Transcript Moisture and Psychrometrics

Pump Affinity Laws
Pump Affinity Laws
P. 100 of text – section 4: vary
only speed of pump
P. 100 of text – section 5:
vary only diameter
P. 106 of text – vary BOTH
speed and diameter of
impeller.
N 1  D1 
(
)

Q2
N 2  D2 
Q1
2
3
 N   D 
  1   1 
W2  N 2   D2 
W1
2
Power out equations
3
 P o 1   N 1   D1 

 
 

 P  N   D 
 o2   2   2 
5
Po  W Q  g
 1 


 
 2 
p. 106
p. 89

A pump is to be selected that is geometrically
similar to the pump given in the performance
curve below, and the same system. What D and
N would give 0.005 m3/s against a head of 19.8
m?
W
D = 17.8 cm
N = 1760 rpm
1400W
900W 9m
0.01 m3/s
What is the operating point of
first pump?
N1 = 1760
D1 = 17.8 cm
Q1 = 0.01 m3/s
W1 = 9m
Q2 = 0.005 m3/s
W2 = 19.8 m
Now we need to “map” to new
pump on same system curve.
Substitute into
Solve for D2
N2 = ?
Try it yourself
If the system used in the previous
example was changed by removing a
length of pipe and an elbow – what
changes would that require you to make?
 Would N1 change? D1? Q1? W1? P1?
 Which direction (greater or smaller)
would “they” move if they change?

Moisture and Psychrometrics
Core Ag Eng Principles Session IIB
Moisture in biological products can
be expressed on a wet basis or dry
basis
m
M
Wm
(W m  W d )
Wm
Wd
wet basis
dry basis (page 273)
Standard bushels
ASAE 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
0.135 
Wm
Wt

Wm
60lb
W m  0.135(60lb )  8.1lb
W d  60lb  8.1lb  51.9lb
0.25 
0.25W
Wm
W m  51.9
m
 13  W m
13  0.75W
Wm 
13
0.75
m
 17.3lb
So water removed =
H2O @ 25% - H2O @ 13.5%
 17.3
 9.2
lb
bu
lb
bu
 8.1
lb
bu
 9.2
lb
bu
* 2000bu  18,400lbH
2
O
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 = 26 BTU/lbda
 Humidity ratio=0.0088 lbH2O/lbda
 Specific volume = 13.55 ft3/lbda
 Dew point temp = 54 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
Δh  22
Energy 
 119kW
kJ
kg da
ΔhQ
V
22

kJ
kg da
0.922
m
3
kg da

300m
min
3

1min
60s
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 four T/RH points from EMCs

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 bu
 20% to 13%


Now what?
Step 1
Std bu = 60 lb @ 0.135
mw = 0.135(60 lb) = 8.1 lb H2O
md = mt – mw = 60 – 8.1 = 51.9 lbdm
@ 13%:
0.13 
mw
m w  51.9
m w  0.13m
m w  7.76lb
w
 6.75
Step 1
mw
0.2 
m w  51.9lb
m w  12.98lb
ΔH
2
O:
12.98lb  7.76lb  5.22
 5.22
lb
bu
lb
bu
 2000bu  10440lb
H 2O
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
0.0056  0.0033
lb H 2 O
lb da
 0.0023
lb H 2 O
lb da
We need to remove 10500 lbH2O.
Each lbda removes 0.0023 lbH2O.
4,565,217l b da

1lb da

 10500lb H 2 O
 0.0023lb

H 2O




Step 3
Determine the cubic feet of air we
need to remove necessary water
Step 3 Calculations
60,260,870 ft
3
air
 4,565,217l b da  13.2
ft
3
lb da
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.
60,260,870 ft
6300
 159hrs
 6.6d
ft
3
min
3
 9565min
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,
(not Q)

m  5.95
m
Q pt 2  6.8
3

s
kg
s

is conserved
m
kg
0.875m

3
kg
1.125m
3
 6.8
kg
 7.65
s
m
s
3
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
Propane  50,000
h 2  140
h1  49.5

kJ
kg fuel
kJ
kg da
kJ
kg da
Δh m  615.4
kJ
s
 44
kg propane
hr
Example 2
4. Determine the amount of fruit
residue dried per hour.
Example 2
@ 85%, 0.15 of every kg is dry
matter
0.22 
wm
w m  0.15
w m  0.0423kg
H 2O
Example 2
Remove 0.85 – 0.0423 = 0.8077
kg H 2 O
kg wetresidue
ΔH  0.038  0.006  0.032
kg H 2 O
kg da
Example 2
 0.032
 970
kg H 2 O
kg da
kg wetfruit
hr

6.8kg
s
da

kg wetresidue
0.8077kg
H2O
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.