Transcript of Core Ag Engineering Principles – Session 1 Bernoulli’s Equation Pump Applications.
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- Core Ag Engineering Principles Session 1 Bernoullis Equation
Pump Applications
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- Bernoullis Equation Hydrodynamics (the fluid is moving)
Incompressible fluid (liquids and gases at low pressures) Therefore
changes in fluid density are not considered
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- 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:
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- For incompressible fluids density remains constant and the
equation becomes: Q is volumetric flow rate in m 3 /s A is
cross-sectional area of pipe (m 2 ) and V is the velocity of the
fluid in m/s
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- 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?
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- Example D 1 = 0.15 mD 2 = 0.3 m V 1 = 0.3 m/sV 2 = ? How do we
find V 2 ?
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- Example D 1 = 15 cm IDD 2 = 30 cm ID V 1 = 0.3 m/sV 2 = ? We
know A 1 V 1 = A 2 V 2
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- Answer V 2 = 0.075 m/s
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- What is the volumetric flow rate?
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- Volumetric flow rate = Q
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- What is the mass flow rate in the larger section of pipe?
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- Mass flow rate =
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- Bernoullis 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.
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- Bernoullis Theorem h = elevation of point 1 (m or ft) P 1 =
pressure (Pa or psi) = specific weight of fluid v = velocity of
fluid
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- Bernoullis Theorem Special Cases When system is open to the
atmosphere, then P=0 if reference pressure is atmospheric (can be
one P or both Ps)
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- When one V refers to a storage tank and the other V refers to a
pipe, then V of tank
- Reynolds numbers: < 2130 Laminar > 4000 Turbulent Affects
what?
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- Reynolds numbers: < 2130 Laminar > 4000 Turbulent Affects
what? The f in Darcys equation for friction loss in pipe Laminar: f
= 64 / Re Turbulent: Colebrook equation or Moody diagram
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- Total F F = F pipe + F expansion + F contraction + F
fittings
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- Darcys Formula
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- Where do you use relative roughness?
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- Relative roughness is a function of the pipe material; for
turbulent flow it is a value needed to use the Moody diagram ( /D)
along with the Reynolds number
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- Example Find f if the relative roughness is 0.046 mm, pipe
diameter is 5 cm, and the Reynolds number is 17312
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- Example
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- Solution / D = 0.000046 m / 0.05 m = 0.00092 Re = 1.7 x 10 4 Re
> 4000; turbulent flow use Moody diagram
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- Find /D, move to left until hit dark black line slide up line
until intersect with Re #
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- Answer f = 0.0285
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- Energy Loss due to Fittings and Sudden Contractions
- Slide 38
- Energy Loss due to Sudden Enlargement
- Slide 39
- 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 m 3 /min. Calculate F.
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- Step 1:
- Slide 41
- Step 1: Calculate Re number
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- Calculate v = ? Calculate v 2 / 2g, because well need this a
lot
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- What is viscosity? What is density?
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- Viscosity = 2.13 x 10 -3 Pa s = 1030 kg/m 3
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- So Re = 154,000
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- f = ? F pipe =
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- F fittings = F expansion = F contraction =
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- F total = 199.7 m
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- Try it yourself Find F for milk at 20.2 C flowing at 0.075 m 3
/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.
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- Pump Applications
- Slide 55
- Power The power output of a pump is calculated by: W = work
from pump (ft or m) Q = volumetric flow rate (ft 3 /s or m 3 /s) =
density g = gravity
- Slide 56
- System Characteristic Curves A system characteristic curve is
calculated by solving Bernoullis theorem for many different Qs and
solving for Ws This curve tells us the input head required to move
the fluid at that Q through that system
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- Example system characteristic curve
- Slide 58
- 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
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- Efficiency Total head Power
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- Pump Operating Point Pump operating point is found by the
intersection of pump performance curve and system characteristic
curve
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- What volumetric flow rate will this pump discharge on this
system?
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- Performance of centrifugal pumps while pumping water is used as
standard for comparing pumps
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- 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
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- Pump Affinity Laws
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- Power out equations
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- 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 m 3 /s against a head of 19.8 m? 900W
9m 1400W W 0.01 m 3 /s D = 17.8 cm N = 1760 rpm
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- What is the operating point of first pump? N 1 = 1760 D 1 =
17.8 cm Q 1 = 0.01 m 3 /s Q 2 = 0.005 m 3 /s W 1 = 9m W 2 = 19.8
m
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- Now we need to map to new pump on same system curve. Substitute
into Solve for D 2
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- N 2 = ?
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- 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 N 1 change? D 1 ? Q 1 ? W 1 ?
P 1 ? Which direction (greater or smaller) would they move if they
change?
- Slide 72
- Bernoullis Theorem for Fans PE Review Session VIB section
1
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- Fan and Bin 1 2 3
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- static pressure velocity head total pressure
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- Power
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- F total =F pipe +F expansion +F floor +F grain F pipe =f (L/D)
(V 2 /2g) for values in pipe F expansion = (V 1 2 V 2 2 ) / 2g V 1
is velocity in pipe V 2 is velocity in bin V 1 >> V 2 so
equation reduces to V 1 2 /2g
- Slide 77
- F floor Equation 2.38 p. 29 (4 th edition) for no grain on
floor Equation 2.39 p. 30 (4 th edition) for grain on floor O f
=percent floor opening expressed as decimal p =voidage fraction of
material expressed as decimal (use 0.4 for grains if no better
info)
- Slide 78
- ASAE Standards graph for F floor
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- F grain Equation 2.36 p. 29 (C f = 1.5) A and b from standards
or Table 2.5 p. 30 Or use Shedds curves (Standards) X axis is
pressure drop/depth of grain Y axis is superficial velocity (m 3
/(m 2 s) Multiply pressure drop by 1.5 for correction factor
Multiply by specific weight of air to get F in m or f
- Slide 80
- Shedds Curve (english)
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- Shedds curves (metric)
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- 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
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- F=F(pipe)+F(exp)+F(floor)+F(grain) F(pipe)=
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- f
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- F exp
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- F floor Equ. 2.39
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- V = V bin =
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- O f =0.1
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- F grain
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- 1599 Pa = _________ m?
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- Using Shedds Curves V=0.2 m/s Wheat
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- F total = 3.2 + 21.2 + 2.3 + 130 = 157 m
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- 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%?