msu-ceco/aec_v1 · Datasets at Hugging Face

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Only a soil test can provide an objective assessment of the need for fertilization and liming, and the appropriate amounts of materials to apply.
For more information regarding collecting and processing a soil sample, please refer to Extension PB1061, Soil Testing.
The nutrient required by fescues in greatest amounts is nitrogen.
Nitrogen is mobile in the soil and is the nutrient most likely to be deficient in fescues.
Fescue lawns deficient in nitrogen appear yellowish to light green, are often thin and grow very slowly.
The application of too much nitrogen often results in rapid growth and reduces the lawn's tolerance of high and low temperature extremes, traffic and drought.
The monthly nitrogen requirement varies among the fescues.
Tall fescues require a medium level of nitrogen fertility compared to the high level of nitrogen fertility most often required by Kentucky bluegrass (from 1/2 to 1 1/2 pounds of nitrogen per 1,000
square feet per growing month).
Red and chewings fescues usually require from 1/5 to 3/5 pound of nitrogen per 1,000 square feet per growing month.
Some nitrogen sources may be very soluble in water and may release nitrogen for plant uptake very quickly.
Others may be relatively insoluble in water and release nitrogen slowly, over an extended period of time.
No more than 1 pound of quickly available nitrogen should be applied per 1,000 square feet of lawn surface at one time.
Ammonium nitrate [34-0-0, containing 34 percent nitrogen ] and urea are examples of quickly available nitrogen sources.
Isobutylidene diurea , sulfur-coated urea , milorganite [6-2-0, containing 6 percent N and 2 percent available phosphate and ureaformaldehyde are sources of slowly available nitrogen.
Both mature fescue plants and developing seedlings need phosphorus.
Plants deficient in phosphorus may have red to reddish-purple leaves, and may grow slowly, due to low energy levels.
Fescues growing in soils testing low in phosphorus should receive more phosphorus each year than those maintained on soils testing medium or high.
Concentrated superphosphate (0-46-0, containing 46 percent P2O5 and diammonium phosphate (18-46-0, containing 18 percent N and 46 percent P2O5 are common sources of phosphorus.
Hydraulic Considerations for Citrus Microirrigation Systems
Brian Boman and Sanjay Shukla
Hydraulics is the study of the behavior of liquids as they move through channels or pipes.
Hydraulic principles govern the flow of water through irrigation pipes.
A basic understanding of these principles is necessary for understanding the design and operation of irrigation systems.
This publication provides an introduction to hydraulics as it relates to citrus microirrigation.
Important Basic Terminologies Head
Head is the energy in water expressed in terms of the equivalent height of a water column above a given reference.
Head at any point in the irrigation line is the sum of three components: static head, velocity head, and friction head.
Static or Elevation Head e
It is the vertical distance between the water inlet and the discharge point.
It represents potential energy per unit mass of the water.
Velocity Head
It is the energy needed to keep irrigation water moving at a given velocity.
It represents the kinetic energy per unit mass of water.
Friction head is the energy needed to overcome friction in the irrigation pipes and is expressed in units of length.
It is the force per unit area.
It is expressed in pounds per square inch , Pascals, or Newtons per square meter.
A few basic hydraulic concepts related to water movement through pipes are particularly important to the design and proper operation of microirrigation systems.
1 ft3 of water = 62.4 pounds
1 gallon of water = 8.341
Water weighs about 62.4 pounds per cubic foot.
This weight exerts a force on its surroundings, which is expressed as force per unit area or pressure.
The pressure on the bottom of a cubical container filled with water is 62.4 lb/ft2.
Conversion to psi can be achieved as follows:
Pressure on 1 ft2 = 62.41
2.
Brian Boman, Emeritus professor, Department of Agricultural and Biological Engineering, UF/IFAS Indian River Research and Education Center; and Sanjay Shukla, professor, UF/IFAS Southwest Florida REC; UF/IFAS Extension, Gainesville, FL 32611.
Pressure in psi for 1 foot of water = 62.4 lb/1 ft2 = 62.4 lb/144 in2 = 0.433 psi.
Therefore to convert feet of water to psi and vice versa we can use:
Head = Pressure /0.433 = 2.31 = X pressure Eq.
1
The column of water does not have to be vertical.
To calculate the static pressure between two points resulting from an elevation difference, only the vertical elevation distance between the two points needs to be known.
However, other factors such as friction affect water pressure when water flows through a pipe.
If the pressure gauge at the bottom of a tank filled with water displays 10 psi, determine the height of water in the tank.
H1 = Pressure X 2.31 = 10 X 2.31 = 23.1 ft
Velocity is the average speed at which water moves through a pipe.
Velocity is usually expressed in units of feet per second.
Water velocity in a pipe is greatest in the middle of the pipe and smallest near the pipe walls.
Normally only the average velocity of water in the pipe is needed for hydraulic calculations.
To avoid excessive pressure losses due to friction and excessive potentially damaging surge pressures, a rule of thumb for pipe sizing for irrigation pipelines is to limit water velocities to 5 ft/sec or less.
Figure 1.
Typical velocity cross-section profile for full-flowing pipe.
The relationship between flow rate and velocity is given by the equation of continuity, a fundamental physical law.
This equation can be used to calculate flow by multiplying the velocity with the cross-sectional area of flow.
Q =AxV Eq.
2 or 7=Q/A Eq.
3 where,
Q = flow rate in ft3/sec,
A = cross-sectional area of flow in pipes in ft2 ,
V = velocity in ft/sec, and
D = pipe diameter.
If pipe diameters change in sections of the pipe without any change in flow rate, the relationship between flow and velocity can be calculated by:
A1 V1 = A2 Eq.
4 where, A1 = cross-sectional area of flow for first section = velocity in first section A2 = cross-sectional area of flow for second section V2 = velocity in second section
If the velocity is the same in a 2and a 4-inch diameter pipe , the flow rate with the 4-inch pipe would be four times as large as the flow rate from the 2-inch diameter pipe.
Note that the cross-sectional area is proportional to the diameter squared: 2 = 4 in., while 2 = 16 in2.
Therefore, doubling the pipe diameter increases the carrying capacity of a pipe by a factor of 4.
Figure 2.
Diameter and velocity relationships for adjoining pipe sections with a constant flow rate.
This principle is known as Bernoulli's Theorem and can be expressed as :
Determine the flow rate in a 4-in Class 160 PVC pipe if the average velocity is 5 ft/s.
Total head = =He+H,+ Eq.
H H Eq.
6 where, H = Static head or elevation head in feet above some reference point
The I.D.
for 4-in.
pipe is 4.154 ir
Q = AxV = 0.094 ft2x5 ft/s = 0.47 ft3/s
H = P/G and G = specific weight of water)
Q = 0.47 cfs X 448 gal/cfs = 211 gpm
H = Velocity head = V2/2g,
V = velocity in ft/s,
g = gravitational constant 32.2 ft/s2,
H = friction head , and subscripts 1 and 2 refer to two points in the fluid system.
The pressure at the pump of an irrigation system is 30 psi.
A microsprinkler is located at another location, which is 10 ft higher than the pump.
What is the static water pressure at the microsprinkler ?
For convenience, the pump can be used as the elevation datum.
The friction head can be calculated by the Hazen-Williams equation discussed in the next section.

Only a soil test can provide an objective assessment of the need for fertilization and liming, and the appropriate amounts of materials to apply.

For more information regarding collecting and processing a soil sample, please refer to Extension PB1061, Soil Testing.

The nutrient required by fescues in greatest amounts is nitrogen.

Nitrogen is mobile in the soil and is the nutrient most likely to be deficient in fescues.

Fescue lawns deficient in nitrogen appear yellowish to light green, are often thin and grow very slowly.

The application of too much nitrogen often results in rapid growth and reduces the lawn's tolerance of high and low temperature extremes, traffic and drought.

The monthly nitrogen requirement varies among the fescues.

Tall fescues require a medium level of nitrogen fertility compared to the high level of nitrogen fertility most often required by Kentucky bluegrass (from 1/2 to 1 1/2 pounds of nitrogen per 1,000

square feet per growing month).

Red and chewings fescues usually require from 1/5 to 3/5 pound of nitrogen per 1,000 square feet per growing month.

Some nitrogen sources may be very soluble in water and may release nitrogen for plant uptake very quickly.

Others may be relatively insoluble in water and release nitrogen slowly, over an extended period of time.

No more than 1 pound of quickly available nitrogen should be applied per 1,000 square feet of lawn surface at one time.

Ammonium nitrate [34-0-0, containing 34 percent nitrogen ] and urea are examples of quickly available nitrogen sources.

Isobutylidene diurea , sulfur-coated urea , milorganite [6-2-0, containing 6 percent N and 2 percent available phosphate and ureaformaldehyde are sources of slowly available nitrogen.

Both mature fescue plants and developing seedlings need phosphorus.

Plants deficient in phosphorus may have red to reddish-purple leaves, and may grow slowly, due to low energy levels.

Fescues growing in soils testing low in phosphorus should receive more phosphorus each year than those maintained on soils testing medium or high.

Concentrated superphosphate (0-46-0, containing 46 percent P2O5 and diammonium phosphate (18-46-0, containing 18 percent N and 46 percent P2O5 are common sources of phosphorus.

Hydraulic Considerations for Citrus Microirrigation Systems

Brian Boman and Sanjay Shukla

Hydraulics is the study of the behavior of liquids as they move through channels or pipes.

Hydraulic principles govern the flow of water through irrigation pipes.

A basic understanding of these principles is necessary for understanding the design and operation of irrigation systems.

This publication provides an introduction to hydraulics as it relates to citrus microirrigation.

Important Basic Terminologies Head

Head is the energy in water expressed in terms of the equivalent height of a water column above a given reference.

Head at any point in the irrigation line is the sum of three components: static head, velocity head, and friction head.

Static or Elevation Head e

It is the vertical distance between the water inlet and the discharge point.

It represents potential energy per unit mass of the water.

Velocity Head

It is the energy needed to keep irrigation water moving at a given velocity.

It represents the kinetic energy per unit mass of water.

Friction head is the energy needed to overcome friction in the irrigation pipes and is expressed in units of length.

It is the force per unit area.

It is expressed in pounds per square inch , Pascals, or Newtons per square meter.

A few basic hydraulic concepts related to water movement through pipes are particularly important to the design and proper operation of microirrigation systems.

1 ft3 of water = 62.4 pounds

1 gallon of water = 8.341

Water weighs about 62.4 pounds per cubic foot.

This weight exerts a force on its surroundings, which is expressed as force per unit area or pressure.

The pressure on the bottom of a cubical container filled with water is 62.4 lb/ft2.

Conversion to psi can be achieved as follows:

Pressure on 1 ft2 = 62.41

2.

Brian Boman, Emeritus professor, Department of Agricultural and Biological Engineering, UF/IFAS Indian River Research and Education Center; and Sanjay Shukla, professor, UF/IFAS Southwest Florida REC; UF/IFAS Extension, Gainesville, FL 32611.

Pressure in psi for 1 foot of water = 62.4 lb/1 ft2 = 62.4 lb/144 in2 = 0.433 psi.

Therefore to convert feet of water to psi and vice versa we can use:

Head = Pressure /0.433 = 2.31 = X pressure Eq.

1

The column of water does not have to be vertical.

To calculate the static pressure between two points resulting from an elevation difference, only the vertical elevation distance between the two points needs to be known.

However, other factors such as friction affect water pressure when water flows through a pipe.

If the pressure gauge at the bottom of a tank filled with water displays 10 psi, determine the height of water in the tank.

H1 = Pressure X 2.31 = 10 X 2.31 = 23.1 ft

Velocity is the average speed at which water moves through a pipe.

Velocity is usually expressed in units of feet per second.

Water velocity in a pipe is greatest in the middle of the pipe and smallest near the pipe walls.

Normally only the average velocity of water in the pipe is needed for hydraulic calculations.

To avoid excessive pressure losses due to friction and excessive potentially damaging surge pressures, a rule of thumb for pipe sizing for irrigation pipelines is to limit water velocities to 5 ft/sec or less.

Figure 1.

Typical velocity cross-section profile for full-flowing pipe.

The relationship between flow rate and velocity is given by the equation of continuity, a fundamental physical law.

This equation can be used to calculate flow by multiplying the velocity with the cross-sectional area of flow.

Q =AxV Eq.

2 or 7=Q/A Eq.

3 where,

Q = flow rate in ft3/sec,

A = cross-sectional area of flow in pipes in ft2 ,

V = velocity in ft/sec, and

D = pipe diameter.

If pipe diameters change in sections of the pipe without any change in flow rate, the relationship between flow and velocity can be calculated by:

A1 V1 = A2 Eq.

4 where, A1 = cross-sectional area of flow for first section = velocity in first section A2 = cross-sectional area of flow for second section V2 = velocity in second section

If the velocity is the same in a 2and a 4-inch diameter pipe , the flow rate with the 4-inch pipe would be four times as large as the flow rate from the 2-inch diameter pipe.

Note that the cross-sectional area is proportional to the diameter squared: 2 = 4 in., while 2 = 16 in2.

Therefore, doubling the pipe diameter increases the carrying capacity of a pipe by a factor of 4.

Figure 2.

Diameter and velocity relationships for adjoining pipe sections with a constant flow rate.

This principle is known as Bernoulli's Theorem and can be expressed as :

Determine the flow rate in a 4-in Class 160 PVC pipe if the average velocity is 5 ft/s.

Total head = =He+H,+ Eq.

H H Eq.

6 where, H = Static head or elevation head in feet above some reference point

The I.D.

for 4-in.

pipe is 4.154 ir

Q = AxV = 0.094 ft2x5 ft/s = 0.47 ft3/s

H = P/G and G = specific weight of water)

Q = 0.47 cfs X 448 gal/cfs = 211 gpm

H = Velocity head = V2/2g,

V = velocity in ft/s,

g = gravitational constant 32.2 ft/s2,

H = friction head , and subscripts 1 and 2 refer to two points in the fluid system.

The pressure at the pump of an irrigation system is 30 psi.

A microsprinkler is located at another location, which is 10 ft higher than the pump.

What is the static water pressure at the microsprinkler ?

For convenience, the pump can be used as the elevation datum.

The friction head can be calculated by the Hazen-Williams equation discussed in the next section.