Sheet (1)

1
1

?Discuss the main type of cold stores

.Seasonal cold store -1
.Distribution cold store -2
.Parts cold store -3
.Transport cold store -4
.Commercial cold store -5
.House hold cold store -6

2 -:Define
-: Volumetric loading rate .It represent the volume in m3 that will occupied
Area use factor :- it is the ratio
between the area occupied by the
product to the total area of the store
3 ? Discuss the method used to decrease both primary and operation cost
Primary cost :- since the building process represent 50 % of the primary
-:cost so we should decrease it is cost by
making all services builds and cold stores and freezing stores in are -1
.build instead of separated builds
.increase the area use factor -2
using of a prefabricated walls and isolating -3
.Panels for cooling stores and an ordinary building for services zones

Sheet 2
o

1) Calculate the Cooling load for a chilling room (4*8*3) m

o
maintained
at 2o C and 90 % R.H and designed to chill 5 ton of beef
from 38 C to 7 C in 24 hours, There are 2 occupants in the room . If
the temperature (Enclose to Solar connection ) opposite the Walls is 32
o
and above the roof is 42 o C. Overall heat transfer coefficient through
o
the room is 0.58 W/m 2 C . Assume the inside temperature is 2 C and
infiltration load is 10 % of the total load ??

Given :

Note : Always

TR = Tin = 2 C

L>W>H

So L = 8 m
W=4 m
H=3 m

, (R.H)R = 0.9

, T P i= 38 C , T Po = 7 C

.
m = 5 ton / 24 hour = (5*1000)/(24*3600) = 0.058 kg/s
No. of person = 2
Q inf

, T wall = 31 C , Troof = 42 C , h = 0.58 W/m .C

= 10 % of the total load , Req. : Calculate The Cooling Load ??

Solution :
Transmission heat gain
Q
Q
Q
Q

Tr

=UA T=Q

wall
roof
ceiling

wall

+Q

roof

+ Q ceiling = Transmission load = Q

=U

A T = 0.58 [(4+8) *2*3] (32-2) = 1252.8

Watt

=U

Watt

=U

W
R

A T
C

= 0.58[(8*4)] (42-2)
= 0.58(8*4) (18-2)

= 742.4

= 296.96 Watt

Wall
(L+W) * 2 * H
Roof
(L * W) = Ceiling ,
T 0 = 18 ( constant) For Beef
QTr=1252.8+742.4+296.96 = 2292.16 watt
kw 2.29=
-:Product load
Where the product "beef" is cool from 38 to7C so the load is
Above freezing
Q prod = mcpat where
Cpa = above freezing specific heat
to 3.4 = 3.14kj/kg.k (from table) 2.9 =
Qprod= (5*1000/24*3600)* 3.14 * (38-7)
Kg/s
k
*kj/kg.k*
kj/S-----=5.646 kw 5.646=

-:people load
from table. Q people=250 w
Assume No of working hours = 4
People = No of person* Q/per * (working hour/24)
(4/24) * 250 * 2=
watt 83.33 =

:Light load
Qlight= w/m2 * A * (working hour/24)
watt 53.33 = 4/24 * (4*8) * 10 =
where : 10 w / m2
Light intensity

-:Air changing load

QAC = N V a CPa T
-:Where
N: No of air change = 8
Cp = 1.005 Kj /kg.k
= 1.19 kg/m3
QACH = (8/(24*3600))(4*8*3)*1.19*1.005(32-2)
watt 318.9 =
Qtotal = Qtr + Qpeople + Qlight + QA.ch
318 + 53 + 83 + 5646 + 2292 =
watt = 8.392 kw 83.92 =
Qinf = 0.1 Qtotal
kw 0.839 = 8.392 * 0.1 =
Cooling Load = (Qtotal + Qinf)* 1.1*(24/16)
-:Where
safty factor = 0.1
working hours of unit = 16
Cooling load = (8.392 + 0.8392)*1.1*(24/16)
kw 15.23 =
C.L = (15.23/3.517) = 4.33 TR
OR
Qrunning = Qtotal * (24/16) =12.588 kw
C.L = Qrunning + Qf = 1.12 Qrunning
Qrunning = 14 kw

Const.

2- Calculate the Evaporator capacity for a Freezing room 1.75 * 2.5 *

o
2.5 m maintained at -18 C and 85 % RH . The room capacity is 2 ton
o
meat where its temperature is cooled from 5 to -12 C (take the daily
loading rate 10 % of the room capacity) Insulation Thickness 15 cm
with thermal conductivity 0.047 W/mo C . Air film coefficients for
o
inside walls is 6 and for outside surface 30 W/m C .Assume the outside
o
temperature is 27 C
Given :
V = (1.75 * 2.5 * 2.5) m
.
m = 2 ton

, TR m = -18 C , (RH) R = 0.85

daily loading rate = 0.1

.
m = (2*0.1*1000)/(3600*24) = 0.0023 kg/s
o

TProd)in = 5 C , TProd)out = -12 C , ins= 0.15 m , K ins = 0.047 W/m .C

2o

h In = 6 W/m

C , h Out = 30 W/m

2o

, To = 27 C

Req : Evaporator Capacity ???

Solution
1

U =

1/h in + ins / Kins + 1/h o

1
1/6 + 0.15/0.047 + 1/30

2o

U = 0.295 W/m C
Transmission heat gain = Q
+Q

+Q

Wall

Roof

Floor

= U A

T = 0.295 (2.5*1.75*4)(27-(-18)) = 232.2 Watt

= U A

T = 0.295(2.5*2.5)(27-(-18))

= 83

= U A

T = 0.295(2.5*2.5)(10-(-18))

= 51.6 Watt

Where TFloor

Wall

Roof

Floor

W
R
F

W
R

= 10 C ( Assume For Freezing Room)

Q

= 232.2 + 83 + 51.6 = 366.8 Watt

Watt

Product Load :
Where the product cooled from 5 to -12 C
5

Sensible

Tf

Tf

Latent

Sensible

-12

Ice

Water

QPr = Qab Fre + Q Fre + Qbelow Fre

= m Cp a ( 5 - T f ) + m L.H + m Cp b (T f -(-12))
From table meat :
Cpa = 3.14 kj/kg.k , Cp b = 1.7 kj/kg.k , L.H = 231.5 kj/kg
Tf

: Freezing temp. = -2.2 C ( From Table )

Qprod = 0.0023(5-(-2.2) * 3.14 + 0.0023*1.7 (-2.2+12) + (0.0023 *

231.5)
Q prod = 0.623 Kw

Light Load :
Q L = 10 ( 2.5 * 2.5 ) * 4/24 = 10.42 Watt
Air changing Load :
QAch = N * V * a * Cp a *T
= 20/(24*3600) * (2.5*2.5*1.75) * 1.16 * 1.005 * (27-(18))
QAch = 0.1328 KW

Qtotal = Q Ev. = 0.336 + 0.623 + 0.132 + 0.10 = 1.133 Kw

= 0.333 w / m2 .k .k
, Cp = 1.67 kj/kg.k
PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
PPPPPPPPPPPPPPPP

3-A cold plant consists of four rooms maintained at-30 C

and four rooms maintained at -5 C. Frozen products at -10 C are
cooled to -25 C in 18 hours at the rate of 50 tons for each room.
Fresh products are cooled from 25 C to 0 C in 18 hours at the rate
of 50 tons for each room. The specific heats of frozen and fresh
products are 2 and 3.35 kJ/kg respectively. The other loads estimated
for each room at -30 C are 6 T.R and 6.5 T.R for each room at -5 C.
Estimate the cooling load for the plant.
-:Given
Room at (-30 C)
No
=4
other load=6 T.R
Room at (-5 C)
No
=4
other load=6.5 T.R
frozen product
Tin = -10 C
,T= -25 C ,18 hr
3
m= (50*10 )/(18*3600) kg/s
,cp=2
fresh product
Tin=25 C ,T=0 C ,cp=3.35
3
m=(50*10 )/(18*3600)
Req cooling load for the plant
Solution
Qfrozen product = m cp T
((25-)-10-)*2*[(3600*18)/(103*50)]*4=
Kw 92.59=
Qfresh prod =4*[(50*103)/(18*3600)]*3.35 (25-0)
Kw 258.49=
Total load for rooms at -30 C
T.R 50.33 = (4*6) + (92.59/3.517)=
Total load for rooms at -5 C
T.R 99.5 = (4*6.5) + (258.49/3.517)=
Total capacity load = 50.33 + 90.5 =149.83 T.R

, Acold store 40 x 50 x 8 m with inside air temperature -20 C -4

outside temperature 40 C and below floor temperature 20 C is used to
store 3000 tons capacity. The insulation thickness (k = 0.02 w/m.k) for
ceiling and wall is 18 cm and for floor is 6 cm . The temperature solar
correction is 10 C for the ceiling and 5 C for the walls. If the daily
loading rate is 10 % of the store capacity and introduced to the store at
-10 C to be freezed to -15 C in 12 hours. Product specific heat 1.67
kJ/kg.k .If the service load and air change load amount to 20 % of the sun
, transmission and products load . Machine working hours 20 per day and
the cold store is epuipped with 8 indentical evaporatos. Find the
. evaporator capacity in kw
-:GIVEN
V = (W*L*H) =(40*50*8)
TRI = -20 C
,

TRo = 40 C

Tfloor = 20 C
M = 3000 * 103 kg
Kins = 0.02 w/m .k
,
Sins)floor = 0.06 m
Sins)ceil,wall 0.18 m
Tceiling = 10 C , Tw = 5 C)
m = 0.1M
Tprod )inlet = - 10 C
12 hr

, Tprod )out = - 15 C
Cp = 1.67 kj/kg.k

Air change = 0.2 ( QTran + Qpr )

Machine working time = 20 hr / day
No of evaporator (coil) = 8
Req : Evaporator Capacity

Solution
1- Transmission heat gain
QTr = U A T
Where:
Uw= Uceiling = (kw/ w) = (0.02 / 0.06) = 0.111 w / m2 .k
Ufloor = (kf / f) = (0.02 /0.06) = 0.333 w / m2 .k

Q = U Wall A Wall ((T R + T o) ST R )

i
Wall
Q = 0.111 * [(40+50) * 2 * 8] [(40+5)-(-20)]
Q Wall= 10389.6 Watt
Wall

Q ceiling = U ceilingA ceiling((T Ro+ T S) T Ri)

Q ceiling = 0.111 * (40*50) * [(40+10)-(-20)]
Qceiling = 15540 Watt
Q floor= Ufloor A floor T floor
Q floor= 0.333 ( 40 * 50 ) ( 20-(-20))
Q floor= 26640 Watt
Q Tr = 10389.6 + 15540 + 26640

= 52570 Watt = 52.57 Kw

Product Load :
Q

= m Cp T
Pro
= (3000 *10 )/(12
3 * 3600) * 1.67 (-10-(-15))
= 57.99 Kw

timum
ad:-PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

QServ = 0.2 ( QTr + QPro )

= 0.2 ( 52.57
+ 57.99 )
.
= 22.1 Kw
Qtotal = QTr + QPro+ Q Serv
Q total = 52.57 + 57.99 + 22.1 = 132.67 Kw

5- Discuss the common type of plate forms and elevators

?
type of plate forms :
1- open platform

2- closed platform

Type of Elevators :
1- outside elevator

2- inside elevator

6) It is Required to construct a prefabricated Cold Store Of

50,000 Ton capacity and 6.5 height For preservation of the
Products at a Port at -20/4 C. Draw The layout of cold store
showing the Location of Platforms , Machine Rooms, Offices,
Store in the Layout ??
Note 1 Ton = 1000 Kg
Given :
6
M=50.000 Ton = 50*10 Kg , H B = 6.5 C A
m = C A * A P , Let C A = 250 Kg/m2
A =
P

m
CA

= 50 * 10 /250 = 200.000 m 2

From The table For Large Store :

 a= 0.8 : 0.85 = 0.82

=
a

AP

AB = 200.000/0.82 = 243903 m

AB

As a Long Area We can divide The Store to 20 Identical Store

A ) Room = 243903/20 = 12195.2 m
B

Store

So Each Store has Area = 12195 m 2

Using Aspect Ratio

3:1

A * W * L = W * 3W = 3W
W=

12195.2

W = 64 m

L=3*W

L = 3 * 64 = 192 m
= 0.333 w / m2 .k .k
, Cp = 1.67
kj/kg.k
PPPPPPPPPPPPPP
PPPPPPPPPPPPPP
PPPPPPPPPPPPPP
PPPPPPPPPPP

A cold store of 800 tons capacity and 8 m height food products in (7

ballets of dimensions (L x W x H) 1.2 x 1 x 1.7 m and 0.75 ton Capacity .
Determine the floor load ( ton / m2 ) and the required area of cold store if
. the area use factor is 0.75
:Given
M = 800 ton
H = 8m
Ballets dimensions ( L x W x H ) = 1.2 x 1 x 1.7 m
mbell = 0.75 ton
area use factor a = 0.75

((Solution
No of ballets in vertical direction = H/Hdirection = 8/1.7 = 4 ballets

Floor load CA = m/Ap =( 4 * 0.75)/(1.2 * 1) = 2.5 ton/m2

M = CA * Ap
:Where
Ap = M/CA = 800/2.5 = 320 m2
a Ap/AB
AB = Ap/a = 320/0.75 = 427 m2
Required area for cold sore = 427 m2
Assume L : W = 3 : 1
A = W * L = W * 3W
W2 3 = 427
W = 12 m
L = 36 m

8) A Cold Store of 3000 Ton Storage Capacity consists of 6 Identical

rooms. If the Loading rate is 2 ton / m For the Floor Area and 1/3
ton/m for the cold store Volume. Determine The main dimensions of the
cold store??
If the Central distance Between Columns is a multiple of 6 m. Find the
Length and Width of each room and the Floor use Factor. Assuming the
corridor width 6 m. Sketch Plan & Side view of the cold store to Protect
the cold store of the Sun and to show the rooms, corridor, Platforms and
service area

Given :
M = 3000 Ton , No. of Rooms = 6 ( Identical)
C = 2 ton / m2 ( Floor)

Machine room
12 m 6 m 12 m

C = 1/3 ton / m3
V

Req. :

N
The Main Dim. Of the cold Store

Platform

Solution

M = CA * AP

AP = M / CA

AP = 3000/2 = 1500 m 2
From table
For Medium Store :
a = 0.75 : 0.8 = 0.75

AP

A = 1500/0.75 = 2000 m 2
B

AB

Area For Each Room = 2000/6 = 334 m

A = 3W 2

W = 11 m , L = 33 m

H P = CA / C V
Where :
Hp : Product height = 2/(1/3) = 6 m
HB

store building height

HB = HP + 0.5

HB = 6 + 0.5 = 6.5 m
CV = cold store volume or volumetric loading
Area use factor = a = area product / Area Building = AP/AB
Building hight HB = product height + 0.5
Capacity
m = CA*AP

AB = LB * WB
where
LB : WB = (1 : 1 , 3 : 1 , 9 : 1)
HP = CA/Cv = loading rate for floor area / loading rate for volume of
rate = (ton/m2) * (m3/ton) = m
Ap = Vp / Hp

Cold Stores :

1 Corck
2- Expanded Polystercne
3- Polyare thane

Discuss The Method used to calculate

the optimum thickness of the thermal
insulation based on the capital and
operation cost ??
First calculate (Cm) the total cost of Insulation
C m = [ A s * * P total/ Z ] * F m

20

30

40

70

3
Kg/m

Where :
Atot : The Summation of all surface area
al :
Insulation thickness
Pm
:
Price of Insulation material per LE / m
Fm
:
Maintenance Factor [ 1.5 - 2 ]
Z
:
No. Of Years

Cm

Q Tr. = U

= ( A * T ) / [ ( 1/h i) + ( 1/h o) + ( L/k ) + ( / k ins) ]

C.O.P = F * [T Ev / ( TCond - TEv ) ]
W = Q Tr. / C.O.P

C = W * H * PE * FE

Where :
H

No . Of hours in years

PE

Price of power

( L.E / Kw.h)

FE

Factor

( 1.5 - 2 )

C total = C E + C m

Optimum

Discuss the heat transfer rate to

Insulate the pipes And Tubes ??
Q =

Where :
do = di + 2

T ,h

Ts

Ts - T_
1
Ln ( do / d i )
+
h
2R in L

, Ao= d o L

From the Graph it Noted that :

When d o Increase
But ( Q ) it also increased due to the
increase in surface area untill reach
the point ( B ) then as ( ) increase
the heat rate ( Q ) decrease

Q
B
A
C
Insulation
Zone

di

o : For insulation of tanks and pipes we use thickness for the

zone between ( B ) And ( C )

do

Discuss the common types of vapor bntillers used in

cold stores ??
1- Asphalt

[ bitumen + sand ]
1 : 5 mm Thickness

2- Bitumen rolls

[ Sack or Aluminum ]
1 : 5 mm Thickness

Bitumen resin :
[ Bitumen
+ Asphalt +
40 - 50%
55 - 60%

Sand

Asphalt

: Bitumen + Sand

Bitumen amulise

: Asphalt + water

Bitumen rolist

: Sack + Aluminum

Bitumen resin

: bitumen + aspestos + sand

PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Discuss the common methods used to construct the wall

ceiling and roof of the used in cold stores wall ??
Jalvaried
Aluminum
Wood
Fiber glass
Stainless Steel

1- Prefabricated

Thermal
Insulation

2- Ceiling :
R.C
V.B
Thermal
Insulation
R .C
Finishing

V.B

:Common Building- 3

Finishing

Finishing

Single Wall

Double Wall
:Roof- 4
Sand
Thermal
Insulation
V.B
R .C
Finishing

Operation Cost:

We can decrease Operation Cost by :

1-Decrease The heat gain by Transmission through Walls and coiling by
use of gamelans and additional aluminum walls
Using Air curtains on doors to reduce the amount of incoming air- 2
3- Use a Closed type Plat forms and simplification of Pipe work

4) Draw The Common type of Cold Store Floor:

1-40 mm Asphalt

2-120 mm Plan
Concrete

3-Vapor Insulation

4-400 mm thermal
Insulation

5-30 mm Sand
6-Vapor Insulation
7-300 mm Concrete
Layer
8-Bvc types
(d=(100:250)mm)
High = Cost

6
d
7
8
(1)

(1) 120 mm Plan

Concrete
(2) Sand

1
2
3

(3) Solid Filling

(4) 100 mm Rein
Forced Concrete Layer
With embedded tubes

4
5

Note : This Type Is Cheap

And Wide use

1
1- 40 mm asphalt
2- 120 mm concrete
3- 600 mm gravel
4- 50 mm concrete
5- 50 mm reinforced
concrete
6- steel tube

1
2
3
4
5

NO(3)in sheet (2)

Givenpart (1)
mp=800toncapacity
identical rooms 4
anti rooms 4
wa=2m-----width of anti room
Floor-loading rate=F.L.R=2ton/m2
Volume-loading-rate=V.L.R=0.5 ton/m2
u= 0.8

floor use factor

?? = Determine the main dimension of cold store

Solutions

highet

H
(
H
4

1
2
1

3
1

wideh

eight=(F.L.R)/
V.L.R)=(2Ton/m2)/(0.5Ton/m2)
eight=4m
APR/AR=
=APR (area of room
product)
AR (area room) /
APR/AR=.8

4
1

0
lon
g

= F.L.R=total capacity (mp)/area store product (Ap)

Room capacity (mpr)/ area of room product (APR)
F.L.R=(mPR)/(APR)
Apr/200=2
Apr=200/2=100m2
Ar =Apr/4=100/980=125m2
AR = L W

*
2

* 125 m

=L W

So We Will To Assume many Value Of W And

When W = 10 m

L = 12.5 m . And H = 4 m

When W = 8m

L = 15.6 m .. And H = 4 m

When W = 9 m

L = 13.8 m .. And H = 4 m

When W = 7 m

L = 17.8 m .. And H = 4 m

Take Any dimension For The Room Say (12.5*10*4) =

(L*W*H)
So The dimension Of Cold Storage = (12.5*10*4)
Dimension Of Cold Storage
H = HR + H anti R
L= LR = 4 Room
W = WR

We can Note that cold Store has 3 dimension (H * W * L) So when we

Find H There Isn't Any problem and also W Because the height Equal to
The height Of Room plus the height of antiroom , to Find the length we
get it and multiple by 4 ( no of rooms )

U = Uc = U = 0.3W/m 2. k
f
w
o
= 40 C
o
o
W.b.T o = 31 C
D.b.T

o
T = 12 C
a
o
T =0 C
1

Orange Storage Type :

Q = 500 W \ room Floor Area
E
Q = 10 W \ room Floor Area
L
N

total

People

= 2 Person / Storage Room

= 275W/ Person , 4 Hr/ day

Operating Time = 16 Hr/ day

5 Evaporator
Required

Evaporator Cooling Capacities = ??

Solution

We Can Solving By Assuming that The Rooms (1) , (4) And the
rooms (2) , (3) Are Identical
how to Get Transmission hear Gain Load For Room (1) ,(4) :

celling

1
ROOM
1 OR 4

w=10m

2
1

H=4 m

W=40m

L=1.25
m

3
1

4
1

Tanti=10C

=Q +Q +Q

1,4

Where :
Q
Q

= U * A W* T W

= U * A C* T C

Q F = U * A F* T

Qw = u * Aw * Tw
Qw = 0.3 * [ (12.5 * 4) * (40 0) + (10 * 4) *(40 0) + (12.5 * 4) (0 -0)
+ (10 *4) * (12 -0) ] = 1224 w
Qc = u * Ac * Tc
Qc = 0.3 *(12.5 * 10) (40 0) = 25.5 w
Qf = 0.3 * (12.5 *10) (18 0) = 32.1 w

QTr1 = QTr4 = Qw + Qc +Qf =1224 + 25.5 + 32.1=1281.6w

How To Get Transimssion heat Ggin Load for room 2 ,3

QTr 2,3 = Qw + Qc + Qf
Qw = u + Aw + Tw
Qc = u + Ac + Tc
Qf = u +Af + Tf
Insulated Side

Q W= 0.3((12.5*4)(0-0)(12.5*4)(0-0)(10*4)(400)(10*4)(12-0))

m4

Q = 624 W
W

Q
Q
Q

= 0.3(12.5*10)(40-0)

= 1500 W

= 0.3(12.5*10)(18-0)

= 675 W

C
F

r2

=Q

r3

10 m
12.5 m

= 624+1500+675 = 2799 W Insulated Side

How to Get Solar heat Gain Load:

From table 4 At dark Cooled

Tse Tss
5

Flat roof

Tsw
5

Tsc
11

Q = U * AC * Tsc
sc
Q = 0.3(10*12.5)(11) = 412.5 W
sc
Q = U * AS * Tss
ss
Q = 0.3(10*4)(3)
sc
Q = U * AE * Tse
se
Q = 0.3(12.5*4)(5)
sc

= 36 W

= 75 W

Q = U * AW * Tsw
sw
Q = 0.3(12.5*4)(5)
sc

75 W