Download Core Ag Engineering Principles – Session 1

Things to grab for this session
(in priority order)
 Pencil
 Henderson,
Perry, and Young text
(Principles of Process Engineering)
 Calculator
 Eraser 
 Scratch paper
 Units conversion chart
 Tables of fluid properties
 Moody diagram
 Pump affinity laws
Core Ag
Engineering
Principles –
Session 1
Bernoulli’s Equation
Pump Applications
Bernoulli’s Equation
 Hydrodynamics
 Incompressible
low pressures)

(the fluid is moving)
fluid (liquids and gases at
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  A2V2  ...  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
D2 = 0.3 m
V2 = ?
How do we find V2?
Example
D1 = 15 cm ID
V1 = 0.3 m/s
D2 = 30 cm ID
V2 = ?
We know A1V1 = A2V2
Answer
π(0.15m)
(0.3m/s)
A1V1
4
V2 

2
π(0.3m)
A2
4
2
V2 = 0.075 m/s
What is the volumetric flow
rate?
Volumetric flow rate = Q
Q  A1 V1  A 2 V2
π(0.15m)
m

(0.3 )
4
s
3
m
 0.0053
s
2
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 = elevationγ 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
 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
 How
are the typical units of energy?
do we start the problem?
Example
Total EnergyA = Total EnergyB
PA v 2A
PB v B2
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'  h C 

2g
γ
125'  25'
ft
v 2  80.2
s
v C2
ft
2(32.2 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 units
 Diameter of pipe
 Velocity
 Density of fluid
DVρ
Re 
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 +
Fexpansion+ 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
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
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
m
2g
2(9.81 2 )
s
3
 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
(v 1  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
 Q1/Q2=(N1/N2)(D1/D2)3
 W1/W2=(N1/N2)2(D1/D2)2
 Po1/Po2=(N1/N2)3(D1/D2)5(ρ1/ρ2)
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?
D = 17.8 cm
N = 1760 rpm
W
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
 Q1/Q2=(N1/N2)(D1/D2)3
 N2=
N1(Q2/Q1)(D1/D2)3
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?