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,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,制 冷 原 理 与 技 术,Chapter 3:THERMAL PROPERTIES OF FOODS,Needed for heat transfer calculations in equipment design and estimation of process times for refrigerating,freezing,heating,or drying,Properties required:density,specific heat,（显热）,enthalpy,thermal conductivity.,Fresh fruits and vegetables also generate heat via respiration,（,呼吸,）,and lose moisture via transpiration,（,蒸发,）,.,1.FOOD CONSTITUENTS,Common constituents:water,protein,fat,carbohydrate,fibre,ash.,2.THERMAL PROPERTIES,Well behaved above initial freezing point.,Vary greatly below due to complex freezing processes.,Initial freezing point somewhat lower than pure water duo to dissolved substances(Fig.1).,At initial freezing point,a portion of water crystallises,remaining solution becomes more concentrated,thus reducing freezing point of unfrozen portion.,Ice and water fractions in frozen food depend on temperature.,Properties vary dramatically with temperature.,Properties above and below freezing point drastically different.,3.WATER CONTENT X,wo,Water is predominant constituent,in most foods,X,wo,significantly inflences properties.,Some values(%by mass):,Cauliflower(,波菜,)91.9%;apple 83.9;salmon,（鲑鱼）,76.4;shrimp 75.9;beef 58.9;pork shoulder 72.6;chicken 66;whole egg 75.3;cheddar cheese 36.8;strawberry ice cream 60;milk 87.8;orange juice 89.,4.INITIAL FREEZING POINT t,f,No foods freeze completely single T but over a range.,Foods with high sugar content or packed in high syrup concentrations may never be completely frozen,even at typical storage T.,No distinct freezing point for foods and beverages.,Initial freezing point indicates start of crystallisation.,Important for determining proper storage conditions and calculation of thermophysical properties.,For storage of fresh fruits and vegetables,T must be above initial freezing point to avoid freezing damage.,As drastic changes in properties when freez,initial freezing point is necessary to accurately model properties.,Some values():,（与前面一一对应）,-0.8;-1.1;-2.2;-2.2;-1.7;-2.2;-2.8;-0.6;-12.9;-5.6;-0.6;-0.4.,5.ICE FRACTION,Food items:water,dissolved solids,undissolved solids.,During freezing,some liquid water crystallises,solids dissolved in remaining liquid become increasingly more concentrated,thus lowering freezing temperature.,Miles(1974)equation for predicting ice fraction:,x,wo,=mass fraction of water in unfrozen food item(moisture content),x,b,=mass fraction of“bound water”=0.4x,p,x,p,=mass fraction of protein,t,f,=initial freezing point of food,t=food temperature,It underestimares x,ice,at T near t,f,and overestimates at low T.,Tchigeov(1979)prosposed an empirical relationship:,Example 1.A 150kg beef carcass is frozen to-20.What is mass of frozen and unfrozen water at-20?,Solution:,Mass fraction of unfrozen water:,Mass of frozen water at-20:,Mass of unfrozen water at-20:,6.DENSITY,Modelling density requires knowledge of food porosity(,孔隙率,),mass fraction and density of food componenets:,Densities(kg/m,3,)of food components(-40 to150):,Water:,Ice:,Protein:,Fat:,Carbohydrate:,Fibre:,Ash:,7.SPECIFIC HEAT c,（比热）,Unfrozen food c slightly lower as T rises from 0 to 20(Fig.2).,For frozen foods,large decrease in c as T decrease.,Some value kJ/(kgK):,（与前面一一对应）,3.89 kJ/(kgK)(above freezing)3.82&1.96.,Unfrozen Food,Specific heats kJ/(kgK)of food components(-40 to 150):,Water:,Ice:,Protein:,Fat:,Carbohydrate:,Fibre:,Ash:,If detailed composition data not available Chen(1985):,x,s,=mass fraction of solids in food item.,Frozen Food,Below freezing point,sensible heat due to T change and latent heat due to fusion,（熔化）,of water must be considered.,Latent heat not released at a constant T,but over a range.,Apparent(,表观,)specific heat used to account for sensible and latent heat effects.,Common method to predict apparent specific heat Schwartzberg(1981):,c,f,=specific heat of fully frozen food item(typically at-40),t,0,=freezing point of water=0,L,0,=latent heat of fusion of water=333.6 kJ/kg,Simpler model by Chen(1985):,Example 2.A 150 kg lamp,（羊肉）,is cooled from 10 to 0.Use specific heat to determine amount of heat removed.,Solution:Composition of lamb:,x,wo,=0.7342;x,f,=0.0525;x,p,=0.2029,x,a,=0.0106,（,ash,）,Evaluate c at average T of(0+10)/2=5,:,Heat to be removed:,T=150 3.59(10-0)=5390 kJ,=0.0525;,8.ENTHALPY H,Above freezing point,enthalpy consists of sensible energy.,Below,include sensible and latent energy.,Unfrozen Food,H,i,=enthalpy of individual food components,Method of Chen(1985):,H,f,=enthalpy of food at initial freezing temperature,kJ/kg,Frozen Foods,Based on Chen(1985)(can be used to estimate):,t,r,=reference temperature(zero enthalpy)=-40,Chang and Tao(1981)developed empirical correlations as functions of water content,initial and final T,food type(meat,juice or fruit or vegetable).,The correlations at a reference T(-45.6):,T,r,=reference temperature=227.6 K(-45.6),Y,z=correlation parameters,For Meat Group:,For Fruit,Vegetable,and Juice Group:,(Fruit/vegetabl
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