2011 Core Ag Engineering Principles - Session II

7/27/2019 2011 Core Ag Engineering Principles - Session II

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Pump Affinity Laws



Pump Affinity LawsP. 100 of text

–section 4: vary

only speed of pump

P. 100 of text – section 5:vary only diameter

P. 106 of text – vary BOTHspeed and diameter ofimpeller.

3

2

1

2

1

2

1)(

D

D

N

N

Q

Q

2

2

1

2

2

1

2

1

D

D

N

N

W

W



Power out equations

2

1

5

2

1

3

2

1

2

1

D

D

N

N

P

P

o

o

g QW P o

p. 106

p. 89



A pump is to be selected that is geometricallysimilar to the pump given in the performancecurve below, and the same system. What D

and N would give 0.005 m3

/s against a headof 19.8 m?

900W 9m

1400W

W

0.01 m3 /s

D = 17.8 cm

N = 1760 rpm



What is the operating point of first pump?

N1 = 1760

D1 = 17.8 cmQ1 = 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 systemcurve.

Substitute into Solve for D2





N2

= ?



Try it yourself

If the system used in the previousexample was changed by removing alength of pipe and an elbow – what

changes would that require you tomake?

Would N1 change? D1? Q1? W1? P1?

Which direction (greater or smaller)would “they” move if they change?





Moisture and PsychrometricsCore Ag Eng Principles Session IIB



Moisture in biological products

can be expressed on a wet basisor dry basis

wet basis

dry basis (page 273)

d

m

dm

m

W

WM

)W(W

Wm



Standard bushels

ASAE Standards

Corn weighs 56 lb/bu at 15% moisturewet-basis

Soybeans weigh 60 lb/bu at 13.5%moisture wet-basis



Use this information todetermine how much water needs to be removed to dry

grain We have 2000 bu of soybeans at 25%moisture (wb). How much water mustbe removed to store the beans at

13.5%?



Remember grain is made up of drymatter + H2O

The amount of H2O changes, but the

amount of dry matter in bu is constant.



Standard bu

51.9l8.1lb60lbW

8.1lb)0.135(60lbW

60lb

W

W

W0.135

d

m

m

t

m



17.3l0.7513W

0.75W13

W130.25W

51.9W

W0.25

m

m

mm

m

m



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

O18,400lbH2000bu*bu

lb9.2

bu

lb9.2bu

lb8.1bu

lb17.3

2



Your turn:

How much water needs to be removedto dry shelled corn from 23% (wb) to15% (wb) if we have 1000 bu?



Psychrometrics

If you know two properties of anair/water vapor mixture you know allvalues because two propertiesestablish a unique point on the psychchart

Vertical lines are dry-bulb temperature



Psychrometrics

Horizontal lines are humidity ratio(right axis) or dew point temp (leftaxis)

Slanted lines are wet-bulb temp andenthalpy

Specific volume are the “other” slantedlines



Your turn:

List the enthalpy, humidity ratio,specific volume and dew pointtemperature for a dry bulb

temperature of 70F and a wet-bulbtemp 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 theright

Sensible cooling – horizontally to the

left

Note that RH changes without

changing the humidity ratio





Example

A grain dryer requires 300 m3/min of 46C air. The atmospheric air is at 24Cand 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



119kW

60s

1min

min

300m

kg

m0.922

kgkJ22

V

ΔhQEnergy

kg

kJ22 Δh

3

da

3

da

da



Equilibrium Moisture Curves

When a biological product is in a moistenvironment it will exchange water with theatmosphere in a predictable way – depending on the temperature/RH of themoist air surrounding the biological product.

This information is contained in the EMC for each product





Equilibrium Moisture Curves

Establish second point on theevaporative cooling line – i.e. can’t

remove enough water from the

product to saturate the air under allconditions – sometimes the exhaustair is at a lower RH because the

product won’t “release” any morewater



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 thestate of the exhaust air



Sample EMC



We are drying corn to 15% wb;with natural ventilation usingoutside air at 25C and 70% RH.What will be the Tdb and RH of the exhaust air?



Drying Calculations



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 staticpressure. 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 beremoved (from moisture content before andafter; 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 dryproduct



Step 1

How much water must beremoved?

2000 bu

20% to 13%

Now what?



Step 1

Std bu = 60 lb @ 0.135mw = 0.135(60 lb) = 8.1 lb H2O

md = mt – mw = 60 – 8.1 = 51.9 lbdm

@ 13%:

7.76lbm

6.70.13mm

51.9m

m0.13

w

ww

w

w





Step 2

How much water can eachpound 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



Change in humidity ratio





We need to remove 10500lbH2O.

Each lbda removes 0.0023

lbH2O.

OH

daOHda

2

2 0.0023lb1lb10500lbb4,565,217l





Step 3 Calculations

da

3

da3air

lb

ft13.2b4,565,217lft60,260,870



Step 4

Determine the fan operatingspeed

How do we approach this?



Step 4

Main term in F is Fgrain

Airflow (cfm/ft2)

5030

15

10

Pressure drop (“H2O/ft)

0.50.23

0.09

0.05

x depth x CF



Step 4

½

Fgrain

6300 cfmQ

PS





6.6d

159hrs

9565mi

min

ft6300

ft60,260,8703

3



Example 2

Ambient air at 32C and 20% RH isheated to 118 C in a fruit residuedryer. The flow of ambient air into

the propane heater is at 5.95m3/sec. The drying is to be carriedout from 85% to 22% wb. The air leaves the drier at 40.5C.

Determine the airflow rate of theheated air.





Example 2

2. Determine the relative humidityof the air leaving the drier.



Example 2

32 40.5 118

78% RH



Example 2

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





Example 2

4. Determine the amount of fruitresidue dried per hour.

E l 2



Example 2

@ 85%, 0.15 of every kg is drymatter

OHm

m

m

2

0.0423kgw

0.15w

w0.22

E l 2



Example 2

Remove 0.85 – 0.0423 = wetresidue

OH

kg

kg0.8077 2

da

OH

kg

kg0.0320.0060.038 ΔH

2



Example 2

hr kg970

0.8077kg

kg

s

6.8kg

kg

kg0.032

wetfruit

OH

wetresidueda

da

OH

2

2



Your Turn:

A grain bin 26’ in diameter has aperforated floor over a plenumchamber.

Shelled field corn will be dried from an

initial mc of 24% to 14% (wb). Batch

drying (1800 std. bu/batch) will beused

with outside air (55F, RH 70%) that hasbeen heated 10F before being passed

through the corn. To dry the corn in 1

week -



1. What is the necessary fandelivery rate (cfm)?



2. What is the approximate totalpressure drop (in inches of water)required to obtain the needed air

flow?



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



4. If the drying air is heated byelectrical resistance elements andthe power costs is $0.065/KWH,

calculate the cost of heatingenergy per standard bushel.

Documents

2011 Core Ag Engineering Principles - Session II