v-c-masstransferbetweenphases-2012_for_distribution

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P.E. Review Session: P.E. Review Session V–C. Mass Transfer between Phases by Mark Casada , Ph.D., P.E. (M.E.) USDA-ARS Center for Grain and Animal Health Research Manhattan, Kansas [email protected] Current NCEES Topics: Primary coverage: Exam % V. C. Mass transfer between phases 4% I. D. 1. Mass and energy balances ~2% Also: I. B. 1. Codes, regs ., and standards 1% Overlaps with: I. D. 2. Applied psychrometric processes ~2% II. A. Environment (Facility Engr.) 3-4% Current NCEES Topics Specific Topics/Unit Operations: Specific Topics/Unit Operations Heat & mass balance fundamentals Evaporation (jam production) Postharvest cooling (apple storage) Sterilization (food processing) Heat exchangers (food cooling) Drying (grain) Evaporation (juice) Postharvest cooling (grain) Mass Transfer between Phases: Mass Transfer between Phases A subcategory of: Unit Operations Common operations that constitute a process, e.g.: pumping, cooling, dehydration (drying), distillation, evaporation, extraction, filtration, heating, size reduction, and separation. How do you decide what unit operations apply to a particular problem? Experience is required (practice). Carefully read (and reread) the problem statement. Principles: Principles Mass Balance Inflow = outflow + accumulation Energy Balance Energy in = energy out + accumulation Specific equations Fluid mechanics, pumping, fans, heat transfer, drying, separation, etc. Illustration – Jam Production: Illustration – Jam Production Jam is being manufactured from crushed fruit with 14% soluble solids. Sugar is added at a ratio of 55:45 Pectin is added at the rate of 4 oz/100 lb sugar The mixture is evaporated to 67% soluble solids What is the yield (lb jam /lb fruit ) of jam? Illustration – Jam Production: Illustration – Jam Production m J = ? (67% solids) m f = 1 lb fruit (14% solids) m s = 1.22 lb sugar m p = 0.0025 lb pectin m v = ? Illustration – Jam Production: Illustration – Jam Production m J = ? (67% solids) m f = 1 lb fruit (14% solids) m s = 1.22 lb sugar m p = 0.0025 lb pectin m v = ? Total Mass Balance: Inflow = Outflow + Accumulation m f + m s = m v + m J + 0.0 Illustration – Jam Production: Illustration – Jam Production m J = ? (67% solids) m f = 1 lb fruit (14% solids) m s = 1.22 lb sugar m p = 0.0025 lb pectin m v = ? Total Mass Balance: Inflow = Outflow + Accumulation m f + m s = m v + m J + 0.0 Solids Balance: Inflow = Outflow + Accumulation m f ·C sf + m s · C ss = m J · C sJ + 0.0 (1 lb) · (0.14 lb / lb ) + (1.22 lb) · (1.0 lb / lb ) = m J · (0.67 lb / lb ) Illustration – Jam Production: Illustration – Jam Production m J = ? (67% solids) m f = 1 lb fruit (14% solids) m s = 1.22 lb sugar m p = 0.0025 lb pectin m v = ? Total Mass Balance: Inflow = Outflow + Accumulation m f + m s = m v + m J + 0.0 Solids Balance: Inflow = Outflow + Accumulation m f ·C sf + m s · C ss = m J · C sJ + 0.0 (1 lb) · (0.14 lb / lb ) + (1.22 lb) · (1.0 lb / lb ) = m J · (0.67 lb / lb ) m J = 2.03 lb Jam /lb fruit m v = 0.19 lb water /lb fruit Illustration – Jam Production: Illustration – Jam Production m J = ? (67% solids) m f = 1 lb fruit (14% solids) m s = 1.22 lb sugar m p = 0.0025 lb pectin m v = ? What if this was a continuous flow concentrator with a flow rate of 10,000 lb fruit /h? Principles: Principles Mass Balance: Inflow = outflow + accumulation Chemical concentrations: Energy Balance: Energy in = energy out + accumulation Principles: Principles Mass Balance: Inflow = outflow + accumulation Chemical concentrations: Energy Balance: Energy in = energy out + accumulation (sensible energy) total energy = m·h Illustration − Apple Cooling: Illustration − Apple Cooling An apple orchard produces 30,000 bu of apples a year, and will store ⅔ of the crop in refrigerated storage at 31°F. Cool to 34°F in 5 d; 31°F by 10 d. Loading rate: 2000 bu /day Ambient design temp: 75°F (loading) decline to 65°F in 20 d … Estimate the refrigeration requirements for the 1 st 30 days. Apple Cooling: Apple Cooling q frig Principles: Principles Mass Balance Inflow = outflow + accumulation Energy Balance Energy in = energy out + accumulation Specific equations Fluid mechanics, pumping, fans, heat transfer, drying, separation, etc. Illustration − Apple Cooling: Illustration − Apple Cooling q frig energy in = energy out + accumulation q in,1 + ... = q out,1 + ... + q a Illustration − Apple Cooling: Illustration − Apple Cooling Try it... Illustration − Apple Cooling: Illustration − Apple Cooling Try it... An apple orchard produces 30,000 bu of apples a year, and will store ⅔ of the crop in refrigerated storage at 31°F. Cool to 34°F in 5 d; 31°F by 10 d. Loading rate: 2000 bu/day Ambient design temp: 75°F (loading) decline to 65°F in 20 d … Estimate the refrigeration requirements for the 1 st 30 days. Apple Cooling: Apple Cooling q r q m q m q b q s q e q so q frig q in Apple Cooling: Apple Cooling Sensible heat terms… q s = sensible heat gain from apples, W q r = respiration heat gain from apples, W q m = heat from lights, motors, people, etc., W q so = solar heat gain through windows, W q b = building heat gain through walls, etc., W q in = net heat gain from infiltration, W q e = sensible heat used to evaporate water, W 1 W = 3.413 Btu/h , 1 kW = 3413. Btu/h Apple Cooling: Apple Cooling Sensible heat equations… q s = m load · c pA · Δ T = m load · c pA · Δ T q r = m tot · H resp q m = q m1 + q m2 + . . . q b = Σ ( A / R T )· ( T i – T o ) q in = ( Q a c pa / v sp )· ( T i – T o ) q so = ... 0 0 Apple Cooling: Apple Cooling definitions… m load = apple loading rate, kg/s (lb/h) H resp = sp. rate of heat of respiration, J/ kg ·s (Btu/ lb·h ) m tot = total mass of apples, kg (lb) c pA = sp. heat capacity of apples, J/kg·°C (Btu/ lb°F ) c pa = specific heat capacity of air, J/kg·°C (Btu/ lb°F ) Q a = volume flow rate of infiltration air, m 3 /s ( cfm ) v sp = specific volume of air, m 3 / kg DA (ft 3 / lb DA ) A = surface area of walls, etc., m 2 (ft 2 ) R T = total R-value of walls, etc., m 2 ·° C/W (h · ft 2 ·°F/Btu) T i = air temperature inside, °C ( °F ) T o = ambient air temperature, °C ( °F ) q m1 , q m2 = individual mechanical heat loads, W (Btu/h) Example 1: Example 1 An apple orchard produces 30,000 bu of apples a year, and will store ⅔ of the crop in refrigerated storage at 31°F. Cool to 34°F in 5 day; 31°F by 10 day. Loading rate: 2000 bu /day Ambient design temp: 75°F (at loading) declines to 65°F in 20 days r A = 46 lb/ bu ; c pA = 0.9 Btu/ lb°F What is the sensible heat load from the apples on day 3? Example 1: Example 1 q r q m q m q b q s q e q so q frig q in Example 1: Example 1 q s = m load ·c pA · Δ T m load = (2000 bu /day · 3 day)·(46 lb/ bu ) m load = 276,000 lb (on day 3) Δ T = (75°F – 34°F)/(5 day) = 8.2°F/day q s = (276,000 lb)·(0.9 Btu/ lb°F )·(8.2°F/day) q s = 2,036,880 Btu/day = 7.1 ton (12,000 Btu/h = 1 ton refrig .) Example 1, revisited: Example 1, revisited m load = 276,000 lb (on day 3) T i,avg = (75 + 74.5 + 74)/3 = 74.5°F Δ T = (74.5°F – 34°F)/(5 day) = 8.1°F/day q s = (276,000 lb)·(0.9 Btu/ lb°F )·(8.1°F/day) q s = 2,012,040 Btu/day = 7.0 ton (12,000 Btu/h = 1 ton refrig .) Example 2: Example 2 Given the apple storage data of example 1, r = 46 lb/ bu ; c pA = 0.9 Btu/ lb°F ; H = 3.4 Btu/ lb·day What is the respiration heat load (sensible) from the apples on day 1? Example 2: Example 2 q r = m tot · H resp m tot = (2000 bu/day · 1 day)·(46 lb/bu) m tot = 92,000 lb q r = (92,000 lb)·(3.4 Btu/lb·day) q r = 312,800 Btu/day = 1.1 ton Additional Example Problems: Additional Example Problems Sterilization Heat exchangers Drying Evaporation Postharvest cooling Sterilization: First order thermal death rate (kinetics) of microbes assumed (exponential decay) D = decimal reduction time = time, at a given temperature, in which the number of microbes is reduced 90% (1 log cycle) Sterilization Sterilization: Sterilization Thermal death time: The z value is the temperature increase that will result in a tenfold increase in death rate The typical z value is 10°C (18°F) ( C. botulinum ) F o = time in minutes at 250°F that will produce the same degree of sterilization as the given process at temperature T Standard process temp = 250°F (121.1°C) Thermal death time: given as a multiple of D Pasteurization: 4 − 6D Milk: 30 min at 62.8°C (“holder” method; old batch method) 15 sec at 71.7°C (HTST − high temp./short time) Sterilization: 12D “Overkill”: 18D (baby food) Sterilization: Sterilization Thermal Death Time Curve ( C. botulinum ) (Esty & Meyer, 1922) t = thermal death time, min z = D T for 10x change in t , °F F o = t @ 250°F (std. temp.) z 2.7 Sterilization: Sterilization Thermal Death Rate Plot (Stumbo, 1949, 1953; ...) D = decimal reduction time 121.1 D r = 0.2 z Sterilization equations: Sterilization equations Sterilization: Sterilization Popular problems would be: Find a new D given change in temperature Given one time-temperature sterilization process, find the new time given another temperature, or the new temperature given another time Example 3: Example 3 If D = 0.25 min at 121°C, find D at 140°C. z = 10°C. Example 3: Example 3 equation D 121 = 0.25 min z = 10°C substitute solve ... answer: Example 4: Example 4 The F o for a process is 2.7 minutes. What would be the processing time if the processing temperature was changed to 100 ° C? NOTE: when only F o is given, assume standard processing conditions: T = 250°F (121.1°C); z = 18°F (10°C) Example 4: Example 4 Thermal Death Time Curve ( C. botulinum ) (Esty & Meyer, 1922) t = thermal death time, min z = D T for 10x change in t , °C F o = t @ 121.1°C (std. temp.) 2.7 Example 4: Example 4 Heat Exchanger Basics: Heat Exchanger Basics Heat Exchangers: Heat Exchangers subscripts : H – hot fluid i – side where the fluid enters C – cold fluid o – side where the fluid exits variables : m = mass flow rate of fluid, kg/s c = c p = heat capacity of fluid, J/kg-K C = m  c , J/s-K U = overall heat transfer coefficient, W/m 2 -K A = effective surface area, m 2 D T m = proper mean temperature difference, K or °C q = heat transfer rate, W F ( Y,Z ) = correction factor, dimensionless Example 5: Example 5 A liquid food (c p = 4 kJ/ kg°C ) flows in the inner pipe of a double-pipe heat exchanger. The food enters the heat exchanger at 20°C and exits at 60°C. The flow rate of the liquid food is 0.5 kg/s. In the annular section, hot water at 90°C enters the heat exchanger in counter-flow at a flow rate of 1 kg/s. Assuming steady-state conditions, calculate the exit temperature of the water. The average c p of water is 4.2 kJ/ kg°C . Example 5: Example 5 Solution 90°C 60°C ? 20°C m f c f D T f = m w c w D T w (0.5 kg/s)·(4 kJ/kg°C)·(60 – 20°C) = (1 kg/s)·(4.2 kJ/kg°C)·(90 – T Ho ) T Ho = 71°C Example 6: Example 6 Find the heat exchanger area needed from example 5 if the overall heat transfer coefficient is 2000 W/m 2 ·° C . Example 6: Example 6 Find the heat exchanger area needed from example 5 if the overall heat transfer coefficient is 2000 W/m 2 ·° C . Data: liquid food, c p = 4 kJ/ kg°C water, c p = 4.2 kJ/ kg°C T food,inlet = 20°C, T food,exit = 60°C T water,inlet = 90°C m food = 0.5 kg/s m water = 1 kg/s Example 6: Example 6 Solution 90°C 60°C 71°C 20°C q = m f c f D T f = (0.5 kg/s)·(4 kJ/kg°C)·(60 – 20°C) = 80 kJ/s D T lm = ( D T max – D T min )/ln( D T max / D T min ) = 39.6°C A e = (80 kJ/s)/{(2 kJ/s·m 2 ·°C)·(39.5°C)} 2000 W/m 2 ·°C = 2 kJ/s·m 2 ·°C A e = 1.01 m 2 D T max = 71°–20°C D T min = 90°–60°C More about Heat Exchangers: More about Heat Exchangers Effectiveness ratio (H, P, & Young, pp. 204-212) One fluid at constant T: R  D T lm correction factors PowerPoint Presentation: Time Out Reference Ideas: Reference Ideas Full handbook The one you use regularly ASHRAE Fundamentals . Processing text Henderson, Perry, & Young (1997), Principles of Processing Engineering Geankoplis (1993), Transport Processes & Unit Operations . Need Mark’s Suggestion Standards ASABE Standards , recent ed. Other text Albright (1991), Environmental Control... Lower et al. (1994), On-Farm Drying and... MWPS-29 (1999), Dry Grain Aeration Systems Design Handbook . Ames, IA: MWPS. Studying for & taking the exam: Studying for & taking the exam Practice the kind of problems you plan to work Know where to find the data See presentation I-C Economics and Statistics, on Preparing for the Exam Mass Transfer Between Phases: Mass Transfer Between Phases Psychrometrics A few equations Psychrometric charts (SI and English units, high, low and normal temperatures; charts in ASABE Standards ) Psychrometric Processes – Basic Components: Sensible heating and cooling Humidify or de-humidify Drying/evaporative cooling Mass Transfer Between Phases cont.: Mass Transfer Between Phases cont. Grain and food drying Sensible heat Latent heat of vaporization T wb “drying” Psychrometrics Moisture content: wet and dry basis, and equilibrium moisture content (ASAE Standard D245.6) Airflow resistance (ASAE Standard D272.3) Mass Transfer Between Phases cont.: Mass Transfer Between Phases cont. Mass Transfer Between Phases cont.: Mass Transfer Between Phases cont. ASAE Standard D245.6 – . Use previous revision (D245.4) for constants or use psychrometric charts in Loewer et al. (1994) Mass Transfer Between Phases cont.: Mass Transfer Between Phases cont. Loewer, et al. (1994) Mass Transfer Between Phases cont.: Mass Transfer Between Phases cont. Deep Bed Drying Process: Deep Bed Drying Process rh e T wb “drying” T G T o rh o Use of Moisture Isotherms: Use of Moisture Isotherms Drying Deep Bed: Drying Deep Bed Drying grain (e.g., shelled corn) with the drying air flowing through more than two to three layers of kernels. Dehydration of solid food materials ≈ multiple layers drying & interacting (single, thin-layer solution is a single equation) Drying Deep Bed vs. Thin Layer: T hin-layer process is not as complex. The common Page eqn. is: (falling rate drying period) Definitions: k, n = empirical constants (ANSI/ASAE S448.1) t = time Deep bed effects when air flows through more than two to three layers of kernels. Drying Deep Bed vs. Thin Layer n t k e MR × - = content moisture basis dry M M M M M MR m equilibriu initial m equilibriu = - - = ; Grain Bulk Density for deep bed drying calculations: Grain Bulk Density for deep bed drying calculations kg/m 3 lb/bu [1] Corn, shelled 721 56 Milo (sorghum) 721 56 Rice, rough 579 45 Soybean 772 60 Wheat 772 60 1 Standard bushel. Source: ASAE D241.4 Basic Drying Process Mass Conservation: Basic Drying Process Mass Conservation Compare: moisture added to air to moisture removed from product Basic Drying Process Mass Conservation: Basic Drying Process Mass Conservation Fan Basic Drying Process Mass Conservation: Basic Drying Process Mass Conservation Try it: Total moisture conservation equation: Basic Drying Process Mass Conservation: Basic Drying Process Mass Conservation Compare: moisture added to air to moisture removed from product Total moisture conservation: kg a s s kg w kg a kg w kg g kg g Basic Drying Process Mass Conservation – cont’d: Basic Drying Process Mass Conservation – cont’d Calculate time: Assumes constant outlet conditions (true initially) but outlet conditions often change as product dries… use “deep-bed” drying analysis for non-constant outlet conditions (Henderson, Perry, & Young sec. 10.6 for complete analysis) Drying Process time varying process: Drying Process time varying process Assume falling rate period, unless… Falling rate requires erh or exit air data Drying Rate Time → Constant Rate Falling Rate erh = 100% a w = 1.0 erh < 100% a w < 1.0 Evaporative Cooling (Thin-layer) Drying Process cont.: Drying Process cont. T wb “drying” erh ASAE D245.6 Example 7: Example 7 Hard wheat at 75°F is being dried from 18% to 12% w.b. in a batch grain drier. Drying will be stopped when the top layer reaches 13%. Ambient conditions: T db = 70°F, rh = 20% Determine the exit air temperature early in the drying period. Determine the exit air RH and temperature at the end of the drying period? Example 7: Example 7 Part II Use Loewer, et al. (1994 ) (or ASAE D245.6) RH exit = 55% T exit = 58°F T wb “drying” emc=13% rh exit T exit Example 7: Example 7 Loewer, et al. (1994) 13% Example 7b: Example 7b Part I Use Loewer , et al. (1994 ) (or ASAE D245.6) T exit = T db,e = T G T wb “drying” emc=18% T db,e Example 7b: Example 7b Loewer, et al. (1994) 18% 53.5 Example 7b: Example 7b Part I Use Loewer , et al. (1994 ) (or ASAE D245.6) T exit = T db,e = T G = 53.5°F T wb “drying” emc=18% T db,e Cooling Process Energy Conservation: Cooling Process Energy Conservation Compare: heat added to air to heat removed from product Sensible energy conservation: Total energy conservation: Cooling Process (and Drying): Cooling Process (and Drying) T wb “drying” erh Airflow in Packed Beds Drying, Cooling, etc.: Airflow in Packed Beds Drying, Cooling, etc. Source: ASABE D272.3, MWPS-29 Aeration Fan Selection: Aeration Fan Selection Pressure drop (loose fill, “ Shedd’s data”): D P = (inH 2 O/ft) LF x M S x (depth) + 0.5 Pressure drop (design value chart): D P = (inH 2 O/ft) design x (depth) + 0.5 Shedd’s curve multiplier (M s = PF = 1.3 to 1.5) Aeration Fan Selection: Aeration Fan Selection Pressure drop (loose fill, “ Shedd’s data”): D P = (inH 2 O/ft) LF x M S x (depth) + 0.5 Pressure drop (design value chart): D P = (inH 2 O/ft) design x (depth) + 0.5 0.5 inH 2 O pressure drop in ducts - Standard design assumption (neglect for full perforated floor) Standards, Codes, & Regulations: Standards, Codes, & Regulations Standards ASABE Already mentioned ASAE D245.6 and D272.3 ASAE D243.3 Thermal properties of grain and… ASAE S448 Thin-layer drying of grains and crops Several others Others not likely for unit operations PowerPoint Presentation: More Examples Evaporator (Concentrator): Evaporator (Concentrator) m S m F m P m V Juice Evaporator: Evaporator Solids mass balance: Total mass balance: Total energy balance: Example 8: Example 8 Fruit juice concentrator, operating @ T =120°F Feed: T F = 80°F, X F = 10% Steam: 1000 lb/h, 25 psia Product: X P = 40% Assume: zero boiling point rise c p,solids = 0.35 Btu/lb ·°F, c p,w = 1 Btu/lb ·°F Example 8: Example 8 m S m F m P = ? m V Juice ( 120°F ) T F = 80°F X F = 0.1 lb/lb T P = 120°F X P = 0.4 lb/lb T V = 120°F Example 8: Example 8 Steam tables: ( h fg ) S = 952.16 Btu/lb, at 25 psia ( T S = 240°F) ( h g ) V = 1113.7 Btu/lb, at 120°F ( P V = 1.69 psia) Calculate: c p,mix = 0.35· X + 1.0· (1 – X) Btu/lb°F c pF = 0.935 Btu/lb·°F c pP = 0.74 Btu/lb·°F Example 8: Example 8 m S m F m P = ? m V Juice ( 120°F ) T F = 80°F X F = 0.1 lb/lb T P = 120°F X P = 0.4 lb/lb T V = 120°F h g = 1113.7 Btu/lb h fg = 952.16 Btu/lb c pF = 0.935 Btu/lb°F c pF = 0.74 Btu/lb°F Example 8: Example 8 Solids mass balance: Total mass balance: Total energy balance: Example 8: Example 8 Solve for m P : m P = 295 lb/h Aeration Fan Selection: Aeration Fan Selection 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft 2 = (0.8) x (depth) x (cfm/bu) 3 . Pressure drop: D P = (inH 2 O/ft) LF x M S x (depth) + 0.5 D P = (inH 2 O/ft) design x (depth) + 0.5 4. Total airflow: cfm = (cfm/bu) x (total bushels) or: cfm = (cfm/ ft 2 ) x (floor area) 5. Select fan to deliver flow & pressure (fan data) Aeration Fan Selection: Aeration Fan Selection Aeration Fan Selection: Aeration Fan Selection Example Wheat, Kansas, fall aeration 10,000 bu bin 16 ft eave height pressure aeration system Example 9: Example 9 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft 2 = (0.8) x (depth) x (cfm/bu) 3 . Pressure drop: D P = (inH 2 O/ft) LF x M S x (depth) + 0.5 4. Total airflow: cfm = (cfm/bu) x (total bushels) or: cfm = (cfm/ ft 2 ) x (floor area) 5. Select fan to deliver flow & pressure (fan data) Example 9: Example 9 * Higher rates increase control, flexibility, and cost. Example 9 Select lowest airflow (cfm/bu) for cooling rate: Example 9 Select lowest airflow ( cfm / bu ) for cooling rate Example 9: Example 9 cfm/ft 2 = (0.8) x (16 ft) x (0.1 cfm/bu) cfm/ft 2 = 1.3 cfm/ft 2 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft 2 = (0.8) x (depth) x (cfm/bu) Example 9: Example 9 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft 2 = (0.8) x (depth) x (cfm/bu) 3 . Pressure drop: D P = (inH 2 O/ft) LF x M S x (depth) + 0.5 4. Total airflow: cfm = (cfm/bu) x (total bushels) or: cfm = (cfm/ ft 2 ) x (floor area) 5. Select fan to deliver flow & pressure (fan data) Pressure drop: DP = (inH2O/ft) x MS x (depth) + 0.5 (note: Ms = 1.3 for wheat): Pressure drop: D P = (inH 2 O/ft) x M S x (depth) + 0.5 (note: M s = 1.3 for wheat) 0.028 1.3 Pressure drop: DP = (inH2O/ft)design x (depth) + 0.5: Pressure drop: D P = (inH 2 O/ft) design x (depth) + 0.5 0.037 1.3 Example 9: Example 9 D P = (0.028 inH 2 O/ft) x 1.3 x (16 ft) + 0.5 inH 2 O D P = 1.08 inH 2 O 1. Select lowest airflow ( cfm / bu ) for cooling rate 2. Airflow: cfm /ft 2 = (0.8) x (depth) x ( cfm / bu ) 3 . Pressure drop: D P = (inH 2 O/ft) LF x M S x (depth) + 0.5 Example 9: Example 9 D P = (0.037 inH 2 O/ft) x (16 ft) + 0.5 inH 2 O D P = 1.09 inH 2 O 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft 2 = (0.8) x (depth) x (cfm/bu) 3 . Pressure drop: D P = (inH 2 O/ft) design x (depth) + 0.5 Example 9: Example 9 cfm = (0.1 cfm/bu) x (10,000 bu) cfm = 1000 cfm 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft 2 = (0.8) x (depth) x (cfm/bu) 3 . Pressure drop: D P = (inH 2 O/ft) LF x M S x (depth) + 0.5 4. Total airflow: cfm = (cfm/bu) x (total bushels) Example 9: Example 9 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft 2 = (0.8) x (depth) x (cfm/bu) 3 . Pressure drop: D P = (inH 2 O/ft) LF x M S x (depth) + 0.5 4. Total airflow: cfm = (cfm/bu) x (total bushels) or: cfm = (cfm/ ft 2 ) x (floor area) 5. Select fan to deliver flow & pressure (fan data) Example 9: Example 9 Axial Flow Fan Data (cfm): Example 9: Example 9 Selected Fan: 12" diameter, ¾ hp, axial flow Supplies: 1100 cfm @ 1.15 inH 2 O (a little extra  0.11 cfm / bu ) Be sure of recommended fan operating range. Final Thoughts: Final Thoughts Study enough to be confident in your strengths Get plenty of rest beforehand Calmly attack and solve enough problems to pass - emphasize your strengths - handle “data look up” problems early Plan to figure out some longer or “iffy” problems AFTER doing the ones you already know

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