Transcript Moisture and Psychrometrics

Bernoulli’s
Theorem for
Fans
PE Review Session VIB –
section 1
Fan and Bin
3
1
2
P3 v 23
P1 v12
h1 

 W  F  h3 

γ 2g
γ 2g
P3
P1
h1 
 h3 
γ
γ
v1  v 3  0
WF
2
2
P2 v
PT
W 

F
γ 2g γ
static
pressure
velocity
head
total
pressure
Power
PTQ
P
eT
or
PsQ
P
es
Ftotal=Fpipe+Fexpansion+Ffloor+Fgrain
 Fpipe=f
(L/D) (V2/2g) for values in pipe
 Fexpansion= (V12 – V22) / 2g




V1 is velocity in pipe
V2 is velocity in bin
V1 >> V2 so equation reduces to
V12/2g
Ffloor
 Equation
2.38 p. 29 (4th edition) for no grain on
floor
 Equation 2.39 p. 30 (4th edition) for grain on floor


Of=percent floor opening expressed as decimal
εp=voidage fraction of material expressed as
decimal (use 0.4 for grains if no better info)
ASABE Standards - graph for
Ffloor
Fgrain
 Equation

A and b from standards or Table 2.5 p. 30
 Or




2.36 p. 29 (Cf = 1.5)
use Shedd’s curves (Standards)
X axis is pressure drop/depth of grain
Y axis is superficial velocity (m3/(m2s)
Multiply pressure drop by 1.5 for correction
factor
Multiply by specific weight of air to get F in
m or f
Shedd’s Curve (english)
Shedd’s curves (metric)
Example
 Air
is to be forced through a grain drying bin
similar to that shown before. The air flows
through 5 m of 0.5 m diameter galvanized iron
conduit, exhausts into a plenum below the
grain, passes through a perforated metal floor
(10% openings) and is finally forced through a 1
m depth of wheat having a void fraction of 0.4.
The area of the bin floor is 20 m2. Find the static
and total pressure when Q=4 m3/s
F=F(pipe)+F(exp)+F(floor)+
F(grain)
 F(pipe)=
L v 

 f     
 D   2g 
2
Fpipe
Vpipe

Q

A pipe
f

D
and Re
Re 
Dv

Re 
 1m 
0.15m m


1000m m 


D
0.5m
 3 10 4
f (m oody) 
f 
Fpipe 
Fexp
Fexp

v
2
1
 v
2 g
2
2

2
m

 20.4    0
s

Fexp 
 21.2m
m

2   9.81 2 
s 

Ffloor
Equ. 2.39
2


2 


Pa

s
v
 1.071




2
 o    

m

 f p  


 g








3
V
= Vbin =
m
4
Q
m
s

 0.2
2
Abin 20m
s
 Of=0.1
 p  0.4
Pa  s  0.2 

1.071 2 

m  0.1 0.4 


 2.3m
kg
m
1.202 2  9.81 2
m
s
2
F floor
Fgrain
Fwheat
a V 2c f
P


L
ln1  bV 


2
m

2.7 104  0.2  1.51m 
s

P 
 1599Pa

m 

ln1  8.77 0.2  
s 


 1599
Pa = _________ m?
N
1599 2
m 
g
N
1599 2
m
 135m
kg 
m
1.202 3  9.81 2 
m 
s 
Using Shedd’s Curves
 V=0.2
m/s
 Wheat
P

L
 Ftotal
= 3.2 + 21.2 + 2.3 + 130

= 157 m
Problem 2.4 (page 45)
 Air
(21C) at the rate of 0.1 m3/(m2 s) is to
be moved vertically through a crib of
shelled corn 1.6 m deep. The area of the
floor is 12 m2 with an opening percentage
of 10% and the connecting galvanized
iron pipe is 0.3 m in diameter and 12 m
long. What is the power requirement,
assuming the fan efficiency to be 70%?
Moisture and
Psychrometrics
Core Ag Eng Principles
Session IIB
Moisture in biological
products can be expressed
on a wet basis or dry basis
Wm
m
(Wm  Wd )
Wm
M
Wd
wet basis
dry basis (page 273)
Standard bushels
 ASABE
Standards
 Corn weighs 56 lb/bu at 15% moisture
wet-basis
 Soybeans weigh 60 lb/bu at 13.5%
moisture wet-basis
Use this information to
determine how much water
needs to be removed to dry
grain
 We
have 2000 bu of soybeans at 25%
moisture (wb). How much water must be
removed to store the beans at 13.5%?
 Remember
grain is made up of dry
matter + H2O
 The amount of H2O changes, but
the amount of dry matter in bu is
constant.
 Standard
bu
Wm
0.135
Wt
Wm 
Wm
0.25 
Wm  51.9
So water removed =
H2O @ 25% - H2O @ 13.5%
Your turn:
 How
much water needs to be removed to
dry shelled corn from 23% (wb) to 15%
(wb) if we have 1000 bu?
Psychrometrics
 If
you know two properties of an
air/water vapor mixture you know
all values because two properties
establish a unique point on the
psych chart
 Vertical lines are dry-bulb
temperature
Psychrometrics
 Horizontal
lines are humidity ratio (right
axis) or dew point temp (left axis)
 Slanted lines are wet-bulb temp and
enthalpy
 Specific volume are the “other” slanted
lines
Your turn:
 List
the enthalpy, humidity ratio, specific
volume and dew point temperature for a
dry bulb temperature of 70F and a wetbulb temp of 60F
 Enthalpy
= ____ BTU/lbda
 Humidity ratio=______ lbH2O/lbda
 Specific volume = ______ ft3/lbda
 Dew point temp = _____ F
Psychrometric Processes
 Sensible
heating – horizontally to the right
 Sensible cooling – horizontally to the left
 Note
that RH changes without changing
the humidity ratio
Psychrometric Processes
 Evaporative
cooling = grain drying (p 266)
Example
A
grain dryer requires 300 m3/min of 46C
air. The atmospheric air is at 24C and 68%
RH. How much power must be supplied
to heat the air?
Solution
@ 24C, 68% RH: Enthalpy = 56 kJ/kgda
@ 46C: Enthalpy = 78 kJ/kgda
V = 0.922 m3/kgda
kJ
Δh  22
kgda
ΔhQ
Energy 
V
Equilibrium Moisture Curves
 When
a biological product is in a
moist environment it will exchange
water with the atmosphere in a
predictable way – depending on
the temperature/RH of the moist air
surrounding the biological product.
 This information is contained in the
EMC for each product
Equilibrium Moisture Curves
 Establish
second point on the evaporative
cooling line – i.e. can’t remove enough
water from the product to saturate the air
under all conditions – sometimes the
exhaust air is at a lower RH because the
product won’t “release” any more water
Establishing Exhaust Air RH
 Select
EMC for product of interest
 On Y axis – draw horizontal line at the
desired final moisture content (wb) of
product
 Find the three T/RH points from EMCs (the
fourth one is typically out of the
temperature range)
Establishing Exhaust Air RH
 Draw
these points on your psych chart
 “Sketch” in a RH curve
 Where this RH curve intersects your drying
process line represents the state of the
exhaust air
Sample EMC
We are drying corn to 15%
wb; with natural ventilation
using outside air at 25C and
70% RH. What will be the Tdb
and RH of the exhaust air?
Drying
Calculations
Example problem
 How
long will it take to dry 2000 bu of
soybeans from 20% mc (wb) to 13% mc
(wb) with a fan which delivers 5140-9000
cfm at ½” H2O static pressure. The bin is
26’ in diameter and outside air (60 F, 30%
RH) is being blown over the soybeans.
Steps to work drying problem





Determine how much water needs to be
removed (from moisture content before and
after; total amount of product to be dried)
Determine how much water each pound of
dry air can remove (from psychr chart;
outside air – is it heated, etc., and EMC)
Calculate how many cubic feet of air is
needed
Determine fan operating CFM
From CFM, determine time needed to dry
product
Step 1
How much water must be
removed?
 2000
 20%
bu
to 13%
 Now
what?
Step 1
Std bu = 60 lb @ 0.135
mw =
md = mt – mw =
Step 1
Step 2
How much water can each
pound of dry air remove?
 How
do we approach this step?
Step 2
Find exit conditions from EMC.
Plot on psych chart.
0C = 32F = 64%
10C = 50F = 67%
30C = 86F = 72%
Step 2
@ 52F – 68% RH
Change in humidity ratio
Each
pound of dry air can
remove
We need to remove 10,500
lbH2O.
Each lbda removes 0.0023
lbH2O.
Step 3
Determine the cubic feet of air
we need to remove necessary
water
Step 3 Calculations
Step 4
Determine the fan operating
speed
 How
do we approach this?
Step 4
Main term in F is Fgrain
Airflow (cfm/ft2)
50
30
15
10
Pressure drop (“H2O/ft)
0.5
0.23
0.09
0.05
x depth x CF
Step 4
Fgrain
½
PS
Q
6300 cfm
From
cfm of fan and cubic
feet of air, determine the
time needed to dry the
soybeans.
Example 2
 Ambient
air at 32C and 20% RH is heated to 118
C in a fruit residue dryer. The flow of ambient air
into the propane heater is at 5.95 m3/sec. The
drying is to be carried out from 85% to 22% wb.
The air leaves the drier at 40.5C.
 Determine the airflow rate of the heated air.
Example 2
With

heated air, m is
conserved (not Q)

m
Q pt 2 
Example 2
2. Determine the relative
humidity of the air leaving
the drier.
Example 2
32 40.5 118
78% RH
Example 2
3. Determine the amount of
propane fuel required per
hour.
Example 2
kJ
P ropane 50,000
kgfuel
h2 
h1 

Δh m 
Example 2
4. Determine the amount of
fruit residue dried per hour.
Example 2
@ 85%, 0.15 of every kg is dry
matter
Example 2
Remove 0.85 – 0.0423 =
ΔH 
kgH 2O
0.8077
kgwetresidue
Example 2
Your Turn:
A grain bin 26’ in diameter has a
perforated floor over a plenum
chamber. Shelled field corn will be
dried from an initial mc of 24% to
14% (wb). Batch drying (1800 std.
bu/batch) will be used
with outside air (55F, RH 70%) that
has been heated 10F before being
passed through the corn. To dry the
corn in 1 week -
1. What is the necessary fan
delivery rate (cfm)?
2. What is the approximate
total pressure drop (in
inches of water) required
to obtain the needed air
flow?
3. The estimated fan HP
based on fan efficiency of
65%
4. If the drying air is heated by
electrical resistance elements
and the power costs is
$0.065/KWH, calculate the
cost of heating energy per
standard bushel.