1 / 108

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%

orlando
Download Presentation

P.E. Review Session

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related