ITP330-Evaporasi 2016-pha-handout ready

ITP330 - Prinsip Teknik Pangan

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EVAPORATION

Baca

• http://www.nzifst.org.nz/unitoperations/evaporation.htm• Landasan Teknik Pangan (Buku Bacaan ITP330)

EVAPORATION

• Frequently in the food industry a raw material or a potential foodstuff contains more water than is required in the final product.

• When the foodstuff is a liquid, the easiest method of removing the water, in general, is to apply heat to evaporate it.

• Evaporation is thus a process that is often used by the food technologist.


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The basic factors that affect the rate of evaporation are the:

• rate at which heat can be transferred to the liquid,

• quantity of heat required to evaporate each kg of water,

• maximum allowable temperature of the liquid,

• pressure at which the evaporation takes place,

• changes that may occur in the foodstuff during the course of the evaporation process.

EVAPORATION

PROSES ………..??

• Evaporasi : 1. Atmosferik

2. Vakum

EVAPORATION

Vakum• Boling Pt lbh rendah• Disain: Vacuum system• Energi: Lebih efisien• Single & Multiple effect• Cocok untuk pangan

sensitif thd panas

Atmosferik :- Boiling pt lb tinggi- Disain: Sederhana- Energi: kurang efisien (- Single & Multiple effect- Tidak cocok untuk pangan

sensitif thd panas : off flavor


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PROSES ………..??

• Evaporasi : 1. Atmosferik

2. Vakum

EVAPORATION

Mass and energy transfer phenomena

An evaporator has 2 functions:

1. Heat exchange the evaporating section, where the liquid boils and evaporates,

and

1. Separate vapor formed from liquid (leaves the liquid and passes off to the condenser or to other equipment)

EVAPORATION


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EVAPORATION

Open kettle or pan evaporator:


EVAPORATION

Concentrate

Condensate

Pan

Boiler

Jacket

Steam



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EVAPORATION



EVAPORATION


• Simplest form of evaporators

• Inexpensive

• Simple to operate

• Very poor heat economy

• In some cases paddles and scrapers for agitation are used


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EVAPORATION

Horizontal-tube evaporator:

Dilute feed

Condensate

Concentratedproduct

Vapour

Steam inlet


EVAPORATION

Horizontal-tube evaporator:

- Relatively cheap

- Used for non-viscous liquids having high heat-transfer coefficients and liquids that do not deposit scales

- Poor liquid circulation (and therefore unsuitable for viscous liquids)


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EVAPORATION

Vertical-type short-tube evaporator:

Dilute feed

Condensate

Concentratedproduct

Vapour

Steam inlet


EVAPORATION



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EVAPORATION

• Liquid is inside the tubes

• Steam condenses outside the tubes

• Used for non-viscous liquids having high heat-transfer coefficients and liquids that do not deposit scales



Factors effecting evaporation:Concentration in the liquid:

• Liquid feed to an evaporator is initially at a very dilute condition;

• Viscosity is low heat-transfer coefficient high.

• As evaporation proceeds, the solution becomes concentrated.

• Viscosity increases heat-transfer coefficient drops.

• Density and the boiling point of solution also increase.


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Factors effecting evaporation:Solubility

As evaporation progress

concentration of the solute increases

when solubility limit of the solute in solution is exceeded, then crystals may form.

Solubility of the solute: determines the maximum concentration of the solute in the product stream.

Generally: the solubility of the solute increases with temperature.

When a hot concentrated solution from an evaporator is cooled to room temperature crystallization may occur.


• Functional products/foods componens may be damaged when heated to moderate temperatures for relatively short times.

• Special techniques are employed to reduce temperature of the liquid and time of heating during evaporation

Factors effecting evaporation:Heat Sensitivity of compounds


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Foaming and frothing:- Solutions like organic compounds tend to foam and froth during vaporization.

- The foam is carried away along with vapor leaving the evaporator.

- Entrainment losses occur.

Factors effecting evaporation:


Pressure and temperature:- The boiling point of the solution is related to the pressure of the

system.

- The higher the operating pressure of the evaporator, the higher the temperature at boiling.

- Also, as the concentration of the dissolved material in solution increases by evaporation, the temperature of boiling may rise (a phenomenon known as boiling point rise/elevation).

- To keep the temperatures low in heat-sensitive materials, it is often necessary to operate under atmospheric pressure (that is, under vacuum).



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Scale deposition:- Some solutions deposit solid materials (called scale) on the

heating surfaces.

- The result is that the overall heat-transfer coefficient (U) may drastically decrease, leading to shut down of the evaporators for cleaning purposes.



Materials of construction:- Evaporators are made of some kind of steel.

- However many solutions attack ferrous metals and are contaminated by them.

- Copper, nickel, stainless steels can also be used.



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Method of operation of evaporators

- When a single evaporator is used ,the vapor from the boiling liquid is condensed and discarded. This is called single effect evaporation.

- It is simple but utilizes steam ineffectively.

- To evaporate 1 kg of water from the solution we require 1-1.3 kg of steam.

- Increasing the evaporation per kg of steam by using a series of evaporators between the steam supply and condenser is called multiple effect evaporation

Single-effect evaporation:

Multiple-effect evaporation:


mf, mp : Aliran cairan (liquid) (kg/jam)mv : Aliran uap (kg/jam)x : Konsentrasi padatan (kg padatan/kg total)T : Suhu (0 C)

MASS & ENERGY BALANCE

Uap (mv, Ti)

Produk (mp, xp, Ti)

Steam (ms,Ts)

Kondensat (ms,Ts)

Heat Exchanger

Feed (mf, xf, Tf)

Interface uap-cairan (liquid)


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ASUMSI :

• Umpan yg masuk (Tf) langsung mencapai titik didih (Ti) dan uap (mv)

serta produk (mp) meninggalkan evaporator pada suhu tersebut

• Uap (steam) masuk heat exchanger sbg uap jenuh (ms) pada suhu Ti

meninggalkan H.E sebagai kondensat cairan (ms)


Uap (mv, Ti)

Produk (mp, xp, Ti)

Steam (ms,Ts)

Kondensat (ms,Ts)

Heat Exchanger

Feed (mf, xf, Tf)



MASS BALANCE

• Material balance keseluruhan :

mf = mv + mp …………………………………………(1)

• Kesetimbangan padatan larut

xf mf = xp mp …………………………………………(2)

• O.K.I :

mv = mf (1- (xf / xp))………………………………….(3)


Uap (mv, Ti)

Produk (mp, xp, Ti)

Steam (ms,Ts)

Kondensat (ms,Ts)

Heat Exchanger

Feed (mf, xf, Tf)



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ASUMSI : Heat loss Negligible

Maka panas yg disuplai steam yang berkondensasi = panas yg diperlukan untuk memanaskan danmenguapkan


Uap (mv, Ti)

Produk (mp, xp, Ti)

Steam (ms,Ts)

Kondensat (ms,Ts)

Heat Exchanger

Feed (mf, xf, Tf)

ENERGY BALANCE



mf Hf + ms Hvs = mv Hvi + mp Hpi + ms Hcs

Dimana : H = entalpi (kJ/kg) : Lihat Tabel UapHvs = entalpi uap jenuh pada Ts

Hvi = entalpi uap jenuh pada Ti

Hcs = entalpi kondensat


Uap (mv, Ti)

Produk (mp, xp, Ti)

Steam (ms,Ts)

Kondensat (ms,Ts)

Heat Exchanger

Feed (mf, xf, Tf)

ENERGY BALANCE



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Uap (mv, Ti)

Produk (mp, xp, Ti)

Steam (ms,Ts)

Kondensat (ms,Ts)

Heat Exchanger

Feed (mf, xf, Tf)

ENERGY BALANCE

Hf = cpf (Tf – 0oC)cpf = panas jenis (kJ/kgoC)

Hpi = cpp (Ti -0oC)

Hvs dan Hvi dari Tabel Uap pada saturated vaporHcs dari Tabel Uap pada saturated liquid



Laju Pindah Panas

q = UA (Ts-Ti) = ms Hvs - ms Hcs

Dimana: U =Overall heat transfer coefficient (W/m2K)A = Luas permukaan pindah panas (m2)q = Laju pindah panas (W)


Uap (mv, Ti)

Produk (mp, xp, Ti)

Steam (ms,Ts)

Kondensat (ms,Ts)

Heat Exchanger

Feed (mf, xf, Tf)



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Example 1:

A continuous single-effect evaporator concentrates 9072 kg/h of a 1.0 wt % salt solution entering at 38ºC to a final concentration of 1.5 wt %.

The vapor space of the evaporator is at 101.325 kPa (1.0 atm abs) and the steam supplied is saturated at 150 kPa. The overall coefficient U = 1704 W/m2.K.

Calculate the amounts of vapor and liquid products and the heat-transfer area required. Assumed that, since it its dilute, the solution has the same boiling point as water.


Uap (mv, Ti)

Produk (mp, xp, Ti)

Steam (ms,Ts)

Kondensat (ms,Ts)

Heat Exchanger

Feed (mf, xf, Tf)



Data provided:

F = 9072 kg/h

xF = 1 wt % = 0.01 kg solute / kg feed

TF = 38ºC

xL = 1.5 wt %

= 0.015 kg solute / kg liquid product

P = 101.325 kPa (1.0 atm abs)

PS = 150 kPa

U = 1704 W/m2.K

T1 = saturated temperature at P (= 101.325 kPa) = 100ºC

TS = saturated temperature at 150 kPa = 111.4ºC

Steam, SPS, TS, HS

Feed, F

xF, TF, hF

Condensate, S

PS, TS, hS

Vapour, V

yV, T1, HV

Concentrate, L

xL, T1, hL

P

T1

Calculation methods for single-effect evaporators(taken from http://www.nzifst.org.nz/unitoperations/evaporation.htm)


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Overall material balance:

F = L + V

Solute balance:

F xF = L xL (no solute in the vapour)

Heat balance:

F hF + S λ = L hL + V HV

where λ = HS – hS

Data provided:

F = 9072 kg/h

xF = 0.01 kg solute / kg feed

TF = 38ºC

xL = 0.015 kg solute / kg liquid product

P = 101.325 kPa; T1 = 100ºC

PS = 150 kPa; TS = 111.4ºC

U = 1704 W/m2.K

q = S λ = U A ∆T = U A (TS – T1

Amounts of vapor and liquid products = ?

F, xF and xL are known, and therefore

L = 6048 kg/h and V = 3024 kg/h

Available equations:



Heat balance:

F hF + S λ = L hL + V HV

where λ = HS – hS

Data known:

F = 9072 kg/h; L = 6048 kg/h, V = 3024 kg/h

TF = 38ºC

P = 101.325 kPa; T1 = 100ºC

PS = 150 kPa; TS = 111.4ºC

U = 1704 W/m2.K

q = S λ = U A ∆T = U A (TS – T

Available equations:

S λ = L hL + V HV – F hF

= (F – V) hL + V HV – F hF

= F (hL – hF) + V (HV – hL )= F Cp (T1 - TF) + V (Latent heat of vapourization at 101.325 kPa )


Heat transfer area A = S λ / U (TS – T1) = ?


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S λ = F Cp (T1 - TF) + V (Latent heat of vapourization at 101.325 kPa)F, T1 , TF and V are already known. Cp = 4.14 kJ/kg.K (assumed)Latent heat of vapourization at 101.325 kPa = 2256.7 kJ/kgTherefore, S λ = (9072) (4.14) (100 – 38) +(3024) (2256.7) kJ/h

= 9152862 kJ/h


Data known:

F = 9072 kg/h; L = 6048 kg/h, V = 3024 kg/h

TF = 38ºC

P = 101.325 kPa; T1 = 100ºC

PS = 150 kPa; TS = 111.4ºC

U = 1704 W/m2.K



S λ = 9152862 kJ/h = 9152862 * 1000 / 3600 WT1 and TS are knownU = 1704 W/m2.KTherefore, A = S λ / U (TS – T1)

= [9152862 * 1000 / 3600] / [1704 * (111.4 – 100)]= 130. 9 m2


Data known:

F = 9072 kg/h; L = 6048 kg/h, V = 3024 kg/h

TF = 38ºC

P = 101.325 kPa; T1 = 100ºC

PS = 150 kPa; TS = 111.4ºC

U = 1704 W/m2.K



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Technical Problems of evaporation :

• Foaming losses, occupation hazards

• Fouling reduced Heat Transfer

• Increase viscosity

• Heat sensitive vs Boiling-point elevation (BPE)


Problems of evaporation : Boiling Point Elevation (BPE)

Duhring plot for boiling point of

sodium chloride solutions


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Contoh:

Duhring Chart : tentukan titik didih awal & akhir suatu cairan pangan dengan komposisi yang menghasilkan tekanan uap air mirip dengan larutan NaCl. Tek.dlm evaporator 20 kPa. Produk dikonsentrasikan dari 5% ke 25 % total solids

Pendekatan :

Penggunaan Duhring Chart perlu titik didih air bisa diperoleh dg Tabel Uap

Next Duhring chart titik didih larutan


Prosedur :

1. Steam Table 20 kPa ttk didih air, yaitu60oC

2. Duhring Plot: Tarik garis vertikal ke atasdari 333oK (60oC) titikdidih air

• 5% TS bp=61oC

• 25 % TS bp=66oC



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Prosedur :

1. Steam Table 20 kPa ttk didih air, yaitu60oC

2. Duhring Plot: Tarik garis vertikal ke atasdari 333oK (60oC) titikdidih air

• 5% TS bp=61oC

• 25 % TS bp=66oC



Diagram of single-effect evaporator


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Susunan MEE:1. Searah (Forward)2. Berlawanan arah (Reverse)

feed product

steam

vapour

Diagram of multiple-effect evaporator


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Susunan MEE:1. Searah (Forward)2. Berlawanan arah (Reverse)

feedproduct

steam

vapour

Diagram of multiple-effect evaporator


Double-effect evaporator-– forward feed


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Baca

http://www.nzifst.org.nz/unitoperations/evaporation.htm


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Selesai ….……….

…….

SELAMAT BERLIBUR

SELAMAT UAS

Prinsip Teknik Pangan (ITP330)

• Review Matematika• Satuan dan Dimensi

• Neraca Massa• Thermodinamika dan Neraca Energi

• Aliran Fluida• Transportasi Fluida

• Pindah Panas• Alat Penukar Panas

• Proses Thermal• Refrigerasi

• Pembekuan• Psikrometri

• Pengeringan• Evaporasi

ITP330-Evaporasi 2016-pha-handout ready

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