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554 CHAPTER 8 Evaporation are presented in a general way. Custom-designed modifications of these types can significantly alter the specific duties. 8.3 DESIGN OF A SINGLE-EFFECT EVAPORATOR In a single-effect evaporator, as shown in Figure 8.11, dilute liquid feed is pumped into the heating chamber, whe...

Description

554

CHAPTER 8 Evaporation

are presented in a general way. Custom-designed modifications of these types can significantly alter the specific duties.

8.3 DESIGN OF A SINGLE-EFFECT EVAPORATOR In a single-effect evaporator, as shown in Figure 8.11, dilute liquid feed is pumped into the heating chamber, where it is heated indirectly with steam. Steam is introduced into the heat exchanger, where it condenses to give up its heat of vaporization to the feed, and exits the system as condensate. The temperature of evaporation, T1, is controlled by maintaining vacuum inside the heating chamber. The vapors leaving the product are conveyed through a condenser to a vacuum system, usually a steam ejector or a vacuum pump. In a batch system, the feed is heated until the desired concentration is obtained. The concentrated product is then pumped out of the evaporator system. Heat and mass balances conducted on the evaporator system allow determination of various design and operating variables. Such variables may include mass flow rates, final concentration of product, and heat-exchanger area. The following expressions can be obtained by conducting a mass balance on flow streams and product solids, respectively. (8.1)

ɺf  m ɺv m ɺp m

■ Figure 8.11

Schematic diagram of a single-effect evaporator.

Condenser · mv, T1 · ms, Ts

T1

System boundary

U, A

· ms, Ts · mf, xf, Tf

m· p, xp, T1

Vacuum ejector

8.3 Design of a Single-Effect Evaporator 555

ɺ f is the mass flow rate of dilute liquid feed (kg/s), m ɺ v is the where m ɺ p is the mass flow rate of conmass flow rate of vapor (kg/s), and m centrated product (kg/s), ɺ f  x pm ɺp xf m

(8.2)

where xf is the solid fraction in the feed stream (dimensionless) and xp is the solid fraction in the product stream (dimensionless). An enthalpy balance conducted on the evaporator system gives the following expression: ɺ f Hf  m ɺ s H vs  m ɺ v H v1  m ɺ p H p1  m ɺ s Hcs m

(8.3)

ɺ s is the mass flow rate of steam (kg/s); Hf is enthalpy of where m dilute liquid feed (kJ/kg); Hp1 is enthalpy of concentrated product (kJ/kg); Hvs is enthalpy of saturated vapor at temperature Ts (kJ/kg); Hv1 is enthalpy of saturated vapor at temperature T1 (kJ/kg); Hcs is enthalpy of condensate (kJ/kg); Ts is temperature of steam (°C); T1 is the boiling temperature maintained inside the evaporator chamber (°C); and Tf is the temperature of dilute liquid feed (°C). ɺ f H f , represents the total enthalpy The first term in Equation (8.3), m associated with the incoming dilute liquid feed, where Hf is a function of Tf and xf. The enthalpy content Hf can be computed from H f  cpf (Tf  0°C)

(8.4)

The specific heat may be obtained either from Table A.2.1 or by using Equation (4.3) or Equation (4.4). ɺ s H vs , gives the total heat content of steam. It is The second term, m assumed that saturated steam is being used. The enthalpy, Hvs, is obtained from the steam table (Table A.4.2) as enthalpy of saturated vapors evaluated at the steam temperature Ts. ɺ v H v1, repOn the right-hand side of Equation (8.3), the first term, m resents total enthalpy content of the vapors leaving the system. The enthalpy Hv1 is obtained from the steam table (Table A.4.2) as the enthalpy of saturated vapors evaluated at temperature T1. ɺ p H p1 , is the total enthalpy associated with the The second term, m concentrated product stream leaving the evaporator. The enthalpy content Hp1 is obtained using the following equation: H p1  cpp (T1  0°C)

(8.5)

where cpp is the specific heat content of concentrated product (kJ/[kg °C]).

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CHAPTER 8 Evaporation

Again, cpp is obtained from Table A.2.1 or by using Equation (4.3) or Equation (4.4). ɺ s Hcs , represents the total enthalpy associated with The last term, m the condensate leaving the evaporator. Since an indirect type of heat exchanger is used in evaporator systems, the rate of mass flow of incoming steam is the same as the rate of mass flow of condensate leaving the evaporator. The enthalpy Hcs is obtained from the steam table (Table A.4.2) as enthalpy of saturated liquid evaluated at temperature Ts. If the condensate leaves at a temperature lower than Ts, then the lower temperature should be used to determine the enthalpy of the saturated liquid. In addition to the mass and enthalpy balances given previously, the following two equations are also used in computing design and operating variables of an evaporator system. For the heat exchanger, the following expression gives the rate of heat transfer: ɺ s H vs  m ɺ s Hcs q  UA(Ts  T1) = m

(8.6)

where q is the rate of heat transfer (W), U is the overall heat transfer coefficient (W/[m2 K]), and A is the area of the heat exchanger (m2). The overall heat-transfer coefficient decreases as the product becomes concentrated, due to increased resistance of heat transfer on the product side of the heat exchanger. In addition, the boiling point of the product rises as the product becomes concentrated. In Equation (8.6), a constant value of the overall heat-transfer coefficient is used and would result in some “overdesign” of the equipment. Steam economy is a term often used in expressing the operating performance of an evaporator system. This term is a ratio of rate of mass of water vapor produced from the liquid feed per unit rate of steam consumed. ɺ v /m ɺs Steam economy  m

(8.7)

A typical value for steam economy of a single-effect evaporator system is close to 1.

Example 8.2

Apple juice is being concentrated in a natural-circulation single-effect evaporator. At steady-state conditions, dilute juice is the feed introduced at a rate of 0.67 kg/s. The concentration of the dilute juice is 11% total solids. The juice

8.3 Design of a Single-Effect Evaporator 557

is concentrated to 75% total solids. The specific heats of dilute apple juice and concentrate are 3.9 and 2.3 kJ/(kg °C), respectively. The steam pressure is measured to be 304.42 kPa. The inlet feed temperature is 43.3°C. The product inside the evaporator boils at 62.2°C. The overall heat-transfer coefficient is assumed to be 943 W/(m2 °C). Assume negligible boiling-point elevation. Calculate the mass flow rate of concentrated product, steam requirements, steam economy, and the heat-transfer area. The system is sketched in Figure E8.1.

Given ɺ f  0.67 kg/s Mass flow rate of feed m Concentration of food xf  0.11 Concentration of product xp  0.75 Steam pressure  304.42 kPa Feed temperature Tf  43.3°C Boiling temperature T1 in evaporator  62.2°C Overall heat transfer coefficient U  943 W/(m2 K) Specific heat of dilute feed cpf  3.9 kJ/(kg °C) Specific heat of concentrated product cpp  2.3 kJ/(kg °C)

Approach We will use the heat and mass balances given in Equations (8.1), (8.2), and (8.3) to determine the unknowns. Appropriate values of enthalpy for steam and vapors will be obtained from steam tables.

· mv, T1 · ms, Ts

T1

System boundary

U, A

· ms, Ts · mf, xf, Tf

· mp, xp, T1

■ Figure E8.1

Schematic diagram of a single-effect evaporator.

558

CHAPTER 8 Evaporation

Solution 1. From Equation (8.2), ɺp ( 0.11)( 0.67 kg/s )  ( 0.75 )m ɺ p  0.098 kg/s m Thus, mass flow rate of concentrated product is 0.098 kg/s. 2. From Equation (8.1), ɺ v  ( 0.67 kg/s )  ( 0.098 kg/s ) m ɺ v  0.57 kg/s m Thus, mass flow rate of vapors is 0.57 kg/s. 3. To use the enthalpy balance of Equation (8.3), the following quantities are first determined: From Equation (8.4), Hf  ( 3.9 kJ/[kg °C] )( 43.3°C  0 °C )  168.9 kJ/kg From Equation (8.5), H p1  ( 2.3 kJ/[kg °C] )( 62.2 °C  0 °C )  143.1 kJ/kg From the steam table (Table A.4.2), Temperature of steam at 304.42 kPa  134°C Enthalpy for saturated vapor Hvs (at Ts  134°C)  2725.9 kJ/kg Enthalpy for saturated liquid Hcs (at Ts  134°C)  563.41 kJ/kg Enthalpy for saturated vapor Hv1 (at T1  62.2°C)  2613.4 kJ/kg ɺ s kg/s )( 2725.9 kJ/kg ) ( 0.67 kg/s )(168.9 kJ/kg )  ( m  ( 0.57 kg/s )( 2613.4 kJ/kg )  ( 0.098 kg/s )(143.1 kJ/kg ) ɺ s kg/s )( 563.41 kJ/kg ) ( m ɺ s  1390.5 2162.49 m ɺ s  0.64 kg/s m 4. To calculate steam economy, we use Equation (8.7). Steam economy 

ɺv 0.57 m   0.85 kg water evaporated/kg steam ɺs 0.67 m

5. To compute area of heat transfer, we use Equation (8.6). A( 943 W/[m2 °C] )(134 °C  62.2 °C )  ( 0.64 kg/s )( 2725.9  563.14 kJ/kg )(1000 J/kJ ) A  20.4 m2

8.4 Design of a Multiple-Effect Evaporator 559

8.4 DESIGN OF A MULTIPLE-EFFECT EVAPORATOR In a triple-effect evaporator, shown in Figure 8.12, dilute liquid feed is pumped into the evaporator chamber of the first effect. Steam enters the heat exchanger and condenses, thus discharging its heat to the product. The condensate is discarded. The vapors produced from the first effect are used as the heating medium in the second effect, where the feed is the partially concentrated product from the first effect. The vapors produced from the second effect are used in the third effect as heating medium, and the final product with the desired final concentration is pumped out of the evaporator chamber of the third effect. The vapors produced in the third effect are conveyed to a condenser and a vacuum system. In the forward feed system shown, partially concentrated product from the first effect is fed to the second effect. After additional concentration, product leaving the second effect is introduced into the third effect. Finally, product with the desired concentration leaves the third effect. Design expressions for multiple-effect evaporators can be obtained in the same manner as for a single-effect evaporator, discussed in Section 8.3.

■ Figure 8.12

Schematic diagram of a triple-effect evaporator.

· mv3,T3

System boundary

· ms, Ts

m· v1

T1

T2

T3

U1, A1

U2, A2

U3, A3

· mf1 xf1 · mf, xf, Tf

m· v2

· mf2 xf2

· mp, T3, xp

560

CHAPTER 8 Evaporation

Conducting mass balance analysis on the flow streams, ɺf  m ɺ v1  m ɺ v2  m ɺ v3  m ɺp m

(8.8)

ɺ f is the mass flow rate of dilute liquid feed to the first effect where m ɺ v2, and m ɺ v3 are the mass flow rates of vapor from the ɺ v1, m (kg/s); m ɺ p is the mass first, second, and third effect, respectively (kg/s); and m flow rate of concentrated product from the third effect (kg/s). Using mass balance on the solids fraction in the flow streams, ɺ  x pm ɺp xf m

(8.9)

where xf is the solid fraction in the feed stream to be consistent with the first effect (dimensionless) and xp is the solid fraction in the product stream from the third effect (dimensionless). We write enthalpy balances around each effect separately. ɺ f Hf  m ɺ s H vs  m ɺ v1H v1  m ɺ f1H f1  m ɺ s Hcs m

(8.10)

ɺ f1H f1  m ɺ v1H v1  m ɺ v2 H v2  m ɺ f2 Hf2  m ɺ v1Hc1 m

(8.11)

ɺ f2H f2  m ɺ v2H v2  m ɺ v3H v3  m ɺ p Hp3  m ɺ v2Hc2 m

(8.12)

where the subscripts 1, 2, and 3 refer to the first, second, and third effect, respectively. The other symbols are the same as defined previously for a single-effect evaporator. The heat transfer across heat exchangers of various effects can be expressed by the following three expressions: ɺ s H vs  m ɺ s Hcs q1  U1 A1(Ts  T1)  m

(8.13)

ɺ v1H v1  m ɺ v1Hc1 q2  U 2 A2(T1  T2 )  m

(8.14)

ɺ v2H v2  m ɺ v2Hc2 q3  U3 A3(T2  T3 )  m

(8.15)

The steam economy for a triple-effect evaporator as shown in Figure 8.12 is given by Steam economy 

ɺ v1  m ɺ v2  m ɺ v3 m ɺ ms

(8.16)

Example 8.3 illustrates the use of these expressions in evaluating the performance of multiple-effect evaporators.

8.4 Design of a Multiple-Effect Evaporator 561

Calculate the steam requirements of a double-effect forward-feed evaporator (Fig. E8.2) to concentrate a liquid food from 11% total solids to 50% total solids concentrate. The feed rate is 10,000 kg/h at 20°C. The boiling of liquid inside the second effect takes place under vacuum at 70°C. The steam is being supplied to the first effect at 198.5 kPa. The condensate from the first effect is discarded at 95°C and from the second effect at 70°C. The overall heat-transfer coefficient in the first effect is 1000 W/(m2 °C); in the second effect it is 800 W/(m2 °C). The specific heats of the liquid food are 3.8, 3.0, and 2.5 kJ/(kg °C) at initial, intermediate, and final concentrations. Assume the areas and temperature gradients are equal in each effect.

Example 8.3

Given

. Mass flow rate of feed  mf  10,000 kg/h  2.78 kg/s Concentration of feed xf  0.11 Concentration of product xp  0.5 Steam pressure  198.5 kPa Feed temperature  20°C Boiling temperature T2 in second effect  70°C Overall heat-transfer coefficient U1 in first effect  1000 W/(m2 °C) Overall heat-transfer coefficient U2 in second effect  800 W/(m2 °C) Specific heat of dilute feed cpf  3.8 kJ/(kg °C) Specific heat of feed at intermediate concentration c pf  3.0 kJ/(kg °C) Specific heat of concentrated food product cpp  2.5 kJ/(kg °C)

■ Figure E8.2

System boundary

· ms, Ts

· mv1

T1

T2

U1, A1

U2, A2

· mf1 xf1 · mf, xf, Tf

m· v, T2

· mp, T2, xp

Schematic diagram of a double-effect evaporator.

562

CHAPTER 8 Evaporation

Approach Since this is a double-effect evaporator, we will use modified forms of Equations (8.8), (8.9), (8.10), (8.11), (8.13), and (8.14). Enthalpy values of steam and vapors will be obtained from steam tables.

Solution 1. From Equation (8.9), ɺp ( 0.11)( 2.78 kg/s )  ( 0.75 )m ɺ mp  0.61 kg/s 2. From Equation (8.8), ɺ v1  m ɺ v2  0.61 2.78  m Thus, the total amount of water evaporating is ɺ v1  m ɺ v2  2.17 kg/s m 3. Steam is being supplied at 198.5 kPa or 120°C, the temperature in the second effect is 70°C, and thus the total temperature gradient is 50°C. T1 T2  50°C Assuming equal temperature gradient in each evaporator effect, T1  T2  25°C 4. The area of heat transfer in the first and second effects are the same. Thus, from Equations (8.13) and (8.14), q1 q2  U1( Ts  T1 ) U2 ( T1  T2 ) or ɺ sHvs  m ɺ sHcs ɺ H m ɺ v1Hc1 m m  v1 v1 U1( Ts  T1 ) U2 ( T1  T2 ) 5. To use Equations (8.10) and (8.11), we need values for enthalpy of product. Hf  c pf ( Tf  0 )  ( 3.8 kJ/[kg °C])(20 °C 0 °C)  76 kJ/kg Hf1 = c pf ( T1 0 )  ( 3.0 kJ / [ kg °C])(95 °C 0 °C)  285 kJ/kg Hf 2  c pp ( T2 0 )  ( 2.5 kJ/[kg °C])(70 °C 0 °C) 175 kJ/kg

8.4 Design of a Multiple-Effect Evaporator 563

In addition, from steam tables, At Ts 120 °C At T1  95 °C At T2  70 °C

Hvs  2706.3 kJ/kg Hcs  503.71 kJ/kg Hv1  2668.1 kJ/kg Hc1  397.96 kJ/kg Hv2  2626.8 kJ/kg Hc2  292.98 kJ/kg

6. Thus, substituting enthalpy values from step (5) in the equation given in step (4), ɺ s kg/s( 2706.3 kJ/kg )  ( m ɺ s kg/s)( 503.71 kJ/kg )](1000 J/kJ ) [( m 2 (1000 W/[m °C] )(120 °C  95 °C ) 

ɺ v1 kg/s( 2668.1 kJ/kg )  ( m ɺ v1 kg/s)( 397.96 kJ/kg )](1000 J/kJ ) [( m 2 ( 800 W/[m °C] )( 95 °C  70 °C )

or ɺs ɺ v1 2205.59 m 2270.14m  25 , 000 20 , 000 7. Using Equations (8.10) and (8.11), ɺ s )( 2706.3 ) ( 2.78 )( 76 )  ( m ɺ v1 )( 2668.1)  ( m ɺ f 1 )( 285 )  ( m ɺ s )( 503.71) (m ɺ f 1 )( 285 )  ( m ɺ v1 )( 2668.1) (m ɺ v 2 )( 2626.8 )  ( m ɺ p )(175 )  ( m ɺ v1 )( 397.96 ) (m 8. Let us assemble all equations representing mass flow rates of product, feed, vapor, and steam. From step (1): From step (2): From step (6): From step (7):

ɺ p  0.61 m ɺ v1  m ɺ v 2  2.17 m ɺ ɺ v1 0.088 ms  0.114m

ɺ s  2668.1m ɺ v1  285 m ɺ f1  211.28 2202.59 m ɺ p  285 m ɺ f1 ɺ v1  2626.8 m ɺ v 2 175 m 2270.14m ɺp , m ɺ v1 , 9. In step (8), we have five equations with five unknowns, namely, m ɺ v2 , m ɺ s , and m ɺ f1 . We will solve these equations using a spreadsheet procem dure to solve simultaneous equations. The method described in the following was executed on Excel™ . 10. The simultaneous equations are rewritten so that all unknown variables are collected on the right-hand side. The equations are rewritten so that the

564

CHAPTER 8 Evaporation

■ Figure E8.3

A spreadsheet to solve simultaneous equations.

A B 1 1.000 2 3 0.000 4 0.000 5 0.000 175.000 6 7 MINVERSE(B2:F6) 8 1.000 9 0.045 10 0.034 11 12 0.034 0.022 13 14

C

D

0.000 0.000 0.000 1.000 0.088 0.114 2202.590 2668.100 0.000 2270.140 0.000 0.670 0.517 0.483 0.336

0.000 4.984 4.925 4.925 84.621

E

F

0.000 0.000 1.000 0.000 0.000 0.000 0.000 285.000 2626.800 285.000 0.000 0.000 0.000 0.000 0.003

G

H

0.61 2.17 0 211.28 0 MMULT(B9:F13,H2:H6)

0.000 0.000 0.000 0.000 0.000

0.61 1.43 1.10 1.07 1.46

coefficients can easily be arranged in a matrix. The spreadsheet method will use a matrix inversion process to solve the simultaneous equations. ɺ p  0m ɺ s  0m ɺ v1  0 m ɺ v2  0 m ɺ f1  0.61 m ɺ p  0m ɺ s m ɺ v2  0 m ɺ f1  2.17 ɺ v1  m 0m ɺ p  0.088 m ɺ s  0.114m ɺ v1  0 m ɺ v2  0 m ɺ f1  0 0m ɺ p  2202.59 m ɺ s  2668.1m ɺ v1  0 m ɺ v2  285 m ɺ f1 211.28 0m ɺ ɺ ɺ ɺ ɺ 175 mp  0 ms  2270.14mv1  2626.8 mv2  285mf1  0 11. As shown in Figure E8.3, enter the coefficients of the left-hand side of the preceding equations in array B2:F6; enter the right-hand side coefficients in a column vector H2:H6. 12. Select another array B9:F13 (by dragging the cursor starting from cell B9). Type  MINVERSE(B2:F6) in cell B9 and press the CTRL, SHIFT, and ENTER keys simultaneously. This procedure will invert the matrix B2:F6 and give the coefficients of the inverted matrix in array B9:F13. 13. Highlight cells H9:H13 by dragging the cursor starting from cell H9. Type  MMULT(B9:F13,H2:H6) into cell H9; press the CTRL, SHIFT, and ENTER keys simultaneously. The answers are displayed in the column vector H9:H13. Thus, ɺ p  0.61 kg/s m ɺ s  1.43 kg/s m ɺ v1  1.10 kg/s m ɺ v2  1.07 kg/s m ɺ f1  1.46 kg/s m 14. The steam requirements are computed to be 1.43 kg/s. 15. The steam economy can be computed as ɺ v1  m ɺ v2 1.10 1.07 m   1.5 kg water vapor/kg steam ɺs 1.43 m

8.5 Vapor Recompression Systems 565

8.5 VAPOR RECOMPRESSION SYSTEMS The preceding discussion on multiple-effect evaporators has shown how energy requirements of the total system are decreased by using exit vapors as the heating medium in subsequent effects. Two additional systems that employ vapor recompression assist in reduction of energy requirements. These systems are thermal recompression and mechanical vapor recompression. A brief introduction to these two systems follows.

8.5.1 Thermal Recompression Thermal recompression involves the use of a steam jet booster to recompress part of the exit vapors, as shown in Figure 8.13. Through recompression, the pressure and temperature of exit vapors are increased. These systems are usually applied to single-effect evaporators or to the first effect of multiple-effect evaporators. Application of this system requires that steam be available at high pressure, and lowpressure steam is needed for the evaporation process.

Water Recycled vapors