last week: heat exchangers refrigeration this week: more on refrigeration combustion
DESCRIPTION
Last Week: Heat Exchangers Refrigeration This Week: More on Refrigeration Combustion Pasteurization Process Control Materials. Refrigeration. Q out. Condenser. Compressor. W in. Evaporator. Q in. Refrigeration. Q out. Hop Storage Cooler. Cond. Air Conditioning. - PowerPoint PPT PresentationTRANSCRIPT
Last Week:Heat ExchangersRefrigeration
This Week:More on RefrigerationCombustionPasteurizationProcess ControlMaterials
Refrigeration
Condenser
Evaporator
Compressor
Qout
Qin
Win
Refrigeration
Cond
Comp
Qout
Win
Fermenting Room
Lagering Cellar Cooler
Hop Storage Cooler
Flash Tank
EvaporatorSecondary Refrigerant
Storage Tank
Wort Cooler
Fermenting Vessels
Green Beer Chiller
Beer Chiller
Pasteurizer
Yeast Tanks
Air Conditioning
Primary RefrigerantsAmmonia, R-12, R-134aSaturation temp < Desired application temp
2 to 8C Maturation tanks0 to 1C Beer Chillers-15 to -20C CO2 liquefaction
Typically confined to small region of brewery
Secondary RefrigerantsWater with alcohol or salt solutionsMethanol/glycol, potassium carbonate, NaClLower freezing temperature of waterNon-toxic (heat exchange with product)Pumped long distances across brewery
ExampleA maturation tank is maintained at 6C 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 18C and the overall heat transfer coefficient between the maturation tank and surroundings is 12 W/m2K. Determine the rate of heat gain to the maturation tank. The glycol water solution is supplied from a storage tank at -5C, it exits the maturation tank at 2C and its specific heat is 3.5 kJ/kg.K. Determine the mass flow rate of secondary refrigerant required.
Wort BoilingImportance
• Flavor development• Trub formation• Wort stabilization• Wort concentration
Time and temperature – color, flavor, sterilization, etc.Turbulence – trub formation and volatile removalRolling boil required.
Temperature above boiling (C)
Hea
t tra
nsfe
r coe
f.
Interface Evaporation
(forced convection) <2C
Film Boiling >25C
Bubbles(nucleate boiling)
2C < T < 25C
Wort BoilingIn wort boiling it is important to maintain a temperature
difference below the critical difference between the wort and heating element surface (25C) If the wort is boiling at 105C, calculate the maximum operational steam pressure you would recommend for an indirect steam heated wort boiler. The wall of the steam heating element is 1.0 mm thick and has a thermal conductivity of 15 W/m.K. The condensing steam’s heat transfer coefficient is 12,000 W/m2.K and the maximum heat flux is 160,000 W/m2.
0.35 MPa 139.0C0.40 MPa 143.5C0.45 MPa 148.0C0.50 MPa 152.0C
0.55 MPa 155.5C0.60 MPa 159.0C0.65 MPa 162.0C
CombustionFuel + Oxidizer Heat + ProductsOxidizer: Air (79% N2, 21% O2 by Volume)Fuels: Typically hydrocarbons
Methane CH4
Ethane C2H6 GasesPropane C3H8 Natural Gas = 95% CH4
Butane C4H10
C6 – C18 LiquidsGasoline (Average C8)Fuel Oil No. 1 (Kerosene)Fuel Oil No. 2 (Diesel)
Fuel Oil No. 3-6 (Heating Oils)
CombustionTo Balance Stoichiometric Combustion Reaction:
1. Balance Carbon (CO2 in products)2. Balance Hydrogen (H2O in products)3. Balance Oxygen (O2 in reactants)4. Balance Nitrogen (N2 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% CH4.
(b) Determine the mass of CO2 emitted per kg of natural gas burned.
CombustionActual combustion process Excess air
Complete combustion (reduce CO, UHC)Reduce flame temperature (reduce NOx)
Example: Determine the composition of CH4 combustion products with 25% excess air.
CombustionFlue gas analysis – Work backwards to find %
excess air.
Example: Determine the excess air used for CH4 combustion when the O2 concentration in the products is 5.5% volume. (Note, for ideal gas mixtures, volume fraction = mole fraction).
Calorific Value of Fuels (= Heating Value)Solids, Liquid: MJ/kgGases: MJ/m3 (at STP) or “Therms”LHV = H2O vapor in products, HHV = liquid
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
Microbiological Load Reduction – LRVSterile 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)
Outletat Organisms of No.Inletat Organisms of No.Reduction Titre
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.”
Factors effecting flow rate and life:
• Pressure Drop• Surface Area
P increases as dirtblocks pores
Increased surfacearea has greatincrease on dirtcapacity
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
(120C, 20 min)
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%
Pasteurization Units• Measure of effect of heat and time on
microorganisms• 1.0 PU corresponds to 1 minute at 60C• PU = t * 1.393(T-60C) (t in minutes)
Rules of Thumb• Increase T by 2C, double PU’s for same time• Increase T by 10C, PU’s increase 10x• 20 PU’s indicates that 1 in 10 Billion
microorganisms surviveEffect of PU’s on specific microorganisms needed
PasteurizationMicroorganisms growing in beer
• Wild yeast strains• Lactic acid bacteria
No – Homogeneous population of microbesN – Remaining number of microbest – time in minutesD – Decimal reduction time at temperature T
Time (min) Number of microbes per Liter0 10,0002 1,0004 1008 110 0.1
Dt
oNN
10
min 260 D
PasteurizationTypically choose D value of most resistant organism1.0 P.U. = “one minute of heating at 60C”
An average Z value of 6.94C is used
ZT
T DD60
6010
TLogDLogDTZ
60
60
tPU TTotal
60394.1
PasteurizationFor the data given below, calculate the total number of
pasteurization units (PU). Assume a Z value of 6.94C.
What type of pasteurizer is this?
Minute Mean Temp (C)
PU’s
21 49.722 53.023 55.924 58.325 60.226 61.527 62.2528 62.65
Minute Mean Temp (C)
PU’s
29-34 62.835 62.636 61.237 58.638 5639 53.740 51.7541 50
Total
Flash PasteurizationTi
me
(min
)0.
1
1
1
0
100
50 60 70Temperature (C)
Over Pasteurization
Under Pasteurization
Minimum Safe Pasteurization
5.6 min
Flash Pasteurization
Beer in = 0C
Pasteurizer60-70C30 sec - 2 min
90-96%regeneration
Flash PasteurizationP
ress
ure
(Bar
)
Tem
pera
ture
(C
)
Time (sec)
Pressure in Pasteurizer
CO2 equilibrium pressure
Temperature in Pasteurizer
Flash PasteurizationTypical Conditions:
Beer inlet: 3COutlet from regenerative heating: 66CHolding tube: 70COutlet from regenerative cooling: 8COutlet from cooling section: 3CHolding Time: 30 sec
AdvantagesLittle space requiredRelatively inexpensive equipment and operationShort time at “intermediate” temperatures where chemical changes occur without pasteurization
Plate/Flash PasteurizationTypical plates: Stainless steel, 0.6 mm thicknessCan withstand 20 bar pressure
Plate Pasteurizer Design• 95% Heat Recovery in regenerator• Product enters Pasteurizer at 4C• Holding temperature 72C• Holding time 25 seconds• Hot water typically used for heating, 2C
warmer than holding temperature
Level of Regeneration
Plate Pasteurizer Control• 0.15C corresponds to 1 PU
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
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)
Tunnel Pasteurization
Simpler system than flash pasteurizationSlow process (may take up to 40 minutes)Energy intensive processBeer near outside of can/bottle over pasteurizedMechanical failure, other stoppage could cause
over pasteurization, effecting beer flavor
Tunnel PasteurizationPasteurized after bottled or cannedBottles or cans move slowly down conveyer
systemHot water sprays heat beer to pasteurization
temperatureCool water sprays cool beer after pasteurization is
completePressure builds in headspace
- Volume of headspace- CO2 concentration in beer
Bottles could break (Typical 1 in 500)CO2 could leak if bottles are not sealed well
Typical temperature regime
Tunnel PasteurizationP
ress
ure
(Bar
)
Tem
pera
ture
(C
)
Time (min)
Spray water temperature
Product Temperature
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
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
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?
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
Definitions• Controlled Variable• Setpoint• Measured Variable• Manipulated Variable
Example
Disturbance?
VariablesControlled?Measured?Manipulated?
More AccurateMore Complicated
On/Off Control• Valve Open or Closed, Heater On or Off• Inexpensive and Simple• Oscillatory, Wear on Switching Device
Sequence Control• Series of Events (Washing Machine)• CIP Sequence, Fermentation Temperature, Keg
Washing and Filling• Achieved with PLC, Pegged Drum (Mechanical)
Closed-Loop Control
Open-Loop Control• Controlled Variable Measured Prior to
Intervention by Manipulated Variable
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
Control Example
Proportional Control
Proportional + Integral Control
Proportional + Integral + Derivative Control
Feedback vs. Feedforward Control
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
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
Stainless Steel• Considered stainless if chromium > 11%• Typical values 11-30% chromium• Cr2O3 oxidation layer gives ss it’s passivity
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 protectedErosion and Cavitation
• Abrasive particles and/or high velocity• Cavitation corrosion (bubbles near pumps)
Sensitisation – Inter-grainal corrosion (415-825C)Pitting – Occurs below surface, chloride ion
Localized weak points in passive surface