(Mass and Energy Balance).pdf

ABEN 4510 FOOD PROCESS ENGINEERING MATERIAL BALANCE (Mass and Energy Balance) MAY ALISBO-CABRAL Department of Agricultural and Biosystems Engineering College of Engineering Central Luzon State University...

ABEN 4510 FOOD PROCESS ENGINEERING MATERIAL BALANCE (Mass and Energy Balance) MAY ALISBO-CABRAL Department of Agricultural and Biosystems Engineering College of Engineering Central Luzon State University 1 TOPIC OUTLINE INTRODUCTION TO MASS AND ENERGY BALANCE ANALYSIS AND COMPUTATION LEARNING OUTCOME  Discuss the principles and concepts of mass and energy balance  Compute the mass balance of a system Introduction Material (mass and energy) balance −Useful tool in the analysis of many engineering problems −Described the quantities of materials in the unit operations in food processing −Governed with the law of conversation of mass Introduction Mass and energy balance −Useful tool the formulation of the product, design of food processes, processing equipment, process utilities, waste treatment utilities, estimation of the cost and calculation of the process efficiency −process of tracking down the input material to a process and the output from the process and the accumulated amount of the product. Introduction Mass and energy balance − selection and sizing of process equipment, the required quantity of formulation, and the information on energy inflow, outflow and conversion can be determine and calculated − engineer can gain the idea on where the loss or conversion of energy takes place and find out means to resolve the problem in the system −The law of conservation of mass/energy states that mass/energy can neither be created nor destroyed Introduction Process − an engineering system that transform raw materials into desired products − comprises of unit operations having specified functions for the completion of the intended transformation. Unit operation –stage in a process that serves to perform a specified job, such as cleaning, mixing, separation, heating, cooling, and extraction. Introduction Mass and energy balance − considered as the fundamental control of processing particularly in the yield of the products Exploratory stage of product Material balance − Planning − Testing − Checking − Changes in the system − Refine − Maintain System and Boundary mass (solid and liquid) balances is based on the fact that in a given system under a steady state condition, the total mass entering a system is equal to the total mass that leaves the system - Carpio (2010) steady state condition of the system or body happens when there is neither accumulation nor decay of mass within the confine of its boundary  System and Boundary System- any region prescribe in a space or a finite quantity of material enclosed by a boundary (Singh and Heldman,2009) Boundary confinement of a given system and it can be real such as wall of the tanks or it can be an imaginary surface that encloses the systems. stationary or movable Surroundings- everything outside the Figure 1. A system containing a tank with a boundary discharge pipe and valve (Source: Singh and Heldman, 2009) System 1. Close system the boundary of any system is impervious to flow of mass which means that it doesn’t exchange mass with its surroundings. − may exchange heat and work with its surroundings that may result in change in energy, volume or other properties of the system, but the mass remains constant 2. Open System − also known as control volume is considered when there is both heat and mass that can flow into and out of a system boundary (control surface). A close and open system (control volume) showing the mass inlet and exit (Source: Singh and Heldman, 2009) Systems 1. Isolated system − when mass, heat, or work does not exchange with its surroundings − no effect on its surroundings 2. Adiabatic System − either closed or an open system − when there is no exchange of heat that takes place with the surroundings 3. Isothermal system − happens when a process occurs at a constant temperature often with an exchange of heat with surrounding ISOLATED VS NON ISOLATED Systems Isolated system Non-isolated system (Source: https://www.hkdivedi.com/2016/08/difference-between-isolated-and-non.html) System Property State Properties − refer to the physical condition of the working substance such as temperature, pressure, density, specific volume, specific gravity or relative density  Transport Properties − refer to the measurement of diffusion within the working medium resulting from the molecular activity like viscosities, thermal conductivities System Property Intensive Properties − size independent such as temperature, pressure and density  Extensive Properties − depends on the size or extent of the system such as mass, volume and total energy Law of Conservation of Mass Mass can be neither created nor destroyed, however, its composition can be altered from one form to another. Even in the case of chemical reaction, the reactant mass composition and the product before and after the reaction may be different, but the mass of the system remains unchanged Rate of mass entering Rate of mass Rate of mass through the boundary exiting through the accumulation within - = of a system boundary of the the system system Inflow = Outflow + Accumulation  Law of Conservation of Mass INFLOW OUTFLOW ACCUMULATION  no accumulation then the inflow is equal to  formation of outflow and the material by  depletion of process is at steady chemical = material through + state reaction or chemical and  there is accumulation microbial biological within the system, it growth is an unsteady state, reactions then the quantity and processes concentration of the components in the system could be change with the time and process Steps in Material Balancing According to Singh and Heldman(2009), the following steps should be advantageous in conducting a material balance. a) Collect all known data on mass and composition of all inlet and exit streams from the statement of the problem. b) Draw a block diagram, indicating the process, with inlet and exit streams properly identified. Draw the system boundary. c) Write all available data on the block diagram. d) Select a suitable basis (such as mass or time) for calculations. The selection of basis depends on the convenience of computations. e) Write the material balances in terms of the selected basis for calculating unknowns. For each unknown, an independent material balance is required. f) Solve material balances to determine the unknowns  ANALYSIS AND COMPUTATION Mass Balance In any unit operation in food processing, the nature of the system as whole can be represented through a diagram (Figure 4). The mass and energy that going into the system must be balance with the mass and energy coming out as stated in the law of conservation of mass. Figure 4. Mass and energy balance ( Source: Earle, R.L. and M.D. Earle, 1966) Mass Balance The law of conservation of mass leads to material balance which can be written using the equation below: 𝑀𝑎𝑠𝑠 𝑖𝑛 = 𝑚𝑎𝑠𝑠 𝑜𝑢𝑡 + 𝑚𝑎𝑠𝑠 𝑠𝑡𝑜𝑟𝑒𝑑 𝑅𝑎𝑤 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 = 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 + 𝑤𝑎𝑠𝑡𝑒𝑠 + 𝑠𝑡𝑜𝑟𝑒𝑑 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 Mass Balance ෍ 𝑚𝑅 = ෍ 𝑚𝑃 + ෍ 𝑚𝑊 + ෍ 𝑚𝑆 ෍ 𝑚𝑅 = 𝑚𝑅1 + 𝑚𝑅2 + 𝑚𝑅3 + ⋯. = 𝑇𝑜𝑡𝑎𝑙 𝑅𝑎𝑤 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 ෍ 𝑚𝑃 = 𝑚𝑃1 + 𝑚𝑃2 + 𝑚𝑃3 + ⋯. = 𝑇𝑜𝑡𝑎𝑙 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 ෍ 𝑚𝑊 = 𝑚𝑊1 + 𝑚𝑊2 + 𝑚𝑊3 + ⋯. = 𝑇𝑜𝑡𝑎𝑙 𝑊𝑎𝑠𝑡𝑒 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 ෍ 𝑚𝑆 = 𝑚𝑆1 + 𝑚𝑆2 + 𝑚𝑆3 + ⋯.. = 𝑇𝑜𝑡𝑎𝑙 𝑆𝑡𝑜𝑟𝑒𝑑 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 Mass Balance If there is unknown losses, it still need to be identified and the material balance will be as follow: 𝑅𝑎𝑤 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 = 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 + 𝑤𝑎𝑠𝑡𝑒𝑠 + 𝑠𝑡𝑜𝑟𝑒𝑑 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 + 𝑙𝑜𝑠𝑠𝑒𝑠𝑠 ෍ 𝑚𝑅 = ෍ 𝑚𝑃 + ෍ 𝑚𝑊 + ෍ 𝑚𝑆 + ෍ 𝑚𝐿 Energy Balance As mass is conserve, so is the energy is conserved too in food processing operations. The energy coming into a unit operation can be balance with the energy coming out and energy stored. This can be written with the equation below: 𝐸𝑛𝑒𝑟𝑔𝑦 𝑖𝑛 = 𝐸𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡 + 𝐸𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑 Energy Balance ෍ 𝐸𝑅 = ෍ 𝐸𝑃 + ෍ 𝐸𝑊 + ෍ 𝐸𝐿 + ෍ 𝐸𝑆 ෍ 𝐸𝑅 = 𝐸𝑅1 + 𝐸𝑅2 + 𝐸𝑅3 + ⋯. = 𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝐸𝑛𝑡𝑒𝑟𝑖𝑛𝑔 ෍ 𝐸𝑃 = 𝐸𝑃1 + 𝐸𝑃2 + 𝐸𝑃3 + ⋯. = 𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝐿𝑒𝑎𝑣𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 ෍ 𝐸𝑊 = 𝐸𝑊1 + 𝐸𝑊2 + 𝐸𝑊3 + ⋯. = 𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝐿𝑒𝑎𝑣𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝑊𝑎𝑠𝑡𝑒 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 ෍ 𝐸𝐿 = 𝐸𝐿1 + 𝐸𝐿2 + 𝐸𝐿3 + ⋯. = 𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝐿𝑜𝑠𝑡 𝑡𝑜 𝑆𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 ෍ 𝐸𝑆 = 𝐸𝑆1 + 𝐸𝑆2 + 𝐸𝑆3 + ⋯. = 𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑆𝑡𝑜𝑟𝑒𝑑 MASS and ENERGY BALANCE (FOOD PROCESSING) - material balance calculations in food processing are based on a key food component, such as total solids, water, sugars, or oil. - energy balances in food processing involve mainly heat energy. Mechanical and electrical energy are of secondary importance in most food processing operations. TWO TYPES OF MATERIAL BALANCES NORMALLY USED IN ENGINEERING 1. The overall material balance of the total material in the system (Total mass in) − (total mass out) = (Total mass accumulated) 2. The component balance of a characteristic food component, such as total soluble solids (TSS), water (moisture), fat, oils, salt, protein and sugars. (Total component in) − (total component out) = (Total component accumulated) *For continuous operations, the accumulated materials (total and component) are zero. ENERGY BALANCES NORMALLY USED IN ENGINEERING ‒ components (i) participate in sensible heating or cooling by (ΔT) degrees (°C or K) ‒ components ( j) are involved in evaporation (condensation) or freezing (fusion) ‒ m (kg) is the mass of each component ‒ ΔH (kJ/kg) is the enthalpy of phase change (evaporation or freezing) ‒ Cp is the specific heat of water normally taken as equal to 4.18 kJ/kg ‒ at atmospheric pressure (P = 1.013 bar and T = 100°C), ΔH = 2.26 MJ/kg ‒ heat of freezing or fusion of water is ΔHf = 0.333 MJ/kg Sample problem Illustrative Problems 1. How much dry sugar must be added in 100 kg of aqueous sugar solution in order to increase its concentration from 20% to 50%? Assumption: Dry sugar is composed of 100 % sugar Process diagram Overall mass balance: 100 + S2 = S3 (eq 1) Soluble solids mass balance: 0.2 (100kg) + S2 = 0.5 (S3) (eq. 2) Solving equation 1 and 2 simultaneously, (0.2*100kg) + S2 = (0.5)(100kg + S2) 20 + S2 = 50kg + 0.5S2 𝑆2 = 30/0.5 = 60 kg ∴ 𝑆3 = 100𝑘𝑔 + 𝑆2 = 100𝑘𝑔 + 60𝑘𝑔 = 𝟏𝟔𝟎 𝒌𝒈 Conclusion: 60 kg of dry sugar per 100 kg of feed must be added to increase its concentration from 20% to 50 %. 2. Tomato juice flowing through a pipe at the rate of 100 kg/min is salted by adding saturated salt solution (26% salt) into the pipeline at a constant rate. At what rate would the saturated salt solution be added to have 2% salt in the product? Total mass balance: 100 + S = P Salt balance : 0.26S= 0.02P 0.26𝑆 From salt balance equation : 𝑃 = 0.02 Therefore : S = P − 100 0.26𝑆 𝑆= − 100 0.02 𝑆 =8.33 kg/ min 3. Fresh orange juice with 12% soluble solids content is concentrated to 60% in a multiple effect evaporator. To improve the quality of the final product the concentrated juice is mixed with an amount of fresh juice (cut back) so that the concentration of the mixture is 42%. Calculate how much water per hour must be evaporated in the evaporator, how much fresh juice per hour must be added back and how much final product will be produced if the inlet feed flow rate is 10, 000 kg/h fresh juice. Assume steady state.   Assumption: steady state  Overall mass balance in Envelop I: 10,000kg/h = W + X (eq 1)  Soluble solids mass balance in Envelop I: 0.12 * 10,000kg/h = 0.60X (eq. 2)  Overall mass balance in Envelop II: X+F=Y (eq 3) Soluble solids mass balance in Envelop II: 0.6 X + 0.12F =0.42Y (eq. 4)  From equation 2 find X, then substituting X in equation 1 and find W. Solve equation simultaneously. From eq 2 0.12(10,000kg/h) =0.6X X= 2000 kg/h Soluble solids mass balance in Envelop II continuation: From eq 1, substituting the value of X W= 10,000kg/h -2,000kg/h W= 8,000kg/h From equation 3, substituting the value of X 2000 kg/h + F =Y (eq. 5) From eq 4 and 5, substituting value of X and solving simultaneously 0.6X + 0.12F=0.42Y 0.6 (2000 kg/h) + 0.12F = 0.42 (2000kg/h + F) 1200 kg/h + 0.12F = 840kg/h + 0.42 F 1200−840 kg/h F= (0.42−0.12) F= 1200 kg/h Substituting the value of F to eq 5 Y= 2000 kg/h + 1200 kg/h Y = 3200 kg/h  Therefore: 8000 kg/h of water will be evaporated, 1200 kg/h of fresh juice will be added back and 3200 kg/h of concentrated orange juice with 42% soluble solids will be produced. 4. What is the energy required to raise the temperature of 3 kg of water from 50°C to 80 °C? The specific heat of water is 4180 J/kg K and latent heat of vaporization of water is 2257.06 kJ/kg Energy required = mcpΔT = 3 kg (4180 J/kg K) (353K - 323K) = 376.2 kJ 5. 1000 kg/h milk is heated in a heat exchanger from 45°C to 72°C. Water is used as a heating medium. It enters the heat exchanger at 90°C and leaves at 75°C. Calculate the mass flow rate of the heating medium, if the heat losses to the environment are equal to 1 kW. The heat capacity of water is given equal to 4.2kJ/kg°C and that of milk is 3.9kJ/kg°C. Assumptions:  The terms of kinetic and potential energy in the energy balance equation are negligible.  A pump is not included in the system (Ws= 0)  The heat capacity of the liquid stream does not change significantly with temperature  The system is at steady state. Energy balance equation: Rate of energy input = mሶ win Hwin + mሶ min Hmin Rate of energy output = mሶ wout Hwout + mሶ mout Hout + q At steady state: mሶ win Hwin + mሶ min Hmin = mሶ wout Hwout + mሶ mout Hout + q Calculating the known terms:  Enthalpy of water stream is : 4.2kJ 378𝑘𝐽 Input: 𝐻𝑤𝑖𝑛 = 𝑐𝑝 𝑇 = − °C 90 °C = kg 𝑘𝑔 Output: 𝐻𝑤𝑜𝑢𝑡 = 𝑐𝑝 𝑇 = 4.2kJ/kg − °C 75 °C = 315𝑘𝐽/𝑘𝑔  Enthalpy of milk stream Input: 𝐻𝑚𝑖𝑛 = 𝑐𝑝 𝑇 = 3.9kJ/kg − °C (45°C ) = 175.5 𝑘𝐽/𝑘𝑔 Output: 𝐻𝑚𝑜𝑢𝑡 = 𝑐𝑝 𝑇 = 3.9kJ/kg − °C 72 °C = 280.8𝑘𝐽/𝑘𝑔 Substituting the values above and taking into consideration that 𝑚ሶ 𝑤𝑖𝑛 = 𝑚ሶ 𝑤𝑜𝑢𝑡 = 𝑚ሶ 𝑤 𝑎𝑛𝑑 𝑚ሶ 𝑚𝑖𝑛 = 𝑚ሶ 𝑚𝑜𝑢𝑡 Then, 𝑚ሶ 𝑤𝑖𝑛 𝐻𝑤𝑖𝑛 + 𝑚ሶ 𝑚𝑖𝑛 𝐻𝑚𝑖𝑛 = 𝑚ሶ 𝑤𝑜𝑢𝑡 𝐻𝑤𝑜𝑢𝑡 + 𝑚ሶ 𝑚𝑜𝑢𝑡 𝐻𝑜𝑢𝑡 + 𝑞 𝑚ሶ 𝑤 (378𝑘𝐽/𝑘𝑔) + 1000(175.5𝑘𝐽/𝑘𝑔) = 𝑚ሶ 𝑤 (315𝑘𝐽/𝑘𝑔) + (1000)(280.8𝑘𝐽/𝑘𝑔) + (1𝑘𝐽/𝑠)(3600𝑠/ℎ) 𝑚ሶ 𝑤 = 1728.6𝑘𝑔/ℎ  References Carpio, 2000. E.V. Engineering for Food Technologist. UPLB Publishing Center. Singh R. P and Heldman, D.R.. 2009. Introduction to Food Engineering Fourth Edition. Elsevier Inc. Retrieved at http://www.ucarecdn.com/fb7332e8-c35a-47b0-9805-051fa171f8fa/. Park, S.H, Lamsal, B. P and Balasubramaniam.2014. Principles of Food Processing. Ohio State University, USA. Retrieved at https://www.academia.edu/35933662/Principles_of_Food_Processing Yanniotis S.. 2008. Solving Problems in Food Engineering. Springer. Retrieved at https://link.springer.com/book/10.1007/978-0-387-73514-6 Earle, R.L. and M.D. Earle. 1966. Unit Operations in Food Processing. Retrieve at https://nzifst.org.nz/resources/unitoperations/matlenerg.htm Mass and Energy Balance. Retrieved athttps://www.cpp.edu/~tknguyen/che302/Notes/chap4-1.pdf Alshanableh, F. 2012. Lecture Notes. Material and Energy Balance. Retrieved at http://old.staff.neu.edu.tr/~filiz/file/FDE201/FDE%20201%20-%20Lecture%20Notes.pdf Thank you for listening Stay safe and healthy God bless