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.
Power out equations 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 Q2 = 0.005 m3/s W1 = 9m W2 = 19.8 m
Now we need to “map” to new pump on same system curve. Substitute into Solve for D2
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 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.
So water removed = H2O @ 25% - H2O @ 13.5%
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 wet-bulb 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
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
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?
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%:
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
We need to remove 10500 lbH2O. Each lbda removes 0.0023 lbH2O.
Step 3 Determine the cubic feet of air we need to remove necessary water
Step 4 Determine the fan operating speed How do we approach this?