Example 4:

Estimate the requirements of steam and heat transfer surface, and the evaporating temperatures in each effect, for a triple effect evaporator evaporating 500 kg h -1 of a 10% solution up to a 30% solution.

Steam is available at 200 kPa gauge and the pressure in the evaporation space in the final effect is 60 kPa absolute. Assume that the overall heat transfer coefficients are 2270, 2000 and 1420 J m -2 s -1 °C -1 in the first, second and third effects, respectively. Neglect sensible heat effects and assume no boiling-point elevation, and assume equal heat transfer in each effect. Source: http://www.nzifst.org.nz/unitoperations/evaporation2.htm

Prof. R. Shanthini 28 May 2012

500 kg h -1 10% solution 60 kPa (abs) 200 kPa (g) 30% solution

Overall heat transfer coefficients are 2270, 2000 and 1420 J m -2 s -1 first, second and third effects, respectively. °C -1 in the Estimate the requirements of steam and heat transfer surface, and the evaporating temperatures in each effect.

Neglect sensible heat effects and assume no boiling-point elevation, and assume equal heat transfer in each effect. Prof. R. Shanthini 28 May 2012

Overall mass balance:

Data: A triple effect evaporator is evaporating 500 kg/h of a 10% solution up to a 30% solution.

Solids Solvent (water) Solution (total)

Feed Concentrated product

10% of total = 50 kg/h 500 kg/h – 50 kg/h = 450 kg/h 500 kg/h 50 kg/h 167 kg/h – 50 kg/h = 117 kg/h (50/ 30 )*100 = 167 kg/h

Vapour from all effects

0 333 kg/h 500 kg/h - 167 kg/h = 333 kg/h Prof. R. Shanthini 28 May 2012

Steam properties:

Data: Steam is available at 200 kPa gauge and the pressure in the evaporation space in the final effect is 60 kPa absolute. (Neglect sensible heat effects and assume no boiling-point elevation)

Steam pressure

200 kPa (g) = 2 bar (g) = 3 bar (abs) 60 kPa (abs) = 0.6 bar (abs)

Saturation temperature

133.5

o C 86.0

o C

Latent heat of vapourization

2164 kJ/kg 2293 kJ/kg Prof. R. Shanthini 28 May 2012

Evaporator layout:

Data: Steam is available at 200 kPa gauge and the pressure in the evaporation space in the final effect is 60 kPa absolute. (Neglect sensible heat effects and assume no boiling-point elevation) Steam temperature Solution temperature Temperature driving force

First effect

133.5

o C T 1 o C ΔT 1 = 133.5 – T 1

Second effect

T 1 o C T 2 o C ΔT 2 = T 1 – T 2

Third effect

T 2 o C 86.0

o C ΔT 1 = T 2 – 86.0 Prof. R. Shanthini 28 May 2012

Heat balance:

Data: Assume that the overall heat transfer coefficients are 2270, 2000 and 1420 J m -2 s -1 °C -1 in the first, second and third effects respectively. Assume equal heat transfer in each effect.

q

1 =

q

2 =

q

3 which gives U 1 A 1 ΔT 1 = U 2 A 2 ΔT 2 = U 3 A 3 ΔT 3 U 1 , U 2 and U 3 are given. A 1 , A 2 and A 3 can be found if ΔT 1 , ΔT 2 and ΔT 3 are known.

Let us assume that the evaporators are so constructed that

A

1 then we have =

A

2 =

A

3 , U 1 ΔT 1 = U 2 ΔT 2 = U 3 ΔT 3 That is, 2270 (133.5 – T 1 ) = 2000 (T 1 – T 2 ) = 1420 (T 2 – 86.0 ) There are two equations and two unknowns in the above expression. The equations can be solved to give the following: T 1 Prof. R. Shanthini 28 May 2012 = 120.8

o C and T 2 = 106.3

o C

Properties in all effects:

Steam temperature

First effect

133.5

o C

Second effect

T 1 = 120.8

o C

Third effect

T 2 = 106.3

o C Solution temperature T 1 = 120.8

o C Temperature driving force ΔT 1 = 12.7

o C Heat transfer coefficient Latent heat of vapourization of steam U λ 1 1 = 2270 J m -2 s -1 °C -1 = 2164 kJ/kg Latent heat of vapourization of solution 2200 kJ/kg Prof. R. Shanthini 28 May 2012 T 2 ΔT = 106.3

2 = 14.4

o o U 2 = 2000 J m -2 s -1 °C -1 λ 2 = 2200 kJ/kg 2240 kJ/kg C C ΔT 86.0

1 o C = 20.3

o U 3 = 1420 J m -2 s -1 °C -1 λ 3 = 2240 kJ/kg 2293 kJ/kg C

Consider the first effect:

Steam temperature Solution temperature Temperature driving force Heat transfer coefficient Latent heat of vapourization of steam Latent heat of vapourization of solution T

First effect

1 ΔT 133.5

1 o C = 120.8

= 12.7

o o C C U 1 = 2270 J m -2 s -1 °C -1 λ 1 = 2164 kJ/kg 2200 kJ/kg Prof. R. Shanthini 28 May 2012

Steam used = ?

Assuming feed enters at the boiling point, S 1 ( λ 1 ) = V 1 (Latent heat of vapourization of solution) where S 1 is the flow rate of steam used in the first effect and V 1 is the flow rate of vapour leaving the first effect. Therefore, S 1 (2164) = V 1 (2200)

Consider the second effect:

Steam temperature Solution temperature Temperature driving force

Second effect

T T 1 2 ΔT = 120.8

= 106.3

2 = 14.4

o o o C C C Heat transfer coefficient Latent heat of vapourization of steam Latent heat of vapourization of solution U 2 = 2000 J m -2 s -1 °C -1 λ 2 = 2200 kJ/kg 2240 kJ/kg Prof. R. Shanthini 28 May 2012

Steam used = ?

- Feed enters at the boiling point - steam used in the second effect is the vapour leaving the first effect Therefore, V 1 ( λ 2 ) = V 2 (Latent heat of vapourization of solution) where V 2 is the flow rate of vapour leaving the second effect. Therefore, V 1 (2200) = V 2 (2240)

Consider the third effect:

Steam temperature Solution temperature Temperature driving force Heat transfer coefficient Latent heat of vapourization of steam Latent heat of vapourization of solution

Third effect

T 2 ΔT = 106.3

86.0

1 o C = 20.3

o o 2293 kJ/kg C C U 3 = 1420 J m -2 s -1 °C -1 λ 3 = 2240 kJ/kg Prof. R. Shanthini 28 May 2012

Steam used = ?

- Feed enters at the boiling point - steam used in the third effect is the vapour leaving the second effect Therefore, V 2 ( λ 3 ) = V 3 (Latent heat of vapourization of solution) where V 3 is the flow rate of vapour leaving the third effect. Therefore, V 2 (2240) = V 3 (2293)

Steam economy:

S 1 (2164) = V 1 (2200) = V 2 (2240) = V 3 (2293) Vapour leaving the system = V 1 balance) + V 2 + V 3 = 333 kg/h (from the mass Therefore, S 1 (2164/2200) + S 1 (2164/2240) + S 1 (2164/2293) = 333 kg/h 2164 S 1 (1/2200 + 1/2240 + 1/2293) = 333 kg/h 2164 S 1 (1/2200 + 1/2240 + 1/2293) = 333 kg/h S 1 = 115 kg/h We could calculate the vapour flow rate as V 1 = 113.2 kg/h; V 2 = 111.2 kg/h; V 3 = 108.6 kg/h

Steam economy

= kg vapourized / kg steam used = 333 / 115 =

2.9

28 May 2012

Heat transfer area:

Steam temperature Solution temperature Temperature driving force Heat transfer coefficient Latent heat of vapourization of steam Latent heat of vapourization of solution T

First effect

1 ΔT 133.5

1 o C = 120.8

= 12.7

o o C C U 1 = 2270 J m -2 s -1 °C -1 λ 1 = 2164 kJ/kg 2200 kJ/kg Prof. R. Shanthini 28 May 2012 A 1 = S 1 λ 1 / U 1 ΔT 1 = (115 kg/h) (2164 kJ/kg) = (115 x 2164 x 1000 /3600 J/s) = Overall heat transfer area required = A 1 = / [2270 J m -2 s -1 / [2270 x 12.7 J m

2.4 m 2

+ A 2

7.2 m 2

+ A 3 °C -1 x (12.7) °C] = 3 * A 1 -2 s -1 ]