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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 casada@ksu.edu. Current NCEES Topics. Primary coverage: Exam % V. C. Mass transfer between phases 4%

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P.E. Review Session

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  1. 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

  2. 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%

  3. 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)

  4. 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.

  5. Principles • Mass Balance Inflow = outflow + accumulation • Energy Balance • Energy in = energy out + accumulation • Specific equations • Fluid mechanics, pumping, fans, heat transfer,drying, separation, etc.

  6. 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?

  7. mv = ? mf = 1 lbfruit (14% solids) ms =1.22 lbsugar mp =0.0025 lbpectin mJ = ? (67% solids) Illustration – Jam Production

  8. mv = ? mf = 1 lbfruit (14% solids) ms =1.22 lbsugar mp =0.0025 lbpectin mJ = ? (67% solids) Illustration – Jam Production Total Mass Balance: Inflow = Outflow + Accumulation mf + ms = mv + mJ + 0.0

  9. mv = ? mf = 1 lbfruit (14% solids) ms =1.22 lbsugar mp =0.0025 lbpectin mJ = ? (67% solids) Illustration – Jam Production 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)

  10. mv = ? mf = 1 lbfruit (14% solids) ms =1.22 lbsugar mp =0.0025 lbpectin mJ = ? (67% solids) Illustration – Jam Production 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

  11. mv = ? mf = 1 lbfruit (14% solids) ms =1.22 lbsugar mp =0.0025 lbpectin mJ = ? (67% solids) Illustration – Jam Production What if this was a continuous flow concentrator with a flow rate of 10,000 lbfruit/h?

  12. Principles • Mass Balance: Inflow = outflow + accumulation Chemicalconcentrations: • Energy Balance: • Energy in = energy out + accumulation

  13. total energy = m·h Principles • Mass Balance: Inflow = outflow + accumulation Chemicalconcentrations: • Energy Balance: • Energy in = energy out + accumulation (sensible energy)

  14. 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.

  15. qfrig Apple Cooling

  16. Principles • Mass Balance Inflow = outflow + accumulation • Energy Balance • Energy in = energy out + accumulation • Specific equations • Fluid mechanics, pumping, fans, heat transfer,drying, separation, etc.

  17. qfrig Illustration − Apple Cooling energy in = energy out + accumulation qin,1+ ... = qout,1+ ... + qa

  18. Illustration − Apple Cooling Try it...

  19. 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.

  20. qfrig qm qb qr qso qe qs qm qin Apple Cooling

  21. 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

  22. 0 0 Apple Cooling • Sensible heat equations… qs = mload· cpA· ΔT = mload· cpA· ΔT qr = mtot· Hresp qm = qm1 + qm2 + . . . qb = Σ(A/RT)· (Ti – To) qin = (Qacpa/vsp)· (Ti – To) qso = ...

  23. 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) Apple Cooling

  24. 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 rA = 46 lb/bu; cpA = 0.9 Btu/lb°F What is the sensible heat load from the apples on day 3?

  25. qfrig qm qb qr qso qe qs qm qin Example 1

  26. 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.)

  27. 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.)

  28. Example 2 Given the apple storage data of example 1, r = 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?

  29. 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

  30. Additional Example Problems • Sterilization • Heat exchangers • Drying • Evaporation • Postharvest cooling

  31. 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)

  32. 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) • 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)

  33. z 2.7 Sterilization Thermal Death Time Curve (C. botulinum)(Esty & Meyer, 1922) t = thermal death time, min z = DT for 10x change in t, °F Fo = t @ 250°F (std. temp.)

  34. z Dr = 0.2 121.1 Sterilization Thermal Death Rate Plot (Stumbo, 1949, 1953; ...) D = decimal reduction time

  35. Sterilization equations

  36. 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

  37. Example 3 • If D = 0.25 min at 121°C, find D at 140°C.z = 10°C.

  38. Example 3 equation D121 = 0.25 min z = 10°C substitute solve ... answer:

  39. 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)

  40. 2.7 Example 4 Thermal Death Time Curve (C. botulinum)(Esty & Meyer, 1922) t = thermal death time, min z = DT for 10x change in t, °C Fo = t @ 121.1°C (std. temp.)

  41. Example 4

  42. Heat Exchanger Basics

  43. 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 = mc, 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

  44. 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.

  45. 90°C ? 60°C 20°C Example 5 • Solution mf cfDTf = mw cwDTw (0.5 kg/s)·(4 kJ/kg°C)·(60 – 20°C) = (1 kg/s)·(4.2 kJ/kg°C)·(90 – THo) THo = 71°C

  46. Example 6 • Find the heat exchanger area needed from example 5 if the overall heat transfer coefficient is 2000 W/m2·°C.

  47. 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

  48. 90°C 71°C 60°C 20°C Example 6 DTmin= 90°–60°C • Solution DTmax= 71°–20°C q = mf cfDTf = (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

  49. More about Heat Exchangers • Effectiveness ratio (H, P, & Young, pp. 204-212) • One fluid at constant T: R • DTlm correction factors

  50. Time Out

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