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Heat balancesenthalpylatent heatsensible heatfreezing
Other forms of energy
mechanical energyelectrical energy

Energy takes many forms such as heat, kinetic energy, chemical energy, potential energy but, because of interconversions, it is not always easy to isolate separate constituents of energy balances. However under some circumstances certain aspects predominate. In many heat balances, other forms of energy are insignificant; in some chemical situations, mechanical energy is insignificant and in some mechanical energy situations, as in the flow of fluids in pipes, the frictional losses appear as heat but the details of the heating need not be considered. We are seldom concerned with internal energies.

Therefore practical applications of energy balances tend to focus on particular dominant aspects and so a heat balance, for example, can be a useful description of important cost and quality aspects of a food process situation. When unfamiliar with the relative magnitudes of the various forms of energy entering into a particular processing situation, it is wise to put them all down. Then after some preliminary calculations, the important ones emerge and other minor ones can be lumped together or even ignored without introducing substantial errors. With experience, the obviously minor ones can perhaps be left out completely though this always raises the possibility of error.

Energy balances can be calculated on the basis of external energy used per kilogram of product, or raw material processed, or on dry solids. or some key component. The energy consumed in food production includes direct energy which is fuel and electricity used on the farm, in transport, in factories, in storage, selling, etc.; and indirect energy which is used to build the machines, to make the packaging, to produce the electricity and the oil and so on. Food itself is a major energy source, and energy balances can be determined for animal or human feeding; food energy input can be balanced against outputs in heat and mechanical energy and chemical synthesis.

In the SI system there is only one energy unit, the joule. However, kilocalories are still used by some nutritionists, and British thermal units (Btu) in some heat-balance work.

The two applications used in this book are heat balances, which are the basis for heat transfer, and the energy balances used in analysing fluid flow.

Heat Balances

The most common important energy form is heat energy and the conservation of this can be illustrated by considering operations such as heating and drying. In these, enthalpy (total heat) is conserved and as with the mass balances so enthalpy balances can be written round the various items of equipment. or process stages, or round the whole plant, and it is assumed that no appreciable heat is converted to other forms of energy such as work.

Fig. 2.2. Heat balance
Figure 2.2. Heat balance.

Enthalpy (H) is always referred to some reference level or datum, so that the quantities are relative to this datum.
Working out energy balances is then just a matter of considering the various quantities of materials involved, their specific heats, and their changes in temperature or state (as quite frequently latent heats arising from phase changes are encountered). Fig. 2.2 illustrates the heat balance.

Heat is absorbed or evolved by some reactions in food processing but usually the quantities are small when compared with the other forms of energy entering into food processing such as sensible heat and latent heat. Latent heat is the heat required to change, at constant temperature. the physical state of materials from solid to liquid, liquid to gas, or solid to gas. Sensible heat is that heat which when added or subtracted from food materials changes their temperature and thus can be sensed.
The units of specific heat (c) are J kg-1 °C-1, and sensible heat change is calculated by multiplying the mass by the specific heat by the change in temperature, m c
DT,    (J).
The units of latent heat are J kg-1 and total latent heat change is calculated by multiplying the mass of the material, which changes its phase, by the latent heat. Having determined those factors that are significant in the overall energy balance, the simplified heat balance can then be used with confidence in industrial energy studies. Such calculations can be quite simple and straightforward but they give a quantitative feeling for the situation and can be of great use in design of equipment and process.

EXAMPLE 2.10. Heat demand in freezing bread
It is desired to establish freezing of 10,000 loaves of bread each weighing 0.75 kg from an initial room temperature of 18°C to a final store temperature of -18°C. If this is to be carried out in such a way that the maximum heat demand for the freezing is twice the average demand, estimate this maximum demand, if the total freezing time is to be 6 h.

If data on the particular bread are unavailable, in the literature are data on bread constituents, calculation methods, enthalpy/temperature tables

(a) Tabulated data (Appendix 7) suggests specific heat above freezing 2.93 kJ kg-1 °C-1, below freezing 1.42 kJ kg-1 °C-1, latent heat of freezing 115 kJ kg-1 and freezing temperature is -2°C.

Total enthalpy change, (DH) = [18 - (-2)] 2.93 + 115 + [-2 - (-18)] 1.42 = 196 kJ kg-1.

(b) Formula (Appendix 7) assuming the bread is 36% water gives
- specific heat above freezing
     4.2 x 0.36 + 0.84 x 0.64 = 2.05 kJ kg-1 °C-1,
- specific heat below freezing
     2.1 x 0.36 + 0.84 x 0.64 = 1.29 kJ kg-1 °C-1,
- latent heat
            0.36 x 335             = 121 kJ kg-1.

Total enthalpy change, (
DH) = [18 - (-2)]2.05 + 121 +[-2-(-18)] 1.29 = 183 kJ kg-1.

(c) Enthalpy/temperature data for bread of 36% moisture (Mannheim et al., 1957) suggest
              H18.3°C =  210.36 kJ kg-1,
              H-17.3°C = 65.35 kJ kg-1.

So from +18°C to -18°C total enthalpy change (DH ) = 210 - 65 = 145 kJ kg-1.

(d) The enthalpy/temperature data in Mannheim et al. can also be used to estimate "apparent" specific heats as
DH /DT = c and so using the data:


T°C =
H kJ kg-1 =

Giving c-18 =
65.35 - 55.88
= 3.4 kJ kg-1 °C-1
20.6 - 17.8

Giving c18 =
210.4 - 203.4
= 2.6 kJ kg-1 °C-1
18.3 – 15.6

Note that the "apparent" specific heat at -18°C, 3.4 kJ kg-1 °C is higher than the specific heat below freezing in (a), 1.42, and in (b) , 1.29 kJ kg-1 °C-1. The reason for the high apparent specific heat at -18°C is due to some freezing still continuing at this temperature. It is suggested that at -18°C only about two-thirds of the water is actually frozen to ice. This implies only two-thirds of the latent heat has been extracted at this temperature. Making this adjustment to the latent-heat terms, estimates for the total emthalph change (a) and (b) give 158kJ kg-1 and 142kJ kg-1 respectively, much improving the agreement with (c) of 145 kJ kg-1
DH = 150 kJ kg-1

Total heat change

= 150 x 10,000 x 0.75  = 1.125 x 106 kJ.
Total time                   = 6h = 2.16 x 104s.

         (DH /Dt) = 52 kJ s-1 = 52 kW on average.

And if the maximum rate of heat removal is twice the average:

    (DH /Dt) max = 2 x 52   = 104 kW.

Example 2.10 illustrates the application of heat balances, and it also illustrates the advisability of checking or obtaining corroborative data unless reliable experimental results are available for the particular system that is being considered. The straightforward application of the tabulated overall data would have produced a result about 30% higher than that finally calculated. On the other hand, for some engineering calculations to be within 30% may be about as close as you can get.

In some cases, it is adequate to make approximations to heat balances by isolating dominant terms and ignoring less important ones. To make approximations with any confidence, it is necessary to be reasonably sure about the relative magnitudes of the quantities involved. Having once determined the factors that dominate the heat balance, simplified balances can then be set up if appropriate to the circumstances and used with confidence in industrial energy studies. This simplification reduces the calculation effort, focuses attention on the most important terms, and helps to inculcate in the engineer a quantitative feeling for the situation.

EXAMPLE 2.11. Dryer heat balance for casein drying
In drying casein the dryer is found to consume 4 m3/h of natural gas with a calorific value of 800 kJ/mole. If the throughput of the dryer is 60 kg of wet casein per hour, drying it from 55% moisture to 10% moisture, estimate the overall thermal efficiency of the dryer taking into account the latent heat of evaporation only.
Basis: 1 hour of operation

60 kg of wet casein contains    60 x 0.55 kg water  = 33 kg moisture
    and                                           60 x (1 - 0.55)  = 27 kg bone dry casein.

As the final product contains 10% moisture, the moisture in the product is 27/9 = 3 kg

          and so moisture removed = (33 - 3) = 30 kg
         Latent heat of evaporation = 2257 kJ kg-1(at 100 °C from Appendix 8)
 so     heat necessary to supply = 30 x 2257 = 6.8 x l04 kJ

Assuming the natural gas to be at standard temperature and pressure at which 1 mole occupies 22.4 litres
  Rate of flow of natural gas = 4 m3 h-1  =
4 x 1000
= 179 moles h-1
Heat available from combustion           = 179 x 800                   = 14.3 x 104 kJ/h

Approximate thermal efficiency of dryer          =
heat needed
= 6.8 x 104 / 14.3 x 104 = 48%
heat used

To evaluate this efficiency more completely it would be necessary to take into account the sensible heat of the dry casein solids and the moisture, and the changes in temperature and humidity of the combustion air, which would be combined with the natural gas. However, as the latent heat of evaporation is the dominant term the above calculation gives a quick estimate and shows how a simple energy balance can give useful information.

Similarly energy balances can be carried out over thermal processing operations, and indeed any processing operations in which heat or other forms of energy are used.

EXAMPLE 2.12. Heat balance for cooling pea soup after canning
An autoclave contains 1000 cans of pea soup. It is heated to an overall temperature of 100°C. If the cans are to be cooled to 40°C before leaving the autoclave, how much cooling water is required if it enters at 15°C and leaves at 35°C?
The specific heats of the pea soup and the can metal are respectively 4.1 kJ kg-1 °C-1 and 0.50 kJ kg-1°C-1. The weight of each can is 60 g and it contains 0.45 kg of pea soup. Assume that the heat content of the autoclave walls above 40°C is 1.6 x 104 kJ and that there is no heat loss through the walls.

Let w = the weight of cooling water required;
and the datum temperature be 40°C, the temperature of the cans leaving the autoclave.

Heat entering
Heat added to cans = weight of cans x specific heat x temperature above datum

                   = 1000 x 0.06 x 0.50 x (100 - 40) kJ = 1.8 x 103 kJ.

Heat added to can contents = weight pea soup x specific heat x temperature above datum
                                         = 1000 x 0.45 x 4.1 x (100 - 40)
                                         = 1.1 x 105 kJ.

Heat added to water = weight of water x specific heat x temperature above datum
                              = w x 4.21 x (15 - 40)       (Appendix 4)                                          
                              =   -104.6 w kJ.

Heat leaving
Heat in cans = 1000 x 0.06 x 0.50 x (40 - 40) (cans leave at datum temperature)
                   = 0.

Heat in can contents = 1000 x 0.45 x 4.1 x (40 – 40)
                               = 0.

Heat in water = w x 4.186 x (35 - 40)
                    = -20.9 w.



Heat entering (kJ) Heat Leaving (kJ)
Heat in cans 1,800   Heat in cans 0
Heat in can contents 110,000   Heat in can contents 0
Heat in autoclave wall 16,000   Heat in autoclave wall 0
Heat in water 105.3w   Heat in water -20.9 w
Total heat entering 127,800 -105.3w   Total heat leaving -20.9 w
Total heat entering
Total heat leaving
127,800 - 105.3w
-20.9 w
And so
1514 kg
Amount of cooling water required = 1514 kg.

Other Forms of Energy

The most common mechanical energy is motor power and it is usually derived, in food factories, from electrical energy but it can be produced from steam engines or water power. The electrical energy input can be measured by a suitable wattmeter, and the power used in the drive estimated. There are always losses from the motors due to heating, friction and windage; the motor efficiency, which can normally be obtained from the motor manufacturer, expresses the proportion (usually as a percentage) of the electrical input energy which emerges usefully at the motor shaft and so is available.

When considering movement, whether of fluids in pumping, of solids in solids handling, or of foodstuffs in mixers, the energy input is largely mechanical. The flow situations can be analysed by recognising the conservation of total energy whether as energy of motion, or potential energy such as pressure energy, or energy lost in friction. Similarly, chemical energy released in combustion can be calculated from the heats of combustion of the fuels and their rates of consumption. Eventually energy emerges in the form of heat and its quantity can be estimated by summing the various sources.

EXAMPLE 2.13. Refrigeration load in bread freezing
The bread-freezing operation of Example 2.10 is to be carried out in an air-blast freezing tunnel. It is found that the fan motors are rated at a total of 80 horsepower and measurements suggest that they are operating at around 90% of their rating, under which conditions their manufacturer's data claims a motor efficiency of 86%.
If 1 ton of refrigeration is 3.52 kW, estimate the maximum refrigeration load imposed by this freezing installation assuming (a) that fans and motors are all within the freezing tunnel insulation and (b) the fans but not their motors are in the tunnel. The heat-loss rate from the tunnel to the ambient air has been found to be 6.3 kW.

Extraction rate from freezing bread (maximum) = 104 kW
Fan rated horsepower = 80

Now 0.746 kW = 1 horsepower (Appendix 2) and the motor is operating at 90% of rating,
and so (fan + motor) power = (80 x 0.9) x 0.746
                                        = 53.7 kW

(a) With motors + fans in tunnel

heat load from fans + motors = 53.7 kW
heat load from ambient          = 6.3 kW
Total heat load                      = (104 + 53.7 + 6.3) kW
                                           = 164 kW
                                           = 164/3.52    =    46 tons of refrigeration (see Appendix 2)

(b) With motors outside, the motor inefficiency = (1 - 0.86) does not impose a load on the refrigeration.

Total heat load                      = (104 + [0.86 x 53.7] + 6.3)
                                           = 156kW
                                           = 156/3.52 = 44.5 tons refrigeration

In practice, material and energy balances are often combined as the same stoichiometric information is needed for both.

Material & Energy Balances > SUMMARY & PROBLEMS

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Unit Operations in Food Processing. Copyright © 1983, R. L. Earle. :: Published by NZIFST (Inc.)
NZIFST - The New Zealand Institute of Food Science & Technology