1. V–C. Mass Transfer between Phases
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

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 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.

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

8. Illustration – Jam Production
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

9. Illustration – Jam Production
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)

10. Illustration – Jam Production
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)
mJ = 2.03 lbJam/lbfruit
mv = 0.19 lbwater/lbfruit

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

12. Principles Mass Balance: Energy Balance:
Inflow = outflow + accumulation
Chemical concentrations:
Energy Balance:
Energy in = energy out + accumulation

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

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. Apple Cooling
qfrig

16. 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.

17. Illustration − Apple Cooling
energy in = energy out + accumulation
qfrig
qin, = qout, 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. Apple Cooling
qfrig qm qb
qso qr qe qs qm
qin

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 = Btu/h, 1 kW = Btu/h

22. 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 = ...

23. 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)

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

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 = ( )/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. 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

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

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 substitute solve ... answer:
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. 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 = 121.1°C (std. temp.)
2.7

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. 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

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. 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

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

50. Time Out

51. 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.

52. 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

53. 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

54. 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)

55. Mass Transfer Between Phases cont.

56. 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)

57. Mass Transfer Between Phases cont.
Loewer, et al. (1994)

58. Mass Transfer Between Phases cont.

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

60. Use of Moisture Isotherms

61. 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)

62. 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 = - ;

63. 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

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

65. Basic Drying Process Mass Conservation
Fan

66. Basic Drying Process Mass Conservation
Try it:
Total moisture conservation equation:

67. Basic Drying Process Mass Conservation
Compare: moisture added to air to
moisture removed from product
Total moisture conservation:
kga s
kgw
kga
kgw
kgg s
kgg

68. 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)

69. 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)
Assume falling rate period, unless…
Falling rate requires erh or exit air data

70. Drying Process cont.
erh
ASAE D245.6
Twb
“drying”

71. 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?

72. Example 7 Part II Use Loewer, et al. (1994 ) (or ASAE D245.6)
RHexit = 55%
Texit = 58°F
Twb
“drying”
emc=13%
rhexit
Texit

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

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

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

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

77. Cooling Process Energy Conservation
Compare: heat added to air to
heat removed from product
Sensible energy conservation:
Total energy conservation:

78. Cooling Process (and Drying)
Twb
“drying”
erh

79. Airflow in Packed Beds Drying, Cooling, etc.
Source: ASABE D272.3, MWPS-29

80. 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)

81. 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)

82. Standards, Codes, & Regulations
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

83. More Examples

84. Evaporator (Concentrator)mS mF mP mV
Juice
Shows the inflow and outflow rates
Add concentrations, XF and XP, to complete picture (X is different nomenclature from before – common, though)
And add enthalpies...

85. Evaporator Solids mass balance: Total mass balance:
Total energy balance:

86. 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

87. 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…

88. 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 has tables for both unit systems.

89. 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

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

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

92. 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) 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)

93. Aeration Fan Selection

94. Aeration Fan Selection
Example
Wheat, Kansas, fall aeration
10,000 bu bin
16 ft eave height
pressure aeration system

95. 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)

96. Example 9
*Higher rates increase control, flexibility, and cost.

97. Example 9 Select lowest airflow (cfm/bu) for cooling rate

98. 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

99. 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)

100. 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

101. Pressure drop: DP = (inH2O/ft)design x (depth) + 0.5
0.037
1.3

102. 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

103. 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

104. 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)
cfm = (0.1 cfm/bu) x (10,000 bu)
cfm = cfm

105. 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)

106. Example 9
Axial Flow Fan Data (cfm):

107. 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.

108. 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