scalding tank design

8/4/2019 Scalding Tank Design

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HEAT TRANSFER

(MBB 3033)

Khairin Ezzaty binti Abdul Rahman 11957

Mumtaz Hayati binti Abdul Rashid 12095

Nurul Syakila binti Iksan 12169Sakinah Haris 12183


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Problem Statement

The ABC Holding Sdn Bhd wants our

consultant group to design a completescalding tank system for slaughteredchicken to loosen their feather before

they are routed to feather pickingmachine with capacity of 1200 chickenper hour under following conditions


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Objectives

To design a complete scalding tanksystem for slaughtered chicken to loosentheir feather before they are routed tofeather picking machine according to the

prescribed parameters.


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Scope of Project

a) The mass flow rate of the make-up water that mustbe supplied to the scalding tank, in kg/sb) The length of the proposed scalding tank, L in mc) The velocity of the conveyer and thus the chicken

through the tank in m/sd) The total rate of the heat transfer from water tochicken, in KWe) The total rate of the heat transfer from steam pipeto the water, in KWf) The rate of heat loss from the exposed surfaces ofthe scalding tank by conduction and by convection, inKW.


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Scalding Tank Design

Detail Dimension

Thickness 2.5cm

Width 3m

Length 9m

Height 1m

No of conveyor 4

No of chicken per conveyor 25


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Isometric view of the scalding tank

Tc,in = 28oC

Th,in = 150oC

Th,out = 90oC


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More View


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Front View


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Top view


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Parameters:

Calculation

Rate: 1200 chickens/hour

Average temperature of water, Tavg:

90C

Thickness of steel tank: 2.5cm

Initial mass of chicken, mi,chicken: 2.2kg

Ambient temperature T

=35C

Final mass of chicken, mf,chicken:(15% 2.2kg )+2.2kg = 2.53kg

Local water supply temperature:

28CBody temperature of chicken,Tchicken: 33C

Tank: width = 3m, height=1m

Time taken to be immersed: 5mins Center-center distance betweenchickens: 30cm


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Assumptions

Diameter of chicken, Dchicken=20cm Tank is divided into 4 equal parts

Chicken carries out the propertiesof meat:

= 1050kg / m2

k = 0.476 W/m.K

= 0.13 x 10-6

Cp= 3.55 kJ/kg.K


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A) Mass flow rate of then make up water thatmust supplied:

t = 5mins = 5 x 60s = 300s

For 1 chicken;

Weight gain by chicken = Weight of waterloss from the tank

Therefore, the weight of water loss from tank= 15% x 2.2kg = 0.33kg

For every 5mins, 0.33kg water is loss fromthe tank

Calculation


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The tank is assumed to have 4 equal parts.Therefore, for every part, 100/4 = 25chickens.

Since the center-to-center distance requiredis 30cm, the length of the tank is 0.3m x 24 +2(0.3m) = 7.8 m

B) The length of the proposed scalding tank, L in m.


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Where the distance is multiplied by 24 is dueto the spacing between the chickens and 2 x

0.3m is due to extra spaces between thechicken and the tank.

The length of the proposed scalding tank, L =9.0 m

C) The velocity of the conveyer thus thechicken through the tank, V

V = length/time = 7.8m / 300s = 0.026 m/s


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D) The total rate of heat transfer from water tochicken.(using the prop of water 1st for thechicken)


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Using Tube bundle calculation, chicken represent thetube bundle.

St=0.3m Tavg= 90oC Cp=4206J/kgoC

A1= 0.1m Pr= 1.96 Ti=90oC

SL=0.6M =956.3kg/m3 Ts =33

oC

D=0.2m k=0.675W/m.K

078.02.03.0

026.03.0max

V

DS

SV

t

t

3.4780510315.0

2.0078.03.965Re5

max

DV

22566.121001.04 mAs


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100425

/466.190

675.0

2.0434.56)

066.5

96.1()196()3.47805.0(27.0

20

25.036.063.0

TL

C

NNN

CmWh

h

k

hLNu

skgT

DSVNmm ttchicken /277.3

3600

2.04.042.2100

CeeTTTT Ps

Cm

hA

isse

04206277.3

466.190566.12

91.80)9033(33)(

C

TTTT

TTTTT

o

is

is

isis 33.52

91.80339033ln

)91.8033()9033(

)()(ln

)()(ln

kWWQ

ThAQ s

19.12568.125192

33.52566.12466.190ln


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E) Total rate of heat transfer from steam pipe towater, kW. (method of heat exchanger)

Using LMTD method, Diameter for the tube=0.06m Assumption made;

Tc,i=28oC Tc,o=90

oC Th,i=150Oc

Th,o=90oC Tavg=(28+90)/2=59

oC

K=0.653 Pr=3.198 =984.8kg/m3

CP=4185kJ/kgoC =1.0969E-8

smEV

VAVm

m

TCmTCmQ

mean

meammeanh

h

hhphccpc

/38763.9

06.08.9841137.0

)90150()2890(11.0

2

,,


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72.5402280969.1

06.038763.9Re

E

EVD

)__(/466.190

/28.119

675.0

2.096.10)5315.0

50802.1()105.8

06.0198.372.54022(86.1

20

0

20

14.036.0

dfromvalueCmWh

CmWh

h

k

Lh

E

ENu

i

Ci

CmWU

hhU i20

0

/35.73

466.190

1

28.119

1111

C

TT

TT

TTTTT

ocih

icoh

ocihicoh 0

,,

,,

,,,,

ln99.60

90150

2890ln

)90150()2890(

)(

)(ln

)()(


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Assume F=1

kWQ

WTUAQ s

5.1997

199704999.605.806.01062.122ln


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F)Rate of heat loss from surface of scaldingtank by conduction and convection

Assume h = 800 W/m2C)

Area of the side = 3 x1 x2+9 x1 x2 =6m2

Heat loss from tank surface on the sides

Rtotal = x/kA + 1/ hoAo

= 0.25/ 14.9 x 24 + 1/ 800 x 24

= 7.5119 x 10-4

Q =T/Rtotal

= (90-35) / 7.5119 x 10-4

= 73217.31W

=73.22kW


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Surface (top) of scalding tank

Qconv =hAsT

= 800 x (9 x3) x (90 35)= 1188000 W

= 1188kW

Qtotal = 73.22+1188

= 1261.22 kW


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Recommendations Reasons

i. Using a higher temperature for the water

ii. Decrease the area of the scalding tank

iii. Reduce the volume of the water in the

scalding tank

iv. Increase the surface area of the tubes

To increase the heat transfer and lessen the time

consume

v. Enclose the tank

To reduce heat loss

vi. Attach an insulator at the tank

vii. The calculation must also consider the

fouling factor

To get a precise results and able to design the

scalding tank correctly

viii. Increase the number of tubes to 16 or more To avoid fouling factor

Recommendations


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Conclusion

The design finally achieved the objective to fulfil theparameters required with :

The working capacity of 1200 chicken per hour.

The length of 9m which is appropriate to be operatedinside a small poultry factory.

The mass flow rate for the make up water which is0.11kg/s

The velocity of the conveyor together with the hungchicken is 0.026m/s

scalding tank design

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