1205 252 Unit Operation of Food Engineering
Introduction: Evaporation Objective: To protect microbiological growth and assist in reducing transportation and storage costs
Dr. Supatpong Mattaraj Instructor Department of Chemical Engineering Faculty of Engineering Ubon Ratchathani University
To remove water from liquid foods to obtain concentrated liquid products, resulting in a smaller differential of temperature between the heating medium and the products, thus reducing rate of heat transfer.
Presented to
Students from Food Engineering (Agricultural )
Evaporation differs from dehydration because the final product of evaporation process remains in liquid state
Evaporation differs from distillation because the vapors are not divided into fractions as in distillation
Semester 2/2545
Contents : Evaporation
Type of evaporation
Design of a single-effect evaporation
Design of a multiple-effec multiple-effectt evaporation
Single-effect evaporator
Heat exchanger-stream used as heating medium for transferring heat from low-pressure stream to the product.
The product inside the evaporator chamber is kept under vacuum, thus introducing temperature difference between stream and product which boils at low temperature (minimize heat damage)
Condenser - affect condense of vapor and separate it out of the system
Separator - separate vapor from concentrated product
The vapors produced are discarded and stream condenses inside the heat exchanger and the condensate is discarded
Types of Evaporator
Multiple-effect evaporator
Stream is used only in the first effect
- Simpl Simplest est and and oldest oldest types types of evapor evaporator ator
The vapors are reused as the heating medium in another evaporator chamber.
- The product product is heated in a stream-ja stream-jacket cketed ed spherical spherical vessel vessel - The heating heating vessel vessel may be open open to the atmosph atmosphere ere or connect connected ed to a condenser and vacuum.
The partially concentrated product leaving the first effect is introduced as feed into the second effect. The product from the second effect becomes feed for the third effect. The product from the third effect leaves at the desired concentration This is called a forward feed system.
1. Batch-Type Pan Evaporator
- Vacuum Vacuum permits permits boiling boiling the product product at temperat temperatures ures lower lower than the boiling point at atmospheric pressure, thus reducing the thermal damage to products. - The heat-trans heat-transfer fer area per unit unit volume is small small and the residence residence time time of the product is longer up to several hours.
2. Natural Circulation Evaporator - Short vertic vertical al types, types, 1-2 m long long and 50-100 50-100 mm in diameter. diameter. - The heat-trans heat-transfer fer area per unit unit volume is small small and the residence residence time time of the product is longer up to several several hours. - The concentrat concentrated ed liquid falls falls back to the base base of the vessel vessel through a central annular section.
Types of Evaporator
3. Rising-film Evaporator - A low-viscosit low-viscosity y liquid food is allowed allowed to boil inside 10-15 m-long m-long vertical vertical tubes. - The upward movement movement of vapors causes a thin liquid film to move rapidly upward. - A temperature temperature different differential ial of at least least 14 oC betwe between en the product and the heating medium is necessary to obtain a well-developed film. - High convective heat-transfer coefficients are are achieved in the system.
4. Falling-film Evaporator - It has a thin liquid film film moving moving downward downward under gravity gravity on the inside of the vertical tubes. - This is more complicated system than rising-film evaporator. - The residence time is about 20-30 seconds, seconds, compared with a residence residence time of 3-4 minutes in a rising-film evaporator.
Types of Evaporator
5. Rising/Falling-film Evaporator - The product product is concentrate concentrated d by circulation circulation through through a rising-film rising-film section followed by a falling-film section.
6. Forced-circulation Evaporator - Consis Consists ts of a noncontact noncontact heat heat exchanger exchanger where liquid liquid food is circulated at high rates. - A hydrostatic hydrostatic head head is used to eliminate eliminate the boiling boiling of the liquid. - The temperature temperature differenc difference e across across the heating surfac surface e in the heat exchanger is usually 3-5 oC. - Axial flow flow pumps are are used to maintain maintain high circulation circulation rates rates with linear velocities of 2-6 m/s, compared with a li near velocity of 0.3-1 m/s in natural-circulation evaporators. -
Low capital capital and operating operating costs costs when when compared compared with with other types types of evaporator.
Types of Evaporator
7. Agitated Thin-Film Thin-Film Evaporator - Suita Suitable ble for viscous viscous fluid foods foods (i.e. (i.e. tabular or plate plate evaporators). evaporators). - Requir Require e higher rates rates for heat heat transfer transfer due to high agitati agitation. on. - The major major disadvantage disadvantages s are the high capital capital and maintenanc maintenance e costs and low processing capacity. - Plate evaporators evaporators use the the principles principles of rising/falling rising/falling-film -film,, and forced-circulator evaporators.
Schematic of a single-effect evaporator evaporator
Design of a single-effec single-effectt evaporator
condenser
Vacuum ejector
mv,T1
Dilute liquid feed is pumped into the heating chamber and stream is introduced into the heat exchanger.
The temperature of evaporation T1 is controlled by maintaining
ms, Ts
vacuum inside the heating chamber.
System boundary
Heat and mass balances conducted on the evaporator system allow determination of various design and operating variables
U, A
(i.e. mass flow rates, final concentration of products, and heat exchanger area.
ms, Ts
mf , Tf , xf mp, T1, xp
Single-effect evaporator: mass balance on flow stream and product solid Overall mass balance:
Single-effect evaporator: enthalpy balance Enthalpy balance:
m f = mv + m p
m f H f (T f , x f ) + m s H v (T s ) = mv H v (T 1 ) + m p H p (T 1 , x p ) + m s H c (T s )
x f m f = x p m p
ms is the mass flow rate of stream (kg/s)
mf is the mass flow rate of dilute liquid feed (kg/s) mv is the mass flow rate of vapor (kg/s) mp is the mass flow rate of concentrated products (kg/s)
Hf (Tf ,xf ) is enthalpy of dilute liquid feed (kJ/kg) Hp(T1,xp) is enthalpy of concentrated product (kJ/kg) Hv(Ts) is enthalpy of saturated vapor at temperature T s (kJ/kg) Hv(T1) is enthalpy of saturated vapor at temperature T 1(kJ/kg)
xf is the solid fraction in feed stream
Hc(Ts) is enthalpy of condensate T1(kJ/kg)
xp is the solid fraction in product stream
Ts is temperature of stream (oC) T1 is the boiling temperature maintained inside the evaporator chamber( oC). Tf is the temperature of dilute liquid feed ( oC)
Single-effect evaporator: enthalpy balance Total enthalpy for dilute liquid feed: Total heat content of stream :
H f (T f , x f ) = c pf (T f −0 oC )
m f H f (T f , x f )
Single-effect Single-effe ct evaporato evaporator: r: Heat exchanger
The rate of heat transfer:
q = UA (T s − T 1 ) = m s H v (T s ) − m s H c (T s )
Where q is the rate of heat transfer (W)
m s H v (T s )
U is the overall heat transfer coefficient (W/m 2.K) Enthalpy of saturated vapor at Ts obtained from stream table A.4.2:
H v (T s )
• The overall heat-transf heat-transfer er coefficient coefficient decreases decreases as product
mv H V (T 1 )
Total enthalpy of the vapor le aving the system :
A is the area of the heat exchanger (m2) becomes concentrated, due to increased resistance of heat
Enthalpy of saturated vapor at T1 obtained from stream table A.4.2:
H v (T 1 )
Total enthalpy for concentrated product stream leaving the evaporator: m p H p (T 1, x p ) H p (T 1, x p ) = c pp(T 1 −0 C ) o
Where cpp is the specific heat content of concentrated product (kJ/kg. oC) Total enthalpy for condensate leaving the evaporator:
m s H c (T s )
H c (T s ) from Table A.4.2
transfer on the product side of the heat exchanger. Stream economy: used to express the operating performance of an evaporator; the ratio of the rate of mass of water vapor produced from the liquid feed per unit rate of stream consumed Stream economy =
mv m s
Example: Single-effect evaporator
Apple juice is being concentrated in natural-circulation single-effect evaporator. At steady-state conditions, 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 is concentrated to 75% total solids. The specific heats if dilute apple juice and concentrate are 3.9 and 2.3 kJ/kg.oC, respectively. The stream pressure is measured to to be 304.42 kPa. The inlet feed temperature is 43.3 oC. The product inside the evaporator boils at 62.2 oC. The overall heat transfer coefficient is assumed to be 943 W/m2. oC. Calculate the mass flow rate of concentrated product, stream requirements, stream economy, and the heat transfer area.
Example: Single-effect evaporator Solution: (1)
x f m f = x p m p The mass flow rate of concentrated product is 0.098 kg/s
0.11× 0.67 kg / s = 0.75 × m p m p = 0.098 kg / s
(2)
m f = mv + mp
mv = (0.67 kg / s) − (0.098 kg / s) = 0.57 kg / s
mv = m f − m p
The mass flow rate of vapors is is 0.57 kg/s
(3) Enthalpy balance Given
mass flow rate of feed mf = 0.67 kg/s concentration of food xf = 0.11 concentration of product xp = 0.75 stream pressure = 304.42 kPa feed temperature Tf = 43.3 oC boiling temperature T1 in evaporator = 62.2 oC overall heat transfer coefficient U = 943 W/m2.K specific heat of dilute feed cpf = 3.9 kJ/kg. oC specific heat of concentrated product cpp = 2.3 kJ/kg. oC
H f (T f , x f ) = c pf (T f − 0 C ) = (3 .9 kJ / kg . C ) × ( 43 .3 C − 0 C ) = 168 .9 kJ / kg o
o
H p (T 1 , x p ) = c pp (T 1 − 0 C ) = ( 2 .3 kJ / kg . C ) × ( 62 .2 C − 0 C ) = 143 .1 kJ / kg o
Example: Single-effect evaporator (3) Enthalpy balance: From table A.4.2
0
o
o
o
o
Example: Single-effect evaporator (4) Stream economy
Temperature of stream at 304.42 oC = 134 134 oC (inte (interpolat rpolate) e)
Stream economy =
oC)
Enthalpy for saturated vapor Hv (Ts = 134 = 2725.9 kJ/kg Enthalpy for saturated liquid H c (Ts = 134 oC) = 563.41 kJ/kg Enthalpy for saturated vapor Hv (T1 = 62.2 oC) = 2613.4 kJ/kg
mv m s
=
0.57 0.64
= 0.89 kg water evaporated / kg stream
(5) Area of heat transfer
m f H f (T f , x f ) + m s H v (T s ) = mv H v (T 1 ) + m p H p (T 1 , x p ) + m s H c (T s )
q = UA(T s − T 1 ) = m s H v (T s ) − m s H c (T s )
(0.67 kg / s)(168.9 kJ / kg ) + (m s kg / s)( 2725.9 kJ / kg ) = (0.57 kg / s )(2613 .4 kJ / kg ) +
A × (943 W / m . C )(134 C − 62.2 C ) = (0.64 kg / s)[2725.9 kJ / kg − 563.14 kJ / kg ] ×1000 J / kJ 2 o
o
o
(0.098 kg / s )(143.1 kJ / kg ) + (m s kg / s)(563.41 kJ / kg )
A = 20.4 m2
2162.49×ms = 1390.5 ms = 0. 0.64 64 kg kg/s /s
Schematic of a triple-effect evaporator evaporator
Design of a multiple-effect evaporator
mv3, T3
Dilute liquid feed is pumped into the first evaporator chamber while stream enters the heat exchanger and condenses, thus discharging
The vapors produced from the first effect are used as the heating
mv2
mv1
heat to the product. The condensate is discarded. ms, Ts
medium in the second effect, where the feed is the partially
T 1
T 2
T 3
U1, A1
U2, A2
U3, A3
concentrated product from the first effect and so on.
The desired final concentration is pumped out of the evaporator chamber of the third effect.
mf , Tf , xf
mf1, xf1
mf2, xf2
mp, T3, xp
Triple-effect evaporator: mass balance on flow stream and product solids
Triple-effect evaporator: enthalpy balance
Overall mass balance:
Enthalpy balance for each effect:
m f = mv1 + mv 2 + mv3 + m p
m f H f (T f , x f ) + m s H v (T s ) = mv1 H v (T 1) + m f 1 H f 1(T 1, x f 1) + m s H c (T s )
x f m f = x p m p
m f 1 H f 1(T 1, x f 1) + mv1 H v (T 1) = mv2 H v (T 2 ) + m f 2 H f 2 (T 2, x f 2 ) + mv1 H c (T 1)
mf is the mass flow rate of dilute liquid feed in the first effect (kg/s)
m f 2 H f 2 (T f 2 , x f 2 ) + mv2 H v (T 2 ) = mv3 H v (T 3 ) + m p H p (T 3, x p ) + mv2 H c (T 2 )
mv1, mv2, mv3 are the mass flow rates of vapor from the first, second, and third effect, respectively. (kg/s)
Subscripts 1, 2, and 3 refer to the first, second, and third eff ect, respectively.
mp is the mass flow rate of concentrated products from the third effect (kg/s) xf is the solid fraction in feed stream to the first effect xp is the solid fraction in product stream from the third effect
Example: double-effect evaporator
Triple-effect evaporator: heat exchange
Heat transfer across heat exchanger: q1 = U 1 A1 (T s − T 1 ) = m s H v (T s ) − m s H c (T s )
q2 = U 2 A2 (T 1 − T 2 ) = mv1 H v (T 1 ) − mv1 H c (T 1 ) q3 = U 3 A3 (T 2 − T 3 ) = mv 2 H v (T 2 ) − mv 2 H c (T 2 ) Stream economy: Stream economy =
mv1 + mv 2 + mv 3 m s
Schematic of a double-effec double-effectt evaporator
Calculate the stream requirements of a double-effect forward-feed evaporator to concentrate a liquid food from 11% total solids to 50% total solids concentrate. concentrate. The feed rate is 10,000 kg/hr at 20 oC. The boiling of liquid inside the second effect takes takes place under vacuum at 70 oC. The stream is being supplied to the first effect at 198.5 kPa. The condensate from the first effect is discarded at 95 oC and from from the the second effect at 70 oC. The overall heat-transfer coefficient in the first effect is 1,000 W/m2. oC; in the second effect it is 800 W /m2. oC. The specific heats of the liquid food are 3.8, 3.0, and 2.5 kJ/kg. oC at initial, intermediate, and final concentrations. concentrations. Assume the areas and temperature gradients are equal in each effect. Given
mass flow rate of feed mf = 10,000 kg/hr = 2.78 kg/s concentration of feed xf = 0.11 concentration of product xp = 0.50 stream pressure = 198.5 kPa feed temperature Tf = 20 oC boiling temperature T2 in second effect = 70 oC overall heat transfer coefficient U1 in the first effect = 1000 W/m2.K overall heat transfer coefficient U2 in the second effect = 800 W/m2.K specific heat of dilute feed cpf = 3.8 kJ/kg. oC specific heat of feed at intermediate concentration c’pf = 3.0 kJ/kg. oC specific speci fic heat heat of conce concentrate ntrated d food produc productt c = 2.5 kJ/k kJ/kg. g. oC
Solution: double-effect evaporator (1) Mass balance:
x f m f = x p m p
mv2, T2
( 0 . 11 )( 2 . 78 kg / s )
m p
mv1
= ( 0 . 5 ) m p
= 0 . 61 kg / s
ms, Ts T 1
T 2
U1, A1
U2, A2
(2) Total amount of water water evaporating: evaporating: System boundary
m f = m v1 + m v 2 + m p 2.78 = m v1 + m v 2 + 0.61 m v1 + m v 2 = 2.17 kg / s
mf , Tf , xf
mf1, xf1
mp, T2, xp
(3) Stream is being supplied at 198.5 kPa or 120 oC, T2 is 70 oC; The total temperature gradient is 50 oC. ∆T 1 + ∆ T 2 = 50 o C
Solution: double-effect evaporator
Solution: double-effect evaporator
(4) The area of heat heat transfer in the first and second second effects are the same: q1 U 1 (T s − T 1 ) m s H v (T s ) − m s H c (T s ) U 1 (T s − T 1 )
=
=
(5) From table A.4.2;
q2
At T s = 120 o C ;
H v (T s ) = 2706 .3 kJ / kg
At T s = 95 o C ;
H v (T 1 ) = 2668 .1 kJ / kg
At T s = 70 o C ;
H v (T 2 ) = 2626 .8 kJ / kg
H c (T s ) = 503 .71 kJ / kg H c (T 1 ) = 397 .96 kJ / kg
U 2 (T 1 − T 2 )
H c (T 2 ) = 292 .98 kJ / kg
mv1 H v (T 1 ) − mv1 H c (T 1 ) U 2(T 1 − T 2 )
(5) Value for enthalpy enthalpy of product product ∆T =25
Solve equation;
oC:
[(m s kg / s) (2706.3 kJ / kg ) − (m s kg / s) (503.71 kJ / kg )]×1000 J / kJ (1000 W / m . oC )(120 − 95) 2
H f (T f , x f ) = c pf (T f − 0) = (3.8 kJ / kg . C )(20 C − 0) =76 kJ / kg o
o
=
H f 1 (T 1 , x f 1 ) = c pf (T 1 − 0) = (3.0 kJ / kg . C )(95 C − 0) = 285 kJ / kg '
o
o
H f 2 (T 2 , x f 2 ) = c pp (T 2 − 0) = (2.5 kJ / kg . C )(70 C − 0) = 175 kJ / kg o
o
[(mv1 kg / s) (2668.1 kJ / kg ) − (mv1 kg / s) (397.96 kJ / kg )] ×1000 J / kJ (800 W / m . o C ) (95 oC − 70 oC ) 2
2202.59 × m s 25,000
Solution: double-effect evaporator
=
2270.14 × mv1 20,000
or
0 . 088 m s = 0 . 114 m v 1
Solution: double-effect evaporator Overall equations:
(6) Solve equations: (6.1)
m f H f (T f , x f ) + m s H v (T s ) = mv1 H v (T 1) + m f 1 H f 1(T 1, x f 1) + m s H c (T s )
1.
mp = 0.61 kg/s
2.
mv1+mv2 = 2.17 kg/s
3. 0.088 ms = 0.114 mv1
(2.78)(76) + (m s )(2706.3) = mv1 (2668.1) + m f 1 (285) + m s (503.71)
4 . 2 20 202.59 ms = 2668.1 mv1+285 mf1 -211.28
2202.59 × m s = 2668.1 × mv1 + 285 × m f 1 − 211.28
5 . 2 27 270.14 mv1 = 2626.8 mv2 + 175 mp – 285 mf1
(6.2) m f H f 1 (T 1, x f 1 ) + mv1 H v (T 1 ) = mv 2 H v (T 2 ) + m p H f 2 (T 2 , x f 2 ) + mv1 H c (T 1 ) 1
Five unknown and five equations to solve: 1.
m f 1 ( 285 ) + m v1 ( 2668 .1) = m v 2 ( 2626 .8) + m p (175 ) + m v1 (397 .96 )
mp + 0 ms + 0 mv1 + 0 mv2 + 0 mf1 = 0.61
2.
0 mp + 0 ms + mv1 + mv2 + 0 mf1 = 2.17
2270 .14 × m v1 = 2626 .8 × m v 2 + 175 × m p − 285 × m f 1
3.
0 mp + 0.088 ms – 0.114 mv1 + 0 mv2 + 0 mf1 = 0
4.
0 mp + 2202.59 ms – 2668.1 mv1 + 0 mv2 – 285 mf1 = -211.28
5.
-175 mp + 0 ms + 2270.14 mv1 – 2626.8 mv2 + 285 mf1 = 0
Solution: double-effect evaporator 0 1 0 0 0 0 . 088 2202 . 59 0 − 175 0
0
0
0 m p
1
1
− 0 .114 − 2668 . 1
0
2270 . 14
− 2626 .8
0
m p m s mv1 = mv 2 m f 1
0.61 1.43 1.10 1.07 1.46
0 m s 0 m v1 = − 285 m v 2 285 m f 1
0 . 61 2 .17 0 − 211 . 28 0
kg / s
kg / s kg / s kg / s
kg / s
Stream economy: Stream economy =
mv1 + mv2 m s
=
1.10 + 1.07 1.43
= 1.5 kg water vapor / kg stream