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
= UA T
= 440 x 12 x 60
= 317 x 103 Js-1
Therefore 2134 x 103x
x
=317 x 103
= 0.149 kgs-1
= 536 kg h-1
Rate of raw juice feed = 536kgh-1
2. 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)
Raw milk
Concentrated milk
Solids
1425
1425
Heat Balance (Js-1)
q1
U1 A1 T1
Liquids
13,375
2,646
Evaporated water
Total
15,000
4,071
10,929
= q2
= U2 A2 T2
U1 = 600 Jm-2s-1 oC-1 U2 = 450 Jm-2s-1 oC-1 A1 = A2
T1 + T2 = (120 –60) oC = 60 oC
T2 = 60 - T1
600 A T1
600 T1
1050 T1
T1
T2
= 450 A (60 - T1)
= 450 (60 - T1)
= 27 x 103 – 450 T1
=27 x 103
= 25.7 26o C
= 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
ws
= w2 x 2388 x 103
= ws x 2202 x 103
= w1 x 2247 / 2388
= w1 + 2247/2202
w1 + w2
w1 + 0.94 w1
1.94 w1
w1
w2
ws
= 0.94 w1
= 1.02 w1
= 10,929kgh-1
= 10,929
= 10,929 kg h-1
= 5,633 kgh-1
= 5,296 kgh-1
= 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-1 T1 = 26oC
q
= U1 A1 T1
(5,633 x 2247 x 103)/3600 =
600 x A1 x 26
3
3516 x 10
=
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
3. Plate evaporator concentrating milk
10% solids in fresh to 30% solids in concentrated milk. Flow rate 1500kgh-1
Mass balance kgh-1
Fresh milk
Concentrated milk
Solids
150
150
Liquid
1350
350
Evaporated water
Total
1500
500
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-1 oC-1
x
= no. of plates
q
= U A T
= 650 A (120-75) Js-1
= 650 A 45
= 29.25 x 103 A Js-1
Heat to evaporate water
=
(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)
=
=
=
=
U2
=
1/U + x/k
1/650 + 0.001/0.1
0.0015 + 0.01
0.0115
87
Therefore capacity of evaporator is reduced by (87)/ 650
= 0.134
Capacity of evaporator is reduced by 13%
4. 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 120oC, latent heat 2202 kJkg-1
Pressure in last effect 55kPa(abs.), condensing temperature 83oC, latent heat 2303kJkg-1
q1
U1 A1 T1
U1 = 600 Jm-2s-1oC-1U2
A1
= q2
= U2 A2 T2
= 500 Jm-2s-1oC-1
=
A2
=
T1 + T2 + T3
T2 = T1 U1/U2
= q3
= U3 A3 T3
A3
U3
=
=
(120-83)
T3 = T1 U1/U3
T1 + T1 U1/U2 + T1U1/U3
T1 + T1 600/500 + T1 600/350
T1 + 1.2 T1 + 1.71T1
3.91T1
T1
T2
= 1.2 x 9.5
T3
= 1.71 x 9.5
T1 = 9.5
T2 = 11.5 T3
Evaporating temperature in first effect
= 120 – 9.5
Evaporating temperature in second effect
= 110.5 – 11.5
Evaporating temperature in third effect
= 99-16
= 350 Jm-2s-1oC-1
A
= 37oC
= 37oC
= 37oC
= 37oC
= 37oC
= 9.5oC
= 11.5oC
= 16oC
= 16oC
= 110.5oC
= 99oC
= 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
3
= ws 2202 x10
w1 + w2 + w3
= 8000kgh-1
w1 + w1 2229 /2260 + w1 2229 / 2301
= 8000
w1 + 0.986w1 + 0.969w1
2.955 w1
w1
w2
w3
= 8000
= 8000
= 2707 kgh-1
= 2669 kgh-1
= 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
5. Boiling Point Elevations
Evaporating temperature in first effect
Evaporating temperature in second effect
Evaporating temperature in third effect
= 110.5 + 0.60
= 99 + 1.50
= 83 + 4
= 111.1oC
= 100.5oC
= 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
ws
2.959 w1
w1
w2
w3
= 2704 x 2227/2202
= 8000kgh-1
= 8000
= 2704
= 2669
= 2628
= 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.
6. (a)Cooling in a jet condenser
Temperature cooling water
Temperature of hot vapour
Mass flow
Fresh milk
Milk concentrate
= 12oC
Max. temperature exit water = 25 oC
= 70oC
Latent heat
= 2334kJkg-1
-1
= 4000kgh
= 9% solids
= 30% solids
Mass balance kgh-1
Fresh milk
Concentrated milk
(a) Jet condenser
Heat Balance
Heat removed from condensate
Solids
360
360
Liquid
3640
840
Evaporated water
Total
4000
1200
2800
= 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
6
7062 x 10
=
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.
7. Mechanical recompression
In the evaporator:
Total water evaporated
Vapours recompressed
Energy used per kg vapour
Steam
=
=
=
=
Temperature of vapours
=
Total steam w
=
=
=
=
=
=
=
=
=
=
=
=
=
Heat available in steam
Heat in returned vapour
Energy used in compressor
Heat energy available
Steam energy saved
2800 kgh-1
1400kgh-1
160 kJkg-1
100 kPA (abs)
Latent heat 2258 kJkg-1 Temp 99.6oC
70oC
Latent Heat 2334 kJkg-1
(2334 x 2800) / 2258
2894kgh-1
2894 x 2258 x103
6535 x 106J
1400 x 2334 x103
3268 x 106J
160 x 103 x 1400J
224 x 106 J
(3268 –224) x 106
3044 x 106 J
(3044 x 106)/(6535 x 106)
0.466
46.6%
8. 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
Ju
=
=
5.216 x105 x 3600
725kgh-1
Rate of evaporation is 725 kgh-1