Bernoulli’s Theorem for FansPE Review Session VIB – section 1

Fan and Bin

γ

Ph

γ

Ph

2g

v

γ

PhFW

2g

v

γ

Ph

33

11

233

3

211

1

12

3

Fγ

P

2g

v

γ

PW

FW

0vv

T222

31

staticpressure

velocityhead

total pressure

Power

s

s

T

T

e

QPP

or

e

QPP

Ftotal=Fpipe+Fexpansion+Ffloor+Fgrain

Fpipe=f (L/D) (V2/2g) for values in pipe

Fexpansion= (V12 – V2

2) / 2g V1 is velocity in pipe

V2 is velocity in bin

V1 >> V2 so equation reduces to

V12/2g

Ffloor

Equation 2.38 p. 29 (4th edition) for no grain on floor

Equation 2.39 p. 30 (4th edition) for grain on floor Of=percent floor opening expressed as decimal

εp=voidage fraction of material expressed as

decimal (use 0.4 for grains if no better info)

ASABE Standards - graph for Ffloor

Fgrain

Equation 2.36 p. 29 (Cf = 1.5) A and b from standards or Table 2.5 p. 30

Or use Shedd’s curves (Standards) X axis is pressure drop/depth of grain Y axis is superficial velocity (m3/(m2s) Multiply pressure drop by 1.5 for

correction factor Multiply by specific weight of air to get F

in m or f

Shedd’s Curve (english)

Shedd’s curves (metric)

Example Air is to be forced through a grain drying bin

similar to that shown before. The air flows through 5 m of 0.5 m diameter galvanized iron conduit, exhausts into a plenum below the grain, passes through a perforated metal floor (10% openings) and is finally forced through a 1 m depth of wheat having a void fraction of 0.4. The area of the bin floor is 20 m2. Find the static and total pressure when Q=4 m3/s

F=F(pipe)+F(exp)+F(floor)+F(grain) F(pipe)=

g2

v

D

LfF

2

pipe

s

msm

4.20

45.0

4

A

QV

2

3

pipepipe

f

Dv

andD

Re

Re

5

5

3

107.6Re

1082.1

202.14.205.0Re

sPamkg

sm

m

)(

103

5.01000

115.0

4

moodyf

mmmm

mm

D

m

smsm

m

mF

f

pipe

2.3

81.92

4.20

5.0

5015.0

015.0

2

2

Fexp

g

vvF

2

22

21

exp

m

sm

sm

F 2.2181.92

04.20

2

2

exp

Ffloor Equ. 2.39

g

ov

msPa

pf

2

2

2

071.1

V = Vbin = s

m

msm

A

Q

bin

2.020

4

2

3

Of=0.1

4.0p

m

sm

mkgmsPa

Ffloor 3.281.9202.1

4.01.02.0

071.1

22

2

2

Fgrain

Pa

sm

msm

P

bV

cVa

L

PF fwheat

1599

2.077.81ln

15.12.0107.2

1ln2

4

2

1599 Pa = _________ m?

m

sm

mkg

mN

gmN

13581.9202.1

15991599

23

22

Using Shedd’s CurvesV=0.2 m/sWheat

mPacmm

Pa

L

Pf 127150011000

Ftotal = 3.2 + 21.2 + 2.3 + 130

= 157 m

Problem 2.4 (page 45) Air (21C) at the rate of 0.1 m3/(m2 s) is

to be moved vertically through a crib of shelled corn 1.6 m deep. The area of the floor is 12 m2 with an opening percentage of 10% and the connecting galvanized iron pipe is 0.3 m in diameter and 12 m long. What is the power requirement, assuming the fan efficiency to be 70%?

Moisture and PsychrometricsCore 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)d

m

dm

m

W

WM

)W(W

Wm

Standard bushels ASABE 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 grainWe 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.

Standard bu

51.9lb8.1lb60lbW

8.1lb)0.135(60lbW

60lb

W

W

W0.135

d

m

m

t

m

17.3lb0.75

13W

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

lb9.2

bu

lb8.1

bu

lb17.3

2

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

119kW

60s

1min

min

300m

kgm

0.922

kgkJ

22

V

ΔhQEnergy

kg

kJ22Δh

3

da

3da

da

Equilibrium 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 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 three T/RH points from EMCs (the fourth one is typically out of the temperature range)

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

Sample EMC

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?

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 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 1Std 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

6.750.13mm

51.9m

m0.13

w

ww

w

w

Step 1

OH

2

w

w

w

210440lb2000bu

bu

lb5.22

bu

lb5.227.76lb12.98lb

:OΔH

12.98lbm

51.9lbm

m0.2

Step 2

How much water can each pound of dry air remove?

How do we approach this step?

Step 2Find 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

Each pound of dry air can remove

da

OH

da

OH

lb

lb0.0023

lb

lb0.00330.0056 22

We need to remove 10,500 lbH2O.

Each lbda removes 0.0023 lbH2O.

OH

daOHda

2

2 0.0023lb

1lb10500lbb4,565,217l

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

Step 3 Calculations

da

3

da3air lb

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

Step 4Determine the fan operating speed

How do we approach this?

Step 4Main term in F is Fgrain

Airflow (cfm/ft2)50301510

Pressure drop (“H2O/ft)0.5

0.230.090.05

x depth x CF

Step 4

½

Fgrain

6300 cfmQ

PS

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

6.6d

159hrs

9565min

minft

6300

ft60,260,8703

3

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 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.

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

m

s

m7.65

1.125m

kg

s

kg6.8Q

s

kg6.8

0.875m

kg

s

m5.95m

3

3pt

3

3

2

Example 2

2. Determine the relative humidity of 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

hr

kg44

s

kJ615.4mΔh

kg

kJ49.5h

kg

kJ140h

kg

kJ50,000Propane

propane

da1

da2

fuel

Example 2

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

Example 2

@ 85%, 0.15 of every kg is dry matter

OHm

m

m

20.0423kgw

0.15w

w0.22

Example 2

Remove 0.85 – 0.0423 = wetresidue

OH

kg

kg0.8077 2

da

OH

kg

kg0.0320.0060.038ΔH 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 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 -

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

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

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

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.