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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 casada@ksu.edu 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% 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 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 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 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 (lbjam/lbfruit) of jam? Illustration – Jam Production mv = ? mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = 0.0025 lbpectin mJ = ? (67% solids) Illustration – Jam Production mv = ? mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar m = 0.0025 lb p pectin mJ = ? (67% solids) Total Mass Balance: Inflow = Outflow + Accumulation mf + ms = mv + mJ + 0.0 Illustration – Jam Production mv = ? mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = 0.0025 lbpectin mJ = ? (67% solids) Total Mass Balance: Inflow = Outflow + Accumulation mf + ms = mv + mJ + 0.0 Solids Balance: Inflow = Outflow + Accumulation mf·Csf + ms·Css = mJ·CsJ + 0.0 (1 lb)·(0.14lb/lb) + (1.22 lb)·(1.0lb/lb) = mJ·(0.67lb/lb) Illustration – Jam Production mv = ? mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = 0.0025 lbpectin mJ = ? (67% solids) Total Mass Balance: Inflow = Outflow + Accumulation mf + ms = mv + mJ + 0.0 Solids Balance: Inflow = Outflow + Accumulation mf·Csf + ms·Css = mJ·CsJ + 0.0 (1 lb)·(0.14lb/lb) + (1.22 lb)·(1.0lb/lb) = mJ·(0.67lb/lb) mJ = 2.03 lbJam/lbfruit mv = 0.19 lbwater/lbfruit Illustration – Jam Production mv = ? mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = 0.0025 lbpectin mJ = ? (67% solids) What if this was a continuous flow concentrator with a flow rate of 10,000 lbfruit/h? Principles • Mass Balance: Inflow = outflow + accumulation Ci Chemical m1 t concentrations: Ci ,1 m2 Ci,2 • Energy Balance: Energy in = energy out + accumulation m mass flow rate, kg/s T m1 T temperature, K t c p specific heat capacity, J/kg K T1 m2 T2 Principles • Mass Balance: Inflow = outflow + accumulation Chemical Ci concentrations: Ci ,1 m1 Ci , 2 m2 V t • Energy Balance: Energy in = energy out + accumulation T m1 c p T1 m2 c p T2 c p V t (sensible energy) total energy = m·h 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 1st 30 days. Apple Cooling qfrig 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 energy in = energy out + accumulation qfrig qin,1+ ... = qout,1+ ... + qa Illustration − Apple Cooling Try it... 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 1st 30 days. Apple Cooling qfrig qm qso qr qb qe qs qm qin Apple Cooling Sensible heat terms… qs = sensible heat gain from apples, W qr = respiration heat gain from apples, W qm = heat from lights, motors, people, etc., W qso = solar heat gain through windows, W qb = building heat gain through walls, etc., W qin = net heat gain from infiltration, W qe = sensible heat used to evaporate water, W 1 W = 3.413 Btu/h, 1 kW = 3413. Btu/h Apple Cooling Sensible heat equations… qs = mload· cpA· ΔT = mload· cpA· ΔT qr = mtot· Hresp qm = qm1 + qm2 + . . . qb = Σ(A/RT)· (Ti – To) 0 qin = (Qacpa/vsp)· (Ti – To) 0 qso = ... Apple Cooling definitions… mload = apple loading rate, kg/s (lb/h) Hresp = sp. rate of heat of respiration, J/kg·s (Btu/lb·h) mtot = total mass of apples, kg (lb) cpA = sp. heat capacity of apples, J/kg·°C (Btu/lb°F) cpa = specific heat capacity of air, J/kg·°C (Btu/lb°F) Qa = volume flow rate of infiltration air, m3/s (cfm) vsp = specific volume of air, m3/kgDA (ft3/lbDA) A = surface area of walls, etc., m2 (ft2) RT = total R-value of walls, etc., m2·°C/W (h·ft2·°F/Btu) Ti = air temperature inside, °C (°F) To = ambient air temperature, °C (°F) qm1, qm2 = individual mechanical heat loads, W (Btu/h) 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 A = 46 lb/bu; cpA = 0.9 Btu/lb°F What is the sensible heat load from the apples on day 3? Example 1 qfrig qm qso qr qb qe qs qm qin Example 1 qs = mload·cpA·ΔT mload = (2000 bu/day · 3 day)·(46 lb/bu) mload = 276,000 lb (on day 3) ΔT = (75°F – 34°F)/(5 day) = 8.2°F/day qs = (276,000 lb)·(0.9 Btu/lb°F)·(8.2°F/day) qs = 2,036,880 Btu/day = 7.1 ton (12,000 Btu/h = 1 ton refrig.) Example 1, revisited mload = 276,000 lb (on day 3) Ti,avg = (75 + 74.5 + 74)/3 = 74.5°F ΔT = (74.5°F – 34°F)/(5 day) = 8.1°F/day qs = (276,000 lb)·(0.9 Btu/lb°F)·(8.1°F/day) qs = 2,012,040 Btu/day = 7.0 ton (12,000 Btu/h = 1 ton refrig.) Example 2 Given the apple storage data of example 1, = 46 lb/bu; cpA = 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 qr = mtot· Hresp mtot = (2000 bu/day · 1 day)·(46 lb/bu) mtot = 92,000 lb qr = (92,000 lb)·(3.4 Btu/lb·day) qr = 312,800 Btu/day = 1.1 ton Additional Example Problems Sterilization Heat exchangers Drying Evaporation Postharvest cooling Sterilization First order thermal death rate (kinetics) of microbes assumed (exponential decay) N No ek D t D = decimal reduction time = time, at a given temperature, in which the number of microbes is reduced 90% (1 log cycle) N t ln N kD t o D Sterilization ( 250 FT ) Thermal death time: t Fo 10 z 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) Fo = 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 z Thermal Death Time Curve (C. botulinum) (Esty & Meyer, 1922) ( 250 FT ) t Fo 10 z t = thermal death time, min z = DT for 10x change in t, °F 2.7 Fo = t @ 250°F (std. temp.) Sterilization 10 Thermal Death Rate Plot (Stumbo, 1949, 1953; ...) z Decimal Reduction Time N No ek D t 1 D = decimal reduction time Dr = 0.2 N t N D ln 0.1 o 0.01 121.1 100 110 120 130 Temperature, °C Sterilization equations ( 250 T ) D To T log DT D250 10 z Do z No N o Fo FT Fo D250 log log N N Do DT (T 250 F ) (T 121.1C ) Fo t 10 z Fo t 10 z 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 If D = 0.25 min at 121°C, find D at 140°C. z = 10°C. Example 3 equation log D To T D121 = 0.25 min Do z z = 10°C substitute D140 121.1C 140C log 0.25 min 10C solve ... answer: D140 0.003 min Example 4 The Fo 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 Fo is given, assume standard processing conditions: T = 250°F (121.1°C); z = 18°F (10°C) Example 4 Thermal Death Time Curve (C. botulinum) (Esty & Meyer, 1922) (121.1 C T ) t Fo 10 z t = thermal death time, min z = DT for 10x change in t, °C 2.7 Fo = t @ 121.1°C (std. temp.) Example 4 (121.1 C T ) t Fo 10 z (121.1 C 100 C ) t100 (2.7 min) 10 10 C t100 348 min Heat Exchanger Basics q U Ae DTm U A DTlm DT DT (T T ) (T T ) (T T ) (T T ) DTlm max DT min Hi Co Ho Ci Hi Ci Ho Co ln DTmax min T T ln THi TCo Ho Ci counter THi T ln THo TCi Co parallel mH cH DTH mC cC DTC q 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 = cp = heat capacity of fluid, J/kg-K C = mc, J/s-K U = overall heat transfer coefficient, W/m2-K A = effective surface area, m2 DTm = proper mean temperature difference, K or °C q = heat transfer rate, W F(Y,Z) = correction factor, dimensionless Example 5 A liquid food (cp = 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 cp of water is 4.2 kJ/kg°C. Example 5 90°C Solution 60°C ? mf cf DTf = mw cw DTw 20°C (0.5 kg/s)·(4 kJ/kg°C)·(60 – 20°C) = (1 kg/s)·(4.2 kJ/kg°C)·(90 – THo) THo = 71°C Example 6 Find the heat exchanger area needed from example 5 if the overall heat transfer coefficient is 2000 W/m2·°C. Example 6 Find the heat exchanger area needed from example 5 if the overall heat transfer coefficient is 2000 W/m2·°C. Data: liquid food, cp = 4 kJ/kg°C water, cp = 4.2 kJ/kg°C Tfood,inlet = 20°C, Tfood,exit = 60°C Twater,inlet = 90°C mfood = 0.5 kg/s mwater = 1 kg/s Example 6 DTmin = 90°–60°C 90°C DTmax = 71°–20°C Solution 71°C 60°C q U Ae DTlm q mC cC DTC 20°C q = mf cf DTf = (0.5 kg/s)·(4 kJ/kg°C)·(60 – 20°C) = 80 kJ/s DTlm = (DTmax – DTmin)/ln(DTmax/DTmin) = 39.6°C Ae = (80 kJ/s)/{(2 kJ/s·m2·°C)·(39.5°C)} 2000 W/m2·°C = 2 kJ/s·m2·°C Ae = 1.01 m2 More about Heat Exchangers Effectiveness ratio (H, P, & Young, pp. 204-212) (Ta1 Ta 2 ) UA Cb Ecooling , NTU , R (Ta1 Tb ,in ) Cm in Ca One fluid at constant T: R DTlm correction factors q U A DTlm F ( Z , Y ) Time Out Reference Ideas Need Mark’s Suggestion 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. 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 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 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. Grain and food drying Twb Sensible heat Psychrometrics Latent heat of vaporization 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. 25 Effect of temperature on 20 moisture isotherms (corn data) Equilibrium Moisture Content, % 15 10 0°C 20°C 5 40°C 0 0 20 40 60 80 100 Relative Humidity, % Mass Transfer Between Phases cont. 25 Equilibrium Moisture Content, % 20 ASAE Standard D245.6 – . 15 Use previous revision (D245.4) for constants 10 or 0°C 20°C use psychrometric charts in Loewer et al. (1994) 40°C 5 0 0 20 40 60 80 100 Relative Humidity, % Mass Transfer Between Phases cont. Loewer, et al. (1994) Mass Transfer Between Phases cont. 25 Effect of temperature on 20 moisture isotherms (corn data) Equilibrium Moisture Content, % 15 10 0°C 20°C 5 40°C 0 0 20 40 60 80 100 Relative Humidity, % Deep Bed Drying Process rhe Twb rho TG To Use of Moisture Isotherms Air Temp. Grain Temp. Equilibrium Moisture Content, % Mo To TG Me rho rhe Relative Humidity, % 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) 1 1 M wb M db 1 M db 1 M wb W1 (1 M wb ,1 ) W2 (1 M wb ,2 ) Drying Deep Bed vs. Thin Layer Thin-layer process is not as complex. The common k t n Page eqn. is: MR e (falling rate drying period) Definitions: k, n = empirical constants (ANSI/ASAE S448.1) t = time M M equilibrium MR ; M dry basis moisture content M initial M equilibrium Deep bed effects when air flows through more than two to three layers of kernels. Grain Bulk Density for deep bed drying calculations kg/m3 lb/bu[1] Corn, shelled 721 56 Milo (sorghum) 721 56 Rice, rough 579 45 Soybean 772 60 Wheat 772 60 1Standard bushel. Source: ASAE D241.4 Basic Drying Process Mass Conservation Compare: moisture added to air to moisture removed from product Basic Drying Process Mass Conservation humidity ratio : a,out Da a ,out a ,in mg total mass of grain DWg change in grain MC ma Fan humidity ratio : a,in Basic Drying Process Mass Conservation Try it: Total moisture conservation equation: Basic Drying Process Mass Conservation Compare: moisture added to air to moisture removed from product Total moisture conservation: kga kgw kgw kgg s s kga kgg ma t Da mg DWg Basic Drying Process Mass Conservation – cont’d Calculate time: mg DWg t ma Da 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 Rate (Thin-layer) Constant Falling Rate Rate Time → Evaporative erh = 100% erh < 100% Cooling aw = 1.0 aw < 1.0 Assume falling rate period, unless… Falling rate requires erh or exit air data Drying Process cont. erh ASAE D245.6 Twb 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: Tdb = 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 Part II Use Loewer, et al. (1994 ) (or ASAE D245.6) RHexit = 55% rhexit emc=13% Twb Texit = 58°F Texit Example 7 13% Loewer, et al. (1994) Example 7b Part I Use Loewer, et al. (1994 ) (or ASAE D245.6) Texit = Tdb,e = TG emc=18% Twb Tdb,e Example 7b 18% 53.5 Loewer, et al. (1994) Example 7b Part I Use Loewer, et al. (1994 ) (or ASAE D245.6) Texit = Tdb,e = TG = 53.5°F emc=18% Twb Tdb,e Cooling Process Energy Conservation Compare: heat added to air to heat removed from product Sensible energy conservation: ma t ca DTa mg c g DTg Total energy conservation: DTg Tinitial TII ma t Dha mg c g DTg Cooling Process (and Drying) erh Twb Airflow in Packed Beds Drying, Cooling, etc. Design Values for Airflow Resistance in Grain 100 Soybeans (PF=1.3) Corn (PF=1.5) 10 2 Airflow, cfm/ft Sunflower (PF=1.5) Barley (PF=1.5) Milo (PF=1.3) 1 Wheat (PF=1.3) 0.1 0.001 0.01 0.1 1 10 Pressure Drop per Foot, inH2O/ft Source: ASABE D272.3, MWPS-29 Aeration Fan Selection Pressure drop (loose fill, “Shedd’s data”): DP = (inH2O/ft)LF x MS x (depth) + 0.5 Shedd’s curve multiplier (Ms = PF = 1.3 to 1.5) Pressure drop (design value chart): DP = (inH2O/ft)design x (depth) + 0.5 Aeration Fan Selection Pressure drop (loose fill, “Shedd’s data”): DP = (inH2O/ft)LF x MS x (depth) + 0.5 Pressure drop (design value chart): DP = (inH2O/ft)design x (depth) + 0.5 0.5 inH2O pressure drop in ducts - Standard design assumption (neglect for full perforated floor) 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 More Examples Evaporator (Concentrator) mV mF Juice mP mS Evaporator Solids mass balance: mF X F mP X P X Concentration, lb lb Total mass balance: mF mV mP Total energy balance: mF c pF TF mS (h fg ) S mV hgv mP c pP TP Example 8 Fruit juice concentrator, operating @ T =120°F Feed: TF = 80°F, XF = 10% Steam: 1000 lb/h, 25 psia Product: XP = 40% Assume: zero boiling point rise cp,solids = 0.35 Btu/lb·°F, cp,w = 1 Btu/lb·°F Example 8 mV TV = 120°F TF = 80°F XF = 0.1 lb/lb TP = 120°F mF Juice (120°F) XP = 0.4 lb/lb mP = ? mS Example 8 Steam tables: (hfg)S = 952.16 Btu/lb, at 25 psia (TS = 240°F) (hg)V = 1113.7 Btu/lb, at 120°F (PV = 1.69 psia) Calculate: cp,mix = 0.35· X + 1.0· (1 – X) Btu/lb°F cpF = 0.935 Btu/lb·°F cpP = 0.74 Btu/lb·°F Example 8 mV TV = 120°F hg = 1113.7 Btu/lb TF = 80°F XF = 0.1 lb/lb TP = 120°F Juice (120°F) XP = 0.4 lb/lb mF cpF = 0.935 Btu/lb°F mP = ? cpF = 0.74 Btu/lb°F mS hfg = 952.16 Btu/lb Example 8 Solids mass balance: mF X F mP X P Total mass balance: mF mV mP Total energy balance: mF c pF TF mS (h fg ) S mV (hg )V mP c pP TP Example 8 Solve for mP: m S ( h fg ) S mP c pP T P R X c pF TF ( R X 1) ( h g )V mP = 295 lb/h Aeration Fan Selection 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu) 3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5 DP = (inH2O/ft)design x (depth) + 0.5 4. Total airflow: cfm = (cfm/bu) x (total bushels) or: cfm = (cfm/ ft2) x (floor area) 5. Select fan to deliver flow & pressure (fan data) Aeration Fan Selection 1.4 1.2 Static Pressure, inH2O 1 0.8 System 0.6 Fan 0.4 0.2 0 0 500 1000 1500 2000 2500 3000 Airflow, cfm Aeration Fan Selection Example Wheat, Kansas, fall aeration 10,000 bu bin 16 ft eave height pressure aeration system Example 9 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu) 3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5 4. Total airflow: cfm = (cfm/bu) x (total bushels) or: cfm = (cfm/ ft2) x (floor area) 5. Select fan to deliver flow & pressure (fan data) Example 9 Recommended Airflow Rates for Dry Grain (Foster & Tuite, 1982): Recommended rate*, cfm/bu Storage Temperate Subtropic Type Climate Climate Horizontal 0.05 0.10 0.10 0.20 Vertical 0.03 0.05 0.05 0.10 *Higher rates increase control, flexibility, and cost. Example 9 Select lowest airflow (cfm/bu) for cooling rate Approximate Cooling Cycle Fan Time: Airflow rate (cfm/bu) Season 0.05 0.10 0.25 Summer 180 hr 90 hr 36 hr Fall 240 hr 120 hr 48 hr Winter 300 hr 150 hr 60 hr Spring 270 hr 135 hr 54 hr Example 9 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu) cfm/ft2 = (0.8) x (16 ft) x (0.1 cfm/bu) cfm/ft2 = 1.3 cfm/ft2 Example 9 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu) 3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5 4. Total airflow: cfm = (cfm/bu) x (total bushels) or: cfm = (cfm/ ft2) 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) Airflow Resistance in Grain (Loose-Fill) 100 Soybeans 10 Corn 2 Airflow, cfm/ft Barley Milo Wheat 1.3 1 0.1 0.0001 0.001 0.01 0.028 0.1 1 10 Pressure Drop per Foot, inH 2O/ft Pressure drop: DP = (inH2O/ft)design x (depth) + 0.5 Design Values for Airflow Resistance in Grain (w/o duct losses) 100 Soybeans 10 Corn 2 Airflow, cfm/ft Barley Milo Wheat 1.3 1 0.1 0.001 0.01 0.037 0.1 1 10 Pressure Drop per Foot, inH 2O/ft Example 9 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu) 3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5 DP = (0.028 inH2O/ft) x 1.3 x (16 ft) + 0.5 inH2O DP = 1.08 inH2O Example 9 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu) 3. Pressure drop: DP = (inH2O/ft)design x (depth) + 0.5 DP = (0.037 inH2O/ft) x (16 ft) + 0.5 inH2O DP = 1.09 inH2O Example 9 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu) 3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5 4. Total airflow: cfm = (cfm/bu) x (total bushels) cfm = (0.1 cfm/bu) x (10,000 bu) cfm = 1000 cfm Example 9 1. Select lowest airflow (cfm/bu) for cooling rate 2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu) 3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5 4. Total airflow: cfm = (cfm/bu) x (total bushels) or: cfm = (cfm/ ft2) x (floor area) 5. Select fan to deliver flow & pressure (fan data) Example 9 Axial Flow Fan Data (cfm): Static Pressure, in H2O Model 0" 0.5" 1" 1.5" 2.5" 3.5" 12" 3/4 hp 1900 1675 1290 815 325 0 12" 1 hp 2308 1963 1460 876 305 0 14" 1.5 hp 3132 2852 2526 2126 1040 0 Example 9 Selected Fan: 12" diameter, ¾ hp, axial flow Supplies: 1100 cfm @ 1.15 inH2O (a little extra 0.11 cfm/bu) Be sure of recommended fan operating range. 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