Evaporation

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CHEMICAL ENGINEERING SERIES

EVAPORATION

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EVAPORATION

A unit operation that involves the concentration of a solution consisting of a non-volatile solute and a volatile solvent

It is conducted by vaporizing a portion of the solvent to produce a concentrated solution of thick liquor.

It differs with other unit operations in such a way that:

1. Distillation: in evaporation vapor is usually a single component

2. Drying: in evaporation, residue is liquid, sometimes a highly viscous one 3. Crystallization: focus is on concentrating a solution rather than forming crystals

Calculations for the Different Methods of Operations of Evaporators:

1. Single Effect Evaporators – used when the required capacity of operation is relatively small and/or cost of steam is relatively cheap compared to the evaporator cost.

where:

- Mass flow rates of feed, vapor, and steam respectively

- Temperatures of feed,

product and vapor, respectively

- Liquid enthalpy of feed

and product, respectively

- Vapor enthalpy

- Operating temperature

- Operating pressure

- Mass fraction of solute in

feed and product respectively

Over-all Material Balance: Solute Balance: Enthalpy Balance: Heat Balance: ( ) must be evaluated at or

If vacuum pressure is given, Feed, F xF TF hF CP, F Steam, S TS λS TI Product, P xP TP hP Vapor, V TV HV P V PI Condensate TS2

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2. Forward Feed Multiple Effect Evaporators – fresh feed is added to the first effect and flows to the next in the same direction as the vapor flow. This is used when the feed is hot or when the final concentrated product might be damaged at high temperatures

3. Backward Feed Multiple Effect Evaporators – fresh feed enters the last and coldest effect and continues until the concentrated product leaves the first effect. This is used when the fresh feed is cold. This type of evaporation would requires liquid pump for each effect since flow is from low to high pressure

4. Mixed Feed Multiple Effect Evaporators – fresh feed enters any of the available effects and continues not necessarily to the effect next to it.

5. Parallel Feed Multiple Effect Evaporators – involves the adding of fresh feed and the withdrawal of concentrated product from each effect. The vapor from each effect is still used to heat the next effect. This method is used mainly when the feed is almost saturated and solid crystals are the product, as in the evaporation of brine to make salt.

P1,T1 TI P2,T2 TII F Steam, S P VI VII LI P3, T3 TIII VIII LII P1,T1 TI P2,T2 TII F Steam, S P VI VII LI P3, T3 TIII VIII LII

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Performance Evaluation of Steam-Heated Evaporators

1. Capacity – number of kilograms of water vaporized per hour

Evaporator Capacity

Where:

- Rate of heat transfer through the heating

surface of an evaporator - Over-all heat transfer coefficient

- Heat transfer surface area

- Over-all temperature drop

2. Steam Economy – number of kilograms vaporized per kilogram of steam fed to the unit

Boiling point Evaluation (BPE) of a solution is the increase in boiling point over that of water

1. Small for dilute solutions and organic colloids solution

2. Large enough for concentrated solutions of inorganic salts; BPE can be estimated using Figure 11-124 (CHE HB 8th edition)

Dϋhring’s Rule – the boiling point of a given solution is a linear function of the boiling point of

pure water at the same pressure. Figure 16.3 (Unit Operations 7th edition by McCabe and Smith)

For solutions with BPE:

( )

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PROBLEM # 01:

A triple effect forward feed evaporator is being used to evaporate a sugar solution containing 5 wt % solids to a concentrated solution of 80 %. The boiling point rise of the solutions (independent of pressure) can be estimated from BPR °C = 1.78x + 6.22x2, where x is wt fraction of sugar in solution. Saturated steam at 205.5 kPa (121.1°C saturation temperature) is being used. The pressure in the vapor space of the third effect is 13.4 kPa. The feed rate is 10,000 kg/h at 26.7°C. The heat capacity of the liquid solutions is 4.19 – 2.35x, kJ/kg·K. The heat of solution is considered to be negligible. The coefficients of heat transfer have been estimated as U1 = 3123,

U2 = 1987, and U3 = 1136 W/m 2

·K. if each effect has the same surface area, calculate the area, the steam rate used and the steam economy.

SOLUTION:

Step 1:

From steam table, at 13.4 kPa (pressure of the vapor space at 3rd effect)

From the given:

For the 3rd effect with x = 0.80

( ) ( ) ( )

( )

Step 2:

Consider solute balance around the system:

( ) ( _*

Over-all material balance around the system

P1,T1 TI P2,T2 TII F

Steam, S

V

1

V

2

L

1 P3, T3 TIII

V

3

L

2

_L

3

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Where:

Assume , initially equal rate of evaporation in each effect

( )

For the 1st effect:

( )( )

For the 2nd effect:

( )( ) Step 3:

To solve for BPR for the 1st and 2nd effects: 1st effect: ( ) ( ) ( ) 2nd effect: ( ) ( ) ( ) ∑ ∑ ∑ ( ) Estimate ΔT for each effect using equation 8.5-6

∑ ( ) * + ∑ ( ) * + ∑ ( ) * +

Calculate actual boiling point of solution for each effect using the estimated ΔT

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Step 4: Heat Balance with 0°C as datum For 1st effect: ( ) ( ) ( )

From steam table, steam at 121.1°C ( )

From steam table, H(vapor enthalpy) at ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) For 2nd effect: ( ) ( )

From steam table, at TV1 = 109°C

( )

From steam table, H(vapor enthalpy) at ( ) ( )( )( ) ( )( ) ( )( )( ) ( ) For 3rd effect: ( ) ( ) ( )

From steam table, at TV2 = 89.89°C

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( )

From steam table, H(vapor enthalpy) at

( ) ( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) Equate and Step 5:

Solve for heat transfer area for each effect: For 1st effect: ( _{* (} _* ( _{* ( )} For 2nd effect: ( _{* (} _* ( _{* ( )} For 3rd effect: ( _{* (} _* ( _{* ( )}

Since areas are not close, then another trial should be done TRIAL 2:

Conduct new material balance using the computed L values

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( )( ) ( )( )

To solve for BPR for the 1st and 2nd effects: 1st effect: ( ) ( ) ( ) 2nd effect: ( ) ( ) ( ) ∑ ∑ ∑ ( ) Average Area from trial 1:

For the new

( ) ( * ( ) ( * ( ) ( * Adjust to attain total ΔT of 63.53

( ) (

*

( ) (

*

Calculate actual boiling point of solution for each effect using the estimated ΔT

Step 4: Heat Balance with 0°C as datum For 1st effect:

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( )

From steam table, steam at 121.1°C ( )

From steam table, H(vapor enthalpy) at ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) For 2nd effect: ( ) ( )

From steam table, at TV1 = 103.94°C ( )

From steam table, H(vapor enthalpy) at ( ) ( )( )( ) ( )( ) ( )( )( ) ( ) For 3rd effect: ( ) ( ) ( )

From steam table, at TV2 = 87.61°C

( )

From steam table, H(vapor enthalpy) at

( ) ( )

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( )( )( ) ( )( ) ( )( )( ) ( )( ) Equate and Step 5:

Solve for heat transfer area for each effect: For 1st effect: ( _{* (} _* ( _{* ( )} For 2nd effect: ( _{* (} _* ( _{* ( )} For 3rd effect: ( _{* (} _* ( _{* ( )}

Areas are almost equal, therefore new assumptions are valid: ANSWERS:

Area of each effect: Steam Requirement Steam Economy:

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PROBLEM # 02:

A solution with a negligible boiling point rise is being evaporated in a triple effect evaporator using saturated steam at 121.1°C. The pressure in the vapor of the last effect is 25.6 kPa abs. The heat transfer coefficients are U1 = 2840, U2 = 1988, and U3 = 1420 W/m

2

·K and the areas are equal. Estimate the boiling point in each of the evaporators.

(Source: Transport Processes and Separation Processes, Geankoplis)

SOLUTION:

Using the heat transfer area equation:

Assume equal heat transfer flux for each effect

( ) ( ) ( ) ∑ From steam table,

∑ T1 TI T2 TII F Steam, S 121.1 C P p3= 25.6 kPa abs VI VII LI T3 TIII VIII LII

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( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

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PROBLEM # 03:

A forced-circulation evaporator is to

concentrate 60,000 kg/h of 44 percent NaOH to 65 percent using steam at 3 atm pressure. The

feed temperature and the condensing

temperature are both 40°C. The density of the feed solution is 1,450 kg/m3. If the over-all heat transfer coefficient is 2,000 W/m2·°C, calculate (a) the steam requirement, in kg/h; (b) the heat transfer area required.

(Source: Unit Operations of Chemical Engineering, 7th edition, McCabe and Smith)

SOLUTION:

Consider solute balance (NaOH balance)

( ) ( *

Consider Over-all material Balance:

Consider enthalpy balance:

From figure 16.6 ( ) From figure 16.3 (McCabe and Smith)

For the corresponding boiling point of solution at 65% conc, From figure 16.6, at 65% concentration and 101.67°C

For the superheated vapor, assume Cp of steam is 0.45 BTU/lb·°F ( )

From steam table at 40°C, HV, To = 1106.72 BTU/lb

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[ ( * ( )] ( * ( * ( * ( * ( * ( *

At 3 atm, from steam table, ( ) ( ) ( _{) ( )}

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PROBLEM # 04:

A triple-effect evaporator of the long-tube is to be used to concentrate 35,000 gal/h of a 17% solution of dissolved solids to 38% dissolved solids. The feed enters at 60°Fand passes through three tube-and-shell heaters, a, b, and c, in series and then through the three effects in order II, III,I. Heater “a” is heated by vapor taken from the vapor line between the third effect and the condenser, heater “b” with vapor from the vapor line between the second and the third effects, and heater “c” with vapor from the line between the first and the second effect. In each heater the warm end temperature approach is 10°F. Other data are given below:

Steam to I - 230°F, dry and saturated

Vacuum on III - 28 in (referred to a 30-in barometer)

Condensates leave steam chests at condensing temperatures

Boiling point elevations - 1°F in II, 5°F in III, 15°F in I

Coefficients, in BTU/h·ft2·°F, corrected for BPE - 450 in I, 700 in II, 500 in III All effects have equal areas of heating surface

Concentration, % solids Specific gravity Specific heat, BTU/lb·°F 10 1.02 0.98 20 1.05 0.94 30 1.10 0.87 35 1.16 0.82 40 1.25 0.75

Calculate (a) the steam required in lb/h; (b) heating surface per effect; (c) economy in lb per lb of steam; and (d) the latent heat to be removed in the condenser

(Source: Unit Operations of Chemical Engineering, 7th edition, McCabe and Smith)

SOLUTION:

c

b

a

F= 35,000 gal/h xF= 0.17 TF= 60 F P xP= 0.38

I

II

III

_CONDENSER

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Consider solute balance: At 17% concentration, sp. gr = 1.041 ( ) ( ) Consider over-all material balance:

Assume equal evaporation in each effect:

(

*

Material Balance for each effect: For second effect:

Over-all material balance: Solute balance: ( ) ( ₎

For third effect:

Over-all material balance:

Solute balance: ( ) ( ₎

For first effect: To check:

Over-all material balance:

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Solute balance: ( ) ( ₎ TEMPERATURE DISTRIBUTION Δ

Temperature of vapor leaving the 3rd effect corresponds to the pressure in the 3rd effect

From steam table, at 2 in Hg

There’s a need to assume values of ΔT1, ΔT2, ΔT3

( )( ) ( )( ) ( )( ) ( )( ) ( ) ( )

STREAM DESIGNATION TEMPERATURE,

°F 1st Effect

Steam Feed to E-I 230

Liquor from E-III 106

Vapor to E-II 172 Product 187 2nd Effect Feed from H- c 162 Vapor to E-III 144 Liquor to E-III 145 3rd Effect Vapor to condenser 101 Liquor to E-I 106 Feed 60 Feed to “b” 91 Feed to “c” 134

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Consider heat balance around 1st effect: ( )

For x = 0.2692, Cp = 0.89156 BTU/lb·°F

From steam table, at T1 = 172 °F, λ = 995 BTU/lb; at Ts = 230°F, λ = 958.8 BTU/lb

( )( )( ) ( ) ( )

Consider heat balance around the second effect and heater “c”

Note that the vapor coming from 1st effect will be used to heat the heater and the 2nd effect ( ) ( )

( )

For x = 0.17, Cp = 0.952 BTU/lb·°F

From steam table, at T2 = 144 °F, λ = 1011.64 BTU/lb

( ) ( )( )( )

Consider heat balance around the third effect and heater “b”

Note that the vapor coming from 2nd effect will be used to heat the heater and the 3rd effect

( ) ( )

For x = 0.2084, Cp = 0.9341 BTU/lb·°F

From steam table, at T3 = 101 °F, λ = 1036.44 BTU/lb ( ) ( ) ( )( )( ) ( )( )( ) Equate and Substitute and ( ) ( )

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From equation ( ) ( ) ( ) ( * ( * Δ ( _*( ) ( ) ( * ( * ( ) ( * ( * ( * Δ ( _*( ) ( ) ( * ( * ( ) ( * ( * ( * ( _*( )

Since surface of each effect is not the same, therefore, previous assumptions need to be re-adjusted To adjust ΔT:

Assume constant q and U ( ) ( ) ( * ( ) ( * ( ) ( * ( ) ( *

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SECOND TRIAL:

Recompute using the adjusted ΔT:

STREAM DESIGNATION TEMPERATURE,

°F 1st Effect

Steam Feed to E-I 230

Liquor from E-III 106

Vapor to E-II 170 Product 185 2nd Effect Feed from H- c 160 Vapor to E-III 143 Liquor to E-III 144 3rd Effect Vapor to condenser 101 Liquor to E-I 106 Feed 60 Feed to “b” 91 Feed to “c” 133

Consider heat balance around 1st effect: ( )

For x = 0.2692, Cp = 0.89156 BTU/lb·°F

From steam table, at T1 = 170 °F, λ = 996.2 BTU/lb; at Ts = 230°F, λ = 958.8 BTU/lb

( )( )( ) ( ) ( )

Consider heat balance around the second effect and heater “c”

Note that the vapor coming from 1st effect will be used to heat the heater and the 2nd effect ( ) ( )

( )

For x = 0.17, Cp = 0.952 BTU/lb·°F

From steam table, at T2 = 143 °F, λ = 1012.23 BTU/lb

( ) ( )( )( )

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Consider heat balance around the third effect and heater “b”

Note that the vapor coming from 2nd effect will be used to heat the heater and the 3rd effect

( ) ( )

For x = 0.2084, Cp = 0.9341 BTU/lb·°F

From steam table, at T3 = 101 °F, λ = 1036.44 BTU/lb ( ) ( ) ( )( )( ) ( )( )( ) Equate and Substitute and ( ) ( ) From equation ( ) ( ) ( ) ( * ( * ( _*( ) ( ) ( * ( * ( ) ( * ( * ( * ( _*( ) ( ) ( * ( * ( ) ( * ( * ( *

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( _*( )

Since surface of each effect is not the same, therefore, previous assumptions need to be re-adjusted To adjust ΔT:

Assume constant q and U ( ) ( ) ( * ( ) ( * ( ) ( * ( ) ( *

Since there will be no change in ΔT’s, therefore, assumptions are correct: ( ) ( ) ECONOMY: ( ) ( ) ( * ( * ( ) From steam table, at T3 = 101 °F, λ = 1036.44 BTU/lb

( * ( *

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PROBLEM # 05:

A single effect evaporator concentrates 1 MT of 10% wt sucrose solution to 50%. The feed enters the evaporator at 20°C and has a specific heat of 1.0. The evaporator is maintained at a vacuum of 800 mm Hg against a barometric reading of 760 mm Hg. The heat is provided by saturated steam at 8.8 kg/cm2 gage. Assuming that no sensible heat is recovered in the evaporator, calculate the weight of heating steam, in kg, needed for concentrating the sucrose solution.

(Source: CHE BP May 1990)

SOLUTION:

Consider sucrose balance

( )( )

Consider heat balance:

Since system involves solution of non-electrolytes, assume negligible BPE

( )

From the steam table, @ 160 mm Hg,

( ) (

* ( )

From steam table at 160 mm Hg,

( ) ( *

From steam table at 8.8 kg/cm2 gage * ( ) + P x_P= 0.50 Sucrose Soln F= 1 MT xP= 0.10 TF= 20 C Sp ht = 1.0 Steam, S 8.8 kg/cm2_gage Vapor, V pvacuum= 600 mm Hg Pbarometric= 760 mm Hg TI T1

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PROBLEM # 06:

A solution of organic colloids is to be concentrated from 20 to 65% wt solids in an evaporator. Saturated steam is available at 172 kPa absolute and the pressure in the condenser is 61.07 vacuum. The feed enters at 25°C and its specific heat is 4.0 J/g·°C. The solution has a negligible elevation in boiling point. OHTC is 1,000 W/m2·°C and the evaporator must evaporate 9,000 kg/h. a) Determine the steam consumption, kg/h

b) How many square meters of heating surface are required?

c) What is the steam economy?

SOLUTION:

Consider sucrose balance

( )

(

*

From the steam table, @ 460 mm Hg, ( * ( * ( )

From steam table at 460 mm Hg,

( ) ( * P xP= 0.65 Organic Colloid F xF= 0.20 TF= 25 C sp ht = 4.0 J/g· C Steam, S 172 kPa abs Vapor, V = 9,000 kg/h pvacuum= 61.07 cm TI T1

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From steam table at 172 kPa abs ( ) Δ ( ) ( ) ( ) ( )

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PROBLEM # 07:

An evaporator is to concentrate 10% wt caustic soda solution to 50% wt. Feed enters at 100°F. Steam is available saturated at 50 psig and the evaporator can be operated at 9.96 psi vacuum. OHTC of the evaporator is 500 BTU/h·ft2·°F.

d) Determine the heating area required for the production of 10,000 lb/h of the 50% wt NaOH. e) What is the steam economy?

SOLUTION:

Consider NaOH balance

( ) ( )

Since solution is an electrolyte, it can be expected that there will be BPE Solve for TI for a vacuum pressure of 9.96 psi vacuum (evaporator pressure)

From steam table,

From figure 16.3 (McCabe and Smith), for 50% NaOH solution and TI of 159.95°F

From steam table at 87°F, ⁄

( ) ( ) ( *

From figure 16.6 (McCabe and Smith), At 100°F and 10% NaOH At 159.95 °F and 50% NaOH ( * ( * ( * ( * ( * ( * P = 10,000 lb/h xP= 0.50 NaOH soln F xF= 0.100 TF= 100 F Steam, S 50 psig Vapor, V pvacuum= 9.96 psi TI T1

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Δ

( )

For steam at 50 psig ( _*( ) ( ) ( )

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PROBLEM # 08:

A 10% wt NaOH at 80 °F is to be concentrated in a single effect evaporator to 40% wt. Steam is supplied at 20 psig and the vacuum pressure of the barometric condenser is 26 in Hg. 100 gpm of water is fed to the condenser and the water leaving the condenser (including the condensate) is at 100 °F. OHTC of evaporator is 200 BTU/h·ft2·°F. Calculate the heating surface required for the evaporator. (Source: CHE BP May 1984)

SOLUTION:

Consider heat balance around the condenser:

( ) ( ) ( * ( ) ( )

Assume barometric pressure of 1 atm or 29.921 in Hg

From steam table, at 3.921 in Hg,

The vapor will be condensed first before lowering to 100°F, thus, Cp of the liquid water should be used ( ) ( _{) (}_{) ( )} For the evaporator Consider NaOH balance

( )

Consider Over-all material Balance: P xP= 0.40 NaOH soln F xF= 0.10 TF= 80 F Steam, S 20 psig Vapor, V pvacuum= 26”Hg TI T1 tB=T = 100 F tA= 70 F 100 gpm

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Since solution is an electrolyte, it can be expected that there will be BPE

From figure 16.3 (McCabe and Smith), for 40% NaOH solution and T1 of 124.37°F

From steam table at 124.37°F, ⁄

( ) ( ) ( *

From figure 16.6 (McCabe and Smith), At 80°F and 10% NaOH At 170 °F and 40% NaOH ( * ( * ( * ( * ( * ( * Δ ( )

For steam at 20 psig ( _*( )

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PROBLEM # 09:

4,500 kg/h of a 10% wt sugar solution is to be concentrated to 30% wt. Feed enters at 21°C. Saturated steam at 110°C is available and the temperature in the condenser is 43°C. Specific heat of the solutions may be taken as constant at 4 J/g·°C. Determine (a) heating surface required, (b) steam consumption, and (c) steam economy, for each of the following cases:

I. Single effect, OHTC = 2,840 W/m2·°C

II. Double effect, forward feed; U1 = 2,270, U2 = 1,700 W/m 2

·°C III. Double effect, backward feed; U1 = 2,270, U2 = 1,700 W/m

2

·°C

SOLUTION:

Consider sugar balance

( ) ( *

CASE I: SINGLE EFFECT

Since solution is non electrolyte, BPE is negligible

Temperature of vapor leaving the evaporator is the same as the condenser temperature ( ) ( * ( * ( ) P xP= 0.30 Sugar soln F=4,500 kg/h xF= 0.10 TF= 21 C Steam, S 110 C Vapor, V TI T1 T1= 43 C

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From steam table, at 43°C ( * ( * Δ ( ) ( _{) ( )} ( )

From steam table, at 110°C, ( ) ( )

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Initial Assumptions:

Assume equal evaporation rates

(

*

Consider material balance around the 1st effect

Solve for the temperature distribution; to assume ΔT’s Assume equal heat flux

( ) ( ) Δ ( Δ ) ( Δ ) T1 TI T2 TII

Sugar soln

F=4,500 kg/h

x

F

= 0.10

T

F

= 21 C

Steam, S

110 C

P

x

P

= 0.30

T

2

= 43 C

V

I

V

II

P

I

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Δ Δ ( _{) Δ (} * Δ Δ ( ) ( * Δ Δ ( ) Δ ( * Δ Δ ( ) ( * ( )

Enthalpy balance around the 2nd effect:

Enthalpy balance around the 1st effect:

( ) Equate and ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) From steam table;

ΔTI ΔTII Ts= 110 C TII= 43 C TI

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( )( ) ( )( ) ( ) ( ) ( )( ) ( ) Equate and ( * ( * Δ ( _{) ( )} Δ ( _{) ( )}

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Since AI = AII, therefore original assumptions are correct ( ) ( ) ( )

CASE III: DOUBLE EFFECT, BACKWARD FEED

Initial Assumptions:

Assume equal evaporation rates

(

*

Consider material balance around the 1st effect T1 TI T2 TII Sugar soln F=4,500 kg/h xF= 0.10 TF= 21 C Steam, S 110 C P xP= 0.30 T2= 43 C VI VII PI

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Solve for the temperature distribution; to assume ΔT’s Assume equal heat flux

( ) ( ) Δ ( Δ ) ( Δ ) Δ Δ ( _{) Δ (} * Δ Δ ( ) ( * Δ Δ ( ) Δ ( * Δ ( ) ( * ( )

Enthalpy balance around the 2nd effect:

Enthalpy balance around the 1st effect:

Equate and ( ) ( ) ΔTI ΔTII Ts= 110 C TII= 43 C TI

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( ) ( ) ( ) ( ) ( ) ( ) From steam table;

( ) ( ) ( )( ) ( )( ) ( )( ) ( ) Equate and ( * ( * ( _{) ( )}

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( _{) ( )}

Since AI = AII, therefore original assumptions are correct

( ) ( ) ( )

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PROBLEM # 10:

Glycerine is to be concentrated from 12% to 72% in a single-effect evaporator. The inlet steam used is at 25 psig and comes out at 170°F. The vapor space in the evaporator has 25 inches Hg vacuum. Ten metric tons of glycerine per hour are fed at 85°F. The concentrated product is at 125°F. Calculate the amount of water evaporated in kg/h.

(Source: MRII Reviewer)

SOLUTION:

Consider solute balance:

( ) ( *

Consider over-all material balance: P xP= 0.72 T_F= 125 F F=10 MT/h xF= 0.12 T_F= 85 F Steam, S 25 psig Vapor, V TI P = 25” Hg vac 170 F

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PROBLEM # 11:

A feed of 4,535 kg/h of a 2.0 % wt salt solution at 311 K enters continuously a single effect evaporator and is being concentrated to 3%. The evaporation is at atmospheric pressure and the area of the evaporator is 69.7 m2. Saturated steam at 383.2 K is supplied for heating. Since the solution is dilute, it can be assumed to have the same boiling point as water. The heat capacity of the feed can be taken as Cp = 4.10 kJ/kg·K. Calculate the amounts of vapor and liquid product and the over-all heat transfer coefficient.

(Source: Principles of Transport Processes and Separation Processes, 4th edition, by Geankoplis)

SOLUTION:

Consider solute balance:

( ) ( *

Consider over-all material balance:

P xP= 0.03 F=4535 kg/h xF= 0.02 TF= 311 K Steam, S 383.2 K Vapor, V TI P = 1 atm

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Since the evaporator operates at 1 atm, operating temperature will be the the temperature corresponding to 1 atm or water boiling point (373 K)

( )

( * (

* ( )

From the steam table at 373 K, ( * ( * ( ) ( )( )

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PROBLEM # 12:

An evaporator having an area of 83.6 m2 and U = 2270 W/m2·K is used to produce distilled water for a boiler feed. Tap water having 400 ppm dissolved solids at 15.6 °C is fed to the evaporator operating at 1 atm pressure abs. Saturated steam at 115.6°C is available for use. Calculate the amount of distilled water produced per hour if the outlet liquid contains 800 ppm solids.

SOLUTION:

Assume that the boiling point of the solution is 100°C (1 atm operating pressure)

( ) ( ) ( )( ) Consider solute balance:

( )

( ) For water, Cp = 4.1868 kJ/kg·°C ( ) ( * ( )

From the steam table at 100°C, ( ) ( * ( ) P xP= 0.08 xF= 0.04 TF= 15.6 C Steam, S 115.6 C Vapor, V T_I P = 1 atm

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PROBLEM # 13:

A single effect evaporator is concentrating a feed of 9,072 kg/h of a 10 wt % solution of NaOH in water to a product of 50 % solids. The pressure of saturated steam used is 42 kPa (gage) and the pressure in the vapor space of the evaporator is 20 kPa (abs). The over-all heat transfer coefficient is 1,988 W/m2·K. Calculate the steam used, the steam economy in kg vaporized/kg steam, and the area for the following feed conditions:

a) Feed temperature of 288.8 K b) Feed temperature of 322.1 K

SOLUTION:

Consider NaOH balance:

( ) ( *

For 20 kPa evaporator vapor space pressure, the temperature of the vapor (from steam table)

Since system is NaOH, it is expected to have BPE

From figure 8.4-2 (Geankoplis), for 50% NaOH concentration and TV = 333.06 K (60.06°C) [ ( )]

From steam table, at 60.06°C, ⁄ , ⁄

P xP= 0.50 NaOH soln F=9,072 kg/h xF= 0.10 TF Steam, S 42 kPa (gage) Vapor, V T_I PV= 20 kPa (abs)

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( * [

( ⁄ ) ( )]

A. For feed temperature of 288.8 K Consider enthalpy balance:

From steam table, at 42 kPa gage, ⁄ , From figure 8.4-3 (Geankoplis)

( * ( * ( * ( * ( ) ( ) ( * ( * ( ) ( ) ( _{) ( )}

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B. For feed temperature of 322.1 K Consider enthalpy balance:

From steam table, at 42 kPa gage, ⁄ , From figure 8.4-3 (Geankoplis)

( * ( * ( * ( * ( ) ( ) ( * ( * ( ) ( ) ( _{) ( )}

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PROBLEM # 14:

In order to concentrate 4,536 kg/h of an NaOH solution containing 10 wt % NaOH to a 20 wt % solution, a single effect evaporator is being used with an area of 37.6 m2. The feed enters at 21.1 °C. Saturated steam at 110 °C is used for heating and the pressure in the vapor space of the evaporator is 51.7 kPa abs. Calculate the kg/h of steam used and the over-all heat transfer coefficient. (Source: Principles of Transport Processes and Separation Processes, 4th edition, by Geankoplis)

SOLUTION:

( ) ( *

For 51.7 kPa evaporator vapor space pressure, the temperature of the vapor (from steam table)

From figure 8.4-2 (Geankoplis), for 20% NaOH concentration and TV = 82.06°C [ ( )]

From steam table, at 82.06°C, ⁄ , ⁄ ( * [ ( ⁄ ) ( )] P xP= 0.20 NaOH soln F=4,536 kg/h x_F= 0.10 T_F= 21.1 C Steam, S 110 C Vapor, V T_I P_V= 51.7 kPa (abs)

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From steam table, at 110°C, ⁄ From figure 8.4-3 (Geankoplis)

( * ( * ( * ( * ( ) ( ) ( * ( * ( ) ( ) ( )( )

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PROBLEM # 15:

An evaporator is concentrating F kg/h at 311 K of a 20 wt % solution of NaOH to 50%. The saturated steam for heating is at 399.3 K. The pressure in the vapor space of the evaporator is 13.3 kPa abs. The over-all heat transfer coefficient is 1,420 W/m2·K and the area is 86.4 m2. Calculate the feed rate F of the evaporator.

SOLUTION:

( )( )

From figure 8.4-2 (Geankoplis), for 50% NaOH concentration and TV = 51.39°C [ ( )]

From steam table, at 51.39°C, ⁄ , ⁄ ( ) [

( ⁄ ) ( )]

P xP= 0.50 NaOH soln F xF= 0.20 TF = 311 K Steam, S 399.3 K Vapor, V TI PV= 13.3 kPa (abs)

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From steam table, at 399.3 K, ⁄

From figure 8.4-3 (Geankoplis)

( ) ( ) ( * ( ) ( * ( ) ( ) ( )( )

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PROBLEM # 16:

A single effect evaporator is concentrating a feed solution of organic colloids from 5 to 50 wt %. The solution has a negligible boiling point elevation. The heat capacity of the feed is Cp= 4.06 kJ/kg·K and the feed enters at 15.6°C. Saturated steam at 101.32 kPa is available for heating, and the pressure in the vapor space of the evaporator is 15t.3 kPa. A total of 4,536 kg/h of water is to be evaporated. The over-all heat transfer coefficient is 1,988 W/m2·K. What is the

required surface area in m2 and the steam

consumption?

SOLUTION:

( )( )

For steam at 101.32 kPa

Since no BPE

( ) P xP= 0.50 F x_F= 0.05 TF = 15.6 C Steam, S 101.32 kPa Vapor, V = 4,536 kg/h T_I PV= 15.3 kPa (abs)

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