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```Example 1: A single effect evaporator is to be used to concentrate a food solution containing
15% (by mass)dissolved solids to 50% solids. The feed stream enters the evaporator at 291 K
with a feed rate of 1.0 kg s−1. Steam is available at a pressure of 2.4 bar and an absolute pressure
of 0.07 bar is maintained in the evaporator. Assuming that the properties of the solution are the
same as those of water, and taking the overall heat transfer coefficient to be 2300 W m−2K−1,
calculate the rate of steam consumption and the necessary heat transfer surface area. Working in
units of kg s−1, the overall material balance becomes
1.0 = V + L
Substituting into the component material balance for xF = 0.15 and xL = 0.50 gives
0.15 × 1.0 = 0.50 L
=>
L = 0.3kg s−1 , V = 0.7kg s−1
from steam tables: (If the steam and condensate remain saturated at 2.40 bar)
hS = 2715kJ kg−1 and hC = 530kJ kg−1
The feed enthalpy is determined by its temperature. Assuming the feed to be pure water, h F is
equal to hf at 291 K and therefore hF = 75.5kJ kg−1.
The enthalpies of the vapour and liquor streams are a function of the pressure within the
evaporator:
hV = 2572 kJ kg−1 (hg at 0.07 bar) and hL = 163kJ kg−1 (hf at 0.07 bar). The
enthalpy balance:
S(2715 − 530) = (0.70 × 2572) + (0.30 × 163) − (1.0 × 75.5)
S = 0.812 kg s−1 and
Q = 0.812(2715 − 530)kW = 1774kW
The temperature of steam at 2.4 bar is TS = 126.1◦C and the temperature of saturated liquid water
at the evaporator pressure of 0.07 bar is TE = 39.0◦C. Thus, from the rate equation, the heat
transfer area is
Example 2: An aqueous solution at 15.5◦C, and containing 4% solids, is concentrated to 20%
solids. A single effect evaporator with a heat transfer surface area of 37.2 m2 and an overall heat
transfer coefficient of 2000 W m−2 K−1 is to be used. The calandria contains dry saturated steam
at a pressure of 200 kPa and the evaporator operates under a vacuum of 81.3 kPa. If the boiling
point rise is 5 K, calculate the evaporator capacity.
1
At 200 kPa the steam and condensate enthalpies are hS = 2707 kJ kg−1, hC = 505 kJ kg−1,
Ts=120.2◦C.
The pressure within the evaporator is 101.3 − 81.3 = 20.0 kPa at which the boiling point of water
is 60.1◦C. The evaporator temperature is now 60.1◦C plus the boiling point elevation and
therefore TE = 65.1◦C.
=> S = 1.862 kg s−1
From steam tables, the feed enthalpy at 15.5◦C is hF = 65 kJ kg−1.
Vapour enthalpy: hV = 2609 kJ kg−1 (hg at 0.20 bar) + (1.91 × 5) kJ kg−1 = 2618.6 kJ kg−1
Cp (water vapour at 60.1◦C)= 1.91 kJ kg−1 K−1
The enthalpy of the concentrated liquor stream at the evaporator temperature is h L = 272 kJ kg−1
(hf at 65.1◦C). The component balance becomes
0.04F = 0.20L => F = 5L , V = 4L
S(hS − hC) = 4LhV + LhL − 5LhF
1.862(2707 − 505) = (4L × 2618.55) + 272L − 325L
L = 0.393 kg s−1 and the evaporator capacity is F = 1.97 kg s−1
EXAMPLE 3. Single effect evaporator: steam usage and heat transfer surface
A single effect evaporator is required to concentrate a solution from 10% solids to 30% solids at
the rate of 250 kg of feed per hour. If the pressure in the evaporator is 77 kPa absolute, and if
steam is available at 200 kPa gauge, calculate the quantity of steam required per hour and the
area of heat transfer surface if the overall heat transfer coefficient is 1700 J m-2 s-1 °C-1.
Assume that the temperature of the feed is 18°C and that the boiling point of the solution under
the pressure of 77 kPa absolute is 91°C. Assume, also, that the specific heat of the solution is the
same as for water, that is 4.186 x 103 J kg-1°C-1, and the latent heat of vaporization of the
solution is the same as that for water under the same conditions.
From steam tables, the condensing temperature of steam at 200 kPa (gauge)[300 kPa absolute] is
134°C and latent heat 2164 kJ kg-1; the condensing temperature at 77 kPa (abs.) is 91°C and
latent heat is 2281 kJ kg-1.
Mass balance (kg h-1)
2
Solids
Liquids
Total
Feed
25
225
250
Product
25
58
83
Evaporation
167
Heat balance
Heat available per kg of steam
= latent heat + sensible heat in cooling to 91°C
= 2.164 x 106 + 4.186 x 103(134 - 91)
= 2.164 x 106 + 1.8 x 105
= 2.34 x 106 J
Heat required by the solution
= latent heat + sensible heat in heating from 18°C to 91°C
= 2281 x 103 x 167 + 250 x 4.186 x 103 x (91 - 18)
= 3.81 x 108 + 7.6 x 107
= 4.57 x 108 Kg J h-1
Now, heat from steam
= heat required by the solution,
Therefore quantity of steam required per hour = (4.57 x 108)/(2.34 x 106)
= 195 kg h-1
Quantity of steam/kg of water evaporated = 195/167
= 1.17 kg steam/kg water.
Heat-transfer area
Temperature of condensing steam = 134°C.
Temperature difference across the evaporator = (134 - 91) = 43°C.
Writing the heat transfer equation for q in joules/sec,
q = UA DT
(4.57 x 108)/3600 = 1700 x A x 43
A = 1.74 m2
Area of heat transfer surface = 1.74 m2
(It has been assumed that the sensible heat in the condensed (cooling from 134°C to 91°C) steam
is recovered, and this might in practice be done in a feed heater. If it is not recovered usefully,
then the sensible heat component, about 8%, should be omitted from the heat available, and the
remainder of the working adjusted accordingly).
EXAMPLE 4. Concentration of tomato juice in a climbing film evaporator
Tomato juice is to be concentrated from 12% solids to 28% solids in a climbing film evaporator, 3 m
3
high and 4 cm diameter. The maximum allowable temperature for tomato juice is 57°C. The juice is
fed to the evaporator at 57°C and at this temperature the latent heat of vaporization is 2366 kJ kg-1.
Steam is used in the jacket of the evaporator at a pressure of 170 kPa (abs). If the overall heattransfer coefficient is 6000 J m-2 s-1 °C-1, estimate the quantity of tomato juice feed per hour. Take
heating surface as 3 m long x 0.04 m diameter.
Mass balance: basis 100-kg feed
Solids
Liquids
Total
Feed
12
88
100
Product
12
31
43
Evaporation
57
Heat balance
Area of evaporator tube pDHL
= p x 0.04 x 3
= 0.38 m2
Condensing steam temperature at 170 kPa (abs) = 115°C from Steam Tables.
Making a heat balance across the evaporator
q = UA DT
= 6000 x 0.38 x (115 - 57)
= 1.32 x 105 J s-1
Heat required per kg of feed for evaporation
= 0.57 x 2366 x 103
= 1.34 x 106 J
Rate of evaporation = (1.32 x 105)/ 1.34 x 106)
= 0.1 kg s-1
Rate of evaporation = 360 kg h-1
Quantity of tomato juice feed per hour = 360 kg
Problem 1. A single-effect evaporator is used to concentrate 7 kg/s of a solution from 10 to 50
per cent of solids. Steam is available at 205 kN/m2 and evaporation takes place at 13.5 kN/m2. If
the overall heat transfer coefficient is 3 kW/m2 K, calculate the heating surface required and the
amount of steam used if the feed to the evaporator is at 294 K and the condensate leaves the
heating space at 352.7 K. The specific heat capacity of a 10 per cent solution is 3.76 kJ/kgK, the
specific heat capacity of a 50 per cent solution is 3.14 kJ/kg K.
Problem 2. 1.9 kg/s of a liquid containing 10 per cent by mass of dissolved solids is fed at 338 K
to a forward-feed double-effect evaporator. The product consists of 25 per cent by mass of solids
and a mother liquor containing 25 per cent by mass of dissolved solids. The steam fed to the first
4
effect is dry and saturated at 240 kN/m2 and the pressure in the second effect is 20 kN/m2. The
specific heat capacity of the solid may be taken as 2.5 kJ/kg K, both in solid form and in
solution, and the heat of solution may be neglected. The mother liquor exhibits a boiling point
rise of 6 deg K. If the two effects are identical, what area is required if the heat transfer
coefficients in the first and second effects are 1.7 and 1.1 kW/m2 K respectively?
Problem 3. An evaporator, working at atmospheric pressure, is used to concentrate a solution
from 5 per cent to 20 per cent solids at the rate of 1.25 kg/s. The solution, which has a specific
heat capacity of 4.18 kJ/kg K, is fed to the evaporator at 295 K and boils at 380 K. Dry saturated
steam at 240 kN/m2 is fed to the calandria, and the condensate leaves at the temperature of the
condensing stream. If the heat transfer coefficient is 2.3 kW/m2 K, what is the required area of
heat transfer surface and how much steam is required? The latent heat of vaporisation of the
solution may be taken as being the same as that of water.
Problem 4. A liquid with no appreciable elevation of boiling-point is concentrated in a tripleeffect evaporator. If the temperature of the steam to the first effect is 395 K and vacuum is
applied to the third effect so that the boiling-point is 325 K, what are the approximate boilingpoints in the three effects? The overall transfer coefficients may be taken as 3.1, 2.3, and 1.1
kW/m2 K in the three effects respectively.
Problem 5. A single-effect evaporator with a heating surface area of 10 m2 is used to
concentrate a NaOH solution flowing at 0.38 kg/s from 10 per cent to 33.3 per cent. The feed
enters at 338 K and its specific heat capacity is 3.2 kJ/kg K. The pressure in the vapour space is
13.5 kN/m2 and 0.3 kg/s of steam is used from a supply at 375 K. Calculate:
(a) The apparent overall heat transfer coefficient.
(b) The coefficient corrected for boiling point rise of dissolved solids.
(c) The corrected coefficient if the depth of liquid is 1.5 m.
Problem 6. A triple-effect backward-feed evaporator concentrates 5 kg/s of liquor from 10 per
cent to 50 per cent solids. Steam is available at 375 kN/m2 and the condenser operates at 13.5
kN/m2. What is the area required in each effect, assumed identical, and the economy of the unit?
The specific heat capacity is 4.18 kJ/kgK at all concentrations and that there is no boiling-point
rise. The overall heat transfer coefficients are 2.3, 2.0 and 1.7 kW/m2 K respectively in the three
effects, and the feed enters the third effect at 300 K.
5
Problem 7. A single-effect evaporator is to produce a 35% solids tomato concentrate from a 6%
solids raw juice entering at 18°C. The pressure in the evaporator is 20 kPa(absolute) and steam is
available at 100 kPa gauge. The overall heat-transfer coefficient is 440 J m-2 s-1 °C-1, the boiling
temperature of the tomato juice under the conditions in the evaporator is 60°C, and the area of
the heat-transfer surface of the evaporator is 12 m2. Estimate the rate of raw juice feed that is
required to supply the evaporator.
[ 536 kgh-1 ]
Problem 8. Estimate (a) the evaporating temperature in each effect, (b) the reirements of steam,
and (c) the area of heat transfer surface for a two effect evaporator. Steam is available at 100 kPa
gauge pressure and the pressure in the second effect is 20 kPa absolute. Assume an overall heattransfer coefficient of 600 and 450 J m-2 s-1 °C-1 in the first and second effects respectively. The
evaporator is to concentrate 15,000 kg h-1 of raw milk from 9.5 % solids to 35% solids.Assume
the sensible heat effects can be ignored, and that there is no boiling-point elevation.
[ (a) 1st. effect 94°C, 2nd. effect 60°C, (b) 5,746 kgh-1, 0.53 kg steam/kg water (c) 450 m2 ]
6
STEAM TABLE - SATURATED STEAM
Temperature
Pressure(Absolute)
(°C)
(kPa)
Enthalpy
(sat. vap.)
(kJ kg-1)
Latent heat
Specific volume
(kJ kg-1)
(m3 kg-1)
2501
2499
2497
2492
2487
2483
2478
2473
2468
2464
2459
2454
2449
2445
2440
2435
2431
2407
2383
2359
2334
2309
2283
2257
2244
2230
2217
2203
2189
2174
2160
2145
2114
2083
2015
1941
206
193
180
157
138
121
106
93.9
82.8
73.3
65.0
57.8
51.4
45.9
40.0
36.6
32.9
19.5
12.0
7.67
5.04
3.41
2.36
1.67
1.42
1.21
1.04
0.892
0.771
0.669
0.582
0.509
0.393
0.307
0.194
0.127
Temperature Table
0
1
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
40
50
60
70
80
90
100
105
110
115
120
125
130
135
140
150
160
180
200
0.611
0.66
0.71
0.81
0.93
1.07
1.23
1.40
1.60
1.82
2.06
2.34
2.65
2.99
3.36
3.78
4.25
7.38
12.3
19.9
31.2
47.4
70.1
101.4
120.8
143.3
169.1
198.5
232.1
270.1
313.0
361.3
475.8
617.8
1002
1554
2501
2503
2505
2509
2512
2516
2520
2523
2527
2531
2534
2538
2542
2545
2549
2553
2556
2574
2592
2610
2627
2644
2660
2676
2684
2692
2699
2706
2714
2721
2727
2734
2747
2758
2778
2793
7
Pressure Table
7.0
9.7
12.0
14.0
15.8
17.5
21.1
24.1
29.0
32.9
40.3
45.8
60.1
75.9
93.5
99.6
102.3
104.8
107.1
109.3
111.4
113.3
115.2
116.9
118.6
120.2
127.4
133.6
138.9
143.6
147.9
151.6
167.8
179.9
1.0
1.2
1.4
1.6
1.8
2.0
2.5
3.0
4.0
5.0
7.5
10.0
20.0
40.0
80.0
100
119
120
130
140
150
160
170
180
190
200
250
300
350
400
450
500
750
1000
2514
2519
2523
2527
2531
2534
2540
2546
2554
2562
2575
2585
2610
2637
2666
2676
2680
2684
2687
2690
2694
2696
2699
2702
2704
2707
2717
2725
2732
2739
2744
2749
2766
2778
8
2485
2479
2473
2468
2464
2460
2452
2445
2433
2424
2406
2393
2358
2319
2274
2258
2251
2244
2238
2232
2227
2221
2216
2211
2207
2202
2182
2164
2148
2134
2121
2109
2057
2015
129
109
93.9
82.8
74.0
67.0
54.3
45.7
34.8
28.2
19.2
14.7
7.65
3.99
2.09
1.69
1.55
1.43
1.33
1.24
1.16
1.09
1.03
0.978
0.929
0.886
0.719
0.606
0.524
0.463
0.414
0.375
0.256
0.194
```
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