Single Effect Evaporation Final [Compatibility Mode]

8/19/2019 Single Effect Evaporation Final [Compatibility Mode]

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Single Effect Evapora

Robi Andoyo, STP., M.Sc., P


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EVAPORATION

Heat is added to a solution to vaporize the solvent, which i

Steam is introduced into HE indirectly

Vapor from a boiling liquid solution is removed and a mor

so u on rema ns.

Example: concentration of aqueous solutions of sugar. In t

crystal is the desired product and the evaporated water is d


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The feed (usually dilute) enters at T F and sasteam at T S enters the heat-exchange sectio

Condensed leaves as condensate or drips.

The solution in the evaporator is assumed to

SINGLE EFFECT EVAPORATORS

T 1.

The pressure is P 1, which is the vapor pressusolution at T 1.

Wasteful of energy since the latent heat of th

leaving is not used but is discarded.

Are often used when the required capacity ois relatively small, but it will wasteful of steam


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SINGLE EFFECT EVAPOR

T1 is controlled by maintaining vacuum insid

chamber

Heat and mass balancees conducted on thesystem allow determination of various desig

Variables : mass flow rate, final concentratioand HE area


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feed, F

T F , x F , h F .

vapor,V to conde

T 1 , yV , H V

P1

T

heat-exch

tubes

steam, S

T S , H S

concentrated liquid, L

T 1 , x L , h L

conde

T S

Simplified Diagram of single-effect evapora


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Components

1. Vapor chamber

Allow separation of vapor fromliquid and prevent carry-over ofsolids by the vapor

2. Condenser

To condensed water vapor, heatexchanger cooled by refrigerantor by cooling water

3. Heat Exchanger

The rate of evaporation in anevaporator is determined by the

amount of heat transferred in theheat exchanger, HE determinedthe type of evaporation


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CALCULATIONS

a) vapor, V and liquid, L flowrates.

b) heat transfer area, A

c) overall heat-transfer coefficient, U .

CALCULATION METHODS FOR

SINGLE-EFFECT EVAPORATO

d) Fraction of solid content, x L.

(1) To calculate V & L and xL,

- solve simultaneously total material balance &solute/solid balance.

mF = m L + m V total material balancemF (x F ) = m L (x L ) solute/solid balance


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(2) To calculate A or U,

- No boiling point rise :calculate hF, hL, Hv and λ

where, λ = (H S – hs) = latent heat of vah F = c PF (T F – T ref )h L = c PL(T 1 – T ref )H V and h s See Steam table

where, T ref = 00 C = (as datum)

c PF

and c PL

= Specific heat of pro H and h = enthalpy at specific

solve for S:m F h F + m S H S = m L h L + m V H V +

solve for A and U :

q = m S λ = m S (H S – h S ) = U A ∆T


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(3) Steam economy

Steam economy is a term often used expressing the operating performance ofevaporator system

This term is the ratio of rate of mass

steam consumed

Steam economy = mV / mS


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Effects of Processing VariaEvaporatos Operatio

• Effect of feed temp . TF↓, Hs↑ to heat feed to the BP. Pre heating the feed cthe size of evaporator HT area neede

• Effect of pressure . Larger ∆T is desira

,cost of evaporator decrease. ↓P BP odecrease

• Effect of steam pressure . Higher pressaturated steam increased ∆T, size andecrease


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Example 2

Apple juice is being concentrated in a natusingle effect evaporator. At steady statedilute juice is the feed introduced at a ratethe concentration of diluted juice is 11 % tot

juice is concentrated to 75 % TS. The sp

,0C), respectively. The steam pressure is m304,42 kPa. The inlet feed temp. Is 43,3 0Cinside evaporator boils at 62,2 0C. Thecoefficient is assumed to be 943 W/(m2

negligible boiling point elevation. Calculate trate of concentrate prod, steam requireeconomy and HT area


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Solution steps

• Draw the schematic diagram of th

• Write known variables

• Find mL mV HF hL Hs

Steam economy A


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Example 1

A continuous single-effect evaporator co9072 kg/h of a 1.0 wt % salt solution enterin(37.8 ºC) to a final concentration of 1.5 wt %space of the evaporator is at 101.325 kPa (and the steam supplied is saturated at 143.3

= . .

Calculate the amounts of vapor and liquid the heat-transfer area required. Assumed thdilute, the solution has the same boiling poin

and Enthalphy of the food products are negl


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P1 = 101.325 kPa

F = 9072 kg/h

T F = 311 K

x F = 0.01

h F .

V =T 1

T 1 A = ?S , T S , H S

PS = 143.3 kPa L = ?

T 1 , h L

x L = 0.015

S, T

Flow Diagram for Example 1


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Solution

Refer to Fig. 1 for flow diagram for this solution.

For the total balance,

F = L + V

9072 = L + V

For the balance on the solute alone,

x F = x L9072 (0.01) = L (0.015)

L = 6048 kg/h of liquid

Substituting into total balance and solving,

V = 3024 kg/h of vapor


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Since we assumed the solution is dilute as wa

c pF = 4.14 kJ/kg. KFrom steam table,

At P 1 = 101.325 kPa, T 1 = 373.2 K (100

H V = 2257 kJ/kg.

At P S = 143.3 kPa, T S = 383.2 K (110

= .

The enthalpy of the feed can be calculated fro

h F = c pF (T F – T 1 )

h F = 4.14 (311.0 – 373.2)

= -257.508 kJ/kg.


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Substituting into heat balance equation;

F h F + Sλ

= L h L + V H V with h L = 0, since it is at datum of 373.2 K.

9072 (-257.508) + S (2230) = 6048 (0) + 3024 (2257

S = 4108 kg steam /h

The heat q transferred through the heating surface a

q = S ( λ )

q = 4108 (2230) (1000 / 3600) = 2 544 000 W

Solving for capacity single-effect evaporator equatio

q = U A ∆T = U A (T S – T 1 )

2 544 000 = 1704 A (383.2 – 373.2

Solving, A = 149.3 m 2 .


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Single Effect Evaporation Final [Compatibility Mode]

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