PLATE HEAT EXCHANGER CALCULATIONS

For T2 calculation of Regenaration Section

C p,pekmez= 4.18*Xwater+ 2.5*Xsugar
C p,pekmez= 4.18*0.82+ 2.5*0.18 = 3.88 kj/kg

Tave = (T3 + T4)/2 = (105 +105)/2 =105 ° C table of A.2-5 and 12 from
Unit operation by J. G.koplis
hfg =enthalpy of saturated – enthalpy of condensate
steam at 105 °C steam at 105 °C
hfg = 2683.8 – 440.15
hfg = 2243 kj/kg
For T7 calculation, in regeneration section;
msteam*hfg= mpekmez*Cp*Δtpekmez
msteam*2243 kj/kg =mpekmez*Cp*(T1-T2)
0.026*2174 kj/kg =0.436 kg/s*3.88kj/kg.°C*(90- T2) °C
T2 = 55 ° C
For T7 calculation, in regeneration section;
Qpekmez = Qpekmez
mpekmez*Cp*ΔTpekmez = mpekmez*Cp*ΔTpekmez
(T2-Tf) = (T1-T7)
(55-25) = (90- T7)
T7 = 60 ° C
For Tp calculation in cooling section ;
C p of pekmez = 3.88 kJ/kg*K
Tave=(T5+T6)/2 = (4+20) /2 =12° C table of A.2-5 from G.koplis
Cp of water = 4.185 kJ/kg*K
Qwater = Qpekmez
mwater*Cp*ΔTwater = mpekmez*Cp*ΔTpekmez
mwater*Cp*(T6-T5) = mpekmez*Cp*(T7-Tp)
0.88 kg /s*4.185 kJ/kg*K*(20-4) = 0.436 kg /s*3.88 kJ/kg*K*(60- Tp)
Tp = 25 ° C
For plate dimensions:
L = 0.50 m
W = 0.25 m
A = 0.125 m2
Thickness of stell = 0.003 m
Gap between plate = 0.005 m
Deq = 2*a*b/(a+b)
Deq = 2*0.25*0.005/(0.25+0.005)
= 0.0098 m
Heating Section
QH = mpekmez*Cp*ΔTpekmez
=mpekmez*Cp*(T1-T2)
=0.436 kg/s*3.88Kj/kg.°C*(90-55) °C
QH =5.92*104 W
Logaritmic mean temperature

ΔT1= 105-55 =50 °C ΔT2 = 105-90 =15 °C
ΔTLM = (ΔT1- ΔT2)/ln(ΔT1/ΔT2)
= (50- 15)/ln(50/15)
=29.07 °C

For pekmez
Tb = (90+55)/2 = 73 °C
ρ pekmez = 1061 kg/m3 at 73 ° C and 18 °
C p of pekmez = 4.18*0.82+ 2.5*0.18 = 3.88 kJ/kg*K
μpekmez = 0.00988 kg/m.s 18 ° at 73 °C
k = (0.565+0.0018T+0.000006T2)(1-0.54Xs)
Xs mass fraction of solids in the fruit(grape) juice.
k = (0.565+0.0018*53+0.000006*532)(1-0.54*0.18)
k = 0.657 W/m.°C
Vpekmez = Q/A Q=0.378 kg/s /(1061 kg/m3)=3.56*10-4 m3/s
= (3.56*10-4 m3/s)/(0.25m*0.005 m)
= 0.28 m/s

Nre = (Deq*V* ρ)/ μ
= (0.0098 m*0.28 m/s*1061 kg/m3)/ 0.00988 kg/m.s
= 295
NPr = (Cp* μ)/ k
= (3880*0.00988)/ 0.657
= 58.35
NRe > 100 so; Nu= 0.4* NRe0.64* NPr0.4 from Plant Design and Economic for chemical engineering.
(Deq*h)/k = 0.4* (295)0.64* (58.35)0.4
(0.0098 m *h)/ 0.657 W/m.°C = 69.51

h pekmez= 5,190 W/m2.°C

For steam
Tb = (105+105)/2 = 105 °C
ρ steam = 0.596 kg/m3 at 105 ° C C p,steam = 1.888 kJ/kg.°C
μ steam = 1.295*10-5 kg/m.s k = 0.02510 W/m.°C
Vsteam = Q/A Q=0.0272 kg/s /(0.596 kg/m3)=0.0456 m3/s
= (0.0456m3/s)/(0.25m*0.005 m)
= 36.48 m/s
NRe = (Deq*V* ρ)/ μ
= (0.0098 m*36.48 m/s*0.596 kg/m3)/ 1.295*10-5 kg/m.s
= 16,453
NPr = (Cp* μ)/ k
= (1888*1.295*10-5)/ 0.02510
= 0.974
NRe > 10 so; flow is turbulent Nu= 0.4* NRe0.64* NPr0.4
(Deq*h)/k = 0.4* (16,453)0.64* (0.974)0.4
(0.0098m *h)/ 0.02510 W/m.°C = 263.64
h steam= 506 W/m2.°C
Thickness of steel = 3*10-3 m
k steel = 16.3 W/m.°C
1/ UH = (1/hpekmez) + (Δx/k) +(1/hsteam)
= (1/5,190) + (0.003/16.3) +(1/506)
UH = 425 W/m2.°C
QH = UH*AH* ΔTLM
5.92*104 W = UH*AH*ΔTLM
5.92*104 W = 425*AH*29.07
AH =4.15 m2
AH =N*a 4.79 m2=N*0.125 m2
N= 38 plates
Length of holding tube;
Nominal pipe size(in.) = 2
Schedula number = 40
Inside diameter = 0.0525 m
where ρ pekmez = 1071 kg/m3 at 18 ° and 90 ° C from…..
μ pekmez = 0.0132 kg/m.s 18 ° at 90 °C

Vave = Q pekmez / A holding tube
= 3.53*10-4 m3/s/ (π*[0.0525]2) / 4
= 0.163 m/s
NRe = (D*V* ρ) / μ
= (0.0525 m*0.163 m/s*1071 kg/m3) / (0.0132)
= 695 NRe < 2100 So laminar flow
Vmax = 2*Vave
Vmax = 2*0.163 = 0.326 m / s
Length of holding tube = Vmax*pasteurization time
= 0.326 m / sec*106.8 sec
= 34.81 m
MULTIPLE-EFFECT EVAPORATOR CALCULATIONS
(78°C)
Enthalpy balance of each effect;
Effect 1;
F*cp,f*(Tout -Tin) +S* hfg,1= L1* cp,1*(T-T) + V1* H1
1,568*3.88(78-25) + S*2,174 = L1*3.78*(78-78) + (1568- L1)*2,314
Effect 2 ;
L1* cp,1*(Tout -Tin) + V1* hfg,2= L2* cp,2*(T-T) + V2* H2
L1*3.78*(72-78) +(1568- L1) *2,314 = L2*3.58*(72-72) +( L1- L2)*2,333
Effect 3 ;
L2* cp,2*(Tout -Tin) + V2* hfg,3= L3* cp,3*(T-T) + V3* H3
L2*3.58*(65-72) +( L1- L2)*2,333 = 392*2.97*(65-65) +(L2-392)*2,333
S + 1.064* L1=1,520 kg / h eq. 1 2L1- L2=1,555 kg / h eq. 2
2.016L2- L1=394 kg / h eq. 3
Sub. 2nd and 3th eq. each other;
L1 =1,949-1.016*L2
L2 is assumed 784 kg / h V2 = 368 kg / h X1= 0.256
L1= 1,152 kg / h
V1 = 416 kg / h X2= 0.36
L3= 392 kg / h
V3 = 392 kg / h X3= 0.72
S = 295 kg / h
Steam Economy = (1,176 kg/h water is evaporated)/(1,078 kg/h is steam is used)
= 1.09
OVERALL HEAT TRANSFER-COEFFICIENT CALCULATIONS
For first effect,U1;
hfg = (enthalpy of saturated steam at 130 °C )-(enthalpy of condensate steam at 130 °C)
hfg = (2720.5 -546.31) = 2174 kj/kg
Q1= mpekmez*Cp*ΔTpekmez
Q1=0.436 kg/s*3.88kj/kg.°C*(78-25) °C
Q1=8.96*104 W
Logarithmic mean temperature
ΔT1= 130-25 =105 °C
ΔT2 = 130-78 =52 °C
ΔTLM = (ΔT1- ΔT2)/ln(ΔT1/ΔT2)
= (105- 52)/ln(105/52)
=75.42 °C
For pekmez
Table 4.1-2 Constant for use for Natural Convection from Transport Process and Unit Operation Christie J.Geankoplis
Mechanisms h values
Boiling Liquids 1700 to 28000 W/m2.°C
h = 2000 W/m2.°C
For steam
Tb = (130+130)/2 = 130 °C
ρsteam = 0.596 kg/m3 at 130 ° C
C p,steam = 1.895 kJ/kg.°C
μ steam = 1.4134*10-5 kg/m.s
k = 0.02786 W/m.°C
Vsteam = Q/A Q= 0.0819 kg/s /(0.596 kg/m3)
= 0.137 m3/s
D is assumed as inner diameter=5.46*10-3 m
Schedula Number = 80 in.
A steam = π*D2/4 A= π*(10.74*10-3)2/4 = 2.34*10-5 m2
= (0.137 m3/s)/(2.34*10-5 m2)
= 5,872 m/s

NRe = (D*V*ρ)/ μ
= (0.00546 m*5,872 m/s*0.596 kg/m3)/ 1.4134*10-5 kg/m.s
= 1,352,052
NPr = (Cp* μ)/ k
= (1895*1.4134*10-5)/ 0.02786
= 0.961
so; flow is turbulent Nu= 0.4* NRe0.64* NPr0.4
(D*h)/k = 0.4* (1,352,052)0.64* (0.961)(0.4)
(5.46*10-3m *h)/ 0.02786 W/m.°C = 263.64
h steam= 16,856 W/m2.°C
k steel= 16.3 W/m.°C Thickness of steel = 3.43*10-3 m

1/ U1 = (1/hpekmez) + (Δx/k) +(1/hsteam)
= (1/2000) + (0.00343/16.3) +(1/16,856)
U1= 1,299 W/m2.°C

Q1 = U1*A1* ΔTLM,1=mpekmez*Cp,pekmez* ΔT
(1,299 W/m2.°C)* A1*(75.42 °C)=(0.436 kg)*(3880 j/kg.K)*(78-25)
A1 = 0.915 m2

THICKNESS OF EVAPORATOR BOILER
Soft water is used for this purposes;
Q = Q(steam) (energy balance)
Density of fuel oil = 890 kg /m3 fuel oil
Steam outlet pressure = 8 bar (170°C )
hfg (steam at 170 °C and 8 bar ) = (2768.7-719.21)= 2480 kj/kg
Required steam for our process = 556 kg steam /h
Water inlet temperature = 18°C
Heat supplied from
148100 btu/gal × 1.05506 kj/btu × 264.1 gal/m3 × 1 m3/890 kg
= 46,367.17 kj / kg fuel oil (46367 kj energy is supplied from 1 kg fuel oil )
Q =msteam × hfg = 556 kg steam /h × 2048 kj/kg = 1,138,688 kj/h

(46,367.17 kj/ kg fuel oil) × mfuel oil = 1,138,688 kj/h
mfuel oil = 24.55 kg /h fuel oil = 24.55 kg/h × 24 h/day
= 590 kg /day fuel oil is used

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