P.E. Review Session III. Process Engineering, Part 1 by

P.E. Review Session III. Process Engineering, Part 1 by
Mark Casada, Ph.D., P.E. (M.E.) USDA-ARS Center for Grain and Animal Health Research Manhattan, Kansas Process Engineering, Part 1 – there is more to the “Processing” area not covered here (not my area of expertise). Outline: will take a break mid-way... at that time, I’ll talk about a few of the general issues – like suggested references...

Current NCEES Topics Knowledge Areas: Questions
Approx. Exam Knowledge Areas: Questions I. Common System Applications 20 II. Natural Resources and Ecology 15 III. Process Engineering 15 IV. Facilities 15 V. Machines 15 Overview of current NCEES topics

Current NCEES Topics III. Process Engineering 15
Approx. Exam Knowledge Areas: Questions I. Common System Applications 20 II. Natural Resources and Ecology 15 III. Process Engineering 15 IV. Facilities 15 V. Machines 15 While this is primarily from "III. Process Engineering,“ there is overlap with "I. Common System Applications." and “I. Facilities.”

Current NCEES Topics Primary coverage (Process Engineering): Exam
I. B. Energy balances ~1% III. A. Bio’l & Chem kinetics ~1.5% III. D. Mass transfer between phases ~1.5% III. E. Phys & Chem props (D-value…) ~1.5% III. I. Applied psychrometric processes ~1.5% III. J. Mass balances ~1.5% Also: I. P. Codes, regulations, and standards ~1% III. E. Properties of biological materials ~1.5% Overlaps with (Facilities): IV. H, I. Ventilation requirements ~3 % You can see it covers parts of "III. Process Engineering" plus overlap with "I. Common System Applications." (Items in green font are just touched on.) Hopefully, this will help you can pick up a lot of points using these Engineering Principles -- plus knowing where to find the data for these problems.

General area: "Unit Operations"
Within process engineering Unit Operations are: 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; these examples). Carefully read (and reread) the problem statement. The process drawing (you make) is invaluable (Today’s examples should refresh you on which operation is which; so you can apply the right one.) If you’ve reviewed this area already, you should recognize these types of problems from one of the following classes: -- Biological/Biological Systems Engineering: Transport Processes -- Agricultural Engineering: Agricultural Processing -- Food Engineering: Food Processing It also overlaps with Fluids and Heat Transfer courses you may have had.

Specific Topics/Unit Operations
Heat & mass balance fundamentals Evaporation (jam production) Postharvest cooling (apple storage) Sterilization, D-values (food processing) Heat exchangers (food cooling) Drying (grain) Evaporation (juice) Postharvest cooling (grain)

Processing Textbooks Henderson, Perry, & Young (1997), Principles of Processing Engineering Geankoplis (2003), Transport Processes and Unit Operations. I’ll talk more about reference books later, but I’ll be referring to these two the most – so here they are. (Geankoplis, 4th ed., rd ed., 1993)

Principles Mass Balance Energy Balance Specific equations
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?

mJ = ? (67% solids) mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = lbpectin mv = ? Show all inflows... all outflows (no accumulation) Simplified by treating as a batch (1 lb of fruit) mf and mj are actually two components (solids and water) – typically shown as one total flow with specified %solids or moisture content

mJ = ? (67% solids) mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = lbpectin mv = ? Total Mass Balance: Inflow = Outflow + Accumulation mf + ms = mv + mJ + 0.0 Here’s the balance with two inflows and two outflows

mJ = ? (67% solids) mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = lbpectin mv = ? Total Mass Balance: Inflow = Outflow + Accumulation mf + ms = mv + mJ + 0.0 -> the arrows show where I got the terms… What’s missing? (pectin is neglected: < 1% of mf) Accumulation is zero Note: this equation alone is not sufficient – both mj and mv are unknowns

mJ = ? (67% solids) mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = lbpectin mv = ? 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) So, let’s look at a mass balance on the solids: two inflows, one outflow…

mJ = ? (67% solids) mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = lbpectin mv = ? 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) -> arrows show the source of terms again… Mass of Solids = (Masstotal) x (Concentration) (pectin neglected here too, of course)

mJ = ? (67% solids) mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = lbpectin mv = ? 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) Solids balance can be solved for mj, then find mv from total mass balance. mJ = 2.03 lbJam/lbfruit mv = 0.19 lbwater/lbfruit

mJ = ? (67% solids) mf = 1 lbfruit (14% solids) ms = 1.22 lbsugar mp = lbpectin mv = ? Follow-up question about treating as a batch instead of continuous flow: what would be different? all the masses, “m”s, would be mass flow rates... minor change (everything is 10,000 times larger and per hour – eqns look the same except m-dots). We’ll try a real continuous flow one later... (extra example). What if this was a continuous flow concentrator with a flow rate of 10,000 lbfruit/h?

Mass Balance: Energy Balance: 𝒎 𝟏 𝑪 𝒊,𝟏 𝒎 𝟐 𝑪 𝒊,𝟐 𝒎 𝟏 𝑻 𝟏 𝒎 𝟐 𝑻 𝟐
Inflow = outflow + accumulation Chemical concentrations (C): 𝒎 𝟏 𝑪 𝒊,𝟏 𝒎 𝟐 𝑪 𝒊,𝟐 Energy Balance: Energy in = energy out + accumulation This is the general form of the mass and energy balance equations. 𝑚 = mass flow rate, kg/s T = temperature, K cp = specific heat capacity, J/kgK 𝒎 𝟏 𝑻 𝟏 𝒎 𝟐 𝑻 𝟐

Principles Mass Balance: Energy Balance:
Inflow = outflow + accumulation Chemical concentrations: Energy Balance: Energy in = energy out + accumulation Summing terms gives these equations (only one term in and out shown, but multiple are more common) This energy balance is for sensible energy - Note: total energy = m·h (mass flow rate x enthalpy) ,= i.e., replaces cp·T -> see next slide (sensible energy)

Principles Mass Balance: Energy Balance:
Inflow = outflow + accumulation Chemical concentrations: Energy Balance: Energy in = energy out + accumulation total energy = m·h (mass flow rate x enthalpy) so enthalpy (not temperature) is analogous to the concentration in mass transfer m1·h1 (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 Energy Balance Specific equations
Inflow = outflow + accumulation Energy Balance Energy in = energy out + accumulation Specific equations Fluid mechanics, pumping, fans, heat transfer, drying, separation, etc.

qfrig

energy in = energy out + accumulation qfrig qin, = qout, qa

energy in = energy out + accumulation qfrig qin, = qout, qa Try it - identify: qin,1 , qin,2 , ... Think "building ventilation" (facilities engineering) as you look for terms….

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 qb qr qso qe qs qm qin
Note the shortcut on qin (infiltration)… it’s a “net” term: need qinto − qout to complete it

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 = Btu/h, 1 kW = Btu/h Note: this is very close to the balance used all the time for ventilation control in S&E Environment problems (Facilities – Environment, Ventilation), except there is no refrigeration term in livestock buildings…. The main difference is how we use it — the basic energy balance is very similar (apples instead of animals; apples across boundary = sensible load; add refrigeration term)

Apple Cooling Sensible heat equations…
qs = mload· cpA· ΔT = mload· cpA· ΔT qr = mtot· Hresp qm = qm1 + qm qb = Σ(A/RT)· (Ti – To) qin = (Qacpa/vsp)· (Ti – To) qso = ... qx are rates (W, Btu/h) Hresp is a rate [W/kg, Btu/(lb-day)]

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 rA = 46 lb/bu; cpA = 0.9 Btu/lb°F What is the sensible heat load from the apples on day 3? Specific problem. (Ambient temperature declines uniformly to 65°F over 20 days (0.5°F/day). Apple temperatures cool from ambient to 34°F in 5 days; 34° to 31°F in next 5 days.) What is day 3 sensible heat load? (I.e., temperature change only, w/o moisture effects.)

Example 1 qfrig qm qb qso qr 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.) Note: Ti,avg = 74.5°F Note the slight overestimate (should be okay for design). Ti was specified to decline 0.5°F each day. So the exact Ti,avg = 74.5°F For calculations later in the cooler season this error would become quite significant.

Example 1, revisited mload = 276,000 lb (on day 3)
Ti,avg = ( )/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.) This is the exact calculation; again, such a slight overestimate should be okay for design (as long as you recognize it and don’t add too many “safety factors”). The correct Ti,avg is 74.5°F (not 75°F).

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?

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) D = decimal reduction time = time, at a given temperature, in which the number of microbes (spores) is reduced 90% (1 log cycle) Sterilization = “canning” process.

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) t = TDT = thermal death time at specified temperature Also, t = F or “F-value” for thermal death time at standard conditions (250°F); However, “F” is sometimes also used with a temperature subscript (i.e., F240) at non-standard conditions. D = decimal reduction time (see upcoming chart) At Std. Conditions: F = Fo = F250 = F-value; D = Do = D250 = D-value

Sterilization Thermal Death Time Curve (C. botulinum) (Esty & Meyer, 1922) t = thermal death time, min

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

Sterilization Thermal Death Rate Plot (Stumbo, 1949, 1953; ...)
D = decimal reduction time

Sterilization Thermal Death Rate Plot (Stumbo, 1949, 1953; ...)
D = decimal reduction time z 121 Dr = 0.2

Sterilization Common 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

Sterilization equations
Out of these many ways to write the relationships, these two equations usually work fine for most problems.

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

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

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°C); z = 18°F (10°C)

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 = 121°C (std. temp.) Because this problem is just a temperature change from standard conditions, you could simply read the new t off the standard C. botulinum graph (at 100°C). We’ll calculate it in the next slide. 2.7

Example 4 At 100°C the new time is 348 min; same as you should have read off the graph.

Heat Exchanger Basics q = U  Ae  DTm Fluid B Parallel flow Fluid A -
Counter flow Parallel flow Definition of total heat transferred (q) in terms of overall heat transfer coefficient. q = U  Ae  DTm

Heat Exchanger Basics q = U  Ae  DTm = U  Ae  DTlm Fluid B
Fluid A - Counter flow Parallel flow For basic exchangers: Tm = Tlm (log mean temperature difference) (and Tlm is always used even if it then has to be modified for more complex exchangers) q = U  Ae  DTm = U  Ae  DTlm

Heat Exchanger Basics Dtmax Dtmin Dtmin Dtmax Fluid B Fluid A -
or Dtmin Fluid B Counter flow - Fluid A Fluid A - Parallel flow Dtmin or Dtmax adding the definition of Tlm and an energy balance between the two fluids gives the main equations required for heat exchanger calculations

Heat Exchanger Basics Dtmax Dtmin Dtmin Dtmax Fluid B Fluid A -
or Dtmin Fluid B Counter flow - Fluid A Fluid A - Parallel flow Dtmin or Dtmax I think Geankoplis uses the dual definitions (based on defining the hot and cold fluids), but I prefer the simplicity of the min and max method from Henderson, Perry, and Young.

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

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 (2003), Transport Processes & Unit Operations. Standards ASABE Standards, recent ed. Other text Albright (1991), Environmental Control... Loewer 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 “PE Exam Study Tips” by Amy Kaleita Also, “Economics & Statistics” (Marybeth Lima) − earlier webinar. Unit ops. questions:

Standards, Codes, & Regulations
ASABE ASAE D245.6 and D272.3 covered in examples 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 Mainly just these ASABE standards…

Heat Exchanger Basics Dtmax Dtmin Dtmin Dtmax q = U  Ae  DTlm
or Dtmin Fluid B Counter flow - Fluid A Fluid A - Parallel flow Dtmin or Dtmax q = U  Ae  DTlm … the equations again ∆𝑇 𝑙𝑚 = ∆𝑇 𝑚𝑎𝑥 − ∆𝑇 𝑚𝑖𝑛 𝑙𝑛 Δ𝑇 𝑚𝑎𝑥 Δ𝑇 𝑚𝑖𝑛 𝑚 𝐻 𝑐 𝐻 ∆𝑇 𝐻 = 𝑚 𝐶 𝑐 𝐶 ∆𝑇 𝐶 = 𝑞

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 Solution See if you can get started with this solution… then I’ll give a pointer…

Example 5 Solution mf cf DTf = mw cw DTw 90°C ? 60°C 20°C
The energy balance between the two fluids: f = food, w = water (Note: subscripts c = cold and h = hot replaced by f and w…)

Example 5 Solution mf cf DTf = mw cw DTw
? 20°C Solution mf cf DTf = mw cw DTw (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 90°C 60°C 71°C 20°C Solution

Example 6 DTmin = 90°–60°C 90°C 60°C 71°C 20°C Solution DTmax = 71°–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 In this case I didn't solve for Ae with the two equations giving an equation to directly calculate the result, but solved it in two small steps: found q first with the second equation, then used that to calculate Ae with the first equation.

Example 6 DTmin = 90°–60°C 90°C 60°C 71°C 20°C Solution DTmax = 71°–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 (subscripts again: “cold” or ”c” = “food” or “f ”)

More about Heat Exchangers
Effectiveness ratio (H, P, & Young, pp ) One fluid at constant T: R DTlm correction factors The effectiveness ratio is another common, useful approach (and can be quick – good for the exam) – see Henderson, Perry, and Young (pp ) for details. The special case of one fluid at constant temperature (i.e., that fluid is evaporating or condensing) is a common special case that is on the effectiveness ratio charts as R   (equivalent to infinite heat capacity for the evaporating/condensing fluid) Note: I consider Tlm correction factors a specialty topic (mechanical engineering); I would not expect them on this 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 Sensible – horizontal Humidify – vertical (theoretical only; no real normal real process follows a vertical line) Drying (adiabatic saturation, evaporative cooling) – constant wet-bulb line (real “humidifying” processes would normally be adiabatic saturation processes)

Mass Transfer Between Phases cont.
Twb “drying” Psychrometrics Grain and food drying Sensible heat Latent heat of vaporization Moisture content: wet and dry basis, and equilibrium moisture content (ASAE Standard D245.6) Airflow resistance (ASAE Standard D272.3)

ASAE Standard D245.6 – Use previous revision (D245.4) for constants or use psychrometric charts in Loewer et al. (1994) If you don’t use Loewer et al. (1994) you can use current charts in ASAE D245.6 (charts are same as old ones – it’s just the equation constants that are hard to use in the newer version of the standard). It just requires two steps, using two charts, instead of one step with Loewer’s psych. charts.

Loewer, et al. (1994)

Deep Bed Drying Process
rhe Twb “drying” TG To rho

Use of Moisture Isotherms
To = inlet air temperature TG = product (grain) temperature

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) Bottom equation is for the weight/moisture change calculations during drying. Note that “W” is used for mass to avoid confusion with moisture. This equation is a “mass balance” – based on the dry matter content being equal before and after a moisture change. Thus it uses wet basis moisture content [because (1 – Mwb) = dry matter fraction]. Recognize that (W2 – W1) = moisture removed (i.e., weight loss)

Drying Deep Bed vs. Thin Layer
Thin-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. n t k e MR × - = content moisture basis dry M MR m equilibriu initial = - ; Drying equations are normally dry basis, while most industries use only wet basis moisture contents.

Drying Process time varying process
Drying Rate Time → Constant Rate Falling Rate erh = 100% aw = 1.0 erh < 100% aw < 1.0 Evaporative Cooling (Thin-layer) … unless something is clearly specified in the problem indicating the drying is not falling rate… Assume falling rate period, unless… Falling rate requires erh or exit air data

Note on water activity aw
Definition: aw = erh expressed as a decimal i.e., 85% erh = 0.85 aw (recall erh and aw increase with increasing temperature)

Note on water activity aw
Definition: aw = erh expressed as a decimal i.e., 85% erh = 0.85 aw (recall erh and aw increase with increasing temperature) Application: food products with aw ≤ are sterilized by the controlled aw level. ( not subject to FDA processing regulations; 21 CFR Parts 108, 113, and 114). if aw ≤ 0.85, it is effectively sterilized.

Grain Bulk Density for deep bed drying calculations
kg/m3 lb/bu[1] Corn, shelled 721 56 Milo (sorghum) Rice, rough 579 45 Soybean 772 60 Wheat 1Standard bushel Source: ASAE D241.4

Basic Drying Process Mass Conservation
Compare: moisture added to air to moisture removed from product

Fan

Basic Drying Process Drying zones… locations
Wet wa,out = wa,wet wa,in Drying Dry wa,out = wa,dry wa,in Drying Wet Drying Dry “Early” and “late” in the deep bed drying process are reasonable well defined conditions. “Mid-way” in the process may still have wet grain at the top (shown), but could have grain already starting to dry at the top (not shown, but essentially the same as “late” but with a smaller “dry” zone). Early Late

Fan

Try it: Total moisture conservation equation: moisture added to air vs moisture removed from product

moisture added to air vs moisture removed from product Total moisture conservation:

Compare: moisture added to air to moisture removed from product Total moisture conservation: kga s kgw kga kgw kgg s kgg

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 for complete analysis) As we saw in a previous slide, the outlet conditions change somewhere in the middle of the process. The sec deep-bed method is useful if this is your specialty area….

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? “early,” but NOT “initially” in the drying process (initially, it is just 75°F – this very brief initial period is usually ignored in drying; but that short period is the entire period of interest in a cooling process)

Example 7a Part I − Texit “early” in the drying period
Use Loewer, et al. (1994 ) (or ASAE D245.6) Texit = Tdb,e = TG Twb “drying” emc=18% Tdb,e

Example 7a 18% 53.5 Loewer, et al. (1994)

Example 7a Part I Use Loewer, et al. (1994 ) (or ASAE D245.6)
Texit = Tdb,e = TG = 53.5°F Twb “drying” emc=18% Tdb,e

Example 7b Part II - exit air RH and temperature at the end
Use Loewer, et al. (1994 ) (or ASAE D245.6) RHexit = 55% Texit = 58°F Twb “drying” emc=13% rhexit Texit Note Texit is required to find erh from emc... Just like finding grain temperature, TG, always needed for calculations prior to the top beginning to dry. [For TG, then, assume TG = Twb + 1° to 3°C (2° to 5°F).]

Example 7b 13% 58 Loewer, et al. (1994)

Cooling Process Energy Conservation
Compare: heat added to air to heat removed from product Sensible energy conservation: A balance of sensible energy only would be easy (it’s based only on temperatures)…

Cooling Process Energy Conservation
Compare: heat added to air to heat removed from product Sensible energy conservation: but there is almost always some moisture transfer involved making a sensible energy balance inaccurate. A total energy balance accounts for the effect of the evaporation/condensation and allows for accurate calculations of time required. Total energy conservation:

Cooling Process (and Drying)

Cooling Process (and Drying)
Twb “drying” erh

Airflow in Packed Beds Drying, Cooling, etc.
These design values take the raw data from ASABE D272.3 and combine with the Shedd’s Curve Multiplier, MS (sometimes called packing factor) to arrive at a typical field value for pressure drop. 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 Pressure drop (design value chart): DP = (inH2O/ft)design x (depth) + 0.5 Shedd’s curve multiplier (Ms = PF = 1.3 to 1.5) This equation is written to use the Shedd’s curve multiplier separately rather than use the “design values” chart from the previous slide.

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 Neglect the duct loss of 0.5 inH2O for a full perforated floor – which you should always have in a drying bin. 0.5 inH2O pressure drop in ducts - Standard design assumption (neglect for full perforated floor)

The basic concept is finding the intersection of the system and fan curves – that is where the fan will operate when installed on that system.

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

More Examples

Evaporator (Concentrator)
mS mF mP mV Juice Shows the inflow and outflow rates Add concentrations, XF and XP, to complete picture (X will be different nomenclature from what was used before – a common nomenclature, though) And this examples adds eh need for enthalpies...

Evaporator Solids mass balance: Total mass balance:
Total energy balance: We saw a similar problem earlier that could be solved with the two mass balances.

Total energy balance: Before looking at these equations, let’s look at a specific problem and, then, how to find the energy balance (to go with the mass balances from before).

Example 8 Fruit juice concentrator, operating @ T =120°F
Feed: TF = 80°F, XF = 10% Steam: 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 FIND: The product rate mP With zero boiling point rise the data can come directly from steam tables at the specified temperatures Note: the boiling point rise would be more important for calculating heat transfer (e.g., heat loss from the unit), which you would need to base on exact operating temperatures.

Example 8 mV Juice (120°F) mF mP = ? mS TV = 120°F TF = 80°F
XF = 0.1 lb/lb TP = 120°F XP = 0.4 lb/lb TV = 120°F Data added to inflows and outflows…

Total energy balance:

Example 8 Steam tables: (hfg)S = Btu/lb, at 25 psia (TS = 240°F) (hg)V = Btu/lb, at 120°F (PV = 1.69 psia) Calculate: cp,mix = 0.35· X + 1.0· (1 – X) Btu/lb°F cpF = Btu/lb·°F cpP = 0.74 Btu/lb·°F If data is not given, hfg and hg can be found is steam tables. Henderson, Perry, and Young (1997) have only SI table; Geankoplis (2003) has tables for both unit systems.

Example 8 mV mF mP = ? mS Juice (120°F) TV = 120°F hg = 1113.7 Btu/lb
TF = 80°F XF = 0.1 lb/lb TP = 120°F XP = 0.4 lb/lb TV = 120°F hg = Btu/lb Energy terms added… cpF = Btu/lb°F cpF = 0.74 Btu/lb°F hfg = Btu/lb

Example 8 Solids mass balance: Total mass balance:
Total energy balance:

Example 8 Solve for mP: mP = 295 lb/h

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) 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)

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 *Higher rates increase control, flexibility, and cost.
Select 0.1 cfm/bu based on this recommendation and… (next slide) *Higher rates increase control, flexibility, and cost.

Example 9 Select lowest airflow (cfm/bu) for cooling rate
Cooling time of 120 h at this airflow rate is reasonable…

cfm/ft2 = (0.8) x (16 ft) x (0.1 cfm/bu)
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 Convert specific airflow (0.1 cfm/bu) to cfm/ft2 needed for charts

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) Calculate pressure drop for the depth from chart value.

Pressure drop: DP = (inH2O/ft) x MS x (depth) + 0. 5 (note: Ms = 1
Pressure drop: DP = (inH2O/ft) x MS x (depth) (note: Ms = 1.3 for wheat) 0.028 1.3 Note -- this raw (loose-fill) chart from ASABE D gives a low value that needs to be adjusted using MS to be accurate

Pressure drop: DP = (inH2O/ft)design x (depth) + 0.5
0.037 1.3 The design value chart here (and in the handout) already includes MS.

DP = (0.028 inH2O/ft) x 1.3 x (16 ft) + 0.5 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 DP = (0.028 inH2O/ft) x 1.3 x (16 ft) inH2O DP = inH2O You can calculate from the loose-fill value multiplied by MS = 1.3 or… (next slide)

DP = (0.037 inH2O/ft) x (16 ft) + 0.5 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) inH2O DP = inH2O or you can calculate directly from the design value chart.

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/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) When converted to total airflow, this is one (key) point on the system curve… for matching with a fan curve…. cfm = (0.1 cfm/bu) x (10,000 bu) cfm = cfm

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) Finally, use fan data to effectively find the intersection of the fan curve and the system curve.

Example 9 Axial Flow Fan Data (cfm):
Theoretically, you could graph the fan data for this fan and add the previous airflow calculation (1000 cfm at 1.09 inH2O) – adding additional airflow points if necessary. The chart above is sufficient to see this fan is adequate – if you needed to know the exact operating conditions, calculating a second point for the system curve would probably be required (but first graph the first point with the fan curve to see where (at what pressure) to calculate the second point…).

Example 9 Selected Fan: 12" diameter, ¾ hp, axial flow
Supplies: cfm @ inH2O (a little extra  cfm/bu) Be sure of recommended fan operating range. Note: don’t specify a fan outside of its recommended 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

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