• Troullinos Yannis

Packaging Report

In this practical class, different films for food packaging were examined as far as their physical properties and their ability to preserve grapes, cheese, meat and potatoes. Appropriate measurements and tests we done on specific time intervals.

Results

Table 1. Results for the rapid tests for the identification of packaging materials

* Materials were identified using the “packaging materials identification chart for films” in the practical booklet.

Table 2. Results of the mechanical and physical properties of the packaging material

Discussion

Calculations and Questions:

1. Calculate the tensile strength of the three packaging materials tested.

Table 3. Physical properties of different packaging materials

* More than 25% difference from the mean

By using N = ±100 g, Force mean values for each of the materials can be found. Also, Area = Width (m) x Gauge (m) = X m². Tensile strength = Force (N) / Area (m²) so for the above materials we have:

Cellulose 340 DMS Tensile strength = 25.6 / 2.125 x 10^-6 = 12.0 x 10⁶ N/m² = 12.0 x 10⁶ Pa = 12.0 MPa, as 1 N/m² = 1 Pa, while 1 MPa = 1,000,000 Pa

Polypropylene Tensile strength = 108.7 / 2.5 x 10^-6 = 43.5 MPa

Polyethylene Tensile strength = 12.5 / 0.75 x 10^-6 = 16.7 MPa

2.Define tensile strength and discuss what factors will affect the tensile strength of the packaging material

Tensile strength is the maximum load that a material can support without fracture when being stretched, divided by the original cross-sectional area of the material. Generally, as tensile strength increases, the tougher the material is considered (Hui, 2008). Factors affecting the tensile strength are (Yam, 2010; Fellows, 2009):

• Plasticiser levels (increased values give less tensile strength and more elasticity
• Degree of crystallinity (crystal structure)
• Density of the material (increasing density gives more tensile strength)
• Manufacturing process (orientation, treatment, coatings)
• Temperature
• Physical properties of the material (branching, side groups, chain length, molecular weight)
• Duration of the time that the force is applied

3.Compare your tensile strength results to those found in literature.

According to Goodfellow Cambridge Ltd. tensile strength for regenerated cellulose is 50 MPa, which, as mentioned, is affected by a lot of different factors. In our experiment, tensile strength of the cellulose used is a lot lower (12MPa).

Paine (1990) gives values of 30 MPa for polypropylene, while in this experiment a value of 43.5 MPa was calculated.

Finally, polyethylene gave an experimental value of 16.7 MPa, while Goodfellow Cambridge Ltd. reports 5-25 MPa for low density polyethylene (LDPE) and 15-40 MPa for high density polyethylene (HDPE). In this experiment it is unknown which exactly was the type of PE used, as there are many different types in market.

As explained, duration of the force applied affects the tensile strength, so different testing machines give different results. There are numerous more factors as noted in question 2, which greatly affect the measurements and results. Thus, comparing values to literature cannot give objective judgement of the experiment.

4.Calculate the moisture vapour transmission rate (g m^-2 day^-1) for each of the films tested

Table 4. Results of the water vapour permeability test

Circle area = π r² = 0.005 m² (r = 40mm = 0.04m)

Number of Days = 4, as Day 1 is the day we started the storage

Cellulose 340 DMS

1^st measurements:

Total moisture gained = Weight of Day 5 – Weight of Day 1 = 84.8 – 83.9 = 0.9 g

Moisture gained per day = Total moisture gained (g) / Nr Days = 0.9/4 = 0.225 g day^-1

Water vapour permeability per 24h = Moisture gained per day / Circle Area = 0.225 / 0.005 = 45 g/m² 24h (1)

2^nd measurements:

Total moisture gained = 87.6 – 87.1 = 0.5 g

Moisture gained per day = 0.5 / 4 = 0.125 g day^-1

Water vapour permeability per 24h = 0.125 / 0.005 = 25 g/m² 24h (2)

Mean value of water vapour permeability per 24h = [(1) + (2)] / 2 = 35 g/m² 24h

Polypropylene

1^st measurements:

Total moisture gained = 86.0 – 85.9 = 0.1 g

Moisture gained per day = 0.1/4 = 0.025 g day^-1

Water vapour permeability per 24h = 0.025 / 0.005 = 5 g/m² 24h

2^nd measurements:

Total moisture gained = 87.1 – 87.1 = 0.0 g

Moisture gained per day = 0.0 / 4 = 0 g day^-1

Water vapour permeability per 24h = 0 g/m² 24h

Mean value of water vapour permeability per 24h = 2.5 g/m² 24h

Polyethylene

1^st measurements:

Total moisture gained = 84.6 – 84.5 = 0.1 g

Moisture gained per day = 0.1/4 = 0.025 g day^-1

Water vapour permeability per 24h = 0.025 / 0.005 = 5 g/m² 24h

2^nd measurements:

Total moisture gained = 84.6 – 84.5 = 0.1 g

Moisture gained per day = 0.1/4 = 0.025 g day^-1

Water vapour permeability per 24h = 0.025 / 0.005 = 5 g/m² 24h

Mean value of water vapour permeability per 24h = 5 g/m² 24h

5.Discuss the results of the water vapour permeability test.

Water vapour permeability is a measure for breathability or for a textile’s ability to transfer moisture. The results show that PP and PE have relatively low water permeability, while cellulose has a lot more. These values agree with literature (Brennan and Grandison, 2012), which states that PP has lower permeability than PE. Cellulose is also stated as a low barrier of water vapour permeability. These results show that using cellulose to pack food sensitive to humidity such as powders is not considered wise.

6.Discuss the results of the packaging and storage of fresh fruit experiment. Explain what is causing the observed changes in the fruit and how the different packaging/storage conditions influence the shelf life of the fruit.

Table 5. Fresh fruit (grapes) 3 days interval observations

Firstly, the tissues of fruits are alive after harvest and they only die through natural senescence, rotting or when they are consumed, cooked or similarly processed. All these tissues “breath”, a phenomenon called respiration with obvious relations to maintenance of the quality and prolonging the shelf life of the product. Specifically, grapes do not respire very intensively and this is the reason they get harvested when they are ripe. Reducing respiration can extend the shelf life but stalling it will make tissues senesce and die. Cooling temperatures can also lower undesirable effects on fruits (Jongen, 2002).

As far as grapes concerned, mould is primarly because of the fungus Botrytis cinerea.

Browning spotted is a chemical process caused by specific enzymes changing the tissues colour to brown, while shrinkage is caused by increased respiration (tissues eventually lose water as shown in the weight measurements causing them to lose volume). Sweating is caused once again because of the respiration in packages where gas permeability is low or very low.

In the above experiments, it is shown that when using MS and heat seal, grapes got sweaty in day 2 and 3, while in the same packaging with 2 holes, sweating was only slight. This makes sense as the 2 holes allowed the air transfer between package and the environment, lowering the humidity because of the respiration in the package.

In PE and heat seal, sweating was even more obvious as PE has lower gas permeability than MS.

Finally, in the open tray, sweating was absent but mould started to show at day 3, as it partially did in the package with 2 holes. This was caused by a microorganism, probably fungus since grapes have low pH. Another change which was spotted in the open tray was the soft, dried and oxidised appearance of the grapes because of the large amounts of respiration. Room temperatures and total contact with the environment allowed this level of respiration, lowering shelf life dramatically.

7.What changes would you make to the packaging/storage conditions to extend the shelf life of the grapes?

The most important change to the storage conditions would be to lower the storage temperature, as it would significantly reduce respiration. The package should not have holes, as they allow environmental air to get in allowing microorganisms to grow faster.

8.Discuss the results of the packaging and storage of cheese experiment. Explain what is causing the observed changes in the cheese and how the different packaging/storage conditions influence the shelf life of the cheese.

Table 6. Cheese 3 days interval observations

Browning of cheese is significant in high storage temperatures (37°C), less in medium (20°C) and absent in low temperatures of 5°C. Light causes the formation of lipid peroxides in medium temperatures, while compounds such as riboflavin are affected by light unrelated to storage temperature (Kristensen et al., 2001).

Cheese tend to produce free oil when they melt and sweats during storage in relatively high temperatures because of the high humidity of it. When in open air sweating is more and drying out occurs (Wang and Sun, 2004).

From the above, it becomes more obvious in ours experiments why cheese dried out during storage in open tray and why this drying out is more than in aluminium foil (which was not folded enough to keep air from contacting cheese). Another way to see the above is the greater loss of weight in open tray rather in aluminium foil. On the other hand, in both MS and cryovac packages no drying out was noted, as can be seen from the differences in initial and final weight (≤0.1g).

Relatively high storage temperatures (about 25°C) caused the oiling and sweating of the cheese.

9.What changes would you make to the packaging/storage conditions to extend the shelf life of the cheese?

The storage temperature should be as low as about 5°C (refrigerator) in dark and should be kept either in MS or cryovac packaging. Ideally, a modified atmosphere packaging should be used (Khoshgozaran et al., 2012), extending shelf life even more than the usual packages.

10.Discuss the results of the packaging and storage of fresh meat experiment. Explain what is causing the observed changes in the meat and how the different packaging/storage conditions influence the shelf life of the meat.

Table 7. Fresh meat 4 days intervals observations