Pump Affinity Laws. P. 100 of text – section 4: vary only speed of pump P. 100 of text – section...
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- Slide 1
- Pump Affinity Laws
- Slide 2
- 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
- Slide 25
- Psychrometric Processes Evaporative cooling = grain drying (p
266)
- Slide 26
- 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?
- Slide 43
- Step 2 Find exit conditions from EMC. Plot on psych chart. 0C =
32F = 64% 10C = 50F = 67% 30C = 86F = 72%
- Slide 44
- Step 2 @ 52F 68% RH
- Slide 45
- Change in humidity ratio
- Slide 46
- Each pound of dry air can remove
- Slide 47
- 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.