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Experiment No. 5

Evaporation

Queenie Gene Gadong

January 27, 2012

February 11, 2012

I. IntroductionWe have discussed in the previous experiments drying, which is the removal of water from food

materials. Drying however generally considers solid products as materials. In this experiment removal of

water from a liquid will be considered, that is Evaporation.

In evaporation the vapor from a boiling liquid solution is removed and a more concentrated solutionremains. (Geankoplis, 1995) The liquid is heated to boiling to enable a change of phase for the water

from liquid to gas and the gas is either left to evaporate or is separated from the original solution by

condensing. (Earle, 1983) Examples of evaporation products are concentrated sugar solutions, milk,

orange juices, and even glue. In some special cases evaporation is used to form crystals, such as in themaking of common table salt, this is referred to as crystallization. (Geankoplis, 1995)

There are several factors that affect the rate of evaporation and are considerations in the processing of

a liquid material through evaporation.

1. Concentration in the liquid.Initially the liquid is diluted and has low viscosity. However as evaporation proceeds the present

water in the liquid evaporates leaving the solution to become more concentrated. The increased

concentration in the solution leads to a reduction in the heat-transfer coefficient. The heat transfer

coefficient is the reciprocal of the thermal resistance of a material; thereby its reduction would indicate an

increase in the boiling point of the solution to evaporate it. (www.spiraxsarco.com) To be able to

accommodate this change in the solution and to prevent local overheating, adequate circulation isrecommended. (Geankoplis, 1995)

2. Solubility.During the course of the evaporation process the concentration of the liquid material increases and the

solubility decreases. Further evaporation may lead to crystallization due to the super saturation of the

solution. To be able to counter this effect an increase in temperature is recommended since the solubility

is directly related to temperature. (Geankoplis, 1995)

3. Temperature sensitivity of materials.There are products which have a low tolerance for temperatures. That is they degrade upon

introduction to high temperatures at a given length of time. Such products that have a compromised

quality when thoughtlessly placed in a high temperature environment are mostly milk, juices andvegetable extract. Also there are nutrients in a food material which are very volatile, it easily evaporates

from the food material. (Geankoplis, 1995)

4. Foaming or frothing.The capability of the liquid to froth or foam can also be a consideration in the evaporation process.

When the liquid froths, the foam can accompany the vapor evaporating and therefore there will be

reduction in the final volume of the product. (Geankoplis, 1995)

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5. Pressure and temperature.There is a direct relationship in the concentration of the solutes in a solution to the boiling point of the

liquid. As the concentration of the solutes increases the boiling point of the solution also increases. Also

the boiling point of the liquid is related to the pressure of the environment. At low pressure of the

evaporator the boiling point of the solution decreases. These characteristics of a solution can be used to

improve the evaporation process and to protect the integrity of the product by using vacuum pressure for

heat sensitive materials. (Geankoplis, 1995)

6. Scale deposition and materials of construction.Scales are deposited products in the solution, caused by fluid impurities, rust formation and other

reactions of the wall which cling to the surface of the heaters. The scales in turn create a thin insulation

between the liquid material and the heating surface, thereby reducing the over-all heat transfer coefficient.

(Geankoplis, 1995)

A typical evaporator has three functional parts, the heat exchanger, evaporating section, and the

separator. A heat exchanger usually transfers heat by coating with pressurized steam the tubes containing

the solution. The evaporating section is the inside of the tubes wherein the solutions are boiled and

subsequently evaporates. The separator is usually found on top of the machine wherein the vapour passes

thru and proceeds to a condenser. (Earle, 1983)

The type of evaporator in the CFOS sued in this experiment is the open kettle or pan evaporator. It is

the simplest form of evaporator which uses coils or in our case a jacket to boil the liquid solution. Pan

evaporators are easy to use and inexpensive although steam economy is very poor. (Geankoplis, 1995)

Machine evaporators generally work in two ways through the Single-effect evaporator and Multiple-

effect evaporator. A single effect evaporator usually only have a single unit wherein all the functional

parts is present, heat exchanger, evaporator, and separator. In this process the feed enters in a separate

pipe from the pressurized steam. The steam would condense as drips or condensate. Then the solution is

boiled and the vapor separates from the solution leaving a concentrated product. In a single effect

evaporating the process ends here. (Geankoplis, 1995)

A multiple effect evaporator on the other hand makes use of the latent heat of the leaving vapor. The

boiling point of a liquid is achieved only when the vapor pressure of the solution is in equilibrium with

the environment. Assuming that the solution is properly mixed we can infer that the temperatures of both

the solution and the leaving vapor are the same. The latent heat of the vapor serves as the heating medium

for another effect of evaporation. (Geankoplis, 1995)

The efficiency of an evaporator in removing the water from a liquid solution can be measured through

Steam Economy. Efficiency is concerned with the maximization of outputs from the inputs provided.

Determining the steam economy follows this equation:

.

Determining for the efficiency of the evaporating system should also consider the following:

a. Model of the evaporation processb. The heat and material balancec. Heat transfer coefficientsd. Flow of solution as a function of pressure

(www.eng.fsu.edu.com)

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Heat transfer coefficient is almost synonymous to the capability of the material to absorb heat. A

high heat transfer coefficient indicates that the material is able to absorb the heat transferred effectively.

Total heat transfer coefficient is the reciprocal of the overall resistance to heat transfer of all individual

resistances. The total heat transfer coefficient (U) is a combination of both conductive and convective

resistance between the solution and the steam separated by the metal wall, and even the solid scales inside

the pipes. An accurate calculation of the total heat transfer coefficient is hard to produce since itsmagnitude depends on several different factors such as nature of heat transfer process, physical property

of the solution, flow rate and the layout of the evaporator. (www.spiraxsarco.com)

A general equation used for determining the total heat transfer coefficient is:

(Earle, 1983)

Sugar, (sucrose C12H22O11) is the term used for the crystalline form of the carbohydrates which

are readily soluble in water and are sweet. Sugar is obtained from plant extracts mainly in sugarcane,

processed through photosynthesis of the plant. (Encarta, 2009) Being a biological material a sugar

solution has a low tolerance for temperature. The compromise in the quality of the product is a function

of high temperature at a lengthy time. Sugar solutions specifically react to heating resulting to color shifts,from colorless to a golden brown, and alteration of flavor, as a result of the breakdown of the glycosidic

bonds in the sugar. (BeMiller, 2010)

The objective of this experiment is to compare and analyze the performance and efficiency of a

single-effect evaporator at normal pressure under different steam pressure through steam economy. Alsothis experiment will teach the students to be able to determine the overall heat transfer coefficient (U)

from experimental data.

II. MethodologyFor each group a 10 Liter 10% w/v sugar solution was dissolved in a plastic pail. This was done by

dissolving 1 kg of sugar crystal, bought from the Miagao Market, in a 10 L of tap water. The temperatureand specific gravity of the solution was determined using a thermometer and hydrometer respectively.

Also the weight of the solution inside the pail was measured. The solution was then divided into 2 equal

portions in plastic pails. The group then proceeded to the CFOS canning laboratory where they have 2pan

evaporators and a boiler system.

The dimensions of the pan evaporators were measured, such as the width of the mouth and its depth

to allow the group to calculate for the surface area of the solution directly exposed to the steam provided.

The solutions were poured inside the evaporators and the height of the solution inside was measured. On

one of the evaporator the steam pressure was set at 5 psi, and on the other was at 10 psi, therefore theboiler was run at a total of 15psi.

Using the Ellab the group then monitored the temperatures of the two (2) steam condensate and the

two (2) the sugar solutions, with a five (5) minute interval, for 30 minutes, only after the desired steam

pressure was obtained. After the 30 minutes was up the boiler was immediately shutdown and the

product and condensate were withdrawn immediately. The temperature of both the products and steamcondensate was taken immediately and consequently weighed.

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III. ResultsTable 1: Tabulated results of evaporation process at 10 psi and 5 psi.

USteam Economy

(%)

q, Heat

gainedT A U=q/AT

10 psi 63.58 4779.34 115.21.4191

249.47

5 psi 50.3 21324.29 8.53 1761.62

Table 2: Amount of water evaporated from evaporation process.

Water evaporated wt. solution, g wt. product, gWater evaporated,

g

10 psi 2475.3 418 2057.3

5 psi 2475.3 1481.3 994

IV. DiscussionEfficiency of an evaporator in removing water from a solution can be determined by the ratio of the

output steam to the input steam. From Table 1 it can be noticed that between the two different pressures

the, 10 psi and 5 psi, there is a significant difference in the efficiency of both pressures. Assuming all

conditions between the two set-up are equal we can assume that the a higher steam pressure would

generate a more effective evaporation process for using a open kettle or pan evaporator. There is a 13.28

difference in the percent yield between the two pressures.

Calculation for the U, Over all Heat Transfer Coefficient seem complicated and too long at first,

however when the values needed have already been found, integration and final calculations are easy tofollow. Over all Heat Transfer Coefficient (U) is the reciprocal of the resistance of a solution to heat

transfer and is therefore an indication of the effectiveness of an evaporation process. The greater the U,

the harder it is to evaporate. From Table 1 it would be noted that there is a significant difference in the U,

at both the 10 and 5 psi steam pressure. There is a 1512.15 difference between the values of U for 10 and

5 psi. From this big difference in the values we can infer that at steam pressure 10 psi evaporation is moreeffective than at 5 psi. And this is manifested by the Steam economy values we have obtained from Table

1. The lower value of U at 10 psi steam pressure may be caused by the agitation brought about by the

more violent boiling of the solution inside of the pan evaporator. Transfer of heat through the solution

was aided by the constant movement of the molecules in the sugar solution. The reduced amount of

agitation at 5 psi steam pressure could have added for the increase in U for this set-up.

V. ConclusionFrom this experiment we can infer that for an open kettle or pan evaporator a greater steam pressure

would yield a greater Steam Economy. The overall heat transfer coefficient can predict the efficiency of

an evaporation set-up.

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VI. References BeMiller, J. N. (2010). Carbohydrate Analysis. In S. S. Nielsen, Food Analysis 4th Ed. . Earle, R. (1983). Unit Operations in Food Processing. The New Zealand Institute of Food Science

and Technology Inc. .

Geankoplis, C. J. (1995). Transport Processes and Unit Operations 3rd Ed. University ofMinnesota: Prentice-Hall International, Inc.

Spirax-Sarco Limited. (2012) Heat Transfer. http://www.spiraxsarco.com/resources/steam-engineering-tutorials/steam-engineering-principles-and-heat-transfer/heat-transfer.asp, accessed

on 02/9/2012.

FAMU.College of Engineering. 2002. Steam Economy.http://eng.fsu.edu/me/senior_design/2002/folder16/steam_economy.html

Microsoft Encarta. 2008. Sugar. Microsoft Corporation .

VII. AnnexTable 3: Temperature raw data from Ellab probes

T (C)

t, min 10 psi 5 psiCondensate

@ 10 psi

Condensate

@ 5 psi

0 28.8 28.8 - -

5 100.4 96.7 79.4 78.4

10 100.5 100.4 77.4 78.1

15 100.7 100.4 72.4 74.7

20 101.1 100.3 73 76

25 102.1 100.4 71.7 77.2

30 105.4 100.4 74.2 82.5

Ave. 101.7 99.77

Table 4: Evaporator Dimensionsevaporator dimensions (cm)

height diameter height of sample

69.342 48.26 5.08

http://www.spiraxsarco.com/resources/steam-engineering-tutorials/steam-engineering-principles-and-heat-transfer/heat-transfer.asphttp://www.spiraxsarco.com/resources/steam-engineering-tutorials/steam-engineering-principles-and-heat-transfer/heat-transfer.asphttp://eng.fsu.edu/me/senior_design/2002/folder16/steam_economy.htmlhttp://eng.fsu.edu/me/senior_design/2002/folder16/steam_economy.htmlhttp://eng.fsu.edu/me/senior_design/2002/folder16/steam_economy.htmlhttp://www.spiraxsarco.com/resources/steam-engineering-tutorials/steam-engineering-principles-and-heat-transfer/heat-transfer.asphttp://www.spiraxsarco.com/resources/steam-engineering-tutorials/steam-engineering-principles-and-heat-transfer/heat-transfer.asp

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Table 5: Weight sugar solution and specific gravity.

Sugar Solutionwt. container,

gwt. cont.+sol'n, g wt. sol'n, g Specific Gravity

10 psi 523.9 3000 2476.11.035

5 psi 525.5 3000 2474.5ave. 2475.3

Table 6: Weight of concentrated product, after evaporation, and specific gravity.

Concentrated Product wt. container, g wt. container+prod., gproduct wt.,

g

specific

gravity

10 psi 166.9 584.9 418 1.31

5 psi 171.6 1752.9 1581.3 1.055

Table 7: Weight of steam used, condensate.

Condensate

wt.

container,

g

wt.

cont+cond,

g

wt.

condensate,

g

10 psi 187.4 3423 3235.6

5 psi 246.7 2223 1976.3

Sample calculations: