Sheet (1)

Discuss the main type of cold stores?1 -Seasonal cold store.

2 -Distribution cold store.3 -Parts cold store.

4 -Transport cold store.5 -Commercial cold store.

6 -House hold cold store.

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

-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-:

1 -making all services builds and cold stores and freezing stores in are build instead of separated builds.

2 -increase the area use factor.3 -using of a prefabricated walls and isolating

Panels for cooling stores and an ordinary building for services zones.

3

11

2

Sheet “ 2 ”

1) Calculate the Cooling load for a chilling room (4*8*3) m maintained at 2 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 and above the roof is 42 C. Overall heat transfer coefficient through the room is 0.58 W/m C . Assume the inside temperature is 2 C and infiltration load is 10 % of the total load ??

o

2 o

oo

oo

Note : Always L > W > H So L = 8 m W = 4 m H = 3 m

Given :

inRo

R

o oP Pi o

.

ooroof

2wall

inf

T = T = 2 C , (R.H) = 0.9 , T = 38 C , T = 7 C

m = 5 ton / 24 hour = (5*1000)/(24*3600) = 0.058 kg/s

No. of person = 2 , T = 31 C , T = 42 C , h = 0.58 W/m .C

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

QTr=1252.8+742.4+296.96 = 2292.16 watt=2.29 kw

Product load -:

Where the product "beef" is cool from 38 to7ºC so the load isAbove freezingQ prod = mcpaΔt whereCpa = above freezing specific heat

=2.9 to 3.4 = 3.14kj/kg.k (from table)Qprod= (5*1000/24*3600)* 3.14 * (38-7)

Kg/s *kj/kg.k* k=5.646 kj/S-----=5.646 kw

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

=2 * 250* ) 4/24( =83.33 watt

Solution :

Transmission heat gain

Q = U A Δ T = Q + Q + Q = Transmission load = Q

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

Q = U A Δ T = 0.58[(8*4)] (42-2) = 742.4 Watt

Q = U A Δ T = 0.58(8*4) (18-2) = 296.96 Watt Wall (L+W) * 2 * H Roof (L * W) = Ceiling , T = 18 ( constant) For Beef

TroofwallTr

wall WWW

RRRroof

ceiling

CCceiling C

0

Light load : Qlight= w/m2

* A * (working hour/24) =10* ) 8*4( * 4/24 = 53.33 watt

Const.where : 10 w / m2 Light intensity

Air changing load-: QAC = N V ρa CPa ΔT

Where-:N: No of air change = 8Cp = 1.005 Kj /kg.k ρ = 1.19 kg/m3

QACH = (8/(24*3600))(4*8*3)*1.19*1.005(32-2) =318.9 watt

Qtotal = Qtr + Qpeople + Qlight + QA.ch

=2292 + 5646 + 83 + 53 + 318 =83.92 watt = 8.392 kw

Qinf = 0.1 Qtotal

=0.1 * 8.392 = 0.839 kw

Cooling Load = (Qtotal + Qinf)* 1.1*(24/16)Where-:

0.1 = safty factor16 = working hours of unit

Cooling load = (8.392 + 0.8392)*1.1*(24/16) =15.23 kw

C.L = (15.23/3.517) = 4.33 TR OR

Qrunning = Qtotal * (24/16) =12.588 kwC.L = Qrunning + Qf = 1.12 Qrunning

Qrunning = 14 kw

2- Calculate the Evaporator capacity for a Freezing room 1.75 * 2.5 * 2.5 m maintained at -18 C and 85 % RH . The room capacity is 2 ton 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/m C . Air film coefficients for inside walls is 6 and for outside surface 30 W/m C .Assume the outside temperature is 27 CGiven :

V = (1.75 * 2.5 * 2.5) m , T = -18 C , (RH) = 0.85

m = 2 ton daily loading rate = 0.1

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

T = 5 C , T = -12 C , δ = 0.15 m , K = 0.047 W/m .C

h = 6 W/m C , h = 30 W/m C , T = 27 C

Req : Evaporator Capacity ???

o

o

o

o

o

o

oo o

ooo2 2In Out o

RR m

.

.

ins insProd)in Prod)out

Solution

U = 1

1/h + δ / K + 1/hinsinsin o

=1

1/6 + 0.15/0.047 + 1/30

U = 0.295 W/m C2 o

Wall Roof Floor

Wall

Roof

Floor

W W W

RR R

FFF

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

5 T T -12ffSensible SensibleLatent

Q = Q + Q + Q = m Cp ( 5 - T ) + m L.H + m Cp (T -(-12))

From table “ meat ” :Cp = 3.14 kj/kg.k , Cp = 1.7 kj/kg.k , L.H = 231.5 kj/kg

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

Q = 0.0023(5-(-2.2) * 3.14 + 0.0023*1.7 (-2.2+12) + (0.0023 * 231.5)

ab FrePr Fre below Fre

a f fb

Q = 0.623 Kwprod

prod

f

O

O

a b

Water Ice

Transmission heat gain = Q + Q + Q

Q = U A ΔT = 0.295 (2.5*1.75*4)(27-(-18)) = 232.2 Watt

Q = U A ΔT = 0.295(2.5*2.5)(27-(-18)) = 83 Watt

Q = U A ΔT = 0.295(2.5*2.5)(10-(-18)) = 51.6 Watt

Where TFloor = 10 C ( Assume For Freezing Room)

Q = 232.2 + 83 + 51.6 = 366.8 Watt

3-A cold plant consists of four rooms maintained at-30 °Cand 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.RRoom at (-5 °C) No =4 other load=6.5 T.R

frozen product Tin = -10 °C ,Tо= -25 °C ,18 hr m˙= (50*103)/(18*3600) kg/s ,cp=2

fresh product Tin=25 °C ,Tо=0 °C ,cp=3.35 m˙=(50*103)/(18*3600)

Req cooling load for the plant

Light Load :Q = 10 ( 2.5 * 2.5 ) * 4/24 = 10.42 Watt

Air changing Load :Q = N * V * ρ * Cp *ΔT

= 20/(24*3600) * (2.5*2.5*1.75) * 1.16 * 1.005 * (27-(-18))

Q = 0.1328 KW

a aAch

Ach

Q = Q = 0.336 + 0.623 + 0.132 + 0.10 = 1.133 Kw = 0.333 w / m2 .k .k , Cp = 1.67 kj/kg.k

total Ev.

L

Solution

Qfrozen product = m cp ∆T =4)]*50*103)/(18*3600*[(2-)*10-)-25((

= 92.59 Kw

Qfresh prod =4*[(50*103)/(18*3600)]*3.35 (25-0) =258.49 Kw

Total load for rooms at -30 °C )=92.59/3.517) + (6*4 = (50.33 T.R

Total load for rooms at -5 °C )=258.49/3.517) + (6.5*4 = (99.5 T.R

Total capacity load = 50.33 + 90.5 =149.83 T.R

4 -Acold store 40 x 50 x 8 m with inside air temperature -20 ºC, outside temperature 40 ºC 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 ºC M = 3000 * 103 kgKins = 0.02 w/m .k , Sins)floor = 0.06 m

Sins)ceil,wall 0.18 m) ΔTceiling = 10 °C , ΔTw = 5 ºC

m = 0.1M

Tprod )inlet = - 10 C , Tprod )out = - 15 C 12 hr , Cp = 1.67 kj/kg.k

Air change = 0.2 ( QTran + Qpr )Machine working time = 20 hr / dayNo of evaporator (coil) = 8Req : Evaporator Capacity

Solution

1- Transmission heat gain QTr = U A ΔT

Where:Uw= Uceiling = (kw/ δw) = (0.02 / 0.06) = 0.111 w / m2 .kUfloor = (kf / δf) = (0.02 /0.06) = 0.333 w / m2 .k

Q = U A ((T + Δ T ) – T )Q = 0.111 * [(40+50) * 2 * 8] [(40+5)-(-20)]Q = 10389.6 Watt

Q = U A ((T + Δ T ) – T )Q = 0.111 * (40*50) * [(40+10)-(-20)]Q = 15540 Watt

Q = U A Δ TQ = 0.333 ( 40 * 50 ) ( 20-(-20)) Q = 26640 Watt

Q = 10389.6 + 15540 + 26640 = 52570 Watt = 52.57 Kw

S

S

R R

RRo

o i

i

Wall Wall Wall

Wall

Wall

ceiling

ceiling

ceiling

floor

floor

floor

floor floor

ceiling ceiling

Tr

floor

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

Product Load : Q = m Cp ∆T = (3000 *10 )/(12 * 3600) * 1.67 (-10-(-15)) = 57.99 Kw

timum ad:-

.Pro

3

Q = 0.2 ( Q + Q ) = 0.2 ( 52.57 + 57.99 ) = 22.1 Kw

Q = Q + Q + Q

Q = 52.57 + 57.99 + 22.1 = 132.67 Kw

.ProServ Tr

Tr Pro Servtotal

total

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 :M=50.000 Ton = 50*10 Kg , H = 6.5 C6

B A

m = C * AA P , Let C = 250 Kg/mA2

A = = 50 * 10 /250 = 200.000 mm

CAP

26

From The table For Large Store :

7 (A cold store of 800 tons capacity and 8 m height food products in 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 = 8mBallets dimensions ( L x W x H ) = 1.2 x 1 x 1.7 mmbell = 0.75 ton area use factor τa = 0.75

Solution) )

No of ballets in vertical direction = H/Hdirection = 8/1.7 = 4 balletsFloor load CA = m/Ap =( 4 * 0.75)/(1.2 * 1) = 2.5 ton/m2

M = CA * Ap Where :

ƞ = 0.8 : 0.85 = 0.82a

ƞ =a

A

AP

B

A = 200.000/0.82 = 243903 mB2

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

A ) = 243903/20 = 12195.2 m 2

B Room Store

So Each Store has Area = 12195 m 2

Using Aspect Ratio 3 : 1

A * W * L = W * 3W = 3W 2

W = 12195.2

3W = 64 m

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

Ap = M/CA = 800/2.5 = 320 m2 τa Ap/ABAB = Ap/τa = 320/0.75 = 427 m2 Required area for cold sore = 427 m2

Assume L : W = 3 : 1 A = W * L = W * 3W

427 = 3 W2

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

HB store building height

HB = HP + 0.5

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

Machine room

Platform

12 m 12 m6 m

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

C = 1/3 ton / m

A

V

2

3

Req. : The Main Dim. Of the cold Store

N

SolutionM = C * AA P A = M / CAP

A = 3000/2 = 1500 m 2P

From table For Medium Store :ƞ = 0.75 : 0.8 = 0.75

a

a APA B

B

ƞ = A = 1500/0.75 = 2000 m

Area For Each Room = 2000/6 = 334 m2

A = 3W

2

2 W = 11 m , L = 33 m

H = C / C

Where : Hp : Product height = 2/(1/3) = 6 m

P VA

AB = LB * WBwhere 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) = mAp = Vp / Hp

Cold Stores : 1 – Corck2- Expanded Polystercne3- Polyare thane

20 30 40 70

K

ρ

Kg/m3

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 = [∑ A * δ * P / Z ] * Fmm s total

Where : ∑ A : The Summation of all surface area δ : Insulation thicknessP : Price of Insulation material per LE / m F : Maintenance Factor [ 1.5 - 2 ]Z : No. Of Years

δ

Cm

mm

total

Q = U A ∆T

= ( A * ∆T ) / [ ( 1/h ) + ( 1/h ) + ( ∑ L/k ) + ( δ / k ) ]

C.O.P = F * [T / ( T - T ) ]

W = Q / C.O.P

EvEv Cond

Tr.

Tr.

i o ins

C = W * H * P * FEE

Where :

H : No . Of hours in years

P : Price of power ( L.E / Kw.h)

F : Factor ( 1.5 - 2 )

E

E

C = C + CE mtotal

δ Optimum

T , h∞

d

Ts

δ

Discuss the heat transfer rate to Insulate the pipes And Tubes ??

Q = T - Ts ∞_

+1h

Ln ( d / d )

2пR Lin

o i

Where :d = d + 2 δ , A = п d Lo i oo

A

B

C

Q

d d oi

Insulation Zone

From the Graph it Noted that : When d Increase

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

δ : For insulation of tanks and pipes we use thickness for the zone between ( B ) And ( C )

o

o

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 + Sand ] 40 - 50% 55 - 60%

Asphalt : Bitumen + Sand

Bitumen amulise : Asphalt + water

Bitumen rolist : Sack + Aluminum

Bitumen resin : bitumen + aspestos + sand

Discuss the common methods used to construct the wall ceiling and roof of the used in cold stores wall ??

1- Prefabricated Jalvaried Aluminum Wood Fiber glass Stainless Steel Thermal

Insulation

2- Ceiling :R.C

Thermal Insulation

Finishing

R .C

V.B

V.B

3 -Common Building:

Double Wall Single Wall

4 -Roof:

Sand

Thermal Insulation

Finishing

R .C

FinishingFinishing

V.B

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

2 -Using Air curtains on doors to reduce the amount of incoming air 3- Use a Closed type Plat forms and simplification of Pipe work

4) Draw The Common type of Cold Store Floor:

1

d

2

3

4

5

6

7

8

(1)

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

1

5

4

3

2

1

5

4

3

2

11- 40 mm asphalt2- 120 mm concrete3- 600 mm gravel4- 50 mm concrete5- 50 mm reinforced concrete6- steel tube

6

(1) 120 mm Plan Concrete

(2) Sand

(3) Solid Filling

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

Note : This Type Is Cheap And Wide use

NO(3)in sheet (2)Given—part (1)mp=800ton—capacity

4 identical rooms4 anti rooms

wa=2m-----width of anti roomFloor-loading rate=F.L.R=2ton/m2

Volume-loading-rate=V.L.R=0.5 ton/m2

ηu= 0.8 floor use factor

Determine the main dimension of cold store?? =

Solutions

Height=(F.L.R)/(V.L.R)=(2Ton/m2)/(0.5Ton/m2)Height=4m4=APR/AR

η =APR (area of room product)

1

21 3

1 41wideh lon

g

highet

/AR (area room)0.8=APR/AR

F.L.R=total capacity (mp)/area store product (Ap)= Room capacity (mpr)/ area of room product (APR)F.L.R=(mPR)/(APR)

2=200/AprApr=200/2=100m2

Ar =Apr/4=100/980=125m2

AR = L W

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

……… 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

2 *

*

When W = 7 m L = 17.8 m …….. And H = 4 mth

gieh

1

2

34

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 = U = U = 0.3W/m . kw c f

2

D.b.T = 40 Co

oW.b.T = 31 C

oo

T = 12 Co

a

T = 0 Co

1

Orange Storage Type :

Q = 500 W \ room Floor AreaE

LQ = 10 W \ room Floor Area

N = 2 Person / Storage Roomtotal

Q = 275W/ Person , 4 Hr/ dayPeople

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) :

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 wQc = u * Ac * ∆Tc Qc = 0.3 *(12.5 * 10) – (40 – 0) = 25.5 wQf = 0.3 * (12.5 *10) – (18 – 0) = 32.1 w

Q = Ʃ Q + Q + Q W C FП 1,4

Where :

WQ = U * A * ∆T

W W

C CC

FFF

Q = U * A * ∆T

Q = U * A * ∆T

121 3

141

W=40m

H=4 mROOM1 OR 4

L=1.25m

w=10m

Tanti=10°C

celling

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 + QfQw = u + Aw + ∆TwQc = u + Ac + ∆TcQf = u +Af + ∆Tf

Insulated Side

12.5 m

10 m

4 m

Q = 0.3((12.5*4)(0-0)(12.5*4)(0-0)(10*4)(40-0)(10*4)(12-0))

W

WQ = 624 W

Q = 0.3(12.5*10)(40-0) = 1500 WC

FQ = 0.3(12.5*10)(18-0) = 675 W

Q = Q = 624+1500+675 = 2799 WП r2 П r3

How to Get Solar heat Gain Load:

From table 4 At dark Cooled E S W Flat roof

∆Tse ∆Tss ∆Tsw ∆Tsc 5 3 5 11

Insulated Side

Q = U * AC * ∆Tscsc

Q = 0.3(10*12.5)(11) = 412.5 Wsc

Q = U * AS * ∆Tssss

Q = 0.3(10*4)(3) = 36 Wsc

Q = U * AE * ∆Tsese

Q = 0.3(12.5*4)(5) = 75 Wsc

Q = U * AW * ∆Tswsw

Q = 0.3(12.5*4)(5) = 75 Wsc