Sheet 6: Cold Stores1.Given

Solution

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2.Given

Solution

2

neglect

3. A cold store of dimensions m has an overall coefficient of heat transfer through

walls, floor, and ceiling of 0.2 . The outside condition is 30 DBT and 50% RH.

Five thousands kilograms of wet mixed vegetables (specific heat of 3.65 and

respiration heat of 0.092 ) are cooled from 25 to the storage temperature of 5

each day for 20 days. Consider that the total miscellaneous loads are 600 W. Neglect the effect of solar radiation. Compute the cooling load based on a 16 hr/day operating time for the equipment.

3- Given:

Dimensions: 10x5x4 m U=0.2 W/m2.K To=30 oC RHo=50%

Product data: Wet mixed vegetables m=5000 kg C1=3.65 kJ/kg. oC

Respiration heat=0.092 W/kg

T1=25 oC T2=Ti=5 oC

QM=600 W (Total miscellaneous load)

Neglect sun effect

Required: Equipment Load (based on n=16 hr/day “ operating time of equipment” )

Solution

1-Transmission Load (QTr): In general Q=U.Ao.∆T

Awalls =[(5+10) x 2] x 4 = 120 m2

Aceiling = Afloor = 10 x 5 = 50 m2

U=0.2x10-3 kW/ m2.K (given)

To=30 oC Ti=5 oC (given)

Tground=Tg=18 oC from Egyptian code (cooling store)

QTr = Qwalls + Qceiling + Qfloor + Qsun(max)

= U.Awall.(To-Ti) + U.Aceiling.(To-Ti)+U.Afloor.(Tg-Ti)

= 0.2 x 10-3 x 120 x(30-5) + 0.2 x 10-3 x 50 x(30-5)+ 0.2 x 10-3 x 50 x(18-5)

= 0.98 kW

2-Air Change Load (QAC) or Infiltration Load:

Interior volume =V = 10 x 5 x 4=200m3 (neglecting the walls, ceiling and floor thickness)

From table 5: (at V=200 m3 and room above 0 o C )

... Infiltration rate=13.9 L/s

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From table 6-A: (At room above 0 o C , Ti=5 oC, To=30 oC and RHo=50%)

... Enthalpy Change=0.0536 kJ/L

... QAC = Infiltration rate (L/s) x Enthalpy Change (kJ/L) = 13.9 x 0.0536 = 0.745 kW

3-Product Load (Qp): Qp,total=Qp + Qrespiration

= + m. (Respiration rate)

= + 5000 x = 4.68 kW

4- Miscellaneos Load (QM) :

QM=600 W= 0.6 kW (given)

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... Qtotal = QTr + QAC + Qp, total + QM

= 0.98 + 0.745 + 4.68 + 0.6 = 7.005 kW

Add S.F=10%

... Q cooling Load = 1.1 x Qtotal = 1.1 x 7.005 = 7.71 kW

... Required Equipment Load= Q Equipment = Q cooling Load x

= 7.71 x = 11.57 kW

= 11.57 x T.R = 3.3 T.R

4. A cold store of dimensions m is designed to cool 5 tons of milk of specific heat

3.77 from to in 18 hours for 1 week. There are 300 W light and 4

persons each produces 250 W. The overall coefficient of heat transfer through the room

is 0.58 , the outside air condition is DBT and 60% RH, and the number of

air changes is 8 per day. Determine the cooling capacity of the refrigerating unit in ton of refrigeration based on 18 hr/day operating time for the equipment.

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The maximum is from roof

4. Given:

Dimensions: 4x8x4 m

Product data: Milk m=5000 kg C1=3.77 kJ/kg. oC

T1=40 oC T2= 2 oC

Light load = 0.3 kW, No. of persons = 4, Equivalent heat per person=0.250 kW

U=0.58 W/m2.K To=40 oC RHo=60%

Number of air changes per day = 8

Required: Equipment Load (based on n=18 hr/day “ operating time of equipment” )

Solution

1-Transmission Load (QTr): In general Q=U.Ao.∆T

Awalls =[(4+8) x 2] x 4 = 96 m2

Aceiling = Afloor = 4 x 8 = 32 m2

U=0.58x10-3 kW/ m2.K (given)

To=40 oC Ti= T2= 2 oC (given)

Tground=Tg=18 oC from Egyptian code (cooling store)

*Sun effect from table (4) assume medium color

... ∆TE= ∆TW= 3 oC & ∆Troof=8 oC

QTr = Qwalls + Qceiling + Qfloor + Qsun(max)

= U.Awall.(To-Ti)+U.Aroof.(To-Ti)+U.Afloor.(Tg-Ti)+ )+U.Aroof.( ∆Troof)

= 0.58 x 10-3 x 96 x(40-2) + 0.58 x 10-3 x 32 x(40-2)

+ 0.58 x 10-3 x 32 x(18-2)+ 0.58 x 10-3 x 32 x(8) = 3.267 kW

2-Air Change Load (QAC) or Infiltration Load:

Interior volume =V = 4 x 8 x 4=128 m3 (neglecting the walls, ceiling and floor thickness)

... Infiltration rate=11.85 L/s

From table 6-A: (At room above 0 o C , Ti=2 oC, To=40 oC and RHo=60%)

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None

... Enthalpy Change = 0.1099 kJ/L

... QAC = Infiltration rate (L/s) x Enthalpy Change (kJ/L) = 11.85 x 0.1099 = 1.3 kW

3-Product Load (Qp):

Qp =

4- Miscellaneos Load (QM) :

QM= Qlight + Qpeople+ Qmotor

Assume (time in use = 3 hr/day)

=======================================

Add S.F=10%

Required Equipment Load

5. A chilling room of dimension m is used to chill 20 tons of meat from 40 to

0 in 18 hours. The overall heat transfer coefficient through the building is 0.29

and the specific heat of meat is . The total cooling load is 1.2 of

the sum of the product load and heat transmission load. Determine the cooling capacity of the refrigerating unit in ton of refrigeration based on 18 hr/day operating time for the equipment.

5- Given: Dimensions: 15x10x4 m

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The maximum is from roof

Chilling = cooling Product data: “meat” m=20 ton =20 x103 kg

T1=40 oC 18 hr T2=0 oC

C1 =3.14 KJ / kg KU=0.29 W/m2.K Qcooling load= = 1.2 [ QTr + Qp]

Required:

Q Equipment (based on n=18 h/day “ operating time of equipment” )

Solution

1-Transmission Load (QTr): In general Q=U.Ao.∆T

Awalls=[(15+10)x2]x4=200 m2

Aroof = Afloor = 15x10=150 m2

U=0.29x10-3 kW/ m2.K (given)

Ti= T2= 0 oC

To=38 oC from table (1) [Cairo, Summer, Tdb (2.5%)]

Tground=Tg=18 oC from Egyptian code (cooling store)

*Sun effect from table (4) assume medium color

... ∆TE= ∆TW= 3 oC & ∆Troof=8 oC

QTr = Qwalls + Qceiling + Qfloor + Qsun(max)

= U.Awall.(To-Ti)+U.Aroof.(To-Ti)+U.Afloor.(Tg-Ti)+ )+U.Aroof.( ∆Troof)

= 0.29 x 10-3 x 200 x(38-0) + 0.29 x 10-3 x 150 x(38-0)

+ 0.29 x 10-3 x 150 x(18-0)+ 0.29 x 10-3 x 150 x(8) = 4.988 kW

2-Product Load (Qp):

Product is (meat ) : m=20 ton =20 x103 kg , T1=40 oC 18 hr T2=0 oC

From table (9) Select Beef Tfreezing= -0.5 oC

... T2 > Tfreezing

C1=3.14 kJ/kg.K (Specific heat before freezing) only

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... Qp =

38.765 kW

=======================================

... Q cooling Load =1.2 [QTr +Qp ]

= 1.2 [4.988 +38.765 ]= 52.5 kW

... Required Equipment Load = Q Equipment = Q cooling Load x

= 52.5 x = 70 kW

= 70 x T.R = 19.91 T.R

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