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| Diagram n.01 | Word |
Mass balances are an important tool that is used to define the dimension of the production, define the amount of raw materials that enter into the process and the amount of products and waste that leave the process. Overall, with a mass balance it is possible to estimate the yield of a process.
Given a process:
The corresponding mass balance is the following:
$$ input = output + accumulation $$Where $accumulation$ è is the amount of material that accumulate into the reactor.
To design a new process, it is important to define a flow diagram. This diagram must be clear, organized in a way that is easily readable. A well-designed flow diagram should allow to define the main operations that characterize the process. Moreover, a flow diagram should define all the raw materials and the products leaving the process, as well as the waste products. In addition, a flow diagram can be enriched with information about the composition of the raw ingredients, the amount of raw materials, or any other processing variable that is important to understand the process.
Given a process:
The overall mass balance is built by writing on the left all the amount of raw materials entering into the process or reactor and, on the right, the amunt of all the materials leaving the process or reactor. In the example, letter "A" is the raw material. Instead, letters "B" e "C" are products leaving the process. Thus, the overall mass balance can be written as:
$$ A = B + C $$The letters correspond to masses (in kg), to volumes (in m$^3$ or rates (kg / s or m$^3$/s). Conversion between mass and volume requires the knowledge of the density. Density of foods as a function of the temperature can be calculated from the information about the composition, by the following equations:
| Density (kg/m$^3$) | Food component | Thermal model |
|---|---|---|
| Protein | $\rho = 1.3299 \cdot 10^3 – 5.1840 \cdot 10^{–1} \cdot t$ | |
| Fat | $\rho = 9.2559 \cdot 10^2 – 4.1757 \cdot 10^{–1} \cdot t$ | |
| Carbohydrate | $\rho = 1.5991 \cdot 10^3 – 3.1046 \cdot 10^{–1} \cdot t$ | |
| Fibers | $\rho = 1.3115 \cdot 10^3 – 3.6589 \cdot 10^{–1} \cdot t$ | |
| Water | $\rho = 9.9718 \cdot 10^2 + 3.1439 \cdot t^{-3} – 3.7574 \cdot 10^{–3} \cdot t^2$ |
For instance, a fruit juice composed by 10% sugars and 90% water, at 25°C, has a calculated density of 1049 kg/m$^3$. This value is in close agreement with A.M Ramos and A. Ibarz (JFE, 1998, 57-63) for different juices.
The chemical composition of the materials "A", "B" e "C", of the previous example, may be known. In such case, it is possible (and very useful) to report the mass fraction of the components. Mass fraction is defined as:
Determine the amount of concentrated fruit juice that contains 65％ of solids that should be mixed with fruit juice with a content of 15％ of solids in order to give 100 kg of resulting product with 45％ of solids.
It follows the flow diagram:
The solution of the problem requires the definition of at least two balance equation. One equation is the overall mass balance. The other equation is the mass balance of the components.
$$ \begin{cases} A + B = C \\ x_A \cdot A + x_B \cdot B = x_C \cdot C \\ \end{cases} $$By combining the two equation, you have the following result:
$$ \begin{cases} A + B = 100 \\ 0.15 \cdot A + 0.65 \cdot B = 0.45 \cdot 100 \\ \end{cases} $$If you substitute $A = 100 - B$ in the mass balance of the components, this gives:
$$ 0.15 \cdot (100 - B) + 0.65 \cdot B = 45 $$If you simplify, you get the solution:
$$ B = 60 $$Thus:
$$ A = 40 $$The process include a washing step. As soon as the fruit enter into the plant,
fruit are initially washed and, subsequently, the move into a conveyor belt, where
trained operators start with the inspection. In this step, 20 kg of water are used per 100 kg of
product. During washing, about l'1％ of the initial mass is lost. During the following
inspection phase, the amunt of lost products is about 2％.
Afterwards, the cleaned and selected fruits move in the hot areas, where the fruit is crushed and
immidiately heated. The heating step is called hot break. The purpose of the hot break process is to
stabilize the fruit by the denaturation of enzymes, like polyphenoloxidase and pectinase.
In details, polyphenoloxidase are rsponsible for enzymatic browning which it leads to
the change of color of the product. Instead, pectinase are a family of enzymes that are
responsible for the degradation of the pectin, the formation of deposit, the change in consistency of
the product, the change in the colloidal stability, etc. Since pectinase are thermostable,
the hot-break process needs very high temperatures (higher than 85℃)
The heated product passes through one or more pulpers, which is composed by a siever
that is able to remove peals and other undesired component of the fruits.
The outcome of this process leads to a loss of products of about 2％.
At this point, the product must be concentrated. This phase occurs with an evaporator, where
the product fruit enters, loose water by boiling under vacuum, and reach a concentration of
solids of about 32％.
Finally, the concentrated fruit can be sterilized and aseptically packed in bag-in-box or
stainless steel tanks.
Th process should take into acount also the management of the waste. Waste can be recoverd.
Typically, waste from fruits can be dried, mechanically pressed (removing about 50％ of water)
and air dried, untill a dry waste with a final moisture of 8％ is achieved.
The most important phases for the mass balance are the following:
During the step of washing and inspection, about 3％ of the initial products are removed. During washing, fruit products (P) and water (W) enter, leading to clean fruit products (PP) and waste water (R). During the inspection phase, cleaned fruit products enter into the process, leading to selected fruit products (PS) and a percentage of removed products (S). It follows the corresponding flow diagram:
Mass balance of the washing step:
$$P + W = PP + R $$Mass balance of the inspection phase:
$$PP = PS + S $$During washing, about 1％ of the initial products is removed. Also, during inspection, about 2％ of the fruits is removed:
$$PP = P \cdot 0.99 $$e
$$PS = PP \cdot 0.98 $$Assume that the initial mass of fruit is 100 kg, then:
$$PP = 99 \si{\kilo\gram}$$e
$$PS = 97 \si{\kilo\gram}$$Now, the flow diagram can be enriched with the following information:
It follows the flow diagram, where all the information related to the mass balance are reported.