Pressure drop in Triple, Quadruple and Quintuple effect evaporator bodies with online calculator
In this article discussed about history of multiple effect evaporator, Rillieux’s principles for multiple effect evaporator bodies and distribution of pressure drops across the multiple effect evaporators like Triple effect, Quadruple effect and Quintuple effect and also given online calculator for pressure drop calculation in multiple effect evaporator with values of latent heat and specific volume of evaporator bodies.
History of multiple effect evaporator:
a) In the early years of 19th century in sugar industry follows the open boiling system. The steam consumption was very high and steam produced from bagasse had not sufficient and large quantities of extra wood had required.
b) E.C. Howard an English man from worked on improvements to the then primitive open boiling process. In 1812-13, he took out patents relating to his invention of the vacuum pan. It was really a single effect evaporator apparatus which was a tremendous break through. Steam of 15-30 psig was used. A vacuum of 24” was achieved and the condensed water was returned to boiler.
c) Norbert Rillieux took his first patent in 1843 shows two Howard Vacuum Pans in series and some what similar to degrands second attempt. His second patent in 1846 was really break through and this showed his three vessels with horizontal tubular heating surface. Rillieux’s first multiple effect evaporators in the sugar industry were triple effects in which the whole concentration was carried on the last body acting as a vacuum pan and process was intermittent as far as evaporation was covered. The last of Rillieux multiple effect evaporator worked up to 1902 at Sin Clair in Louisiana and disused plant was standing in Cuba up to 1914 at armonia and Santa lucia.
d) Monsieur Jules Robert a Frenchman Director of Seelowitz sugar factory in Austria in 1852 constructed apparatus according to Rillieux’s real system. He actually first changed the horizontal tubes to vertical and probably therefore the evaporator bodies are called as Robert type evaporator bodies.
e) LA Tramp in his book (1936) says “Perhaps the most out standing feature in steam economy in the sugar industry has been the invention of Rillieux of Lousiana about a century ago of the multiple effect evaporator, which is now used all over the Globe.
In a multiple effect evaporator of “ N “ effects, one kg of steam will evaporate N kg of water. Thus in a double effect evaporator, one kg steam will evaporate two kg of water, in a triple effect, three kg, and so on.
The second principle deals with the practice of vapour bleeding. If vapour is withdrawn from the ith effect of a multiple-effect evaporator of N effects and used outside the evaporator system in place of steam, the steam saving will be i/N times the quantity of steam used in this duty.
Example: one kg of vapour withdrawn from the second effect of a four effect evaporator would save ( 2/4 x 1 = 0.5) 0.5 kg
Similarly, five kg of vapour withdrawn from the third effect of the equitable evaporator set would save ( 3/5 x 5 = 3 ) 3 kg steam.
Chen and Chou (1993) give a third principle although only the first two were the subject of Rillieux’s patents.
Wherever steam or vapor is condensed, provision must be made to withdraw in-condensable gases continuously.
Distribution of pressure drop across the multiple effect evaporator:
It is important to distribute the total pressure drop of a multiple effect between the vessels in such a way that the individual pressure drops under which the different vessels are working are approximately equal, but decrease slightly from the first to the last vessel.
The objects of this equalization of pressure drops are to give uniform heights of siphon between vessels, and avoid risks of entertainment which would be caused by giving the earlier vessels too high a pressure drop.
Take one example for pressure drop calculation:
Data to be required:
1st body calendria pressure ( i.e exhaust may be) = 1.2 kg/cm2
Lost body vacuum = 630 mm of Hg
Calculation part :
1st body calendria pressure = 1.2 kg/cm2 = 1.2 + 1.03323 = 2.23323 abs kg/cm2 (Note : Atmospheric pressure = 1.03323 abs kg/cm2 )
Lost body vacuum = 630 mm of Hg =1.03323 x [(760-630)/760] = 0.1767 abs kg/cm2
Total Pressure drop across the evaporator set = 23323 – 0.1767 = 2.05653 abs kg/cm2
|Body Name||Pressure drop ( ΔP) in each body in abs Kg/cm2||Pressure of the body in abs Kg/cm2|
|1st Body Calendria||1.2 + 1.03323 = 2.23323|
|1st body||2.05653 x (11/50) = 0.4524||2.23323 – 0.4524 = 1.7808|
|2nd body||2.05653 x (10.5/50) = 0.4319||1.7808 – 0.4319 = 1.3489|
|3rd body||2.05653 x (10/50) = 0.4113||1.3489 – 0.4113 = 0.9376|
|4th body||2.05653 x (9.5/50) = 0.3907||0.9376 – 0.3907 = 0.5469|
|5th body||2.05653 x (9/50) = 0.3702||0.5469 – 0.3702 = 0.1767|
According to the above absolute pressure of the bodies we will get temperature, latent heat and specific volume from steam table from the following link.
Steam Table for Saturated Steam
Online Calculator for Multiple Effect Evaporator Pressure Drop calculation
From this calculator also get Latent heat & Specific volume of the that vapors
Accessories of Rotary Vacuum Filter and its operation in sugar industry.
Juice Defecator and Juice Sulphitor Design Criteria | Online Calculator.
Flash Vapour Calculation | Flash Vapour Recovery Vessel Design Calculation.
Rillieux’s principles and pressure drops calculation for Multiple Effect Evaporator.
Steam Percent cane and heating surface of the individual evaporator bodies in multiple effect evaporator.
Hi friends Thanks for reading. I Hope you liked it. Give feed back, comments and please don’t forget to share it
rgun007(April 9, 2017 - 4:37 am)
Vijay K(July 8, 2022 - 7:30 am)
Umakant Paul(January 26, 2018 - 7:14 pm)
Thank you for Sugar industry all details with calculation
siva alluri(January 27, 2018 - 5:17 pm)
Thank you Mr. patil
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.
UDAI VEER(May 29, 2018 - 1:41 pm)
Welcome , Thanks to all off seniar persons help to juniars given knowledge of this website
Vry thks shiv ram krishna sir
siva alluri(May 30, 2018 - 1:28 pm)
Thank you Ms.UDAI VEER
BHARATKUMAR MAGDUM(August 22, 2018 - 5:18 pm)
Good calculation of pressure drop calculation of multiple effect evaporator. But how calculate pressure drop of DEVC+Quad
siva alluri(August 23, 2018 - 2:52 pm)
It is very simple take two types of sets one is double effect and another is Quad effect then you will calculate like above.
We will publish one example on DEVC+Quad soon
Ajit Tigadi(July 28, 2021 - 5:04 pm)
How to calculate boiling point rise for different tube length of evaporater
BHARATKUMAR MAGDUM(August 22, 2018 - 5:24 pm)
How calculate ethanol production from final molasses?
siva alluri(August 23, 2018 - 2:48 pm)
We will publish that article soon
Arun Kumar Bhati(December 30, 2018 - 5:15 pm)
How to produce 85% of M30. What is grain or seed size and what are the grader meshes sizes. What are the pan boiling scheme for producing 85% of M30.
siva alluri(January 4, 2019 - 4:25 pm)
Ok we will provide soon
Sanjeev Tiwari(October 5, 2019 - 11:09 am)
Thank you sir, I an safety engineer and not having deep knowledge of sugar industry equipment. I was interested in the pressure in the Body of Pan which is vacuum. We need to test the vessel by providing hydraulic pressure, which is 1.5 times the safe working pressure of the vessel.I am pleased to get the answer from your presentation, once again thanks sir.
siva alluri(October 10, 2019 - 3:52 pm)
Welcome Mr.Sanjeev Tiwari
Ram veer(June 30, 2021 - 3:32 am)
How to calculate pressure drop in evaporator such series given below.
1.Five bodies in a series.
2.first body juice goes to 2nd body after that 3rd and so on.but steam used in (2nd and 3rd) body is of same 1st body.
In this case pressure drop calculation by quard or quintuple manner ?