Transcript Core Ag Engineering Principles – Session 1

Core Ag Engineering Principles
– Session 1
Bernoulli’s Equation
Pump Applications
Bernoulli’s Equation

Hydrodynamics (the fluid is moving)

Incompressible fluid (liquids and gases at
low pressures)
◦ Therefore changes in fluid density are not
considered
Conservation of Mass

If the rate of flow is constant at any point and
there is no accumulation or depletion of fluid
within the system, the principle of conservation
of mass (where mass flow rate is in kg/s)
requires:



m1  m2  ...  mi
For incompressible fluids – density
remains constant and the equation
becomes:
A1V1  A 2 V2  ...  Q
Q is volumetric flow rate in m3/s
A is cross-sectional area of pipe (m2) and
V is the velocity of the fluid in m/s
Example

Water is flowing in a 15 cm ID pipe at a
velocity of 0.3 m/s. The pipe enlarges to
an inside diameter of 30 cm. What is the
velocity in the larger section, the
volumetric flow rate, and the mass flow
rate?
Example
D1 = 0.15 m
V1 = 0.3 m/s
How do we find V2?
D2 = 0.3 m
V2 = ?
Example
D1 = 15 cm ID
V1 = 0.3 m/s
D2 = 30 cm ID
V2 = ?
We know A1V1 = A2V2
Answer
π(0.15m) 2
(0.3m/s)
A1V1
4
V2 

2
π(0.3m)
A2
4
V2 = 0.075 m/s
What is the volumetric flow rate?
Volumetric flow rate = Q
Q  A1V1  A 2 V2
2
π(0.15m)
m

(0.3 )
4
s
3
m
 0.0053
s
What is the mass flow rate in the
larger section of pipe?

Mass flow rate = m

m  Qρ
3
m
kg
 0.0053 (1000 3 )
s
m
kg
 5.3
s
Bernoulli’s Theorem

Since energy is neither created nor
destroyed within the fluid system, the
total energy of the fluid at one point in
the system must equal the total energy at
any other point plus any transfers of
energy into or out of the system.
Bernoulli’s Theorem
2
1
2
2
P1
v
P2
v
h1 

 W  F  h2 

γ
2g
γ
2g
h = elevation of point 1 (m or ft)
P1 = pressure (Pa or psi)
= specific weight of fluid
v = velocity of fluid
γ
Bernoulli’s Theorem Special Cases

When system is open to the atmosphere,
then P=0 if reference pressure is
atmospheric (can be one P or both P’s)
When one V refers to a storage tank and
the other V refers to a pipe, then V of
tank <<<< V pipe and assumed zero
 If no pump or fan is between the two
points chosen, W=0

Example

Find the total energy (ft) at B; assume flow is
frictionless
A
B
125’
75’
C
25’
Example

Why is total energy in units of ft?

What are the typical units of energy?

How do we start the problem?
Example
Total EnergyA = Total EnergyB
PA v 2A
PB vB2
hA  
 W  F  hB  
γ
2g
γ 2g
Total EnergyB
hA = 125’ = Total EnergyB
Example
Find the velocity at point C.
2
C
v
PC
125'  hC 

2g γ
125'  25'
v C2
ft
2(32.2 2 )
s
ft
v 2  80.2
s
0
Try it yourself:
Water is pumped at the rate
of 3 cfs through piping
system shown. If the pump
has a discharge pressure of
150 psig, to what elevation
can the tank be raised?
Assume the head loss due to
friction is 10 feet.
pump
9’
1’
x’
1’
Determining F for Pipes and
Grain
Step 1
Determine Reynolds
number
 Dynamic viscosity Re
units
 Diameter of pipe
 Velocity
 Density of fluid

D Vρ

μ
Reynolds numbers:
<
2130 Laminar
 > 4000 Turbulent
 Affects
what?
Reynolds numbers:
< 2130 Laminar
 > 4000 Turbulent

Affects what?
 The f in Darcy’s equation for friction loss in
pipe
 Laminar: f = 64 / Re
 Turbulent: Colebrook equation or Moody
diagram

Total F
F = Fpipe + Fexpansion + Fcontraction + Ffittings
Darcy’s Formula
 L  v 
 f   
 D  2g 
2
Fpipe
 Where
do you use relative roughness?
 Relative
roughness is a function of the
pipe material; for turbulent flow it is a
value needed to use the Moody
diagram (ε/D) along with the Reynolds
number
Example
 Find
f if the relative roughness is
0.046 mm, pipe diameter is 5 cm,
and the Reynolds number is 17312
Example
 Find
f if the relative roughness is
0.046 mm, pipe diameter is 5 cm,
and the Reynolds number is 17312
Solution
ε / D = 0.000046 m / 0.05 m = 0.00092
 Re = 1.7 x 104

Re > 4000; turbulent flow – use Moody
diagram
Find ε/D , move to left until hit dark black
line – slide up line until intersect with Re
#
Answer

f = 0.0285
Energy Loss due to Fittings and Sudden
Contractions
v 
F  K 
2g
 
2
Energy Loss due to Sudden
Enlargement
(V1  V2 )
F
2g
2
Example
 Milk
at 20.2C is to be lifted 3.6 m
through 10 m of sanitary pipe (2 cm ID
pipe) that contains two Type A elbows.
Milk in the lower reservoir enters the
pipe through a type A entrance at the
rate of 0.3 m3/min. Calculate F.

Step 1:
 Step
1: Calculate Re number
DVρ
Re 
μ

Calculate v = ?

Calculate v2 / 2g, because we’ll need this a lot
3
m 1min 
0.3


Q
m
min  60s 
v 
 15.9
2
(0.02m) π
A
s
4
m 2
(15.9 )
2
v
s

 12.9m
2g 2(9.81 m )
2
s

What is viscosity? What is density?
Viscosity = 2.13 x 10-3 Pa · s
ρ = 1030 kg/m3
 So
Re = 154,000
m
kg
0.02m(15.9 )(1030 3 )
s
m
Re 
3 Ns
2.13 10
2
m
kgm
N 2
s

f=?
 Fpipe
=
ε
0.046mm 0.000046m


 0.0023
D
0.02m
0.02m
5
Re  1.5 10
Moody' s :
f  0.026
 L  v 
 f   
 D  2g 
2
Fpipe
 10m 
 0.026
12.9m
 0.02m 
 167.5m
 Ffittings =
 Fexpansion
=
 Fcontraction=
v 
Ffittings  (0.5  0.5)   12.9m
 2g 
2
2
(v1  v 2 )
v1
Fexp 

 12.9m
2g
2g
2
2
Fcontr
v
 0.5  6.45m
2g
 Ftotal
= 199.7 m
Try it yourself

Find F for milk at 20.2 C flowing at 0.075
m3/min in sanitary tubing with a 4 cm ID
through 20 m of pipe, with one type A
elbow and one type A entrance. The milk
flows from one reservoir into another.
Pump Applications
Power

The power output of
a pump is calculated
by:
W = work from pump (ft or m)
Q = volumetric flow rate (ft3/s or m3/s)
ρ = density
g = gravity
System Characteristic Curves

A system characteristic curve is
calculated by solving Bernoulli’s theorem
for many different Q’s and solving for W’s

This curve tells us the input head
required to move the fluid at that Q
through that system
Example system characteristic
curve
Pump Performance Curves

Given by the manufacturer – plots total
head against volumetric discharge rate

Note: these curves are good for ONLY
one speed, and one impeller diameter –
to change speeds or diameters we need
to use pump laws
Total head
Power
Efficiency
Pump Operating Point

Pump operating point is found by the
intersection of pump performance curve
and system characteristic curve
What volumetric flow rate will this
pump discharge on this system?
 Performance
of centrifugal pumps
while pumping water is used as
standard for comparing pumps

To compare pumps at any other speed
than that at which tests were conducted
or to compare performance curves for
geometrically similar pumps – use affinity
laws
Pump Affinity Laws
Power out equations

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
W1 = 9m
Q2 = 0.005 m3/s
W2 = 19.8 m
Now we need to “map” to new
pump on same system curve.
Substitute into
Solve for D2
N2 = ?
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?

Bernoulli’s Theorem for Fans
PE Review Session VIB – section 1
Fan and Bin
3
1
2
P3 v 32
P1 v12
h1    W  F  h3  
γ 2g
γ 2g
P3
P1
h1   h3 
γ
γ
v1  v 3  0
W F
2
2
P2 v
PT
W 

F
γ 2g γ
static
pressure
velocity
head
total
pressure
Power
PT Q
P
eT
or
PsQ
P
es
Ftotal=Fpipe+Fexpansion+Ffloor+Fgrain
 Fpipe=f (L/D) (V2/2g)
 Fexpansion= (V12 – V22)
◦
◦
◦
◦
for values in pipe
/ 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)
ASAE 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
m^2. Find the static and total pressure when
Q=4 m^3/s
F=F(pipe)+F(exp)+F(floor)+F(grain)

F(pipe)=
L v 
 f      
D  2g 
2
Fpipe
Vpipe
Q

A pipe
m3
4
s

2
π 0.5
4
m
 20.4
s
f
ε
and Re
D
Dvρ
Re 
μ
m 
kg 

0.5m 20.4 1.202 3 
s 
m 

Re 
5
1.82 10 Pa  s
Re  6.7 10
5
 1m 
0.15mm

ε
1000mm 


D
0.5m
 3 10  4
f(moody) 
f  0.015
2


m

  20.4  
s 
 5m  
Fpipe  0.015


 0.5m  2   9.81m  
2 
 
s


 3.2m
Fexp
Fexp

v
2
1
 v
2g
2
2

2
m

 20.4    0
s

Fexp 
 21.2m
m

2   9.81 2 
s 

Ffloor
Equ. 2.39

2 
 1.071Pa  s  v
2 

m
o f ε




ρ g







p 
2








3

V = Vbin =
m
4
Q
m
s


0.2
2
A bin 20m
s

Of=0.1
ε
p
 0.4
2
Ffloor
Pa  s  0.2 

1.071 2 

m  0.1 0.4 


 2.3m
kg
m
1.202 2  9.81 2
m
s
Fgrain
Fwheat
ΔP
a  V 2cf


L
ln1 bV 


2
m
4 
2.7 10  0.2  1.51m
s

ΔP 
 1599Pa

m 

ln1 8.77 0.2  
s 



1599 Pa = _________ m?
N
1599 2
m 
ρ g
N
1599 2
m
 135m
kg 
m
1.202 3  9.81 2 
m 
s 
Using Shedd’s Curves
V=0.2 m/s
 Wheat

ΔP
Pa
 1000 1mc f  1500Pa  127m
L
m
 Ftotal
= 3.2 + 21.2 + 2.3 + 130
= 157 m
Problem 2.4 (page 45)

Air (21C) at the rate of 0.1 m^3/(m^2 s)
is to be moved vertically through a crib of
shelled corn 1.6 m deep. The area of the
floor is 12 m^2 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%?