# Chapter 8 Unit Operations Problems Tomato juice concentration in a single effect evaporator

 Дата канвертавання 22.04.2016 Памер 51.81 Kb.
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 120oC and the latent heat is 2202kJkg-1

Condensing temperature in evaporator is 60oC, and the latent heat is 2358kJkg-1

Temperature of entering juice is 18oC

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 103 (60-18)

= 2358 x 103 x 0.83x + x x 4.186 x103 x 42

= 1957 x103 x + 177 x103 x

= 2134 x 103xJs-1

Heat transfer equation

Temperature of condensing steam = 120oC

Temperature difference across the evaporator = (120 – 60) = 60oC

U = 440 Jm-2s-1 oC-1 A = 12m2

q = UAT

= 440 x 12 x 60

= 317 x 103 Js-1

Therefore 2134 x 103x =317 x 103

x = 0.149 kgs-1

= 536 kg h-1

Rate of raw juice feed = 536kgh-1

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 120oC and the latent heat is 2202kJkg-1

Condensing temperature in second effect is 60oC, 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)

q1 = q2

## U1= 600 Jm-2s-1oC-1 U2= 450 Jm-2s-1oC-1A1 = A2

T1 + T2 = (120 –60) oC = 60 oC

T2 = 60 - T1

## 600 A T1= 450 A (60 - T1)

600 T1 = 450 (60 - T1)

= 27 x 103 – 450 T1

1050 T1 =27 x 103

T1 = 25.7 26o C

T2 = 60 – 26 = 34oC
(a) Evaporating temperatures:

In first effect: (120 –26) = 94oC, latent heat = 2247kJkg-1

In second effect (94 – 34) = 60oC, latent heat = 2358kJkg-1
(b) Steam requirement

ws = steam condensed per hour in effect 1

w1 = water evaporated per hour in effect 1

w2 = water evaporated per hour in effect 2
w1 x 2247 x 103 = w2 x 2388 x 103

= ws x 2202 x 103

w2 = w1 x 2247 / 2388 = 0.94 w1

ws = w1 + 2247/2202 = 1.02 w1

w1 + w2 = 10,929kgh-1

w1 + 0.94 w1 = 10,929

1.94 w1 = 10,929 kg h-1

w1 = 5,633 kgh-1

w2 = 5,296 kgh-1

ws = 5,746 kgh-1

It required 5,746 kgh-1of 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:

U1 = 600 Jm-1s-2 oC-1T1 = 26oC

q = U1 A1T1

(5,633 x 2247 x 103)/3600 = 600 x A1 x 26

3516 x 103 = 15.6 x 103 x A

A1 = 225 m2

As the areas are the same, the heat transfer area in each effect is 225m2

The total area for the two effects = 450 m2

1. 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 120oC, latent heat 2202 kJkg-1

Evaporating temperature 75oC, latent heat 2322 kJkg-1

Heating surface per plate is 0.44m2

## U= 650 Jm-2s-1oC-1

x = no. of plates

q = U AT

= 650 A (120-75) Js-1

= 650 A 45

= 29.25 x 103 A Js-1

= (1000 x 2322 x 103)/3600

= 6.45 x 105

A = (6.45 x 105)(29.25 x 103)

= 22 m2

Each plate = 0.44 m2

#### Number of plates = 50

(b) With a film on the plates:

(1/U2) = 1/U + x/k

= 1/650 + 0.001/0.1

= 0.0015 + 0.01

= 0.0115

U2 = 87
Therefore capacity of evaporator is reduced by (87)/ 650

= 0.134

Capacity of evaporator is reduced by 13%

1. 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

Steam at 200kPa (abs.), condensing temperature 120oC, latent heat 2202 kJkg-1

Pressure in last effect 55kPa(abs.), condensing temperature 83oC, latent heat 2303kJkg-1

q1 = q2 = q3

## U1 A1T1= U2 A2T2 = U3 A3T3

U1 = 600 Jm-2s-1oC-1 U2 = 500 Jm-2s-1oC-1 U3 = 350 Jm-2s-1oC-1

A1 = A2 = A3 = A
T1 + T2 + T3 = (120-83) = 37oC

T2 = T1 U1/U2T3 = T1 U1/U3

T1 + T1 U1/U2 + T1U1/U3 = 37oC

T1 + T1 600/500 + T1 600/350 = 37oC

T1 + 1.2 T1 + 1.71T1 = 37oC

3.91T1 = 37oC

T1 = 9.5oC

T2 = 1.2 x 9.5 = 11.5oC

T3 = 1.71 x 9.5 = 16oC

T1 = 9.5 T2 = 11.5 T3 = 16oC

Evaporating temperature in first effect = 120 – 9.5 = 110.5oC

Evaporating temperature in second effect = 110.5 – 11.5 = 99oC

Evaporating temperature in third effect = 99-16 = 83oC
Latent heat in first effect = 2229 kJkg-1

Latent heat in second effect = 2260 kJkg-1

Latent heat in third effect = 2301 kJkg-1

w1 = water evaporated in first effect per hour

w2 = water evaporated in second effect per hour

w3 = water evaporated in third effect per hour

ws = quantity of steam condensed

w1 x 2229 x 103 = w2 2260 x 103 = w3 2301 x 103

= ws 2202 x103

w1 + w2 + w3 = 8000kgh-1

w1 + w1 2229 /2260 + w1 2229 / 2301 = 8000

w1 + 0.986w1 + 0.969w1 = 8000

2.955 w1 = 8000

w1 = 2707 kgh-1

w2 = 2669 kgh-1

w3 = 2623 kgh-1
Evaporation in each effect: 1st Effect 2707kgh-1, 2nd Effect 2669 kgh-1, 3rd Effect 2623 kgh-1
(b) Input of steam

ws = 2707 x 2229/2202 = 2740 kgh-1
Quantity of steam per kg water = 2740/8000 = 0.343kgkg-1

1. ## Boiling Point Elevations

Evaporating temperature in first effect = 110.5 + 0.60 = 111.1oC

Evaporating temperature in second effect = 99 + 1.50 = 100.5oC

Evaporating temperature in third effect = 83 + 4 = 87oC

Latent heat in first effect = 2227 kJkg-1

Latent heat in second effect = 2256 kJkg-1

Latent heat in third effect = 2291 kJkg-1

w1 = water evaporated in first effect per hour

w2 = water evaporated in second effect per hour

w3 = water evaporated in third effect per hour

ws = quantity of steam condensed

w1 x 2227 x 103 = w2 2256 x 103 = w3 2291 x 103

= ws 2202 x103

w1 + w2 + w3 = 8000kgh-1

w1 + w1 2227 /2256 + w1 2227 / 2291 = 8000

w1 + 0.987w1 + 0.972w1 = 8000kgh-1

2.959 w1 = 8000

w1 = 2704

w2 = 2669

w3 = 2628

ws = 2704 x 2227/2202 = 2735
Evaporation in each effect: 1st Effect 2704kgh-1, 2nd Effect 2669 kgh-1, 3rd Effect 2628 kgh-1

Quantity of steam per kg water = 2735/8000 = 0.342kgkg-1

No change in input steam required.

1. (a)Cooling in a jet condenser

Temperature cooling water = 12oC Max. temperature exit water = 25 oC

Temperature of hot vapour = 70oC 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

### Heat Balance

Heat removed from condensate = 2334 x 103 + (70 –25) x 4.186 x 103

= 2334 x 103 + 188 x 103

= 2522 x 103 Jkg-1

Heat taken in by cooling water = (25-12) x 4.186x103

= 54.4 x x103 Jkg-1

Quantity of heat removed from condensate

= 2800 x 2522 x 103 Jh-1

= 7062 x 106 J h-1

Quantity of cooling water needed = 7062 x 106 / 54.4 x x103

= 130 x 103kgh-1

## (b) Cooling in a surface condenser

U = 2200 Jm-2s-1 oC-1

Mean temperature difference T = (70 – 12)/2 + (70 –25)/2

= 29 + 22.5

= 51.5oC

Quantity of heat removed = 7062 x 106 Jh-1

= UA T

= 2200 x A x 51.5 x 3600

A = 7062 x 106 /(2200 x 51.5 x 3600)

= 17.3m2

Necessary heat transfer area is 17.3m2

Quantity of water needed

q = wt. x spec.heat x T

7062 x 106 = w x 4.18 x 103 x ( 25 –12)

w = 130 x 103 kgh-1

The water needed per hour is 130 x 103 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.

1. 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 99.6oC

Temperature of vapours = 70oC

Latent Heat 2334 kJkg-1

Total steam w = (2334 x 2800) / 2258

= 2894kgh-1

Heat available in steam = 2894 x 2258 x103

= 6535 x 106J

Heat in returned vapour = 1400 x 2334 x103

= 3268 x 106J

Energy used in compressor = 160 x 103 x 1400J

= 224 x 106 J

Heat energy available = (3268 –224) x 106

= 3044 x 106 J

Steam energy saved = (3044 x 106)/(6535 x 106)

= 0.466

= 46.6%

1. 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 = 93.5oC Latent heat 2274 kJkg-1

Specific heat of juice = 4.19kJkg-1oC-1

Pressure of steam = 100kPa(abs) Temperature = 99.6oC Latent heat 2258kJkg-1

A = DL

= 3.14 x 0.05 x 1

= 0.157 m2

Total area for 100 tubes = 15.7 m2

##### Heat Balance

Heat taken in by juice per kg = 2274 x 103 + (93.5 –18) x 4.19 x103

= 2274 x 103 + 316 x103

= 2590 x 10 3 Jkg-1
Heat transferred from steam jacket

q = UA T

= 440 x 15.7 x (93.5 - 18)

= 5.216 x105Js-1
If Ju = wt of water evaporated from juice per hour
2590 x 103 Ju = 5.216 x105 x 3600

Ju = 725kgh-1

Rate of evaporation is 725 kgh-1

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