Pump Affinity Laws. P. 100 of text – section 4: vary only speed of pump P. 100 of text – section...
69
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
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.
Slide 3
Power out equations p. 106 p. 89
Slide 4
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 m 3 /s against a head of 19.8 m? 900W
9m 1400W W 0.01 m 3 /s D = 17.8 cm N = 1760 rpm
Slide 5
What is the operating point of first pump? N 1 = 1760 D 1 =
17.8 cm Q 1 = 0.01 m 3 /s Q 2 = 0.005 m 3 /s W 1 = 9m W 2 = 19.8
m
Slide 6
Now we need to map to new pump on same system curve. Substitute
into Solve for D 2
Slide 7
Slide 8
N 2 = ?
Slide 9
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 N 1 change? D 1 ? Q 1 ? W 1 ?
P 1 ? Which direction (greater or smaller) would they move if they
change?
Slide 10
Slide 11
Moisture and Psychrometrics Core Ag Eng Principles Session
IIB
Slide 12
Moisture in biological products can be expressed on a wet basis
or dry basis wet basis dry basis (page 273)
Slide 13
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
Slide 14
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%?
Slide 15
Remember grain is made up of dry matter + H 2 O The amount of H
2 O changes, but the amount of dry matter in bu is constant.
Slide 16
Standard bu
Slide 17
Slide 18
So water removed = H 2 O @ 25% - H2O @ 13.5%
Slide 19
Your turn: How much water needs to be removed to dry shelled
corn from 23% (wb) to 15% (wb) if we have 1000 bu?
Slide 20
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
Slide 21
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
Slide 22
Your turn: List the enthalpy, humidity ratio, specific volume
and dew point temperature for a dry bulb temperature of 70F and a
wet- bulb temp of 60F
Slide 23
Enthalpy = 26 BTU/lb da Humidity ratio=0.0088 lb H2O /lb da
Specific volume = 13.55 ft 3 /lb da Dew point temp = 54 F
Slide 24
Psychrometric Processes Sensible heating horizontally to the
right Sensible cooling horizontally to the left Note that RH
changes without changing the humidity ratio
Example A grain dryer requires 300 m 3 /min of 46C air. The
atmospheric air is at 24C and 68% RH. How much power must be
supplied to heat the air?
Slide 27
Solution @ 24C, 68% RH: Enthalpy = 56 kJ/kg da @ 46C: Enthalpy
= 78 kJ/kg da V = 0.922 m 3 /kg da
Slide 28
Slide 29
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
Slide 30
Slide 31
Equilibrium Moisture Curves Establish second point on the
evaporative cooling line i.e. cant 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 wont release any
more water
Slide 32
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
Slide 33
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
Slide 34
Sample EMC
Slide 35
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?
Slide 36
Drying Calculations
Slide 37
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 H 2 O static pressure. The bin is 26 in diameter
and outside air (60 F, 30% RH) is being blown over the
soybeans.
Slide 38
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
Slide 39
Step 1 How much water must be removed? 2000 bu 20% to 13% Now
what?
Slide 40
Step 1 Std bu = 60 lb @ 0.135 m w = 0.135(60 lb) = 8.1 lb H 2 O
m d = m t m w = 60 8.1 = 51.9 lb dm @ 13%:
Slide 41
Step 1
Slide 42
Step 2 How much water can each pound of dry air remove? How do
we approach this step?
We need to remove 10500 lb H2O. Each lb da removes 0.0023 lb
H2O.
Slide 48
Step 3 Determine the cubic feet of air we need to remove
necessary water
Slide 49
Step 3 Calculations
Slide 50
Step 4 Determine the fan operating speed How do we approach
this?
Slide 51
Step 4 Main term in F is F grain Airflow (cfm/ft 2 ) 50 30 15
10 Pressure drop (H 2 O/ft) 0.5 0.23 0.09 0.05 x depth x CF
Slide 52
Step 4 F grain 6300 cfm Q PSPS
Slide 53
From cfm of fan and cubic feet of air, determine the time
needed to dry the soybeans.
Slide 54
Slide 55
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 m 3 /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.
Slide 56
Example 2 With heated air, is conserved (not Q)
Slide 57
Example 2 2. Determine the relative humidity of the air leaving
the drier.
Slide 58
Example 2 32 40.5 118 78% RH
Slide 59
Example 2 3. Determine the amount of propane fuel required per
hour.
Slide 60
Example 2
Slide 61
4. Determine the amount of fruit residue dried per hour.
Slide 62
Example 2 @ 85%, 0.15 of every kg is dry matter
Slide 63
Example 2 Remove 0.85 0.0423 =
Slide 64
Example 2
Slide 65
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
-
Slide 66
1. What is the necessary fan delivery rate (cfm)?
Slide 67
2. What is the approximate total pressure drop (in inches of
water) required to obtain the needed air flow?
Slide 68
3. The estimated fan HP based on fan efficiency of 65%
Slide 69
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.