Moisture and Psychrometrics

Pump Affinity Laws1Pump 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.

2Power out equations

p. 106p. 893A 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?900W 9m1400WW0.01 m3/sD = 17.8 cmN = 1760 rpm4What is the operating point of first pump?N1 = 1760D1 = 17.8 cmQ1 = 0.01 m3/s Q2 = 0.005 m3/sW1 = 9m W2 = 19.8 m5Now we need to map to new pump on same system curve.

Substitute intoSolve for D2

6

7N2 = ?8Try it yourselfIf 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?If a length of pipe is removed, then the friction loss through the pipe will be reduced for a given Q, so system curve will be lower. That will result in the operating point being a larger Q against a smaller head (see next figure).

N1 would not change, nor would D but Q, W, and P would.

Q would be larger, W smaller, and P smaller.9

10Moisture and PsychrometricsCore Ag Eng Principles Session IIB11Moisture in biological products can be expressed on a wet basis or dry basis wet basis

dry basis (page 273)

12Standard bushelsASAE StandardsCorn weighs 56 lb/bu at 15% moisture wet-basisSoybeans weigh 60 lb/bu at 13.5% moisture wet-basis13Use this information to determine how much water needs to be removed to dry grainWe have 2000 bu of soybeans at 25% moisture (wb). How much water must be removed to store the beans at 13.5%?14Remember grain is made up of dry matter + H2OThe amount of H2O changes, but the amount of dry matter in bu is constant.15Standard bu

16

17So water removed =H2O @ 25% - H2O @ 13.5%

18Your turn:How much water needs to be removed to dry shelled corn from 23% (wb) to 15% (wb) if we have 1000 bu?19PsychrometricsIf you know two properties of an air/water vapor mixture you know all values because two properties establish a unique point on the psych chartVertical lines are dry-bulb temperature20PsychrometricsHorizontal lines are humidity ratio (right axis) or dew point temp (left axis)Slanted lines are wet-bulb temp and enthalpySpecific volume are the other slanted lines

21Your 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 60F22Enthalpy = 26 BTU/lbdaHumidity ratio=0.0088 lbH2O/lbdaSpecific volume = 13.55 ft3/lbdaDew point temp = 54 F23Psychrometric ProcessesSensible heating horizontally to the rightSensible cooling horizontally to the left

Note that RH changes without changing the humidity ratio24Psychrometric ProcessesEvaporative cooling = grain drying (p 266)25ExampleA 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?26Solution@ 24C, 68% RH: Enthalpy = 56 kJ/kgda@ 46C: Enthalpy = 78 kJ/kgdaV = 0.922 m3/kgda27

28Equilibrium Moisture CurvesWhen 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 product29

30Equilibrium 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 water31Establishing Exhaust Air RHSelect EMC for product of interestOn Y axis draw horizontal line at the desired final moisture content (wb) of productFind the four T/RH points from EMCs32Establishing Exhaust Air RHDraw these points on your psych chartSketch in a RH curveWhere this RH curve intersects your drying process line represents the state of the exhaust air

33Sample EMC

34We 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?35Drying Calculations36Example problemHow 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 H2O static pressure. The bin is 26 in diameter and outside air (60 F, 30% RH) is being blown over the soybeans.37Steps to work drying problemDetermine 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 neededDetermine fan operating CFMFrom CFM, determine time needed to dry product38Step 1How much water must be removed?2000 bu20% to 13%

Now what?39Step 1Std bu = 60 lb @ 0.135mw = 0.135(60 lb) = 8.1 lb H2Omd = mt mw = 60 8.1 = 51.9 lbdm@ 13%:

40Step 1

41Step 2How much water can each pound of dry air remove?How do we approach this step?42Step 2Find exit conditions from EMC.Plot on psych chart.

0C = 32F = 64%10C = 50F = 67%30C = 86F = 72%43Step 2@ 52F 68% RH44Change in humidity ratio45Each pound of dry air can remove

46We need to remove 10500 lbH2O.Each lbda removes 0.0023 lbH2O.

47Step 3Determine the cubic feet of air we need to remove necessary water48Step 3 Calculations

49Step 4Determine the fan operating speedHow do we approach this?50Step 4Main term in F is FgrainAirflow (cfm/ft2)50301510Pressure drop (H2O/ft)0.50.230.090.05x depth x CF51Step 4Fgrain6300 cfmQPS52From cfm of fan and cubic feet of air, determine the time needed to dry the soybeans.53

54Example 2Ambient 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.55Example 2With heated air, is conserved (not Q)

56Example 22. Determine the relative humidity of the air leaving the drier.57Example 232 40.5 11878% RH58Example 23. Determine the amount of propane fuel required per hour.59Example 2

60Example 24. Determine the amount of fruit residue dried per hour.61Example 2@ 85%, 0.15 of every kg is dry matter

62Example 2Remove 0.85 0.0423 =

63Example 2

64Your 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 -

651. What is the necessary fan delivery rate (cfm)?662. What is the approximate total pressure drop (in inches of water) required to obtain the needed air flow?673. The estimated fan HP based on fan efficiency of 65%684. 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.69