Traditional Dry Fruits

•NAME OF COMPANY :TRADITIONAL DRIED FRUIT
•DATE OF ESTABLISH :16.05.2013
•ADDRESS :Yeni sanayi sitesi 34/2 blok no=14 Kahramanmaraş
• WEB :www.tdf.com
•PHONE NUMBER :03442361232
Sketchup and plant layout=Nurcan İzol
Pipe and pump=Zehra Gökalp
Conveying = F.Zehra Doğan
Dryer = Şeyma Narlıoğlu
Tank = Fatma Koç
Cold storage= Nursel Bilici
•INTRODUCTION
Four kinds of raw materials for dried fruits.
•Apricot
•Persimmon
•Mulberry
•Grape
•RAW MATERIALS PROPERTIES
Fruits must be :
Mature
Good
Unspotted
No buggy
Also fruits must be dried at true season
•FINISHED PRODUCT PROPERTIES
Lower moisture content
Bright color
High nutritonal value
Texture
•MASS BALANCE
v For Apricot
280kg/hr
39.2kg/hr
240,8kg/hr 13.48kg/hr

85.3% 25%m.c.
227.32 A
kg/hr
W
•APRICOT
v227.32kg/year×85.3/100=W + 0.25A
v227,32kg/year=W + A
v193.9 kg/hr = 227.32-A+0.25A
A=44,56kg/hr
•GRAPE
vFor grape
560kg/hr 56kg/hr
77.8% m.c. 16.5%m.c.
504kg/hr A
W

133.98kg/hr 0.2679 kg/hr

•GRAPE
v504kg/hr x 0.778=A x 0.165 + W
v
v504=W + A
v392.112=0.165A+504xA
vA=133.98 kg/hr
v133.98kg/hr – 0.2639 kg/hr =133.716kg/hr
•MULBERRY
vFor mulberry
370 37.03kg/hr
kg/hr
72.2% m.c. 12%m.c.
332.9kg/hr A
W
v332.9x 0.722= W+ Ax0.12
v332.9kg/hr= W + A
v240.39=332.9 – 0.12 A
vA=105.12kg/hr
•PERSIMMON
vFor persimmon
833.33 141.6kg/hr
kg/hr
75.2%m.c. 20%m.c.
691.6 kg/hr A
W
v691.6×0.722 = W + Ax0.2
v691.6=W + A
v 520.08= 691.6- 0.8 A
v A=214.396kg/hr
•GANTT CHART
•PIPE AND PUMP
•Zehra gökalp
area= ΠD2/4
area= Π(0,152)/4=0,018 m2
D=0,15 m
V=q/A
q=2,77×10^-4
V= 0,015m/s
•Nre= DgV/µ
• g=density=997,54 kg/m3 at 20-25ºC from gean coplis
Nre=(0,15m*997,54 kg/m3 *0,015m/s)/0,936×10^-3=2397.9
Nre>2100 for turbulent flow
E/D=4,6×10^-5/0,15=3.067×10^-4 L=3,32 m
f = 0,0115 from gean coplis
•From tank to pump
Friction
Σ= 4fΔLV2/D2 +KeXV2 1 /2 + KcV2/2a+KfV2/2
=(4×(0,0115) ×(3,32 m) × (0,015 m/s)2 )/(0,15m×2 )
= 1,14j/kg
3 Elbow in 0,15 diameter
hf= KfV2/2= 3× 0,75 × (0,015 m/s)2 /2=2,53 × 10^-4 j/kg
From tank to pump
Contraction
Kc= 0,55×(1-A2/A1)=0,55
hc= KcV2/2=0,55× (0,015 m/s)2 /2a= 6,1875 ×10^-5 j/kg
From 0,15 pipe to 0,07 m pipe(contraction)
Kc= 0,55×(1-A2/A1)=0, 43
A1= Π (0,15m)^2/4=0,0176m2
A2= Π(0,07 m)^2/4=3,84×10^-3 m2
hc= KcV2/2=0,43 × (0,072m/s)^2/2=1,1×10^-3 j/kg
From pipe to washing
area= ΠD2/4
area= Π(0,072)/4= 3,84×10^-3 m2
D=0,07 m
V=q/A
q=2,77×10^-4
V= 0,072m/s
•Nre= DgV/µ
• g=density=997,54 kg/m3 at 20-25ºC from gean coplis
µ= 0,936×10^-3 from gean coplis
Nre=(0,07m*997,54 kg/m3 *0,072m/s)/0,936×10^-3=5371.4
Nre>2100 for turbulent flow
E/D=4,6×10^-5/0,07=6,6×10^-4
f = 0,008 from gean coplis
Friction in 0,07 diameter pipe
Σ= 4fΔLV2/D2 +KeXV2 1 /2 + KcV2/2a+KfV2/2
=(4×(0,008) ×(2,3 m) × (0,072m/s)2 )/(0,07m×2 )
=2,72×10^-3 j/kg
Exponential from pipe to air
KeX= (1-A1/A2)^2=(1-0)^2=1
hex= KeXV2 1 /2 =(1) ×(0,072m/s)2 /2=2,592×10^-3 j/kg
ΣF=1,15 j/kg
Pump
H= 1/2a×(V2 2 – V2 1 )+g(z2-z1)+(P2-P1)/d+ΣF+Ws=0
P1=dgh+Pa
=997,54 kg/m3 ×(9,8 m/s2) ×(8,5 m) +101,32 pa
=83196 pa
P2=101,32 pa
H=1/2×(0,072^2-0)+9.81(0m)+(101,32-83196pa)/997,54 kg/m3 +1,15j/kg+Ws=0
Ws=82,15 j/kg
Ws=-nWp Wp=117,35j/kg
pump w=(0,277 kg/s*117,35j/kg)=32,50 watt
•CALCULATION OF CONVEYING
Fatma Zehra Doğan
•Conveyor for raw materıals
•Carrying capacity of the conveyor
•T = a*b*v
•T: carrying capacity, (mass/time, t/h, kg/s….)
•a:bo2/11m2 =(0,85)2 /11 = 0,066 m2
•b: 540.82 kg/m3
•v:0,11 (m/s)
T=0,066m2 × 540,82 kg/m3 × 0,11m2
T=3,93 kg/s
The force required to move the empty belt, Ne:
Ne = total weight of belt on idlers x friction coefficient
Ne = mi*lt* g* µe
mi= 3,4 kg/m2 × 0,85 m =2,89 kg/m
Ne =2,89 kg/m × 6 m × 9,81 m/s2 × 0,3 × 0,11m/s=51,03
We=Ne × V=51,03 × 0,11 m/s =5,6133 W
Wm = mm*lt*g*µm *v
mm = T/v = (3,93 kg/s )/ (0,11 m/s)=35,72 kg/m
Wm=35,72 kg/m × 6 m × 9,81 m/s2 × 0,35 × 0,11m/s =80,96W
•WT = We + Wm ± Wr
•Wr=0
•Wt =86,57 W
Eff=90%
MC or Power = Wt/ Eff
=86,57/0,9 =96,19 W
•for seed separation of apricot
•T = a*b*v
•T: carrying capacity, (mass/time, t/h, kg/s….)
•a:bo2/11m2 =(0,85)2 /11 = 0,066 m2
•b: 533,545kg/m3
T= 3tone/hr × hr/3600s × 1000 kg/ton=0,85 kg/s
0,85 kg/s =0,066 m2 × 533,545kg/m3 × V
V=0,0236 m/s
Ne = total weight of belt on idlers x friction coefficient
Ne = mi*lt* g* µe
mi= 3,4 kg/m2 × 0,85 m =2,89 kg/m
Ne =2,89 kg/m × 6 m × 9,81 m/s2 × 0,3 =51,03 kg/ms2
We=Ne × V=51,03 ×0,0236 m/s =1,2 W
Wm = mm*lt*g*µm *v
mm = T/v = (0,85 kg/s)/ (0,0236 m/s)=36,01 kg/m
Wm=36,01kg/m × 6 m × 9,81 m/s2 × 0,35 × 0,0236m/s =17,5W
Wt=18,7 W
Power = 18,7/0,9=23,1 W
•For white light belt
•T = 666kg/13,32min = 0,83 kg/s
•T = a*b*v
•a:bo2/11m2 =(1)2 /11 = 0,09 m2
•b: 449kg/m3
0,83 kg/s =0,09 m2 × 449kg/m3 × V
V=0,02 m/s
Ne = mi*lt* g* µe
mi= 1kg/m2 × 1 m =1 kg/m
Ne =1 kg/m × 5,5 m × 9,81 m/s2 × 0,3 =16,2 kg/ms2
We=Ne × V=16,2 ×0,02 m/s =0,324 W
Wm = mm*lt*g*µm *v
mm = T/v = (0,83 kg/s)/ (0,02 m/s)=41,5kg/m
Wm=41,5kg/m × 5,5 m × 9,81 m/s2 × 0,35 × 0,02m/s =15,7W
Wt=16,02W
•for light =50 W
•Wt=66,02 W
•Power = 66,02/0,9=73,35 W
•CALCULATIONS OF ELEVATOR
330 packet/min = 165 kg/min × min/60s=2,75 kg/s
T = (c*b*v)/p
c: volume of material in each bucket, (m3)
b: bulk density of material
v: speed of elevator
p: bucket spacing, space between buckets, (m)
2,75 kg/s = (0,02m3 ×449 kg/ m3 ×V )/ 0,35m
V = 0,107 m/s
WT = 2*T*g*h*(1.5) = 2 ×2,75kg/s ×9,81 m/s2 ×3,8m ×1,5 = 307,5 W
* Drive efficiency of motor is assumed as 75 %.
Motor power = WT/Efficiency = 307,5W/0,75 = 410W
•DRYER
Şeyma Narlıoğlu
CALCULATION OF DRYING
exit air inlet air
T2=42oC T1=70oC H=0.008kg H2O/kg DA
H=0.021kg H2O/kgDA V=12m/s
Feed=1280kg DS/hr Product=280kg DS/hr
T=25oC T=42oC
Xwt=kg H2O/kg DS xp=0.8kg H2O/kg DS
RH %=40% at 25oC
1120 kg fresh persimmon *(75/100) = 840 kg H2O in persimmon
840 kg H2O/280 kg DS = 3
(842kg H2O*0.2)/0.75 = 224 kg H2O in final product
Basis:
feed: 1120 kg wet solid/hr
1120 kg wet solid/hr*(3 kg H2O/(1+3kg wet solid))
1120 kg wet solid/hr-840 kg H2O = 280 kg DS
MOISTURE BALANCE AROUN DRYER
G1 = ? Kg DA
Ls = 280 kg DS/hr
Moisture in inlet air:
G*Hi = G*0.008 kg H2O/ kg DA = 0.008G H2O/hr
Moisture in inlet feed:
Ls*xwf = 280 kg DS/hr*3 kg H2O/kg DS = 840 kg H2O/hr
Mouisture in outlet air:
G*Ho = G*0.0021 kg H2O/hr = 0.021G kg H2O/hr
Moisture in outlet feed:
Ls*xwp = 280 kg DS/hr*0.8 kg H2O/hr
MOISTURE IN = MOISTURE OUT
0.008G + 840 = 0.021G + 224 G = 47384.6 kg DA/hr
HEAT BALANCE AROUND DRYER
Tref = 0oC and ∆ToC = ∆TK by using eqn. :
H1*g = c1*(Tg-Td)+H*λo
AT datum temp. T = 0 λo = 2501kj/kg (from Gean koplis)
For inlet air :
Hi = (1.005+1.88*0.008)*(70-0)+(0.008*2501) = 91.41kj/kg DA
For exist air :
Ho = (1.005+1.88*0.021)*(42-0)+(0.008*2501) = 63.87 kj/kg DA
ENTHALPY OF WET SOLID
cp of dried food 3.8kj/kgK
Hıs = cps*(Ts-To) +xA*cpg(Tg-To)
= 3.8kj/kgK(25-0)K+(3 kg H2O/kg DS)*(4.187kj/kgK)*(25-0)
H1feed : 409.025 kj/kg DS
H1feed : 409.025 kj/kg DS
H1product : 3.8kj/kgK*(42-0)+(0.8 kg H2O/kg DS)*(4.187kj/kgK)*(42-0)
= 300.283 kj/kg DS
HEAT IN = HEAT OUT
G*Hi+Ls*Hifeed = G*Hıo+Ls*Hısproduct+Qloss
47384 kg DA/hr*(91.41 kj/kg DA)+(280 kgDS/hr)*(409.025 kj/kg DS) =
47384 kg DA/hr*(63.87 kj/kg DA)+(280 kgDS/hr)*(300.283 kj/kg DS)+Qloss
Qloss = 1090.73 kW
η = Qout/Qin = 0.7 = 70%
CALCULATİON OF HEAT EXCHANGER
m= 47384 kgdryair/h
Cp for air= 1.0048 kj/kgK
Q= m*cp*∆T=(47384*1.0048*(70-420C))=
= 1.3×10^6 kj/h
Q= m*cp*∆T = m* ∆hfg
∆hfg= 2188.51 kj/kg for 250C
1.3×10^6 kj/h = ms* 2188.51 kj/kg
ms= 609.15 kg/h
NATURAL GAS
1 kcal= 4.18 kj
1.3×10^6 kj/h * 0.24 kcal/kj
= 319952 kcal/h
Price= 0.397$/m3
= (0.397*1000/8250)*(0.93)= 0.045$/1000 kcal
(3.2×10^5 kcal/hr)*(0.045$/kcal)= 14.4 $/h
14.4 $/h * 720h/month = 10366.5$
•CALCULATİON OF DRYİNG TİME
Inlet air = 70oC and H= 0.008kgH2O/kgdryair
From psychometric chart;
T= 28.9OC
Twetbulb= 0.026 kgH2O/kgdryair
Vh= (2.83×10^-3+4.56×10^-3*H)*T
= 2.83×10^-3+4.56×10^-3*0.008)*(273+70)
= 0.983 m3/kgdryair
Density for 1 kgdryair + 0.008 kgH2O
density= (1+0.008)/(0.983)= 1.025kg/m3
The mass velocity= 12 m/s
G= Velocity*Density
= 12*1.025*3600= 44,280 kg/hm2
h= 0.0204*(G^0.8)= 106.315 W/m2K
Twetbulb= 28.9 0C
From steam table; ƛw= 2433 kj/kg
Rc= (h/ƛw)*(T-Tw)*3600=
= (106.315/2433)*(70-30)*3600= 6.3kg/hm2
Area= 8m*1.25m= 10.02 m2
Total rate= Rc*A= 6.3*10.02= 63.04 kg H2O/h
t=(Ls/A*Rc)*(x1-x2)=
= (280/63.04)*(3-0.8)= 9.7 h
•Tank CALCULATIONS
Fatma Koç
•capacity:90 tone
•90 tone=90000 kg
•Volume of cylinder=П(D/2)2×h h=4/3 D
• (D2/4)×(4/3)D= П D3/3
•90000 kg*m3/920 kg=97.83 m3
•97.83 m3= П D3/3 D=6.4m h=8.5 m
•Workıng pressure
PTOTAL=Poptimum+Hp
H=8.5 m=28 ft
Hp=28 ft (ft2/144 in2)(62.4 lbs/ft3)=12.1 psi
PTOTAL =12.1 psi+(14.7 lbs/ in2 )=26.8 psi
•Shell thıckness of tank
•Ts=(PD+C)/(2Se-P)
•S=Su× Fa×Fr×Fs×Fm
•Su=9000 psi
•Fu=%25, Fm=1 , Fr=1 , Fa=1
•S=9000psi×0,25×1×1×1=2250 psi
•E=0,80 c=1/16 in
Ts=
(26,81bs/in2×28ft×12in/ft +1/16 )/(2×22501bs/in2 ×0,8-26,81bs/in2)
Ts=2,5 in=6,25 cm
•Bottom thickness
• th= PLW/2Se
•Di=251,9in
•Do=Di+2×s
•Do=251,9+2×2,5
=256,9 in
L=Di-6(in in)
L=245,9 in
Kr= 0,06*Do kr=15,4
Th= (26,81bs/in2×245,9 in×1,8)/3600bs/in2=3,3 in=8,37 cm
CALCULATION OF COLD STORAGE
Nursel Bilici
•COLD STORAGE
kytong= 0.13 W/mK
kpolyurethane= 0.023 W/mK
∆Xytong= 0.15m
∆Xpolyurethane= 0.20m
Area= (15 x 16)= 240m2
Toutside= 180C
Tinside= 20C
hi = 9.3 W/m2K
ho= 22 W/m2K
Q=UA∆T
U= 1 / (1/hc + 1/ hi + ∆Xy / ky + ∆Xp / kp )
U= 1 / ( 1/22 + 1/9.3 + 0.15/0.13 + 0.20/0.023 )
= 0.1 W/m2K
EAST AND WEST WALLS
Toutside= 18+3= 210C
Q=UA∆T
Q= 0.1 x 60 x ( 21-2 )
Q= 114W
SOUTH WALL
Toutside= 18+2=200C
Q=UA∆T
Q= 0.1 x 64 x (20-2) = 115,2W
NORTH WALL
Toutside= 180C (no solar effect)
Q= 115,2W
ROOF
Toutside= 18+5=230C
Q= 504 W
FLOOR
Tground= 5+18=230C
U= 0.309 W/m2K
Q= 1557,36 W
Total heat loss by walls = 2405.76 W
• INFILTRATION LOAD
Q air change load = m x houtside – hinside
‘h’ are specific enthalpy values of inside air.
houtside = ha + x * hw
X = humidity ratio of air in K.Maraş
ha = entalpy of dry air
hw = enthalpy of water vapor
ha = Cpa * T
Cpa = specific heat of dry air
T = Temperature of air in working area
ha =( 1.006 kj/kg0C)*250C
=25.15 kj/kg
•hw = cpw *T+hwe
•hwe = evaporation heat of water
•cpw = specific heat of water vapor
•hw = (1.84 kj/kg0C*250C)+2.532kj/kg
• = 48.502 kj/kg
houtside =25.15kj/kg+(0.58*48.502 kj/kg)
=53.28kj/kg
hinside =ha+x*hw (xinside=0.9)
= 25.15+0.9*48.502 = 68.80 kj/kg
•Vroom = 15*16*4 = 960 m3
•V = 5*960 = 4800
•D = m/V
•Density of air at 250C 1.1915 kg/m3
•m = 5719.2 kg/day = 0.0662 kg/s
•=8.08 kW
•= 8081 W
3.Product Load
1000 kg/24h*1 h/3600
m = 0.116 kg/s
Q = m*cp*∆T
Q = 0.116*3.6*(25-2) = 9604.8 W
•Miscallaneous Load
•a ) heat given off by light
number of fluoresan lamp = 20
watt of each lamp = 36W
(36W*24h)/(24h*20) = 720 W
b ) heat given off electrical motor
power of motor = 0,25 kw
power of electric = 1.62 kW
0,25kW*1,62*1000 W/kW = 405 W
•c ) heat gain from people
number of people = 6 person
Q = (272-6*20C)*6 = 1560W
Forklift
(2hr*6150 W)/(0.85*24) = 603 W
Respiration
10000kg*17.5 W/kg =175000
From door (0.7*V+2)DT
V(2m*2.20m*0.02) = 0.088 m3
((0.7*0.088)+2)*(25-2) = 47.4 W
Total =198,4 kw
For safety 198,4 +(198,4*0,1)=218,24 Kw
•COP
•WHY WE USE CHLORINE DIOXIDE
üit can be used stainless steel
üKilling the microorganism
üİt is useful for environmental surface , floor and other areas.
•WASTE TREATMENT
We obtain the core of apricot.
12517 kg core of apricot in year.
We are going to sell $1,9/kg to cosmetics industry.
$24,000 in year.
We obtain stem from grape
520000 kg stem in year
We are going to cell $ 0,4/kg for animal feed
$25000 in year
•OUR SPECIFIC PROCESS
Sulphurization =

1 tone apricot 1600-1800 gr SO2
Immersing =
v 5-8% potassium carbonate
v 0,6-0,9% olive oil
COST ANALYSIS
expense
Slicing machine $500
Water Tank $6204
Pump $208
Income
Core of apricot $24000
Stem of grape $25000
THANKS FOR LISTENING AND Prof.Dr.A.Coşkun Dalgıç

Bir cevap yazın