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Pump Affinity Laws

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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.

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Power out equations p. 106 p. 89

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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

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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

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Now we need to map to new pump on same system curve. Substitute into Solve for D 2

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N 2 = ?

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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?

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Moisture and Psychrometrics Core Ag Eng Principles Session IIB

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Moisture in biological products can be expressed on a wet basis or dry basis wet basis dry basis (page 273)

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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

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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%?

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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.

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Standard bu

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So water removed = H 2 O @ 25% - H2O @ 13.5%

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Your turn: How much water needs to be removed to dry shelled corn from 23% (wb) to 15% (wb) if we have 1000 bu?

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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

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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

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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

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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

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Psychrometric Processes Sensible heating horizontally to the right Sensible cooling horizontally to the left Note that RH changes without changing the humidity ratio

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Psychrometric Processes Evaporative cooling = grain drying (p 266)

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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?

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Solution @ 24C, 68% RH: Enthalpy = 56 kJ/kg da @ 46C: Enthalpy = 78 kJ/kg da V = 0.922 m 3 /kg da

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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

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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

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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

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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

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Sample EMC

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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?

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Drying Calculations

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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.

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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

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Step 1 How much water must be removed? 2000 bu 20% to 13% Now what?

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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%:

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Step 1

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Step 2 How much water can each pound of dry air remove? How do we approach this step?

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Step 2 Find exit conditions from EMC. Plot on psych chart. 0C = 32F = 64% 10C = 50F = 67% 30C = 86F = 72%

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Step 2 @ 52F 68% RH

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Change in humidity ratio

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Each pound of dry air can remove

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We need to remove 10500 lb H2O. Each lb da removes 0.0023 lb H2O.

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Step 3 Determine the cubic feet of air we need to remove necessary water

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Step 3 Calculations

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Step 4 Determine the fan operating speed How do we approach this?

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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

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Step 4 F grain 6300 cfm Q PSPS

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From cfm of fan and cubic feet of air, determine the time needed to dry the soybeans.

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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.

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Example 2 With heated air, is conserved (not Q)

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Example 2 2. Determine the relative humidity of the air leaving the drier.

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Example 2 32 40.5 118 78% RH

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Example 2 3. Determine the amount of propane fuel required per hour.

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Example 2

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4. Determine the amount of fruit residue dried per hour.

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Example 2 @ 85%, 0.15 of every kg is dry matter

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Example 2 Remove 0.85 0.0423 =

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Example 2

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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 -

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1. What is the necessary fan delivery rate (cfm)?

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2. What is the approximate total pressure drop (in inches of water) required to obtain the needed air flow?

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3. The estimated fan HP based on fan efficiency of 65%

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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.