Food Engineering Part II - Energy Balance, Mass Balance - PDF

1  PART II MATERIALS/ 2 MASS AND ENERGY MASS AND ENERGY BALANCE The law of conservation led to material/ energy balance No chemical reaction; the law of conservation will apply to each component 3 then some part of A left unnoticed, w...

1  PART II MATERIALS/ 2 MASS AND ENERGY MASS AND ENERGY BALANCE The law of conservation led to material/ energy balance No chemical reaction; the law of conservation will apply to each component 3 then some part of A left unnoticed, which can be = L, OPERATIONS Types of Process Operations Batch Process Continuous Process Semi batch Process 4 TYPES OF OPERATIONS Continuous process: also known as continuous flow process has all the materials whether dry or fluid processed continuously by subjecting the materials to mechanical or heat treatment Batch process: as the name implies, processing is done in batches. Semi Batch process: this is basically the combination of 5 continuous and batch processes TYPES OF PROCESS OPERATIONS 6 TYPES OF PROCESS OPERATIONS 7 WORKED EXAMPLE - QUESTION CONSTITUENT BALANCE OF MILK 1. Skim milk of 100kg is prepared by the removal of some of the fat from whole milk. This skim milk is found to contain 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original milk contained 4.5% fat, calculate its composition assuming that fat only was removed to make the skim milk and that there are no losses in8 processing. WORKED EXAMPLE - SOLUTION Basis: 100 kg of skim milk. This contains, therefore, 0.1 kg of fat. Let fat which was removed from it to make skim milk be x kg. Total original fat = (x + 0.1) kg Total original mass = (100 + x) kg The original fat content was 4.5% so 9 Basis: 100 kg Skim milk FLOW DIAGRAM OF BASIS Separatio Whole milk (W) kg = Skim milk (B) = 0.045 fat n 100kg, 0.001 fat Process Water – 0.905 Protein – 0.035 Carbohydrate – 0.051 Fat (F) kg Ash – 0.008 10  TOTAL BALANCE (1) Fat Component Balance (2) 11 Determine the composition for protein, carbohydrate and fat 12 TRIAL QUESTION In a process for the production of sugar from cane sugar, 4000 kg/h of a solution containing 10 wt% sugar is evaporated in the 1st evaporator given an 18 wt% sugar solution. This is then fed to a 2nd evaporator, which gives a product of 50 wt% sugar. Calculate the water removed from each evaporator, the feed to the 2nd evaporator and the amount of the product. Assume steady state. 13 CLASS ACTIVITY A mixture containing 45% benzene (B) and 55% toluene (T) by mass is fed to a distillation column. An overhead stream of 95wt% benzene is produced and 8wt% of the benzene fed to the column leaves in the bottom stream. The feed rate is 2000kg/hr. Determine: 1. the overhead flow rate and 2. the mass flow rates for benzene and toluene in the bottom stream. 14 RECYCLE When 100% conversion is not achieved in the process, recycling is done Recyc le R Feed sepa Produc F rator tP The Recycle Process 15 WORKED EXAMPLE - QUESTION In a process producing salt (NaCl) 1000 kg/h of a feed solution containing 20wt % NaCl is fed to an evaporator, which removes some water (H2O) at 455K to produce a 50wt % NaCl solution This is then fed to a crystallizer at 311k where crystals containing 96wt % NaCl are removed. The saturated solution containing 37.5wt % NaCl is recycled to the evaporator. Calculate the amount to recycled stream and the product stream of crystals 16 WORKED EXAMPLE - SOLUTION 17 Assuming a steady state, the Total Balance for the process is (1 ) (2 ) Component balance on salt (NaCl) Solving for W using equation 1 18  Balance on the Crystallizer (3 ) (4 Component balance on salt (NaCl) ) Solving eqn. (3) and (4) simultaneously 19 TRIAL QUESTION In a process producing salt (NH₄)₂SO₄, 1000 kg/h of a feed solution containing 32wt % (NH₄)₂SO₄ is fed to an evaporator, which removes some water (H2O) at 450K to produce a 62wt % (NH₄)₂SO₄ solution. This is then fed to a crystallizer at 355k where crystals containing 88wt % (NH₄)₂SO₄ are removed. The saturated solution containing 44.5wt % (NH₄)₂SO₄ is recycled to the evaporator. Calculate the amount to recycled stream and the product stream of crystals. 20 BYPASS A fraction of the feed diverts and gets into the output stream Feed sepa Produc F rator tP Bypas sB The Bypass Process 21 WORKED EXAMPLE - QUESTION Fresh orange juice contains 12wt % solids and concentrated orange juice contains 42wt % solids. Initially a single evaporation process was used for the concentration but the volatile constituents of the juice escaped with the H2O leaving the concentrate a flat taste. The present process overcomes the problem by bypassing the evaporator with a fraction of the fresh juice. The juice that enters the evaporator is concentrated to 58% solids and the product is mixed with the bypassed fresh juice to achieve the desired final concentration of the solids. Calculate the amount of concentrated juice produced per 100 kg of a fresh juice fed to the process and the fraction of the feed that bypasses the evaporator. 22 WORKED EXAMPLE - SOLUTION 23 Total Balance on the whole process is (1 ) Component balance on solids (2 ) from equation (1) 24  Total Balance (3 C P ) 58% 42% Solids Solids Component balance on solids (4 ) B 12% Solving eqns. (3) and (4) Solids simultaneously For Percentage Bypassed (%B): 25 TRIAL QUESTION 1 Fresh watermelon juice contains 16wt % solids and concentrated watermelon juice contains 52wt % solids. Initially a single evaporation process was used for the concentration but the volatile constituents of the juice escaped with the H2O leaving the concentrate a flat taste. The present process overcomes the problem by bypassing the evaporator with a fraction of the fresh juice. The juice that enters the evaporator is concentrated to 66% solids and the product is mixed with the bypassed fresh juice to achieve the desired final concentration of the solids. Calculate the amount of concentrated juice produced per 100 kg of a fresh juice fed to the process and the fraction of the feed that bypasses the evaporator 26 TRIAL QUESTION 2 In a process for producing tomato paste 5500 kg/h of the juice containing 10% solids is evaporated in the 1st evaporator giving a solution containing 32% total solids. This is then fed to a 2nd evaporator, which gives a product of 75% total solids. Calculate the amount of H2O removed from each evaporator, the feed to 2nd evaporator and the amount of product. 27  CHAPTER II ENERGY BALANCE Energy takes many forms. Which do you remember? In industrial processing, energy interconversions, do not allow for the separation of different energy constituents for balances. Hence, the most predominant energy form in the process flow is used for the energy balances, whiles the other minor energy forms that may exist are considered insignificant and possibly ignored 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 components - Units of Energy = Joule (SI) or kilocalories (used by nutritionist) 28 ENERGY BALANCE Energy consumed in Food 1. Production includes: DIRECT ENERGY This is the fuel and electricity used on theThis is used to actually farm, and in transportbuild the machines, to 2. and in factories and inmake the packaging, to INDIRECT storage selling, etc. produce the electricity ENERGY and the oil and so on. 29  Farm(Fo ENERGY BALANCE od producti on system) so energy balances can Energy be Balance: used to Food is determined examine the also a on both energy animal and involved in source Individu the Whole of human al proces processe energy, feeding s s 30 HEAT BALANCES Heat is the most important energy form Operations such as Heating and Drying are means of illustrating the conservation of heat, where enthalpy (total heat) is conserved Hence, similar to mass balances, 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 31 HEAT BALANCES Heat to Surroundings Heat from Electricity Heat out in Heat from Fuel Combustion Heat Products Heat from mechanical Sources Heat in raw stored Heat out in Waste materials Heat from 32 Surroundings HEAT BALANCE Sensible heat is that heat which when added or subtracted from food materials changes their temperature and thus can be sensed. Latent heat (L) 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 33 (1 ) HEAT BALANCE 34 WORKED EXAMPLE - QUESTION Cows milk at 4.4 is being heated in a heat exchanger to 54.4 hot water. How much heat is needed for a feed flow of 4536kg/h. (Take Heat Capacity of Milk = 3.85kJ/kg.K) 35 WOREKD EXAMPLE - SOLUTION 36 HEAT BALANCE EQUATIONS (2 ) 37 HEAT BALANCE EQUATIONS (3 ) (4 ) (5 ) 38 USE OF STEAM TABLES Water is widely used in the process industries both as feed to many process and as heating medium in the form of steam The physical properties of water, have been tabulated in what is called the steam table The reference state for these tables is liquid water at the triple point (0.01 and 0.00611bars) where the specific internal energy is defined as zero Properties tabulated include specific volume, internal energy, enthalpy and entropy At specified temperatures and pressures, these properties can easily be read from these tables. It is possible to interpolate within the39table to estimate values WORKED EXAMPLE - QUESTION Water at 85°C is being stored in a large insulated tank at atmospheric pressure. It is being pumped at steady state from this tank at point 1 by pump at a rate of 0.567 m3/min. The motor driving the pump supplies energy at a rate of 7.45 kW. The water passes through a heat exchanger where it gives up 1408kW of heat. The cooled water is then delivered to a second large open tank (atmospheric pressure) at point 2 which is 20 m above the first tank. Calculate the final temperature of the final water delivered to the second tank. Neglect any kinetic energy changes since the initial and final velocities in the tanks are essentially 40 zero. WOREKD EXAMPLE - SOLUTION Work done by the fluid is. In this case, Work done is on the fluid 41 WOREKD EXAMPLE - SOLUTION The heat added (Q) to the fluid is also negative, since it gives up heat: From the Steam Table: Substitute into equation: 42 WOREKD EXAMPLE - SOLUTION Interpolating from the steam table: 43 TRIAL QUESTION Water enters a boiler at 22.00oC at atmospheric pressure through a pipe at an average velocity of 3.20m/s. The exist steam is at a height of 22 m above the liquid inlet steady state heat added was 2689.81 J/kg, determine the final temperature steam (take g =9.81) 44 45 46  UHUH…! THAT’S ALL ON MATERIAL AND ENERGY BALANCES WHAT DO YOU REMEMBER? GOT QUESTIONS? 47