12. chilling, freezing and boning - woolwise...notes – lecture 12 – chilling, freezing and...

MEAT418/518 Meat Technology - 0 - 1 ©2009 The Australian Wool Education Trust licensee for educational activities University of New England

12. Chilling, Freezing and Boning

Mike North

Learning objectives At the end of this topic you should be able to: • demonstrate a thorough understanding of heat transfer mechanisms and how they affect product

cooling in the meat industry. • discuss relevant literature relating to grazing management and sustainability. • predict chilling and freezing times for meat products. • understand hot and cold boning and the advantages and disadvantages of each of these

processing methods.

Key terms and concepts Heat, Conduction, Convection, Radiation, Chilling. Heat transfer coefficient. Biot number, Thermal properties, Predicting chilling times, Heat loads, Freezing. Latent heat, Predicting freezing times, Boning, Hot boning, Warm boning, Cold boning.

12.1 Introduction The meat industry uses refrigeration for food preservation. In many countries, animals slaughtered for meat were, and in some cases still are, immediately distributed, sold and consumed. Preservation was therefore unnecessary. As producers began to produce surplus meat, however, preservation methods were required so that excess product could be held and used at a later time, or in a distant location. Chilling meat was an early form of preservation that could be used without changing the form or state of the products. Freezing meat was a logical progression from chilling that gave longer preservation times. The main factors that affect the storage life of meat are the microbial growth and chemical reactions. Microbial spoilage can make the meat less pleasant to eat and it can also make the consumer ill if microbial numbers are too high and enough toxins are produced. If microbial growth is limited, the product may also become unacceptable due to flavour and textural changes, or due to colour deterioration before microbial spoilage becomes apparent. Bacteria and moulds, like all living things, have a range of temperature in which they prefer to live. As the temperature moves away from the most preferred level, their growth rate slows. Once they get too far away from that level, the growth stops completely and the bacteria may even die. Figure 12.1 shows how the growth rate of a commonly-studied bacterium varied with temperature. By cooling the meat from the initial body temperature of the animal (about 38°C) to 10°C, the growth rate drops by about 95%. Below 7°C, this particular bacterium (E.coli) does not grow at all. The growth rate at a given temperature varies between different microorganisms, but the principle remains the same. Cooling the meat is a very good way to slow or even stop microbial growth and thereby reduce the rate of microbial spoilage.

Notes – Lecture 12 – Chilling, Freezing and Boning

12 - 2 – MEAT418/518 Meat Technology ©2009 The Australian Wool Education Trust licensee for educational activities University of New England

Figure 12.1 Growth rate of E. coli vs. temperature. Source: North, (2005).

Figure 12.2 How the rate of a typical chemical reaction varies with temperature. Source: North, (2005).

Chemical reaction rates are also affected significantly by temperature. Figure 12.2 shows how the rate of a typical chemical reaction varies with temperature. Unlike the microbial growth rate, this reaction will still take place at low temperatures, albeit very slowly. Even if the meat is cooled to a very low temperature indeed, some chemical reactions will continue to occur, and the meat will slowly deteriorate, thereby setting a finite shelf life for the meat regardless of other factors. What is heat? Heat is a form of energy and it is measured in Joules (J). It is associated with the movement of atoms and molecules and it may be transferred from molecules and atoms by various mechanisms.

12.2 Product cooling Firstly, let us look at the process by which a food product such as meat is cooled. It is important to understand the fundamental mechanisms through which heat is transferred. There are three of these: conduction, convection and radiation. All of the possible methods for cooling are based on one or more of these mechanisms.



Heat transfer mechanisms Conduction is the simplest of the heat transfer mechanisms. The molecules of a solid, liquid or gas are in contact with each other, either all the time (in a solid), or some of the time as they periodically collide with each other (in a liquid or a gas). When two molecules come into contact, the one with the higher energy level (or at a higher temperature) transfers some of its energy to the one with the lower energy level (or at a lower temperature). Therefore, conduction is responsible for transferring heat from the warm parts of an object to the cool parts by contact. Convection, unlike conduction, does not transfer heat by relying on molecules to pass energy to and from each other. Instead, convection relies on actually moving the molecule with the higher energy level into areas occupied by lower energy molecules. This raises the average energy level of the molecules in that particular area and thereby raises the average temperature of that area. Since convection relies on moving the molecules to transfer heat, it is a mechanism that can only occur in liquids or gases because the molecules in a solid are fixed in position. Radiation is the last of the fundamental heat transfer mechanisms. Like conduction, radiation also transfers heat by passing energy from higher energy molecules to lower energy molecules. However, radiation does not require that the molecules contact one another in order to transfer energy. Instead, the energy is transferred by electromagnetic radiation, which is emitted by all molecules. For radiation to occur the molecules need only be in view of one another and since the amount of radiation emitted by a molecule increases with increasing temperature, a net transfer of energy from the higher temperature molecule to the lower temperature molecule results. Heat transfer by radiation is generally only important when you have very hot surfaces from which the heat is to be transferred. Typical examples include boiler surfaces and electrical resistance heaters. In the chillers and freezers of a meat plant, heat transfer by radiation is usually negligible. Surface heat transfer In the meat industry, product cooling is most commonly achieved by placing the product into a cold liquid or gas. When the lower energy molecules of the fluid come into contact with the higher energy molecules at the surface of the meat, heat is transferred from the meat to the fluid by conduction. This simultaneously cools the surface of the meat and warms the fluid. The warm fluid molecules are then moved away from the surface (either naturally or by an external means such as a fan) and replaced by lower energy fluid molecules. The amount of heat energy that is transferred to these new fluid molecules is less than for the first group because the meat surface is now at a lower temperature or has less energy. Hence, as the meat cools, heat cannot be removed as quickly by the cold fluid. Although the amount of heat removed by the fluid is continually decreasing, the process will continue until the energy level of the meat surface reaches the energy level of the fluid (i.e. until the meat is cooled to the same temperature as the fluid). As implied above, there are two ways to make the fluid molecules move away from the meat surface – let it occur naturally or force them away. Natural convection (sometimes called ‘free convection’) occurs naturally during the cooling process. When the fluid at the meat surface is heated, it becomes less dense than the surrounding fluid. The warm fluid therefore rises due to its buoyancy and the surrounding cold fluid sweeps in from the sides to replace it. Forced convection occurs when the fluid is forced by an external force. For example, if a fan blows air or a pump pushes liquid across a surface. Forced convection results in a higher rate of heat removal than natural convection because the molecules are moving at a higher velocity and therefore carry the heat away from the surface more rapidly. The rate of heat removal in either forced or natural convection depends on “Newton’s Law of Cooling”, which Sir Isaac Newton discovered in 1701. This is described in equation (1)

( )fluidsurface TTAhQ −= (1) Where: Q is the rate of heat removed from the surface (W)

h is the surface heat transfer coefficient (W/m2K) A is the area over which heat is being transferred (m2)

Tsurface is the temperature of the surface (°C) Tfluid is the temperature of the fluid (°C)



“Newton’s Law of Cooling” mathematically demonstrates which variables affect the rate of heat removal from the product surface (Q). Note that Q is measured in Watts (W) and that one Watt is equal to one Joule per second. So removing 100W of heat means that we are removing 100 Joules per second. The temperature difference between the surface and the fluid (Tsurface - Tfluid) affects the rate of heat removal, and as the fluid and surface approach the same temperature the rate of heat removal approaches zero. The rate of heat removal also relies on the area (A) over which the heat transfer is occurring – if we can expose more of the meat surface area to the cold fluid then we can increase the rate of heat removal. The surface heat transfer coefficient (h) also has an impact – but what is it? The surface heat transfer coefficient (h) is an interesting parameter. It depends on the characteristics of the fluid and the level of turbulence. Liquids are denser than gases because their molecules are much closer together. Therefore, when meat is immersed in a cold liquid, its surface is in contact with many more cold fluid molecules than if it were immersed in a cold gas. This higher level of molecular contact means that liquids remove heat much faster than gases, and therefore produce a higher surface heat transfer coefficient. This is a phenomenon that you probably know already, since if you place your hand in ice water it cools much more quickly than if you place your hand into air at 0°C (i.e. it feels “colder” even though it is at the same temperature). The heat transfer coefficient also depends on the level of turbulence in the fluid and the easiest way to increase turbulence is to increase the fluid velocity using forced convection. Again, from your own practical experience you probably know that your body cools down much faster in a cool breeze than it does in still air at the same temperature. Table 12.1 shows some typical ranges for h in a variety of situations. Methods for estimating h for common product cooling cases will be discussed shortly. In the meantime, if you have no better information to rely upon than Table 12.1, it is important to design conservatively. For instance, to have a heat transfer coefficient of 500 W/m2K in air with forced convection, you would need to have the air moving at a very high velocity. Table 12.1 Approximate surface heat transfer coefficients in various convective situations. Source: Welty, (1978)

Mechanism Surface heat transfer coefficient - h (W/m2K) Forced convection, water 250-15,000 Forced convection, air 25-500 Natural convection, air 5-50

Internal heat transfer When the surface of a food product such as meat is cooled, it creates a temperature difference between the surface and internal parts of the product. This temperature difference produces internal heat transfer by conduction, where the warmer molecules inside the product transfer their heat to the cooler surface molecules. For a slab-shaped piece of food cooled from the top surface only, the rate of internal heat flow Q (W) from the bottom to the top, is given by the equation:

( )bottomtop TTxAkQ −= (2)

Where: k is the thermal conductivity of the material (W/mK) A is the area over which heat is being conducted (m2)

x is the thickness of the slab (m) T1 is the temperature of the top slab surface (°C)

T2 is the temperature of the bottom slab surface (°C) Equation (2) is strictly correct only when the heat flow is steady, which is not the case during cooling of meat because the temperatures within the meat are constantly changing. Despite this, equation (2) indicates the factors that are important to internal heat transfer. The slab thickness, slab surface area and the top and bottom surface temperatures are parameters that most of us understand, however, the thermal conductivity may not be. The thermal conductivity of a material specifies how easily heat can be transferred through itself by conduction. Highly conductive materials such as metals, which have a high thermal conductivity, will easily transfer heat from one side of a slab to the other. Whereas, materials that have a low thermal conductivity will not allow as much heat to be transferred. Insulation is an example of a material with a low thermal conductivity, which we often use in buildings to restrict the flow of heat through walls, ceilings and floors.



Surface vs internal heat transfer In practice, meat cooling involves a combination of conduction and convection, where internal conduction carries the product's heat content to the surface, and then convection (natural or forced) carries that heat away from the surface. Because the whole process is a combination of these two factors, you do not necessarily get the results that you might expect when you change some of the chilling parameters. If you already have a high rate of surface heat transfer, increasing the heat transfer coefficient will increase rate of heat removal from the surface but may not significantly increase the total rate of heat removal from the product because it will be limited by the rate of internal conduction. On the other hand, if you have a high rate of internal heat conduction through an object, reducing its thickness will not increase the heat flow rate by very much either, because the total rate of heat removal will be limited by the convection at the surface. The way in which the external resistance to heat transfer (due to convection) is balanced against the internal resistance (due to conduction) is indicated by the Biot number. The Biot number is the ratio of the conductive (internal) resistance (X/k) to heat transfer to the convective (external) resistance (1/h). The Biot number is abbreviated Bi and it is defined by equation (3).

khXBi = (3)

Where: X is the distance from the centre of the meat product to the surface (m) The Biot number can range from 0 to infinity. If Bi is low (much less than 1.0), this means that the external resistance controls the process. In this case, changes to h will have a large effect on the cooling time and changes to X will have little effect. If Bi is high (much more than 1.0) then the internal resistance controls the process. In this case, changes to h will have very little effect on the cooling time, but changes to X will have a large effect.

12.3 The chilling process The chilling process is important to the meat export industry for two reasons. Firstly, an increasing proportion of our meat products are exported in chilled form, using vacuum packaging, or controlled-atmosphere packaging. Secondly, if we are going to freeze the product, then the meat must pass through a chilling process first. Even if we are hot-boning, the meat must always chill before it starts to freeze. Definition Chilling is the process of cooling meat while the meat remains above its freezing temperature. The temperature of the cooling medium (air or water, for instance) doesn't matter. As soon as the meat starts to freeze, it is no longer considered a chilling process. Quality considerations A well designed and controlled chilling process is very important to producing meat of satisfactory quality. An appropriate chilling regime is critical to producing tender meat, and it is just as critical to producing meat with low levels of bacterial growth, and therefore a long shelf life. For tenderness, chilling must be slow enough to prevent too much “cold shortening”. Once the meat has passed through rigor, chilling must continue relatively slowly to give the meat time to “age”. Since the aging rate depends greatly on the meat temperature, meat that is cooled too quickly after rigor can fail to age as much as you might like, and it will be tough. Fortunately, the cuts that you most need to age are generally quite deep in the carcass, and they therefore cool more slowly than the surface. To minimise bacterial growth, meat should be chilled as quickly as tenderness considerations allow. Bacterial growth depends on the meat surface temperature, but it also depends on the “water activity” on the surface of the meat. Fortunately, the surface temperature drops quite quickly during chilling. Air chilling can also dry the surface of the meat, thereby reducing the water activity, and bacteria do not grow as well in low water activity environments. These factors combine to keep bacterial growth to an acceptable level in a well-designed and operated process.



Thermal properties The amount of heat that has to be removed from a meat product during chilling, and the rate at which that heat can be removed both depend on the thermal properties of the meat. The important thermal properties are the thermal conductivity (W/mK), the heat content or enthalpy (J/kg) and the density (kg/m3) of the meat. The enthalpy of a meat product above its freezing temperature increases at an almost constant rate with temperature. This constant rate is called the heat capacity (J/kgK). Values of thermal properties for various meat products may be found in Pham & Willix (1989), Pham et al. (1994), Miles et al. (1983), Willix et al. (1998) and ASHRAE handbooks (e.g. ASHRAE, 2002). If you do not have thermal properties for the meat product that you are interested in, you can estimate the properties from its composition using equations (4) to (6).

850F +

1300S +

1000W

1 = ρ (4)

K W/m 4722

F + 5306

S + 1695

W = kl ⎟⎠

⎞⎜⎝

⎛ρ (5)

K J/kg F1900 +S 1400 + W4180 = Cl (6) Where: ρ is the meat density (kg/m3)

kl is the unfrozen thermal conductivity (W/mK) Cl is the unfrozen heat capacity (J/kgK)

W, F, S are the mass fractions of water, fat and other solids respectively (ie. for a typical piece of lean beef, W = 0.70, F = 0.10, S = 0.20).

Choosing a chilling rate How should we choose a chilling rate? There are three process factors that can be changed in order to change the chilling rate: object size or shape, external resistance, and driving force. Changing the object size or shape changes the length of the internal pathway along which the heat must be conducted. This means that it changes the X in the Biot number [equation (3)]. You can make the pathway shorter by chopping the object up into smaller pieces. This is certainly very successful in reducing the chilling time for a meat product, but it has several disadvantages if it is taken to extremes. First, the exposed meat surface is multiplied many times over the original object's surface area. This can result in much greater evaporation rates and greatly increased weight loss, even given the shorter time spent in the chiller. It can also encourage “oxidative rancidity”. This is perceived as a "warmed-over" flavour. Second, re-packing densities after chilling will be lower than the original undivided meat. Third, the smaller the meat pieces are, the harder it is to identify the original form of the meat. Customers do not want to have to do a genetic fingerprint test to find out whether they are buying beef, lamb, or horse! You can make the pathway longer, and increase the chilling time, by making the product larger. This is achieved by packing many smaller products inside one larger package. Trying to chill meat in very large blocks (e.g. several hundred millimetres thick) inevitably means that the centre temperature will remain high for quite a long time, with a consequent high growth of micro-organisms in that part of the package. The tenderness of meat in the centre of the package will also vary considerably from the tenderness at the surface, particularly if a high rate of heat transfer is used at the surface in an attempt to reduce the overall chilling time. Even in more moderate package sizes, for example a 27 kg carton that is 160 mm thick, the variation between the centre and the surface can be considerable. The amount of external resistance to heat transfer in meat cooling can be changed by changing the heat transfer medium (e.g. using forced convection instead of natural convection, or water instead of air), or by changing the packaging. The effect of altering the heat transfer medium was discussed in the section above on ‘Surface Heat Transfer’ and the effect of packaging will be discussed below. One heat transfer medium not yet discussed is plate freezing. In a plate freezer, the external resistance is negligible, except for any packaging that may surround the product. The final parameter that affects the cooling rate is the temperature driving force. We are limited in how cold we can make the cooling medium during chilling if we want to avoid the surface of the



meat freezing during the process. In principle, we should not have the cooling medium any colder than the freezing temperature of the meat if we want to avoid all risk of freezing. In practice, even if the cooling medium is a few degrees below the freezing temperature, there will be virtually no risk of freezing the surface. Product tenderness Muscles are made up of fibres that contain protein filaments of actin and myosin. In a live animal, these protein filaments move across one another to allow muscles to contract and relax. The energy for contraction is usually generated by a process involving oxygen, which is supplied to muscles by circulating blood. After an animal is slaughtered, blood circulation stops and muscles use up their oxygen supply. Without oxygen the muscle turns to anaerobic glycolysis to generate energy, a process that breaks down glycogen (a sugar stored in muscle) without oxygen. Anaerobic glycolysis produces energy to contract the muscle and it also produces lactic acid, which causes the muscle pH to fall. As glycogen supplies are depleted, the energy for muscle contraction is lost and the actin and myosin filaments lock together in a permanent contraction called rigor mortis. When meat is chilled so that its temperature falls below about 10°C while its pH is still above 6.2, the muscle will contract. This is called cold shortening, and the meat becomes tough in proportion to the amount of shortening that takes place. If the muscle pH is less than about 6.0, it can be cooled fairly rapidly without shortening, although if it is then frozen it may shorten when thawed and again become tough. The rate of pH fall can be increased by holding the meat at a high temperature during the early part of the cooling process, or by electrical stimulation. Electrical stimulation quickly reduces the pH of the meat by speeding up the rate of anaerobic glycolysis considerably and results in a substantial reduction in shortening for a given cooling rate. Electrical stimulation requires sending an electrical current with a specified voltage and particular waveform through the meat. The required parameters have been well-established for lamb and sheep in order to give a maximum pH drop during stimulation (MIRINZ RM 54, RM 135 and RM 141). For other animals, the parameters are not so well-established and they are often finalised for a particular plant by carrying out carefully-controlled trials. However the stimulation is carried out, it is essential to subsequently chill the meat in such a way that the temperature vs. time specifications are met in order to produce meat that is of the desired tenderness. This ensures that there is enough time for aging to take place before the meat is frozen or consumed. Aging of meat is a tenderisation caused by the activity of enzymes that are already present in the muscle when the animal is killed. These enzymes act on the muscle and break down some of the proteins within the meat. Electrical stimulation does not accelerate aging as such, but because stimulated meat goes into rigor more quickly, it starts to age sooner and it does so while the meat is still quite warm. Since aging is accelerated by higher meat temperatures, stimulation usually results in faster aging. More detail on meat quality issues is provided by Devine et al. (1996). Although meat quality may not always be mentioned during the rest of these notes, it should always be borne in mind when discussing chilling or freezing of meat.



Predicting chilling times To design a chilling process, it is important to be able to calculate the length of time that the process will take and the heat loads that will have to be removed. We looked at several surface heat transfer coefficients in the examples above, but we did not look at how to calculate them for real meat chilling and freezing situations. For air chilling and freezing, the surface heat transfer coefficient (h, in W/m2K) in forced convection can be estimated from the following equations:

surfaces flat for 8.0 v7.3 = h (7)

surfaces oval for 6.0 v12.5 = h (8) Where: v is the air velocity (m/s)

The heat transfer coefficients predicted by these equations are shown in Figure 12.3. Oval surfaces include lamb carcasses and beef sides. The flat surface equation should be used for cartoned product.

Figure 12.3 Estimated heat transfer coefficients for flat and oval shapes from air velocities of 0.4 m/s to 6.0 m/s. Source: North, (2005).

These equations are valid down to the lower air velocities where natural convection starts to have a significant influence on the heat transfer coefficient. In general, they should be used in air velocities of more than 0.4 m/s. It is possible to predict heat transfer coefficients below 0.4 m/s but the equations to do so are much more complicated and they involve many other factors. It is generally safe to assume that the heat transfer coefficient at the surface of a meat product due to natural convection is approximately the value shown at the lower end of the applicable line in Figure 12.3. The equations to estimate heat transfer coefficients for water chilling are also more complicated than equations (7) and (8), so I will not give them here. Estimating equations for both the air (natural convection) and water heat transfer coefficients can be found in heat transfer textbooks such as that by Welty (1978) if you require them. Often, you can just use Table 12.1 as a guide for rough calculations, however. The chilling process reduces the temperature of a product from some initial temperature (Ti) towards the ambient temperature of the surrounding cooling medium (Ta). The best way of expressing how far through the chilling process we have gone is to say what fraction of the temperature difference the meat still has to go through. This is called the fraction unaccomplished temperature change, and it has the symbol Y. Y can be calculated for any point in a meat product, but we are most often interested in either the average temperature or the thermal centre temperature. The thermal centre is the slowest-cooling point in the object. This is often the same as the geometric centre of the object, though that is not always the case. The fraction unaccomplished centre temperature change, Yc, is therefore:



T- TT- T = Y

ai

acc (9)

The average temperature is calculated in the same way, but Tc is replaced with Tav:

T- TT- T = Yai

aavav (10)

Alternatively, if we know what the value of Yc or Yav is, we can rearrange equations (9) or (10) to find the centre or average temperature:

( ) T + Y T- T = T acaic (11) ( ) T + Y T- T = T aavaiav (12)

As we discussed above, chilling requires a temperature driving force which is the difference in temperature between the surface of the object and the surrounding medium. If the difference in temperature between the object and the surrounding medium is small, then chilling will be very slow. In fact, as the object gets cooler and cooler, its rate of cooling gets slower and slower. As a result, if you chill a piece of meat initially at 35°C in cold air at a temperature of 0°C, the meat will never quite reach 0°C. It will just keep getting closer and closer to 0°C forever. The meat will never get all the way to the ambient temperature. Although we can never talk about how long the meat will take to reach the ambient temperature, we can talk about how long it would take for the meat to get part-way there. The time that it takes for Yc to change from 1.0 to 0.5 is the time required for the centre of the object to go half way from its initial temperature to the ambient temperature. This is called the half-cooling time and denoted t1/2. This should not be confused with half of the total cooling time. The total cooling time depends upon the final temperature that you set (some temperature above Ta) at which you will take the meat out of the chiller. The half-cooling time, on the other hand, depends only on the physical conditions of the meat in the chiller. For a given set of conditions, it is possible to calculate the half-cooling time in several different ways. We will use the method of Cleland & Earle (1982). For a simple shape such as a slab, or a sphere, the calculation is straightforward. First, calculate the Biot number (Bi) for the situation and use Table 12.2 to find the Fourier Number at the half-cooling time (Fo1/2) for that value of Bi.



Table 12.2 Fo1/2 for slab and sphere shapes. Source: North, (2005). Bi Fo1/2 Slab Fo1/2 Sphere

0.01 69.54 23.15

0.02 34.88 11.59

0.04 17.56 5.82

0.06 11.78 3.89

0.08 8.89 2.93

0.1 7.16 2.35

0.15 4.85 1.58

0.2 3.69 1.2

0.25 3 0.971

0.3 2.54 0.817

0.4 1.96 0.625

0.5 1.62 0.51

0.6 1.39 0.433

0.7 1.23 0.378

0.8 1.1 0.338

0.9 1.01 0.306

1 0.936 0.28

1.1 0.874 0.26

1.2 0.822 0.243

1.3 0.779 0.228

1.4 0.741 0.216

1.5 0.709 0.205

1.6 0.681 0.196

1.7 0.656 0.187

1.8 0.634 0.18

1.9 0.615 0.174

2 0.597 0.168

2.2 0.567 0.158

2.4 0.542 0.15

2.6 0.521 0.143

2.8 0.503 0.137

3 0.487 0.132

3.5 0.456 0.122

4 0.433 0.114



Bi Fo1/2 Slab Fo1/2 Sphere

4.5 0.415 0.109

5 0.401 0.104

6 0.38 0.098

7 0.365 0.093

8 0.354 0.09

9 0.346 0.088

10 0.339 0.086

15 0.319 0.0804

20 0.309 0.0777

30 0.299 0.0751

40 0.295 0.0738

50 0.292 0.0731

60 0.29 0.0726

80 0.287 0.072

100 0.286 0.0716

1000 0.281 0.0703

Then use equation (13) to calculate the half-cooling time:

kX c Fo = t

l

2l1/2

1/2ρ

(13)

The only parameter from equation (13) that we have not looked at is X, the depth of the thermal centre. For a carton, that is easy. If the carton is 160 mm thick, X = 0.08 m (half the thickness). For a carcass or side, it's a bit more complicated. Figures 12.4 and 12.5 show the values of X for various weights of beef sides and lamb carcasses. You can also use these figures for beef quarters (double the weight of the quarter to get the weight of the equivalent side) and for venison, etc. (treat small venison like a sheep carcass and a large venison like a beef side). Now that we can use equation (13) to estimate the half-cooling time, we can calculate the time required to cool to any temperature quite easily.



Figure 12.4 Deep butt depth for various beef side weights. Source: North, (2005).

Figure 12.5 Deep leg depths for various and sheep carcass weights. Source: North, (2005).



Figure 12.6 Number of half-lives, N, required to cool to a given fractional unaccomplished temperature change, Y. Source: North, (2005).

Finding the average temperature. For an average temperature, Tav, that we want to cool to, we can use equation (10) to calculate the fractional unaccomplished temperature change, Yav. From Yav, we can find the number of half-lives, N, required to cool to that Y value from Figure 12.6. Finally, we can find the total time to cool to the specified average temperature using equation (14).

t N = t 1/2 (14) For shapes other than a slab or a sphere, we cannot just read the Fourier number from Table 12.2. Instead, for a flat-sided shape such as a carton, find the Fo1/2 for a slab from Table 12.2, and for an oval shape such as a carcass or side, find the Fo1/2 for a sphere from Table 12.2. Once you have the Fo1/2 for a slab or sphere, calculate the Fo1/2 for the shape using either equation (15) or (16).

EFo

= Foslab 1/2

carton1/2,

, (15)

EFo

= Fosphere 1/2

carcass1/2,

,3

(16)

The factor E is called the Equivalent Heat Transfer Dimensionality. This is a factor that relates the cooling time for a complicated shape to the cooling time for a simple shape such as a slab or a sphere. It is possible to calculate E for a complicated shape, but the procedure is quite difficult. Fortunately, other people have done these calculations for us for many of the shapes that we are interested in. Some of these values are shown in Table 12.3. Where a range is shown in Table 12.3, the exact value depends on the precise shape. In general, the larger the E value is, the thicker the object is relative to its width and length. This means that a deep carton will have an E value of 1.5 while a shallow carton will have an E value of 1.3. Table 12.3 E values for products of interest to the meat industry. Source: North, (2005).

Product E Lamb (shoulder) 1.4 Lamb (deep leg) 2.2 Ewe (deep leg) 2.0 Beef carton (plate freezer) 1.0 Beef carton (air blast freezer) 1.3 - 1.5 Beef side or quarter (deep leg) 1.3



Finding the centre temperature. To calculate the chilling time for the thermal centre of an object, we have to take into account the difference between the thermal centre temperature profile and the other temperature profiles in the object. The thermal centre is the slowest cooling part of the object, so it lags behind the other temperature profiles. Rather than using Figure 12.6 to find the number of half-lives, N, if we are estimating the chilling time for the centre, we need to use Figure 12.7 instead. Simply find the value of E (or EHTD on Figure X) at the bottom left, and go straight up until you reach the value of Bi for this situation. Then go straight across to the right until you are above the value of Y that you need. This will give you the number of half-lives that you should use.

Figure 12.7 Alignment chart to calculate N, the number of half-lives for the centre temperature

from E, Yc and Bi. Source: North, (2005).

Effect of packaging on heat transfer coefficient Up to now, we have dealt with unpackaged objects where you could estimate the heat transfer coefficient from Figure 12.3 or equations (7) and (8). When you have a packaged object, the effective heat transfer coefficient is reduced by the thermal resistance of the packaging and by any trapped air layer between the package and the product. Some thermal conductivities for packaging materials are shown in Table 12.4. Table 12.4 Typical thermal conductivities of packaging materials and still air. Source: North, (2005).

Solid cardboard 0.06 - 0.10 W/mK Plastic film 0.08 - 0.15 W/mK Corrugated cardboard 0.04 - 0.06 W/mK Still air 0.03 W/mK

Although the impact of the packaging material on heat transfer can be very important, often the most important factor is any layer of still air that is trapped in the packaging. You can see from Table 12.4 that still air has a lower thermal conductivity than any of the packaging materials. In fact, most methods of preventing heat transfer (such as household wall insulation, polystyrene panels and peoples' warm clothes) have their main effect by trapping a layer of still air within them. In many cartoned products, there is often an air layer of 0.5 to 2 mm thickness around most of the product in the package, with a larger air gap on top. In the case of stockinet and poly-bagged lamb carcasses,



the effect of the trapped air can reduce the effective heat transfer coefficient predicted by Figure 12.3 by about 20% (for stockinet) or up to 50% (for loose polybags). The overall effect of the packaging on the heat transfer coefficient can be calculated from:

kx +

kx +

h1 =

h1

air

air

package

package

surface (17)

Where: h is the heat transfer coefficient to use in your calculations

hsurface is the surface heat transfer coefficient predicted from Figure 12.3 kpackage is the thermal conductivity of the package from Table 12.4 xpackage is the packaging material thickness (m) kair is the thermal conductivity of still air from Table 12.4 xair is the thickness of the still air layer (m)

In practice, the surface heat transfer coefficient varies significantly between different surfaces of the product, and even between different parts of the same surface. The most important example of this is in a cartoned product with an air gap under the top surface. The heat transfer coefficient under that surface can be much lower than it is under surfaces with little or no air gap. The simple methods for estimating cooling times only use one heat transfer coefficient, however, so you cannot account for this sort of variation. The solution is to use a weighted average heat transfer coefficient rather than any single value. A rough guideline is to estimate the upper and lower surface heat transfer coefficients and use an average of the two, ignoring the coefficients of the carton sides and ends. Product heat loads The product chilling heat load is the total amount of heat that is to be removed from the product spread over the length of the operation. In practice, however, because the temperature difference between the product and the cooling medium is high at the start of chilling and low at the end, the heat load will also be greater at the start than at the end. There are methods that can be used to take all of this into account and produce a smooth heat load profile, but the simplest technique for hand calculation is to use the average-temperature cooling times as shown in the previous section. As an example, to chill a beef side with a heat capacity of about 4000 J/kgK in air at 2°C from an average temperature of 40°C to 5°C, we would have to remove a total amount of heat of (T1 -T2) x Cl , i.e. (40°C-5°C) x 4000 J/kgK = 140000 J/kg. Now let us assume this heat is to be removed over a period of about 24 hours (86400 seconds). The average heat load would be 140000 J/kg / 86400 s = 1.62 W/kg. If this 24-hour process took three half lives (N = 3). It would mean that the average temperature dropped half-way from 40°C to 0°C (i.e. to 20°C) after 8 hours; it then dropped three quarters of the way to 0°C (i.e. to 10°C) after 16 hours; and then dropped seven eighths of the way to 0°C (i.e. to 5°C) after 24 hours. So, the amount of heat removed during the first 8 hours would be (40°C-20°C) x 4000 J/kgK = 80000 J/kg. This period took 28800 seconds, so the average heat load during the first 8 hours would be 80000 J/kg / 28800 s = 2.78 W/kg. During the second 8 hours the temperature drops from 20°C to 10°C, so the heat removed during this time would be (20°C-10°C) x 4000 J/kgK = 40000 J/kg. The average heat load during this period would be 40000 J/kg / 28800 s = 1.39 W/kg. During the last 8 hours the temperature drops from 10°C to 5°C, so the heat removed during this time would be (10°C-5°C) x 4000 J/kgK = 20000 J/kg. The average heat load during this period would be 20000 J/kg / 28800 s = 0.69 W/kg. We can check that this calculation is correct by adding up the amounts of heat removed in each third of the process: 80000 W/kg + 40000 J/kg + 20000 J/kg = 140000 W/kg. This is the same as the total amount of heat removed that we calculated for the whole process, so we have done the calculation correctly. Although the results are given in terms of W/kg, they can be calculated for the whole chiller by multiplying by the number of sides in the chiller (perhaps 200) and by the weight per side (perhaps 150 kg). For the first third of the process, therefore, the average heat load would be 150 kg/side x 200 sides x 2.78 W/kg = 83400 W, or 83.4 kW in this chiller.



In the example above, we have calculated the heat load in the chiller at three points in time and we used 1, 2 and 3 half-lives because it happened to be convenient. You can, of course, calculate the average heat over any period during a chilling process as long as you know the average temperature at the beginning and end of that period. To estimate the average temperature part way through a half-life you can use the following equation to find Yav:

Nav = Y 5.0 (18)

To calculate the instantaneous heat load at the start of the cooling process, you just need to know the product surface area in addition to the other factors that have been covered above, and you can calculate it from: ( ) T- TAh = Q asurfacesurface (19)

Where: Q is the initial heat load (W) Asurface is the meat surface area (m2)

Tsurface is the surface temperature of the meat (°C) Ta is the ambient temperature (°C)

So for the example above, let us assume an air velocity of 1 m/s (and hence a heat transfer coefficient of 12.5 W/m2K using equation (8)), a surface area of 2 m2/side and 200 sides in the chiller, then the initial heat load Q = 12.5 W/m2K x 200 sides x 2 m2/side x (40°C-0°C) = 200000 W, or 200 kW for this chiller. In addition to the product load in a chiller, there will also be loads due to other factors. These include: • Heat infiltration through the walls, floor and ceiling • Hot air entering through open or unsealed doors • Heat load due to fans, lights and machinery (e.g. forklifts) • Heat loads due to people in the chiller • Heat loads due to the structures in the chiller The total heat load to be removed from a chiller is the sum of the product heat load (which varies with time as the chilling process progresses) and these additional loads, most of which do not vary over time. I will not go into detail about how these additional loads may be calculated [Cleland (1990) gives more detail if required] however, it is important to minimise these additional heat loads to ensure that the refrigeration system has as much capacity as possible to handle the product heat load and thereby achieve the desired result.

12.4 The freezing process Definition Freezing is the process of removing heat so that the water content of meat is converted into ice. To make this change from liquid water to ice, we must remove a large amount of heat in addition to the amount that is required to change the temperature. This extra amount is called the latent heat of freezing. As a pure substance freezes, this latent heat is removed without any change in temperature. Meat is not a pure substance, so the latent heat is not all removed at one fixed temperature. You have to remove quite a lot of heat, however, until the temperature really starts to drop again.



Figure 12.8 The variation in heat content (enthalpy) of meat with temperature. Source: North, (2005).

You can see from Figure 12.8 that the amount of heat removed during freezing is quite a lot more than the amount removed during chilling. This means that the freezing part of the process takes longer than the chilling part if the same cooling conditions are used. Eventually, freezing can be regarded as finished and the meat moves into the final, sub-cooling, stage of the process. At this point, the latent heat is mostly gone, and the meat just chills down to its storage temperature. The freezing front Meat does not start to freeze until its surface temperature drops to the initial freezing temperature of the meat. At that time, the piece of meat will start to freeze from the outside towards the inside. The surface layer freezes first and then the frozen layer starts to get thicker over time. The boundary of the frozen layer is called the freezing front. This freezing front moves gradually inwards towards the centre of the meat, with frozen meat on the outside and unfrozen meat on the inside (as shown in Figure 12.9).

Figure 12.9 The Freezing Front. Source: North, (2005).



Once the freezing front has started moving, the temperature of the unfrozen region has usually dropped to the freezing temperature of the meat. This means that the temperature at the centre of the meat remains constant for most of the freezing process — just a little below zero. One consequence of this behaviour is that you can't tell how well frozen a piece of meat is by measuring its centre temperature, unless the meat is almost completely frozen and the centre temperature has already started to drop below the freezing temperature. Further, if you measure the temperature of a point that is not actually at the thermal centre, the plateau temperature that you measure will still be the freezing temperature of the meat. However, the difference is that the temperature at a non-central point will start to drop from the freezing temperature sooner than the centre point and the rate at which it drops will be slower. A non-central point drops in temperature more slowly than the centre point because it is not just releasing its own heat content outwards into the rest of the meat (and thence into the surroundings), but it has other heat passing through it from deeper in the meat. As a consequence, the closer your temperature measurement is to the centre of the meat, the sharper is the "knee" in the temperature plot, e.g. Figure 12.10.

Figure 12.10 Centre temperature during freezing. Source: North, (2005).

The bottom part of the tail on Figure 12.10 indicates that the meat temperature is starting to get close to the air temperature, so the rate of cooling and therefore the rate of temperature decrease has slowed. As you can see, however, there is no doubt about the freezing time for a piece of meat if you measure its centre temperature correctly. Quality considerations When it freezes, water is a crystalline substance. Under normal freezing conditions, it does not form an amorphous solid like glass, but instead forms discrete ice crystals. The sizes of these crystals depend upon the rate of freezing. They always start very small, clustered around an impurity or microscopic flaw in the meat called a nucleation site. If the freezing is done quickly, then the ice crystals will stay small. If the freezing is done more slowly, the crystals merge and grow (Figure 12.11).



Figure 12.11 Ice crystal growth during freezing. Source: North, (2005).

Small ice crystals can exist within and between meat cells, in which case they do not cause any problem. Once ice crystals reach about the size of the meat cells, however, their expansion breaks up the cell structure and penetrates the cell membranes. This has little effect until the meat is defrosted, but then the fluids within the cell can leak out. This is known as drip loss and it is unattractive to customers and causes the defrosted meat to lose weight. The process of freezing aged meat, if sufficiently rapid, generally does not in itself, have any demonstrable effect on the cooked colour, flavour, odour or juiciness of that meat. Indeed, the process of thawing can result in a certain amount of tenderisation, and small amounts of additional moisture released by cell damage can increase the perceived juiciness. Most nutrients are retained during freezing and subsequent storage. Soluble proteins can be lost in the drip during thawing, but the fluid lost as drip (as long as it is not excessive) is similar to the amount of fluid lost when fresh meat is cooked, anyway. Once the meat is frozen, however, it is not completely inert. The undesirable changes in meat during freezing are associated with formation of large ice crystals in extracellular locations, mechanical damage by the ice crystals to cellular structures through distortion and volume changes and chemical damage arising from changes to concentrations of solutes. The fastest freezing rates are associated with the least damage because they result in small ice crystal sizes and they do not provide an opportunity for the chemicals dissolved in the moisture content of the meat to move from their original locations. On the other hand, a slow freezing rate does allow the solutes to move away from the growing ice crystals. This results in ice crystals that are composed of almost pure water rather than the solution of water and various chemicals that normally makes up the moisture in the meat. Even if the meat is frozen quickly to start with, a long period of cold storage will allow the ice crystals to grow to a large size. Ice crystal growth occurs more quickly at higher storage temperatures, and is particularly encouraged by temperature fluctuations. When thawed, the meat will again become mushy due to cell damage. Small ice crystals also make the appearance of the meat surface look better because a surface containing small ice crystals reflects more light than the surface of slowly frozen meat, so if the meat is to be sold while frozen, it is important to retain a small ice crystal size. Thermal properties As for chilling, it is important to know the thermal properties of any meat for which you want to calculate a freezing time or heat load. Thermal property data can be obtained from the same sources as for chilled meat data. If you do not have the required data then it can be calculated in the same way as for chilled meat from W, S, F (the fractional water, solids and fat contents) and I, the ice fraction. However, with freezing, there are more parameters to calculate.



The initial freezing temperature, Tf (°C) can be estimated from:

W +1.8 - = Tf (20) Most unsalted meats have an initial freezing temperature of -0.8 to -1.2°C. The ice fraction when fully frozen, I, can be estimated from:

( ) ⎟⎠

⎞⎜⎝

⎛20-

T- 1 S 0.25- W = I f (21)

The density of frozen meat, ρf (kg/m3) is a little less than that of unfrozen meat:

850F +

1300S +

900I +

1000I- W

1 = fρ (22)

The thermal conductivity of frozen meat, kf (W/mK) can be estimated from:

⎟⎠

⎞⎜⎝

⎛4722

F + 5306

S + 433

I + 1783

I- W = k ff ρ (23)

The heat capacity of frozen meat, Cf (J/kgK) from: F1900 +S 1400 + I1940 + ) I- (W4180 = Cf (24)

The latent heat of freezing of a meat product is released over a range of temperatures, rather than at a single temperature as is the case with a pure substance (such as water). The total amount of latent heat change due to freezing, L (J/kg) can be predicted by:

I600 333 = L (25) Note that the ice fraction when the meat is fully frozen is not equal to the moisture content. Some of the water within the meat is bound so tightly to the protein content that it never freezes no matter how cold you make the meat. This is called the bound water content of the meat. Predicting freezing times Predicting the chilling time of meat is made quite difficult by the changing temperature profile within the product. However during freezing, the temperature profile in the meat is much simpler. Inside the freezing front, the meat is unfrozen, but at its freezing temperature. Outside the freezing front, the meat is already frozen and its temperature is gradually dropping towards the temperature of the surroundings. Most of the heat to be removed is latent heat and this is released from a given part of the meat as the freezing front passes through that part. The task of predicting the freezing time is therefore simplified to the task of predicting how long it will take for the freezing front to get from the surface of the meat to the centre, and then adding factors to account for the initial chilling period and the final sub-cooling period. Both of these periods are typically short compared to the freezing period, as can be seen from Figure 12.10, so there is no need to predict their lengths very accurately as long as the length of the freezing period is well-predicted. Equation (26) was developed by Pham (1986).

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

2kX +

hX

TH +

TH

E1 = t

f2

2

1

1f

2

ΔΔ

ΔΔ

(26)

where: ( ) T- T C ρ = H fminll1Δ (27)

( ) T- T C ρ + L ρ = H cfmffl2Δ (28)

T- 2

T T = T afmin

1+

Δ (29)

T- T = T afm2Δ (30)

T 0.105 + T0.263 +1.8 = T acfm (31)



Where: tf is the freezing time, s E is the Equivalent Heat Transfer Dimensionality

X is the shortest distance from the thermal centre to the surface (m) h is the surface heat transfer coefficient (possibly including packaging resistance) (W/m2K)

kf is the frozen meat thermal conductivity (W/mK) ΔH1 is the heat released during chilling (J/m3)

ΔH2 is the heat released during freezing and subcooling (J/m3) ΔT1 is the temperature driving force during chilling (°C)

ΔT2 is the temperature driving force during freezing and subcooling (°C) ρl, ρf are the densities of unfrozen and frozen meat (kg/m3)

Cl,, Cf are the specific heat capacities of unfrozen and frozen meat (J/kgK) Tin is the initial temperature of the product (°C)

Tfm is the mean freezing temperature of the product (°C) Tc is the final centre temperature of the product (°C)

Ta is the temperature of the cooling medium (°C) L is the latent heat of freezing the ice fraction (J/kg)

This calculation can be expected to be accurate within ±15%, though it is usually a bit better than that. One problem that arises more frequently in freezing time prediction than in chilling time prediction is that of asymmetric heat transfer. Often, the heat transfer coefficient at the top of a carton is different from that at the bottom of the carton, due to the air gap that is intentionally left at the top to accommodate the meat as it freezes and expands. Pham (1987) found a straightforward solution to this problem for slab-shaped food products. It applies quite well to products that are almost slab-shaped too, such as meat cartons in plate or air-blast freezers. When different h values apply at each side of a slab, the freezing fronts progress inwards at different rates. Instead of meeting in the geometric centre of the meat product, the freezing fronts will meet at a point X(1 - a) from the top and X(1 + a) from the bottom. Now if:

hh = r

top

bottom (32)

where: hbottom and htop are the heat transfer coefficients at the top and bottom of the carton respectively. ... then it turns out that the value of a is given by equation (33).

bottomBi2 + 1 + r1- r = a (33)

where: Bibottom is the Biot number calculated with the heat transfer coefficient hbottom and the frozen meat thermal conductivity kf. Having calculated a, you can then calculate the freezing time for the product using equation (26) but setting h = hbottom and X = (1 + a)Xactual, where Xactual is the depth of the thermal centre for the actual product. Product heat loads The total amount of heat released during freezing can be calculated from equation (34).

fl ρH

+ ρH

= H 21 ΔΔΔ (34)

Where: ΔH1 is the heat released during chilling (J/m3) from equation (27) ΔH2 is the heat released during freezing and subcooling (J/m3) from equation (28)

ρl, ρf are the densities of unfrozen and frozen meat (kg/m3) ΔH is the heat released during the whole process (J/kg)

The average product heat load is therefore given by:

tH = φf

fΔ (35)

Where: φf is the average product heat load per kilogram of meat in the freezer (W/kg) The average total heat load can be calculated by multiplying φf by the total weight of meat in the freezer.



Although we have been able to calculate the average product heat load during freezing, we have yet to see how the load varies with time during the process. While product freezing times are definitely easier to calculate than product chilling times, unfortunately the reverse is true for product heat loads, at least for hand calculation. To estimate the freezing product heat load at any given time during the process, you have to trace the position of the freezing front all of the way from the surface to whatever position it has reached by that time. This can be done in a computer program using one of several different techniques (e.g. that of Lovatt et al., 1992), but it would be very laborious to do by hand, so we will not look at those methods. Although it is difficult to calculate accurately, the product heat load during freezing does follow a similar sort of pattern to the heat load during chilling. That is, it starts off very high and drops sharply. Then it tails off towards the end of the process. The main difference between the freezing and chilling heat load profiles is that while the chilling heat load profile forms a smooth curve when you calculate it all the way through the process, the freezing heat load profile flattens while the freezing front is moving and then drops away after that. If you need to predict freezing heat load variation over time accurately, you should use a computer program. The sources of additional heat loads in freezers are the same as for chillers. However, because of the lower temperature inside a freezer the heat load due to infiltration through the walls, floor and ceiling is generally greater than for a chiller. Higher air velocities in air-blast freezers compared with chillers also result in fan power being a more important heat load in freezers than in chillers. If the air velocity in a chiller is 0.5 m/s and that in a freezer of similar design is 2 m/s, the freezer fans will actually use (2.0/0.5)3 = 64 times as much power as the chiller fans! Heat loads from lights, machinery and people in a freezer are generally negligible because of the large heat loads presented by the product, heat infiltration and fans. Software to estimate cooling times and heat loads The lecture notes above give you some background on the theory of chilling and freezing and enable you to carry out calculations for simple situations. However, due to the complexity and limitations of the equations for predicting cooling times and heat loads, it is often more practical to use a software program. Food Product Modeller™ and Refrigeration Loads Analyser™ are two software products that are available from the AgResearch MIRINZ Centre in Hamilton, New Zealand. These programs allow more complicated situations to be calculated and the temperature profiles in all parts of the product to be displayed.

12.5 The boning process Definition Boning refers to the process of removing the meat from the bones and cutting the carcass down into smaller and more manageable pieces. This process is often performed in an area called a ‘boning room’ or ‘cutting room’ where other operations, such as vacuum-wrapping the meat in plastic films and/or packing into cartons, are also carried out. Boning can be carried out before, during or after cooling depending on the requirements of the processor. If boning is carried out before any significant cooling occurs, it is called ‘hot boning’ because the product is still near to the body temperature of the animal. If boning is carried out after chilling (usually to below 7°C), it is called ‘cold boning’. If boning is carried out at any temperature between about 7 and 30°C it is often called ‘warm boning’ or ‘boning on the curve’ in reference to the temperature curve that the product is experiencing. Product tenderness A description of how rigor, cold shortening and aging affects the tenderness of meat is given in the section on ‘Product Tenderness’ above. Please revise this section before continuing.



Hot boning vs cold boning The advantages and disadvantages of hot boning as opposed to cold boning are discussed below. The choice of whether a processor uses hot or cold boning depends on which process characteristics are most important to them. Advantages of hot boning: Lower refrigeration costs When a carcass is hot boned, the meat is removed from the bones before any significant cooling occurs. Therefore, the processor need only cool the useable meat (about two-thirds of the carcass weight) and the bones and unusable meat do not need to be cooled. Faster processing Rigor progresses more quickly at a higher temperature, therefore the delay in product cooling that occurs when boning is carried out directly after slaughter allows the meat to reach rigor more quickly. This often means that a more rapid chilling regime may be used after boning (as long as the meat is not cooled too quickly to avoid cold shortening). Hot-boned meat cuts are not as thick as carcasses, so heat can be removed from the centre of hot-boned meat faster than for carcasses. Often processors do not take advantage of this fact, instead preferring to pack hot-boned meat into cartons before chilling. Since most cartons have a critical depth only slightly lower than beef sides and higher than lamb carcasses, this practice does not aid chilling. If hot-boned meat pieces are vacuum-wrapped, the meat may be chilled by immersion in a cold liquid, which is faster, more energy efficient and more consistent than air cooling. The faster cooling rate achieved by immersion chilling can also reduce drip loss and microbial growth, which may lead to a better quality product with a longer shelf-life. Less inventory and lower capital costs for buildings Hot-boning can significantly reduce the on-site inventory for a meat plant, since animals can be slaughtered, cooled and ready for shipment within a shorter period of time. The requirement of less chiller space means that the capital cost of buildings is also lower. This is particularly the case for meat that is shipped chilled and is not to be consumed within a short period of time. The shipping time and the time spent on the shelf allow the meat to age to acceptable levels of tenderness. If product is to be shipped from the plant frozen or if it is to be consumed very shortly after leaving the plant, it will be important to provide enough chiller space for the product to age to acceptable levels of tenderness. Less weight loss Cold-boned meat loses about 2% of its weight by evaporation of moisture from the meat into the air (about 1.8% during carcass chilling and about 0.2% during boning after chilling). By comparison, hot-boned meat loses only about 0.6% of its weight and almost all of this occurs during boning because the product is usually wrapped or packaged before it is cooled. Disadvantages of hot boning: Bad shape retention/more difficult to cut Since hot boning occurs quite soon after slaughter, the meat is usually in a pre-rigor state. This means that the meat is extremely soft and flexible and is at near-to-body temperature, which also causes the fat to be very soft and sticky. Due to these factors, it is usually more difficult (and potentially more dangerous) to cut and trim hot-boned meat because the meat can easily slip in the boner’s hands and the hot fat can easily smear. Further, as the cuts are no longer held in position by the bones of the carcass, the muscles easily lose their characteristic shape, unless they are packaged in such a way as to maintain their shape until they reach rigor, at which stage they will be able to hold their own shape. Tenderness/electrical stimulation Many cold-boned muscles are held in a stretched state by the bones of the carcass when they go into rigor. Since hot-boned muscles are removed from the carcass before reaching rigor, they are not held in a stretched state and may contract more easily resulting in a rise in toughness. Furthermore, because hot-boned meat cuts are not as thick as carcasses, they can be cooled much quicker which, if not done carefully, this may result in cold shortening.



As mentioned previously, electrical stimulation can be used to reduce the risk of cold shortening. The electrical current forces the muscles to contract, which provides a considerable increase in the rate of glycolysis. This results in a much faster fall in pH while the muscle temperature remains high. For example, in an unstimulated beef animal it may take 10 to 12 hours to reach a pH of 6.2, whereas with electrical stimulation it may only take about one hour to reach this pH. Consequently, with regard to cold shortening, there is very little limit on the cooling rate of electrically stimulated meat since it would be difficult to cool hot meat below about 10°C before a pH of 6.0 was reached. Electrical stimulation may not be necessary in cold boning operations. However, many cold boning plants also use electrical stimulation to advance the rigor process. As mentioned previously, when a carcass reaches rigor it immediately begins to age. If this aging process begins at a higher temperature (i.e. earlier on in the chilling process), the aging rate will be faster and an acceptable tenderness level will be reached sooner.

Microbiological quality The microbiological quality of any food depends on its initial contamination and the subsequent handling and storage processes, which may or may not allow microbial growth to occur. A slaughter and dressing process provides an initial contamination of the carcass surfaces due to handling. However, the slaughter and dressing processes are similar for both hot and cold boning so there will be little difference between hot and cold boning processes at this stage.

As we learned earlier, the chilling process is critical for controlling microbial growth. If boning occurs before chilling (as with hot boning) the meat may become further contaminated (due to handling) at a time when the product is still at a high enough temperature for microbial growth to continue. Whereas, if the meat is already chilled when boned (as with cold boning) the further contamination during boning will not result in as much subsequent growth because the product is already at a lower temperature.

A further consideration is that during cold boning the whole carcass is cooled in air. This provides a certain amount of surface drying that inhibits microbial growth. With hot boning, the product is generally packaged during the chilling step, which means that the product surface is moist, providing a better environment for microbial growth.

Synchronising slaughter floor and boning room In a hot boning plant it is important that the slaughter floor and boning room operations are carefully synchronised. This often means that the slaughter floor needs to start slightly earlier than the boning room in order to supply enough animals for the boning room to start processing. In a cold boning operation the carcasses processed in the boning room are usually from the previous shift on the slaughter floor, therefore it is unnecessary to delay the boning room operation unless the carcasses from the previous slaughter floor shift have not yet been cooled to an acceptable temperature.

Readings The following readings are available on web learning management systems 1. Cleland, A.C., 1990 Food Refrigeration Processes: Analysis, Design and Simulation,

Chapter 1, Elsevier Science Publishers, London. In his opening chapter, Cleland (1990) describes the importance of food refrigeration and the major issues involved in designing refrigeration processes. He goes on to discuss the benefits that can arise from further research into refrigerating operations, using the example of a beef carcass in an air chiller. The chapter ends with the statement of two areas for study: heat transfer within the food product and dynamics of the refrigeration system that is providing the cooling effect. The rest of this book covers these two areas in great detail.

2. ASHRAE 2002, ASHRAE Handbook - Refrigeration, SI Edition, Chapter 15, American Society of Heating, Refrigeration and Air Conditioning Engineers, Atlanta, Georgia, USA. An outline of different freezing methods and food freezing equipment is covered in chapter 15 of the 2002 ASHRAE Handbook (Refrigeration). Most of the equipment outlined in this chapter could be used within the meat industry, making the chapter quite useful to industry members. It contains many diagrams that allow the reader to better understand how the freezing equipment operates. Consideration is also given to selection criteria of freezer systems such as cost, quality, reliability, hygiene, etc.

3. Taylor, A.A., 1995. Carcase boing. Meat Focus International, July pp 280-285. This article is well-written and gives a thorough analysis of the advantages and disadvantages of hot boning. It covers aspects such as weight loss, colour, tenderness, electrical stimulation and microbiology.



Summary Summary slides are available on web learning management systems The importance of chilling and freezing in the meat industry is outlined and the lecture notes provide a basic introduction to the concepts and mechanisms involved in these two heat transfer operations. The material covers prediction of chilling and freezing times for a variety of different products that are of relevance to the meat industry. Also discussed is the importance of heat loads; the possible sources are listed and methods for estimating these are considered. Hot and cold boning is also discussed in the lecture material, with particular emphasis on the advantages and disadvantages of each with respect to cost and quality. References Australian Model Codes of Practice for the welfare of animals: Livestock at Slaughtering

establishments 22 No readings provided ASHRAE 2002. ASHRAE Handbook - Refrigeration, SI edition. American Society of Heating,

Refrigerating and Air-conditioning Engineers, Atlanta, Georgia, USA. Cleland, A.C., 1990. Food Refrigeration Processes: Analysis, Design and Simulation. Elsevier

Science Publishers, London. Cleland, A.C. and Earle, R.L., 1982. A simple method for prediction of heating and cooling rates in

solids of various shapes. International Journal of Refrigeration, vol 5, pp 98-106. Devine, C.E., Bell, R.G., Lovatt, S.J., Chrystall, B.B. and Jeremiah, L.E., 1996. Chapter 2: Red Meats.

In ‘Freezing Effects on Food Quality’, ed. L. Jeremiah, Marcel Dekker, New York, pp 51-84. Gilbert, K.V., Davey, C.L. and Newton, K.G., 1977. Electrical stimulation and the hot boning of beef.

N.Z. Journal of Agricultural Research, vol 20, pp 139-143. Lovatt, S.J., Pham, Q.T., Cleland, A.C., and Loeffen, M.P.F., 1992. A new method of predicting the

time-variability of product heat load during food cooling – Part 1: Theoretical considerations. Journal of Food Engineering, vol 18, pp 13-36.

Miles, C.A., van Beek, G. and Veerkamp, C.H., 1983. Chapter 16: Calculation of the Thermal Properties of Foods. In: ‘Physical Properties of Foods’, eds. R. Jowitt, F. Escher, B. Hallström, H. Meffert, E. Spiess, G. Vos. Applied Science Publishers, London and New York, pp 269-311.

MIRINZ RM54 1977. Specification for accelerated conditioning of lambs after dressing, MIRINZ, Hamilton, New Zealand.

MIRINZ RM135 1981. AC Seminar 1981, MIRINZ, Hamilton, New Zealand. MIRINZ RM141 1982. Guidelines for an AC quality control programme, MIRINZ, Hamilton, New

Zealand. Pham, Q.T., 1986. A simplified equation for predicting the freezing time of foodstuffs. Journal of Food

Technology, vol 21, pp 209-219. Pham, Q.T., 1987. A converging-front model for the asymmetric freezing of slab-shaped food, Journal

of Food Science, vol 52, pp 795-800. Pham, Q.T., Wee, H.K., Kemp, R.M., and Lindsay, D.T. 1994. Determination of the Enthalpy of Foods

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Acknowledgements Thanks to Dr Simon Lovatt at AgResearch Ltd, New Zealand for drawing many of the figures in these notes.

Glossary of terms

Biot number. The Biot number (Bi) is the ratio of the conductive (internal) resistance to heat transfer to the convective (external) resistance, and indicates the way in which the external resistance to heat transfer is balanced against the internal resistance.

Boning. Boning refers to the process of removing the meat from the bones and cutting the carcass down into smaller and more manageable pieces.

Chilling. Chilling is the process of cooling a food while it remains above its freezing temperature. The temperature of the cooling medium (air or water, for instance) doesn't matter, as soon as the food starts to freeze, it can no longer be considered a chilling process.

Cold boning. If boning is carried out after chilling (usually to below 7°C), it is called ‘cold boning’.

Conduction. A mechanism for heat transfer in solids, liquids or gases, where a molecule at a higher energy level will transfer some of its energy to a molecule at a lower energy level when they come into contact with each other.

Convection. A mechanism for heat transfer in liquids or gases, where a molecule at a higher energy level is moved into an area occupied by lower energy molecules, which raises the average energy level of the molecules in that particular area.

Freezing. Freezing is the process of removing heat from a food so that the water content of the food is converted into ice. A food starts to freeze when it is cooled to below its freezing temperature and is considered fully frozen when no more of the water in the food can be converted to ice.

Heat load. Heat load describes the rate at which heat energy enters the cooling medium. The heat load may come from the food product or from some other source of energy such as fans, lights or warm air entering through an opening. To maintain a constant temperature it is important that the refrigeration system is able to remove heat from the cooling medium at the same rate as it enters from the various heat sources in the cooler.

Heat transfer coefficient. The surface heat transfer coefficient (h) indicates the amount of heat that can be transferred from the surface of a food to the surrounding cooling medium for a given surface area and temperature difference. It depends on the characteristics of the cooling medium and its level of turbulence.

Heat. Heat is a form of energy and it is measured in Joules (J). It is associated with the movement of atoms and molecules and it may be transferred from molecules and atoms by various mechanisms.

Hot boning. If boning is carried out before any significant cooling occurs, it is called ‘hot boning’ because the product is still near to the body temperature of the animal.

Latent heat. Latent heat is heat that, when added or removed from a food, does not result in an observable change in temperature. Instead, the heat addition or removal causes the food to undergo a change in state (e.g. it may change the water in the food from a liquid to a solid).



Radiation. A mechanism for heat transfer where energy is transferred from molecules at a higher energy level to molecules at a lower energy level by electromagnetic radiation (i.e. without coming into contact with each other).

Sensible heat. Sensible heat is heat that, when added or removed from a food, brings about an observable change in temperature.

Thermal properties. The thermal properties of a food describe its ability to hold and transfer heat. These include the thermal conductivity (W/mK), the heat content or enthalpy (J/kg) and the density (kg/m3).

Warm boning. If boning is carried out at any temperature between about 7 and 30°C it is often called ‘warm boning’ or ‘boning on the curve’ in reference to the temperature curve that the product is experiencing.

12. chilling, freezing and boning - woolwise...notes – lecture 12 – chilling, freezing and...

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