without exchanger and with evaporator.
We have an alternative compressor with an electric power of 3.58 kW and an
evaporator that evaporates at -5º and a condenser that condenses at 30ºC,
assuming a 5ºC of superheating and 2ºC of subcooling. Our energy operation
coefficient (COP) is 4.33.
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Image 12. Diagram P-h R450A
Image 13. Points of the diagram P-h
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❖ COMPRESSOR SELECTION
For the choice of the compressor a program developed by the Bitzer has been used,
it makes a comparison between two possible devices. Therefore, one of the two will
be appropriate. In our case, we have chosen the one with the greatest cooling
capacity since 15,8 kW are required in the FRIO program, so the 14,48 kW capacity
compressor would be discarded.
However, a general summary of the technical specifications of each compressor is
attached to visualize this difference. The final decision is of the 4TES-9Y-40P
compressor.
Image 14.Technical table of both compressors
The graphic representation of work that we would have for this compressor would be
the following. Taking into account the parameters of overheating, compressor
discharge, etc. It should be noted that 0,319 kg/m³ of the selected refrigerant
(R450A) is required for this circuit. Hence, with the volume we have in the camera
the result would be the following.
M ass of ref rigerant = 0, 319·6·4·3.5 = 26, 796 kg
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Image 15. Steam Cycle
Finally, the planes of the compressor are attached and subsequently those of the
necessary bench to avoid vibrations, since it is a piston compressor. This base is
advisable to be made with concrete.
Image 16. Compressor plane
Image 17. Clamping in transport and bench
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❖ EVAPORATOR SELECTION
Now the selection of the evaporator is going to be discussed, because the
refrigerating chamber is not very large and will not contain excessive load, the most
common is an air-cooled tube and fin evaporator.
This type of evaporator is the most used in the industry thanks to its easy
construction, modulation and wide range of cooling capacities, as well as the
temperature and pressure range, efficiency and easy cleaning. The COP of the
evaporators depends on the exchange surface, the velocity of the refrigerant and the
air as well as it temperatures differences.
Image 18. ​Technical table of differents evaporators
An evaporator with 4 mm fin distance has been selected, what is achieved here is to
increase its capacity, within these ​BSL121A has been selected which is the first one
that meets the evaporator power calculated with FRIO.
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Once an evaporator has been selected, the planes of the same are attached:
Image 19:Evaporator Planes
Image 20: Evaporator Dimensions
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❖ CONDENSER SELECTION
The Bitzer software has been used for the selection of the capacitor, which with the
parameters obtained in FRIO, the program obtains several possible capacitors for
the necessary characteristics and from the 3 possibilities, we select the one that we
believe best adapts to the circuit.
The ​K203H ​(horizontal multitubular water condenser K) compressor will be selected
and these two one is seawater and the other the standard one. We select the one
with a higher volumetric flow and with a maximum capacity of 36,1 kW.
Image 21.Technical table
Here we have a condenser scheme, with both the input and output temperatures:
Image 22. Condenser Diagram
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Finally, in the part of the budget will be the cost of each element, the plans of the
condor are attached:
Image 23. Condenser plane I
Image 24. Condenser plane II+]]
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❖ EXPANSION VALVE
The expansion valve that we are going to use is oversized for this installation
because the manufacturer we have searched for does not have smaller valves for
this application, since it is designed for installations with 30 kW of cooling capacity.
For this reason, the most economical one that meets our needs has been sought.
That is, it can work between 30ºC of condensation and -5ºC of evaporation.
Image 25. Parts of expansion valve
The characteristics of the valve are attached, in our case it is ETS 12.5.
Image 26. General characteristics of the valve
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Finally, the planes of the valve are attached to know their dimensions:
Image 27. Valve plane
❖ PIPES SELECTION
In the calculation of pipes, the software FRIO has been used.
The input data used was: evaporating temperature of -5°C, condensing temperature
of 30°C, roughness of 0,0015 mm, actual length of 15m and cooling capacity of
15,8kW.
Following the data of the book ​Fundamentos de Refrigeración,​ 6 elbows of 45º, 15
large elbows of 90º and 1 siphon have been considered.
The type of pipe will be calculated depending on the line of the refrigeration system.
First we have the suction line. The suction line is the most critical in pipe selection.
The speed should not exceed 20 m/s to avoid rudeness, vibrations and wear that
can provide leaks.
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In this case, the pipe with diameter 1 3/8(1,25)" has been chosen, since it would not
exceed the speed of 20m/s and it is also the most economical than the other allowed
diameters.
Image 28. Recommended (red) and selected (green) pipe diameters in suction line.
Secondly, you have the discharge line. The discharge line has similar considerations
to the suction line in question to oil return and speeds.
In this case the choice we make is the 1(1)" diameter pipe. Option 7/8(1)" would also
be possible, but compared to pressure losses it is much higher (0,3758 kg/cm²)
compared to 1(1)" which is 0,1984 kg/cm².
In the case of the 1 1/8(1)" option, the difference is in the speed of the fluid, which is
lower.
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Image 29. Recommended (red) and selected (green) pipe diameters in discharge line.
Finally, we have the liquid line. The liquid line is not calculated in the same way as
the suction and discharge line. In this case, the maximum pressure loss criterion that
is adopted is to avoid the formation of bubbles at the inlet of the expansion valve and
the maximum speed criterion to avoid water hammer that can cause ruptures in the
pipes and valves of the system.
The diameters in the liquid line will be considerably smaller than those of the gas
line. Therefore, our choice of diameter has been 5/8(0,8)".
Image 30. Recommended (red) and selected (green) pipe diameters in liquid line.
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Suction
Discharge
Liquid
Material
Copper
(standard bar)
Copper
(standard bar)
Copper
(standard bar)
D. Nominal (“)
1 ⅜(1,25)”
1(1)”
⅝(0,8)”
Real length (m)
15
15
15
D. Inside (mm)
32,43
23,40
14,28
Fluid velocity (m/s)
11,520
7,209
0,605
P. Total (kg/cm²)
0,1470
0,1984
-0,3762
Flow (kg/h)
400
400
400
L. equivalent (m)
42,14
34,59
26,95
Viscosity (Pa)
1,09.10​-5
1,26.10​-5
17,83-10​-5
Density (kg/m​3​)
11,72
35,95
1151
Roughness (mm)
0,0015
0,0015
0,0015
Friction factor
0,01426
0,01409
0,02079
Table 7. Data and pipe selection.
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Mechanical Engineering Area
5. BUDGET OF THE MAIN COMPONENTS
_____________________________________________________________________________________________________
The compressor that has been chosen comes from the manufacturer Bitzer and the
Pecomark distributor. The price included is without possible discounts to negotiate
with the distributor
​Image 31. Price of compressor
Image 32. Price of condenser
Due to the difficulty of the evaporator selected, another evaporator has been
selected with similar characteristics but slightly higher than the one that has been
easier to find the sale price.
32
Mechanical Engineering Area
Image 33. Price of evaporator
This valve comes from the manufacturer Danfoss and our distributor is Electric
Automation Network. We provide this item with the price of 220,39 € without counting
possible discounts.
Image 34. Price of expansion valve
As the book ​Fundamentos de Refrigeración​ indicates, we have obtained the price of
pipes from the Pecomark commercial catalog.
Image 35. Price of pipers
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Mechanical Engineering Area
Main components
Type
Units
Unit Price (€/ud)
Total price
(€)
Compressor
4TES-9Y Bitzer
1
4.662,00
4.662,00
Condenser
K203-H Bitzer
1
1.381,00
1.381,00
034G4212 Danfoss
1
220,39
220,39
BSL112A
1
4.067,00
4.067,00
Expansion valve
Evaporator
Garcia Camara
​10.330,39
Table 8. Total cost of main components.
Pipes
Type
Length (m)
Unit Price (€/m)
Total price
(€)
Pipe Cu ⅝”
TB - ⅝”
15
7,80
117,00
Pipe Cu 1”
TB - 1”
15
15,00
225,00
TB - 1-⅜”
15
25,00
375,00
Pipe Cu 1 ⅜”
​717,00
Table 9. Total cost of pipes.
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Mechanical Engineering Area
Enclosure
Type
Surface (m²)
Unit Price (€/m²)
Total price
(€)
Insulation
Expanded
polyurethane
118
11,94
1.408,92
Roof
Precast concrete
hollow core panels
24
22,53
540,72
Wall
Precast concrete
hollow core panels
70
22,53
1.577,10
Floor
Precast concrete
24
22,53
540,72
hollow core panels
​4.067,46
Table 10. Total cost of enclosure.
Others
Door
Refrigerant
Type
Quantity
Quantity Price (€/x)
Total price
(€)
PF 01 034 Salvador
Escoda
1 (ud.)
1.125,00
1.125,00
R450A
28,796 (kg)
10,94
315,03
​1.440,03
Table 11. Total cost of others.
35
Mechanical Engineering Area
Summary budget:
Budget
1. Main components ​10.330,39 ​€
2. Pipes
717,00 €
3. Enclosure
4.067,46 €
4. Others
1.440,03 €
-------------------------------------------------Total: 16.554,88 €
-------------------------------------------------
The total cost of our cold store is 16.554,88€ + staff workforce that will make it.
36
Mechanical Engineering Area
6. BIBLIOGRAPHY
_____________________________________________________________________________________________________
➢ WEBSITES
○ https://store.danfoss.com/es/es/Refrigeraci%C3%B3n/V%C3%A1lv
ulas/V%C3%A1lvulas-de-expansi%C3%B3n/V%C3%A1lvulas-de-e
xpansi%C3%B3n-el%C3%A9ctricas/V%C3%A1lvula-de-expansi%C
3%B3n-electr%C3%B3nica%2C-ETS-12-5/p/034G4212
○ https://www.electricautomationnetwork.com/es/danfoss-refrigerati
on/034g4212-danfoss-refrigeration-valvula-de-expansion-electric
○ https://www.pecomark.com/es/c/p/122376
○ https://www.bitzer.de/websoftware/
○ http://www.calculaconatecyr.com/bpfrio.php
○ http://www.generadordeprecios.info/obra_nueva/Fachadas_y_particio
nes/Fachadas_pesadas/Paneles_prefabricados_de_hormigon/FPP030
_Fachada_pesada_de_paneles_alveolare.html
○ http://www.generadordeprecios.info/obra_nueva/Aislamientos_e_imp
ermeabilizaciones/Aislamientos_termicos/Camaras_frigorificas/NAG
010_Aislamiento_termico_de_suelo_de_cam.html
○ https://www.salvadorescoda.com/tarifas/Puertas_Camaras_Tarifa_P
VP_SalvadorEscoda.pdf
○ http://www.sc.ehu.es/nmwmigaj/CartaPsy.htm
○ http://www.refrigerantes.mobi/herramientas/R450A_bubble.php
➢ BOOKS
○ ATECYR, Fundamentos de Refrigeración, 2015.
○ Apuntes de clase de EM 1044
37