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
Core Principles
Conservation
of mass
Conservation of energy
Assumption/Conditions
 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  m 2  ...  mi
For incompressible fluids –
density remains constant and
the equation becomes:
A1V1  A2 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
D2 = 0.3 m
V2 = ?
How do we find V2?
Example
D1 = 0.15 m
V1 = 0.3 m/s
D2 = 0.3 m
V2 = ?
How do we find 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  A1V1  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
Questions?
Bernoulli’s Theorem
(conservation of energy)
 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
W = work done to the fluid
h = elevationγ of point 1 (m or ft)
P1 = pressure (Pa or psi)
= specific weight of fluid
v = velocity of fluid
F = friction loss in the system
γ
Bernoulli’s Theorem
Special Conditions
(situations where we can simplify the equation)
Special Condition 1
 When
system is open to the atmosphere,
then P=0 if reference pressure is
atmospheric (gauge pressure)
 Either one P or both P’s can be zero
depending on system configuration
Special Condition 2
 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
Special Condition 3
 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’
Preliminary Thinking
 Why
is total energy in units of ft? Is that a
correct measurement of energy?
 What
 How
are the typical units of energy?
do we start the problem?
Preliminary Thinking
 Feet
is a measure of pressure; it can be
converted to more traditional pressure
units.
 Typical units of energy: energy = work =
Nm or ft-lb. If we multiply feet or meter by
the specific weight of the fluid, we obtain
units of pressure.
 How 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 =
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.
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 energy of the
water leaving the pump is
equivalent to the 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’
Answer
 150

psig *62.4 lb/ft3 * 144 in2 / ft2 = 346 ft
1ft + 346 ft + 0 + 0 – 10 = 11 + x
X
= 326’
Bernoulli’s Equation
Adding
on – how do we
calculate F (instead of
having it given to us or
assuming it is negligible like
in the previous problems)
Bernoulli’s Theorem
P1
v12
P2
v 22
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
F = friction loss in the system
γ
Determining F
for Piping
Systems
Step 1
 Determine
Reynolds number
 Dynamic
viscosity units
 Diameter of pipe
 Velocity
 Density of fluid
DVρ
Re 
μ
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 Re.
Step
1: Calculate Re number
DVρ
Re 
μ
Calculate
v=?
Calculate
v2 / 2g, because we’ll need
this a lot
m  1m in
0.3


Q
m
m in  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
m2
k gm
N 2
s
Reynolds numbers:
<
2130 Laminar
> 4000 Turbulent
Affects
what?
Reynolds numbers:
 To
calculate the f in Darcy’s
equation for friction loss in pipe;
need Re
 Laminar: f = 64 / Re
 Turbulent: Colebrook equation or
Moody diagram
Total F in piping sytem
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 
2
g


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
m
kg
0.02m(15.9 )(1030 3 )
s
m
Re 
3 Ns
2.13 10
m2
kgm
N 2
s
Re
= 154,000
f
=?
 Fpipe
=
ε 0.046m m 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

 L  v 
 f   
 D  2g 
2
Fpipe
 10m 
 0.026
12.9m
 0.02m
 167.5m
 L  v 
 f   
 D  2g 
2
Fpipe
 10m 
 0.026
12.9m
 0.02m
 167.5m
 Ffittings =
 Fexpansion
=
 Fcontraction=
Ffittings 
Fexp 
2
Fcontr
v
 0.5  6.45m
2g
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
How is W determined for a
pump?
 Compute
all the terms in the Bernoulli
equation except W
 Solve
for W algebraically
Why is W determined for a
pump?
W
is used to compute the size
of pump needed
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
Remember:
 Po
= the power delivered
to the fluid (sometimes
referred to as hydraulic
power)
 Pin
= Po/pump efficiency
(sometimes referred to
as brake horsepower)
To calculate Power(out)
What we know
What we need
To calculate Power(out)
What we know
W

g
What we need
Q
Flow rate is variable
Depends
pressure”
on “back
Intersection
of system
characteristic curve and
the pump curve
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 power needed to be
supplied 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 Q: 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
Pump Performance Curve
Total head
Power
Efficiency
N = 1760 rpm
D = 15 cm
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 (for ex.
different impeller diameters or different
speeds) – use pump affinity laws (or pump
laws)
Pump Affinity Laws (p. 106)
 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)
 NOTE:
For changing ONLY one
property at a time
Example
 Size
a pump that is geometrically similar
to the pump given in the performance
curve below, for the same system. Find
D and N to achieve Q= 0.005 m3/s
against a head of 19.8 m?
0.01 m3/s
(Watt)
N = 1760 rpm
D = 17.8 cm
Procedure
 Step
1: Find the operating point of the
original pump on this system (so you’ll
have to plot your system characteristic
curve (calculated) onto your pump curve
(given) or vice-versa.
P(W)
N = 1760 rpm
D = 17.8 cm
P(W)
882.9 W
N = 1760 rpm
D = 17.8 cm
0.01 m3/s
What is the operating point
of first pump?
N1 = 1760
D1 = 17.8 cm
Q1 = 0.01 m3/s
W1 =
Q2 = 0.005 m3/s
W2 = 19.8 m
How do we convert Po to W?
How do we convert Po to W?
W
= Po/Qg
 W=(882.9 Nm/s)/(0.01 m3/s *1000 kg/m3 *
9.81 m/s2)
 W= 9 m
Find D that gives both new
W and new Q
 (middle
of p. 109 – can’t use p. 106 for
two conditions changing)
 D2=D1(Q2/Q1)1/2(W1/W2)1/4
 D2 =
 D2 =
Find D that gives both new
W and new Q
 (middle
of p. 109)
 D2=D1(Q2/Q1)1/2(W1/W2)1/4
 D2=0.178m(.005
m)1/4
 D2=0.141
m
m3/s/0.01m3/s)1/2(9 m/19.8
Find N that corresponds to
new Q and D (p. 109 equ)
 N2=N1(Q2/Q1)(D1/D2)3
 N2=
Find N that corresponds to
new Q and D
 N2=N1(Q2/Q1)(D1/D2)3
 N2=
1770 rpm
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?
Answers
 Removing
pipe + elbow would reduce F
and therefore reduce W (increase Q).
System curve would move to the right.
 N would likely change. D- no. Q- yes. P –
depends on the shape of the power
curve but likely it will change.
 N would increase (N2=N1(Q2/Q1)(D1/D2)3)
 Q – increase; P – depends.