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Condensate Receiving & Condensate Flash Recovery tank design
Abstract : Condensate receiving tank or condensate mound is one of the simple equipment in sugar industry and other process industries. If we provide common condensate flash recovery tank then no need to provide individual mounds for each evaporation or heating bodies like juice heaters, evaporator bodies, pans .. etc.
Here explain about calculation of sizing requirement for condensate receiving tank (Condensate mound). Also explain the design concepts of single flash recovery condensate receiving tank and its flash line sizing.
Concepts to find the sizing of the condensate receiving tank:
a) First we conform the purpose of the the condensate mound. I.e It is either condensate receiving tank or condensate flash vapour recovery tank.
b) Find the condensate quantity to handling the condensate mound.
c) Parameters of the condensate water and its flash vapour like temperature, density, velocities, latent heat, specific volume … etc.
d) Calculate the flash vapour velocity at the time of generation from condensate. Its velocity calculated by the “The Souders – Brown equation”
Velocity of flash vapour in flash tank(m/sec) = U_{max} = C x [ (ρ _{L} – ρ_{V})/ ρ_{V}]^{0.5}
Condensate Water Density = ρ _{L }in kg/M^{3}
Flash vapour Density = ρ _{V }in kg/M^{3}
Here C = The value of the constant.
C is a measure of the droplet size that will be carried over and depends on the degree of separation of liquid and vapor required.
According to Cane sugar engineering by Peter Rein , coefficient C recommended 0.01 m/s.
Now calculate flash vapour quantity and its velocity in condensate mound. From this we can easily calculate the dia required for condensate receiving tank.
For common condensate flash recovery system purpose please go through the following link
Flash Vapour Calculation | Flash Vapour Recovery Vessel Design Calculation.
For simple condensate receiving tank purpose, flash vapour to be taken only for 2 to 3^{o}C of the inlet water temperature.
a) Generally the height of the condensate receiving tank required for 1.5 to 2.0 meters with the calculated dia.
b) Provide partition plate in the middle of this condensate mound. Insert the condensate inlet line in partitioned portion(opposite to the condensate outlet line).
c) Bottom of the condensate line provide minimum 200mm gap from the bottom plate of condensate mound. The partition plate height to be provide minimum 500mm less than the total height of the condensate mound.
Now take one example to calculate the sizing of the condensate receiving tank and condensate flash vapour recovery tank.
Condensate Receiving Tank ( Mound )Calculation |
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S.NO | DESCRIPTION | Symbol / Formula | VALUES | UOM |
1 | Condensate Quantity | Qc | 50 | M^{3}/hr |
0.013889 | M^{3}/sec | |||
2 | Condensate inlet temperature | Ti | 125 | ^{o}C |
3 | Condensate outlet temperature (To be take 2 to 3^{o}C diff.) | To | 123 | ^{o}C |
4 | Condensate water outlet velocity from mound | Vc | 0.5 | m/sec |
5 | Flash vapour velocity in vent line or equalizing line | Vf | 50 | m/sec |
U= Flash Vapour maximum velocity in condensate mound | ||||
U = C ((ρ_{L} – ρ_{V})/ρ_{V})^{0.5} | ||||
Condensate water density | ρ_{L} | 939.00 | kg/M^{3} | |
Flash Vapour Density | ρ_{V} | 1.30 | kg/M^{3} | |
Coefficient | C | 0.01 | m/sec | |
Specific volume of flash vapour | µ | 0.77 | M^{3}/kg | |
Latent Heat of flash vapour | λ | 522.50 | Kcal/Kg | |
U | 0.27 | m/sec | ||
Flash vapour Quantity | Qv = Qc x (To – Ti) / λ | 0.19 | T/hr | |
191.39 | Kg/hr | |||
Qv x 1000 x µ | 147.38 | M^{3}/hr | ||
0.04 | M^{3}/sec | |||
Condensate Mound Dia | ||||
Flash vapour Quantity | Qv | 0.04 | M^{3}/sec | |
Flash vapour velocity in mound | U | 0.27 | m/sec | |
Cross sectional area of mound | Am = Qv/U | 0.15 | m^{2} | |
Condensate mound Dia | Dm = sqrt ( Am /0.785) | 0.44 | mtr | |
450 | mm | |||
Condensate mound vent or equalizing line Dia | ||||
Flash vapour Quantity | Qv | 0.04 | M^{3}/sec | |
Flash vapour velocity in vent line | Vf | 50 | m/sec | |
A | Av = Qv / Vf | 0.00082 | ||
Flash vapour Line Dia | Dv = sqrt ( Av /0.785) | 0.032 | mtr | |
i .e | 50 | mm | ||
Condensate mound outlet water line Dia | ||||
Condensate Quantity | Qc | 0.0139 | M^{3}/sec | |
Vc | 0.50 | m/sec | ||
Cross sectional | Ac = Qc / Vc | 0.0278 | m^{2} | |
Condensate water oulet line dia | Dc = sqrt ( Ac /0.785) | 0.1881 | mtr | |
200 | mm | |||
^{ Cross check the dia by this formula }D > 1.116 x (Qc)^{0.4} | ||||
Condensate mound outlet water line Dia | 201.71 | mm |
Individual Flash Recovery Condensate Mound Calculation |
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S.NO | DESCRIPTION | Symbol / Formula |
VALUES | UOM |
1 | Condensate Quantity | Qc | 50 | M^{3}/hr |
0.013889 | M^{3}/sec | |||
2 | Condensate inlet temperature | Ti | 125 | ^{o}C |
3 | Condensate outlet temperature.(After Flash) | To | 105 | ^{o}C |
4 | Condensate water outlet velocity from mound | Vc | 0.5 | m/sec |
5 | Flash vapour velocity in flash outlet line | Vf | 50 | m/sec |
U= Flash Vapour maximum velocity in condensate mound (At the time of generation) | ||||
U = C ((ρ_{L} – ρ_{V})/ρ_{V})^{0.5} | ||||
Condensate water density | ρ_{L} | 939.00 | kg/M^{3} | |
Flash Vapour Density | ρ_{V} | 1.30 | kg/M^{3} | |
Coefficient | C | 0.01 | m/sec | |
Specific volume of flash vapour | µ | 1.42 | M^{3}/kg | |
Latent Heat of flash vapour | λ | 535.75 | Kcal/Kg | |
U | 0.27 | m/sec | ||
Flash vapour Quantity | Qv = Qc x (To – Ti) / λ | 1.87 | T/hr | |
1866.54 | Kg/hr | |||
Qv x 1000 x µ | 2648.33 | M^{3}/hr | ||
0.74 | M^{3}/sec | |||
Condensate Mound Dia | ||||
Flash vapour Quantity | Qv | 0.74 | M^{3}/sec | |
Flash vapour velocity in mound | U | 0.27 | m/sec | |
Cross sectional area of mound | Am = Qv/U | 2.74 | m^{2} | |
Condensate mound Dia | Dm = sqrt ( Am /0.785) | 1.87 | mtr | |
1870 | mm | |||
Condensate mound flash outlet line Dia | ||||
Flash vapour Quantity | Qv | 0.74 | M^{3}/sec | |
Flash vapour velocity in vent line | Vf | 50 | m/sec | |
Cross sectional area | Av = Qv / Vf | 0.01471 | ||
Flash vapour Line Dia | Dv = sqrt ( Av /0.785) | 0.137 | mtr | |
i.e | 150 | mm | ||
Condensate mound outlet water line Dia | ||||
Condensate Quantity | Qc | 0.0139 | M^{3}/sec | |
Vc | 0.50 | m/sec | ||
Cross sectional area | Ac = Qc / Vc | 0.0278 | m^{2} | |
Condensate water outlet line dia | Dc = sqrt ( Ac /0.785) | 0.1881 | mtr | |
200 | mm | |||
^{ Cross check the dia by this formula }D > 1.116 x (Qc)^{0.4} | ||||
Condensate mound outlet water line Dia | 201.71 | mm |
Online calculation for Condensate receiving tank and flash vapour recovery condensate mound
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5 thoughts on “Condensate Receiving Tank Design Calculation | Condensate Mound”
VijayKumar Dure
(October 31, 2017 - 4:15 pm)It’s usefull site I learn much more thanks
siva alluri
(November 1, 2017 - 1:01 pm)Thank you
The “www.sugarprocesstech.com” invites to all sugar technologists to share your knowledge, achievements in your working organization and new developments and technologies in sugar industry and its concerned units. It is very much helpful to show your identity to the world at the same time it will helpful to another technologist to enhance their insight and enhance great execution in there working. This website also provides the basic knowledge in sugar industry technologies and equipment design calculation with online calculators.
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Pranav
(April 22, 2021 - 10:04 pm)Very informative website. Nicely explained the content. I have few queries regarding sizing of condensate storage cum flash tank.
siva alluri
(April 26, 2021 - 3:58 am)What queries
Ram veer
(June 28, 2021 - 8:54 am)Dear sir my question is about steam consumption.
If any pan dropped at same time steam consumption reduced i.e. syrup brix effected. In this situation how much evaoporation should be increased in evaporators that syrup brix will maintain