Chapter 8 Unit Operations Problems
1. Tomato juice concentration in a single effect evaporator
Raw juice 6% solids, tomato concentrate 35% solids.
Mass Balance, basis 100kg fresh juice
Solids Water Total
Raw juice 6 94 100
Concentrate 6 11 17
Water evaporated 83
i.e. 0.83kg water evaporated per kg fresh juice
Heat Balance
Pressure in evaporator 20kPa (abs)
Steam pressure = 100 kPa(gauge) 200kPa(abs)
From steam tables (Appendix 8)
Condensing temperature of steam is 120^{o}C and the latent heat is 2202kJkg^{1}
Condensing temperature in evaporator is 60^{o}C, and the latent heat is 2358kJkg^{1}
Temperature of entering juice is 18^{o}C
Heat required to evaporate water from x kg juice per second
= latent heat + sensible heat
= latent heat x 0.83 x + x x 4.186 x 10^{3} (6018)
= 2358 x 10^{3} x 0.83x + x x 4.186 x10^{3} x 42
= 1957 x10^{3} x + 177 x10^{3} x
= 2134 x 10^{3}xJs^{1}
Heat transfer equation
Temperature of condensing steam = 120^{o}C
Temperature difference across the evaporator = (120 – 60) = 60^{o}C
U = 440 Jm^{2}s^{1} ^{o}C^{1 }A = 12m^{2}
q = UA T
= 440 x 12 x 60
= 317 x 10^{3} Js^{1}
Therefore 2134 x 10^{3}x =317 x 10^{3}
x = 0.149 kgs^{1}
= 536 kg h^{1}
Rate of raw juice feed = 536kgh^{1}

Steam usage in two effects evaporator
Steam at 100kPa (gauge) = 200kPa (abs.)
Pressure in second effect = 20kPa (abs.)
From steam tables (Appendix 8)
Condensing temperature of steam is 120^{o}C and the latent heat is 2202kJkg^{1}
Condensing temperature in second effect is 60^{o}C, and the latent heat is 2358kJkg^{1}
Raw milk 9.5% solids, concentrated milk 35% solids. Flow raw milk 15,000 kg h^{1}
Mass Balance (kgh^{1})
Solids Liquids Total
Raw milk 1425 13,375 15,000
Concentrated milk 1425 2,646 4,071
Evaporated water 10,929
Heat Balance (Js^{1})
q_{1 } = q_{2}
U_{1} A_{1} T_{1} = U_{2} A_{2} T_{2} U_{1} = 600 Jm^{2}s^{1} ^{o}C^{1 }U_{2} = 450 Jm^{2}s^{1} ^{o}C^{1} A_{1 } = A_{2}
T_{1} + T_{2 } = (120 –60)^{ o}C = 60^{ o}C
T_{2 }= 60  T_{1 }
600 A T_{1} = 450 A (60  T_{1})
600 T_{1} = 450 (60  T_{1})
= 27 x 10^{3} – 450 T_{1}
1050 T_{1} =27 x 10^{3}
T_{1} = 25.7 26^{o} C
T_{2 }= 60 – 26 = 34^{o}C
(a) Evaporating temperatures:
In first effect: (120 –26) = 94^{o}C, latent heat = 2247kJkg^{1}
In second effect (94 – 34) = 60^{o}C, latent heat = 2358kJkg^{1}
(b) Steam requirement
w_{s }=_{ }steam condensed per hour in effect 1
w_{1} = water evaporated per hour in effect 1
w_{2} = water evaporated per hour in effect 2
w_{1 }x 2247 x 10^{3} = w_{2} x 2388 x 10^{3}
= w_{s }x 2202 x 10^{3}
w_{2 }= w_{1} x 2247 / 2388 = 0.94 w_{1}
_{ }w_{s }= w_{1} + 2247/2202 = 1.02 w_{1 }
w_{1 }+ w_{2} _{ } = 10,929kgh^{1}
w_{1 }+ 0.94 w_{1} _{ }= 10,929
1.94 w_{1 }= 10,929 kg h^{1}
w_{1 }= 5,633 kgh^{1}
w_{2 }= 5,296 kgh^{1}
w_{s }= 5,746 kgh^{1}
It required 5,746 kgh^{1}of steam to evaporate a total of 10,929 kgh^{1} of water i.e. 0.53 kg steam/kg water
(c) Heat exchange surface for first effect:
U_{1} = 600 Jm^{1}s^{2} ^{o}C^{1} T_{1 }= 26^{o}C
q = U_{1} A_{1} T_{1}
(5,633 x 2247 x 10^{3})/3600 = 600 x A_{1} x 26
3516 x 10^{3} = 15.6 x 10^{3} x A
A_{1} = 225 m^{2 }
As the areas are the same, the heat transfer area in each effect is 225m^{2}
The total area for the two effects = 450 m^{2}

Plate evaporator concentrating milk
10% solids in fresh to 30% solids in concentrated milk. Flow rate 1500kgh^{1}
Mass balance kgh^{1}
Solids Liquid Total
Fresh milk 150 1350 1500
Concentrated milk 150 350 500
Evaporated water 1000
(a) Number of plates
Steam at 200kPa(abs.), condensing temperature 120^{o}C, latent heat 2202 kJkg^{1}
Evaporating temperature 75^{o}C, latent heat 2322 kJkg^{1}
Heating surface per plate is 0.44m^{2}
U = 650 Jm^{2}s^{1} ^{o}C^{1}
x = no. of plates
q = U A T
_{ }= 650 A (12075) Js^{1}
= 650 A 45
= 29.25 x 10^{3 }A Js^{1}
Heat to evaporate water
= (1000 x 2322 x 10^{3})/3600
= 6.45 x 10^{5}
A = (6.45 x 10^{5})(29.25 x 10^{3})
= 22 m^{2}
^{ }Each plate = 0.44 m^{2}
Number of plates = 50
(b) With a film on the plates:
(1/U_{2}) = 1/U + x/k
= 1/650 + 0.001/0.1
= 0.0015 + 0.01
= 0.0115
U_{2} = 87
Therefore capacity of evaporator is reduced by (87)/ 650
= 0.134
Capacity of evaporator is reduced by 13%

Triple effect evaporator
Feed 5% solids, product 25% solids. Input 10,000kgh^{1}
Mass Balance kgh^{1}
Solids Liquid Total
Feed 500 9500 10,000
Product 500 1500 2,000
Water evaporated 8,000
(a) Evaporation in each effect
Steam at 200kPa (abs.), condensing temperature 120^{o}C, latent heat 2202 kJkg^{1}
Pressure in last effect 55kPa(abs.), condensing temperature 83^{o}C, latent heat 2303kJkg^{1}
q_{1 } = q_{2} = q_{3}
U_{1} A_{1} T_{1} = U_{2} A_{2} T_{2}_{ } = U_{3} A_{3} T_{3}
U_{1} = 600 Jm^{2}s^{1o}C^{1} U_{2} = 500 Jm^{2}s^{1o}C^{1} U_{3} = 350 Jm^{2}s^{1o}C^{1}
A_{1} = A_{2} = A_{3 }= A
T_{1} + T_{2 } + T_{3 }= (12083) = 37^{o}C
T_{2 } = T_{1} U_{1}/U_{2} T_{3} = T_{1 }U_{1}/U_{3}
T_{1} + T_{1 }U_{1}/U_{2} + T_{1}U_{1}/U_{3 }= 37^{o}C
T_{1} + T_{1 }600/500 + T_{1} 600/350_{ }= 37^{o}C
T_{1} + 1.2 T_{1 } + 1.71T_{1} _{ }= 37^{o}C
3.91T_{1} = 37^{o}C
T_{1 }= 9.5^{o}C
T_{2} = 1.2 x 9.5 = 11.5^{o}C
T_{3} _{ }= 1.71 x 9.5 = 16^{o}C
T_{1} = 9.5 T_{2 } = 11.5 T_{3 }= 16^{o}C
Evaporating temperature in first effect = 120 – 9.5 = 110.5^{o}C
Evaporating temperature in second effect = 110.5 – 11.5 = 99^{o}C
Evaporating temperature in third effect = 9916 = 83^{o}C
Latent heat in first effect = 2229 kJkg^{1}
Latent heat in second effect = 2260 kJkg^{1}
Latent heat in third effect = 2301 kJkg^{1}
w_{1} = water evaporated in first effect per hour
w_{2} = water evaporated in second effect per hour
w_{3} = water evaporated in third effect per hour
w_{s} = quantity of steam condensed
w_{1} x 2229 x 10^{3} = w_{2} 2260 x 10^{3} = w_{3} 2301 x 10^{3}
= w_{s} 2202 x10^{3}
w_{1} + w_{2} + w_{3} = 8000kgh^{1}
w_{1} + w_{1} 2229 /2260 + w_{1} 2229 / 2301 = 8000
w_{1} + 0.986w_{1} + 0.969w_{1} = 8000
2.955 w_{1} = 8000
w_{1} = 2707 kgh^{1}
w_{2} = 2669 kgh^{1}
w_{3} = 2623 kgh^{1}
Evaporation in each effect: 1^{st} Effect 2707kgh^{1}, 2^{nd} Effect 2669 kgh^{1}, 3^{rd} Effect 2623 kgh^{1}
(b) Input of steam
w_{s} = 2707 x 2229/2202 = 2740 kgh^{1}
Quantity of steam per kg water = 2740/8000 = 0.343kgkg^{1}

Evaporating temperature in first effect = 110.5 + 0.60 = 111.1^{o}C
Evaporating temperature in second effect = 99 + 1.50 = 100.5^{o}C
Evaporating temperature in third effect = 83 + 4 = 87^{o}C
Latent heat in first effect = 2227 kJkg^{1}
Latent heat in second effect = 2256 kJkg^{1}
Latent heat in third effect = 2291 kJkg^{1}
w_{1} = water evaporated in first effect per hour
w_{2} = water evaporated in second effect per hour
w_{3} = water evaporated in third effect per hour
w_{s} = quantity of steam condensed
w_{1} x 2227 x 10^{3} = w_{2} 2256 x 10^{3} = w_{3} 2291 x 10^{3}
= w_{s} 2202 x10^{3}
w_{1} + w_{2} + w_{3} = 8000kgh^{1}
w_{1} + w_{1} 2227 /2256 + w_{1} 2227 / 2291 = 8000
w_{1} + 0.987w_{1} + 0.972w_{1} = 8000kgh^{1}
2.959 w_{1} = 8000
w_{1} = 2704
w_{2} = 2669
w_{3} = 2628
w_{s} = 2704 x 2227/2202 = 2735
Evaporation in each effect: 1^{st} Effect 2704kgh^{1}, 2^{nd} Effect 2669 kgh^{1}, 3^{rd} Effect 2628 kgh^{1}
Quantity of steam per kg water = 2735/8000 = 0.342kgkg^{1}
No change in input steam required.

(a)Cooling in a jet condenser
Temperature cooling water = 12^{o}C Max. temperature exit water = 25^{ o}C
Temperature of hot vapour = 70^{o}C Latent heat = 2334kJkg^{1}
Mass flow = 4000kgh^{1}
Fresh milk = 9% solids
Milk concentrate = 30% solids
Mass balance kgh^{1}
Solids Liquid Total
Fresh milk 360 3640 4000
Concentrated milk 360 840 1200
Evaporated water 2800
(a) Jet condenser
Heat Balance
Heat removed from condensate = 2334 x 10^{3} + (70 –25) x 4.186 x 10^{3}
= 2334 x 10^{3} + 188 x 10^{3}
= 2522 x 10^{3 }Jkg^{1}
Heat taken in by cooling water = (2512) x 4.186x10^{3}
= 54.4 x x10^{3} Jkg^{1}
Quantity of heat removed from condensate
= 2800 x 2522 x 10^{3 }Jh^{1}
= 7062 x 10^{6 }J h^{1}
Quantity of cooling water needed = 7062 x 10^{6 }/ 54.4 x x10^{3}
^{ }= 130 x 10^{3}kgh^{1}
(b) Cooling in a surface condenser
U = 2200 Jm^{2}s^{1} ^{o}C^{1}
Mean temperature difference T = (70 – 12)/2 + (70 –25)/2
= 29 + 22.5
= 51.5^{o}C
Quantity of heat removed = 7062 x 10^{6 }Jh^{1}
^{ }= UA T
= 2200 x A x 51.5 x 3600
A = 7062 x 10^{6 } /(2200 x 51.5 x 3600)
= 17.3m^{2}
Necessary heat transfer area is 17.3m^{2}
Quantity of water needed
q = wt. x spec.heat x T
7062 x 10^{6 }= w x 4.18 x 10^{3} x ( 25 –12)
w = 130 x 10^{3} kgh^{1}
The water needed per hour is 130 x 10^{3} kg
Thi is the same as in the jest condenser, but in practice the jet condenser does not have 100% efficiency in use of water, and water used would be greater.

Mechanical recompression
In the evaporator:
Total water evaporated = 2800 kgh^{1}
Vapours recompressed = 1400kgh^{1}
Energy used per kg vapour = 160 kJkg^{1}
Steam = 100 kPA (abs)
Latent heat 2258 kJkg^{1} Temp 99.6^{o}C
Temperature of vapours = 70^{o}C
Latent Heat 2334 kJkg^{1}
Total steam w = (2334 x 2800) / 2258
= 2894kgh^{1}
Heat available in steam = 2894 x 2258 x10^{3}
= 6535 x 10^{6}J
Heat in returned vapour = 1400 x 2334 x10^{3}
= 3268 x 10^{6}J
Energy used in compressor = 160 x 10^{3} x 1400J
= 224 x 10^{6 }J
Heat energy available = (3268 –224) x 10^{6}
= 3044 x 10^{6 }J
Steam energy saved = (3044 x 10^{6})/(6535 x 10^{6})
= 0.466
= 46.6%

Calandria type evaporator
No. of tubes = 100
Length of tube = 1 metre
Diameter of tube = 5cm = 0.05m
Area of tube = A
Pressure in evaporator = 80kPa(abs,) Temperature = 93.5^{o}C Latent heat 2274 kJkg^{1}
Specific heat of juice = 4.19kJkg^{1o}C^{1}
Pressure of steam = 100kPa(abs) Temperature = 99.6^{o}C Latent heat 2258kJkg^{1}
A = DL
= 3.14 x 0.05 x 1
= 0.157 m^{2}
Total area for 100 tubes = 15.7 m^{2}
Heat Balance
Heat taken in by juice per kg = 2274 x 10^{3} + (93.5 –18) x 4.19 x10^{3}
= 2274 x 10^{3} + 316 x10^{3}
= 2590 x 10 ^{3} Jkg^{1}
Heat transferred from steam jacket
q = UA T
= 440 x 15.7 x (93.5  18)
= 5.216 x10^{5}Js^{1}
If Ju = wt of water evaporated from juice per hour
2590 x 10^{3 }Ju^{ }=^{ }5.216 x10^{5} x 3600
Ju = 725kgh^{1}
Rate of evaporation is 725 kgh^{1} 