1. Last Week: Heat Exchangers Refrigeration This Week: More on Refrigeration Combustion Pasteurization Process Control Materials

2. Refrigeration Condenser Evaporator Compressor Q out Q in W in

3. Refrigeration Cond Comp Q out W in Fermenting Room Lagering Cellar Cooler Hop Storage Cooler Flash Tank Evaporator Secondary Refrigerant Storage Tank Wort Cooler Fermenting Vessels Green Beer Chiller Beer Chiller Pasteurizer Yeast Tanks Air Conditioning

4. Primary Refrigerants Ammonia, R-12, R-134a Saturation temp < Desired application temp 2 to 8  C Maturation tanks 0 to 1  C Beer Chillers -15 to -20  C CO 2 liquefaction Typically confined to small region of brewery Secondary Refrigerants Water with alcohol or salt solutions Methanol/glycol, potassium carbonate, NaCl Lower freezing temperature of water Non-toxic (heat exchange with product) Pumped long distances across brewery

5. Wort Boiling Importance Flavor development Trub formation Wort stabilization Wort concentration Time and temperature – color, flavor, sterilization, etc. Turbulence – trub formation and volatile removal Rolling boil required. Temperature above boiling (  C) Heat transfer coef. Interface Evaporation (forced convection) <2  C Film Boiling >25  C Bubbles (nucleate boiling) 2  C <  T < 25  C

6. Combustion Fuel + Oxidizer  Heat + Products Oxidizer: Air (79% N 2, 21% O 2 by Volume) Fuels: Typically hydrocarbons MethaneCH 4 EthaneC 2 H 6 Gases PropaneC 3 H 8 Natural Gas = 95% CH 4 ButaneC 4 H 10 C 6 – C 18 Liquids Gasoline (Average C 8 ) Fuel Oil No. 1 (Kerosene) Fuel Oil No. 2 (  Diesel) Fuel Oil No. 3-6 (Heating Oils)

7. Combustion To Balance Stoichiometric Combustion Reaction: 1. Balance Carbon (CO 2 in products) 2. Balance Hydrogen (H 2 O in products) 3. Balance Oxygen (O 2 in reactants) 4. Balance Nitrogen (N 2 in products) Example: (a) Determine the theoretical quantity of air required for combustion of natural gas. Give results in kg of air per kg of natural gas. Assume that natural gas is 100% CH 4. (b) Determine the mass of CO 2 emitted per kg of natural gas burned.

8. Sterile Filtration Alternative to pasteurization for microbiological stabilization Avoid heat treatment, flavor deterioration Occurs before packaging (could be contaminated after filtration, before package) Process Requirements Feedstock microbiological and non-mb loads (concentration and particle size) Filtrate concentration, product spoilage concentration allowed Product viscosity, density, flow characteristics

9. Microbiological Load Reduction – LRV Sterile Filters = 99.9999999999% LRV Filtration Mechanisms Direct Interception – Pore smaller than particle Charge Effects – Particles (-), so filter (+) Inertial Impactation – Particles want straight path, fluid curves (different densities required) Diffusional Impactation – Random motion (gas)

10. Key Features Effecting Filter Performance Pore geometry Membrane thickness Surface Charge Removal Ratings Nominal – “An arbitrary micron value assigned by the filter manufacturer, based upon removal of some percentage of a given size or larger.” Absolute – “The diameter of the largest hard spherical particle that will pass through the filter under a specified test condition.”

11. Factors effecting flow rate and life: Pressure Drop Surface Area  P increases as dirt blocks pores Increased surface area has great increase on dirt capacity

12. Surface area can be increased with pleats Filter sizes: Pre-filter: 1.5  m Sterile: 0.45  m Cleaning Backwash (high V) Hot Liquor Sodium Hydroxide Steam Sanitized (120  C, 20 min)

13. Pasteurization Inactivate all microorganisms Inactivate undesired enzymes (chem. changes) Five Key Factors for Effective Pasteurization Temperature Time Types of microorganisms present Concentration of microorganisms present Chemical composition of the product Pasteurization Level Decimal reduction time, D – Time required to inactivate 90% of microorganisms present Temperature dependence value, Z – Increase in temp. require to increase D value by 90%

14. Pasteurization Units Measure of effect of heat and time on microorganisms 1.0 PU corresponds to 1 minute at 60  C PU = t * 1.393 (T-60  C) (t in minutes) Rules of Thumb Increase T by 2  C, double PU’s for same time Increase T by 10  C, PU’s increase 10x 20 PU’s indicates that 1 in 10 Billion microorganisms survive Effect of PU’s on specific microorganisms needed

15. Pasteurization Microorganisms growing in beer Wild yeast strains Lactic acid bacteria N o – Homogeneous population of microbes N – Remaining number of microbes t – time in minutes D – Decimal reduction time at temperature T Time (min)Number of microbes per Liter 010,000 21,000 4100 81 100.1

16. Pasteurization Typically choose D value of most resistant organism 1.0 P.U. = “one minute of heating at 60  C” An average Z value of 6.94  C is used

17. Pasteurization For the data given below, calculate the total number of pasteurization units (PU). Assume a Z value of 6.94  C. What type of pasteurizer is this? MinuteMean Temp (  C) PU’s 2149.7 2253.0 2355.9 2458.3 2560.2 2661.5 2762.25 2862.65 MinuteMean Temp (  C) PU’s 29-3462.8 3562.6 3661.2 3758.6 3856 3953.7 4051.75 4150 Total

18. Flash Pasteurization Time (min) 0.1 1 10 100 50 60 70 Temperature (  C) Over Pasteurization Under Pasteurization Minimum Safe Pasteurization 5.6 min

19. Flash Pasteurization Beer in = 0  C Pasteurizer 60-70  C 30 sec - 2 min 90-96% regeneration

20. Flash Pasteurization Pressure (Bar) Temperature (  C) Time (sec) Pressure in Pasteurizer CO 2 equilibrium pressure Temperature in Pasteurizer

21. Flash Pasteurization Typical Conditions: Beer inlet:3  C Outlet from regenerative heating:66  C Holding tube:70  C Outlet from regenerative cooling:8  C Outlet from cooling section:3  C Holding Time:30 sec Advantages Little space required Relatively inexpensive equipment and operation Short time at “intermediate” temperatures where chemical changes occur without pasteurization

22. Plate/Flash Pasteurization Typical plates: Stainless steel, 0.6 mm thickness Can withstand 20 bar pressure

23. Plate Pasteurizer Design 95% Heat Recovery in regenerator Product enters Pasteurizer at  4  C Holding temperature  72  C Holding time  25 seconds Hot water typically used for heating, 2  C warmer than holding temperature Level of Regeneration

24. Plate Pasteurizer Control 0.15  C corresponds to 1 PU

25. Flow Control Options Fixed Flow Range of Pre-set Flows Fully Variable Flow Most Suitable Option Depends Upon Size of Outlet Buffer Tank Importance of No Recirculation of Product PU Variation Desired Product Quality Type of Filler Minimum Flow typically 1/3 of maximum Pressure drop 1/9 of max flow (must be adjusted downstream to avoid overpressure) Heat transfer coefficient decreases, residence time increases

26. Best Practice - Full flow to 1/3 of full in 15 min while maintaining PU’s within  2.0 Control Loops Holding Cell Temperature Critical for PU Control Must be varied with changes in flow Final Product Outlet Flow – Upstream and downstream influences Pressure – Varied with changes in flow Interrelationships of many variables requires use of sophisticated control (PLC)

27. Tunnel Pasteurization Simpler system than flash pasteurization Slow process (may take up to 40 minutes) Energy intensive process Beer near outside of can/bottle over pasteurized Mechanical failure, other stoppage could cause over pasteurization, effecting beer flavor

28. Tunnel Pasteurization Pasteurized after bottled or canned Bottles or cans move slowly down conveyer system Hot water sprays heat beer to pasteurization temperature Cool water sprays cool beer after pasteurization is complete Pressure builds in headspace - Volume of headspace - CO 2 concentration in beer Bottles could break (Typical 1 in 500) CO 2 could leak if bottles are not sealed well

29. Typical temperature regime

30. Tunnel Pasteurization Pressure (Bar) Temperature (  C) Time (min) Spray water temperature Product Temperature

31. Factors Effecting Tunnel Pasteurization Materials of Construction Structure and weight – lighter stronger matl Corrosion – chemical attack metal, cracking Transport System – typically conveyor Spray System – Votex or spray pan Temperature Heating PU Control

33. Plate/Flash vs. Tunnel Pasteurization Plate uses significantly less floor space 15% reduction in operating cost Reduced capitol costs Beer tastes fresher (approx 92% less TIU) Cleaning and contamination downstream

34. Why is Process Control Needed? Safety Quality Specifications, Consistency Environmental Regulation, Environmental Impact Optimum Operation of Equipment Cost Effectiveness Aims of Control System Suppress Influence of External Disturbances Ensure Stability of a Process Example: External Disturbance on Shower Flow rate of hot water increases? Temperature of hot water decreases? Flow rate of hot water decreases?

35. Basic Control Elements Sensor – Receives Stimulus, Outputs Signal Controller – Receives Signal, Compares to Desired Value, Sends Control Signal Actuator – Receives Control Signal, Makes Corrective Action on Process Process – “The Organized Method of Converting Inputs to Outputs Functions of Control System Measure Compare to Desired Value Compute Error Corrective Action

36. Definitions Controlled Variable Setpoint Measured Variable Manipulated Variable Example Disturbance? Variables Controlled? Measured? Manipulated?

38. More Accurate More Complicated

39. On/Off Control Valve Open or Closed, Heater On or Off Inexpensive and Simple Oscillatory, Wear on Switching Device

40. Sequence Control Series of Events (Washing Machine) CIP Sequence, Fermentation Temperature, Keg Washing and Filling Achieved with PLC, Pegged Drum (Mechanical) Closed-Loop Control

41. Open-Loop Control Controlled Variable Measured Prior to Intervention by Manipulated Variable

42. Definitions Overshoot – Ratio of maximum amount by which response exceeds steady state to final steady state value Rise Time – Time required for response to reach final value for first time Response Time – Time it takes for response to settle at its new steady state value

43. Control Example

44. Proportional Control

45. Proportional + Integral Control

46. Proportional + Integral + Derivative Control

47. Feedback vs. Feedforward Control

48. Carbon and Low Alloy Steels Carbon Steel – Iron alloys with 0.05 to 1% C Low Carbon Steel – aka mild steel Low Alloy Steels – alloying elements with <2% Advantages Inexpensive and readily available Easily worked and welded Good tensile strength and ductility Disadvantages Corrosion Protective coatings often required

49. Copper Pure copper traditionally used Brass – alloyed with zink Bronze – alloyed with tin Advantages Soft and easily worked Readily available for small pipes/tubes Resists corrosion well Resistant to caustic and organic acids/salts Disadvantages Strong acids and oxidizing acids attack Cost

50. Stainless Steel Considered stainless if chromium > 11% Typical values 11-30% chromium Cr 2 O 3 oxidation layer gives ss it’s passivity

53. General Corrosion Covers entire surface “Best” kind of corrosion to have Measurable and predictable (design for) Galvanic Corrosion Two metals in contact in same electrolyte Less noble, less passive, more active metal corroded, other metal protected Erosion and Cavitation Abrasive particles and/or high velocity Cavitation corrosion (bubbles near pumps) Sensitisation – Inter-grainal corrosion (415-825  C) Pitting – Occurs below surface, chloride ion Localized weak points in passive surface

54. Example A maturation tank is maintained at 6  C using a secondary refrigerant (glycol/water solution). The cylindrical tank has a diameter of 3 m and a length of 6 m. The air temperature in the room is 18  C and the overall heat transfer coefficient between the maturation tank and surroundings is 12 W/m 2 K. Determine the rate of heat gain to the maturation tank. The glycol water solution is supplied from a storage tank at -5  C, it exits the maturation tank at 2  C and its specific heat is 3.5 kJ/kg.K. Determine the mass flow rate of secondary refrigerant required.

55. Example A simple vapor compression refrigeration system uses R-134a as the refrigerant. The evaporator pressure is 120 kPa and the condenser pressure is 1.2 MPa. The cooling load is 1000 kW. Determine the power consumed by the compressor and the COP of the system. Enthalpy of sat. vapor at 120 kPa = 239 kJ/kg Enthalpy of sat liquid at 1.2 MPa = 115 kJ/kg Enthalpy of vapor leaving comp. = 290 kJ/kg

56. Example A boiler, using propane fuel, produces 10 kg/s of steam at 1 MPa and 220 deg C from saturated liquid at the same pressure. Assuming that the boiler efficiency is 100%, determine: 1.The mass flow rate of fuel 2.The mass flow rate of air 3.The mass fraction of O2 in the products Enthalpy at 1 MPa and 220 C = 3000 kJ/kg Enthalpy of sat liquid at 1 MPa = 700 kJ/kg