scalding tank design
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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
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