ttp practical wireline electronics turning point student rev2
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TTP - Practical Wireline Electronics
Learning Objectives
• Perform calculations using Ohm’s Law
• Perform calculations for equivalent resistance, capacitance,
and inductance
• Describe the basics of electromagnetism and its relationship to
relays, transformers, and phase shifting.
• Perform basic transformer calculations
• Identify problems faced in basic wireline transmission
• Define the relationship between resistance,resistivity, and
conductivity
• Identify the purpose of transformers, diodes, rectifiers,
capacitors, resistors, and inductors in electronic circuits
The fundamental
equation for electronic
calculations
Voltage = Resistance
times Current = V = I *
R
Current = Voltage
divided by Resistance
= I = V / R
Resistance = Voltage
divided by Current =
R = V / I
OHMS Law
What is the equivalent resistance of the circuit below?
R1 R2 R3
Series Resistors Exercise
1. 0.03 ohms
2. 2.73 ohms
3. 25 ohms
4. 30 ohms
Series Resistors
Equivalent Resistance = 5 + 10 + 15 = 30 ohms
Current through each resistor will be the same. Voltage at each
resistor will vary
Equivalent Resistance will be R1 + R2 + R3
R1 R2 R3
What is the equivalent resistance of the circuit below?
R1
R2
R3
Parallel Resistors Exercise
1. 0.03 ohms
2. 2.73 ohms
3. 25 ohms
4. 30 ohms
Parallel Resistors
Voltage at each resistor will be the same.
Current will vary
Equivalent Resistance will be 1/ (1/R1 + 1/R2 + 1/R3)
Equivalent Resistance=1/(1/5+1/10+1/15) = 2.73 ohms
R1
R2
R3
Currents in a Circuit
The total current entering a junction equals the total current
leaving that junction.
Current flows from positive to negative of battery
(electrons flow the opposite direction)
Voltages in a Circuit
5 + 10 + 15 = 30 OHMS
I = V/R = 30/30 = 1 amp
Voltage Drop = IR = 1x5 = 5 Volts
5 Volts 10 Volts 15 Volts
What is the equivalent resistance of the circuit below?
Complex Circuits Exercise
1. 15 ohms
2. 22.5 ohms
3. 52.5 ohms
4. 60 ohms
Power
P = V x I
P = I2 x R
P= V2/R
Two Identical Electrical Bulbs 110V/60W rating, are connected
in the manner shown. What is the resistance of a single Bulb?
110V Bulb1 Bulb2
A
Series Versus Parallel Power Exercise
1. 0.55 ohms
2. 1.8 ohms
3. 60 ohms
4. 202 ohms
Two Identical Electrical Bulbs 110V/60W rating, are connected
in the manner shown. What is the equivalent resistance of the
circuit?
110V Bulb1 Bulb2
A
1. 1.83 ohms
2. 101 ohms
3. 202 ohms
4. 404 ohms
Series Versus Parallel Power Exercise
Two Identical Electrical Bulbs 110V/60W rating, are connected
in the manner shown. What is the current flow in the circuit?
110V Bulb1 Bulb2
A
1. 0.27 amps
2. 1.84 amps
3. 3.67 amps
4. 6.73 amps
Series Versus Parallel Power Exercise
Two Identical Electrical Bulbs 110V/60W rating, are connected
in the manner shown. What is the power output of the circuit?
110V Bulb1 Bulb2
A
1. 29.5 watts
2. 60 watts
3. 120 watts
4. 202 watts
Series Versus Parallel Power Exercise
Two Identical Electrical Bulbs 110V/60W rating, are connected
in the manner shown. What is the current flow in the circuit?
110V Bulb1 Bulb2
B
1. 0.27 amps
2. 1.084 amps
3. 3.67 amps
4. 6.73 amps
Series Versus Parallel Power Exercise
Two Identical Electrical Bulbs 110V/60W rating, are connected
in the manner shown. What is the power output of the circuit?
110V Bulb1 Bulb2
B
1. 29.5 watts
2. 60 watts
3. 120 watts
4. 202 watts
Series Versus Parallel Power Exercise
Equivalent Capacitance for Series Circuits
Capacitance Effect can be seen when checking resistance. The
resistance will start out at some value and build toward an open
circuit. Reversing polarity will start the charging sequence over.
Capacitors can store high voltages so care must be taken to
insure they are discharged!
Equivalent Capacitance = 1/(1/C1 + 1/C2 + 1/C3 + 1/C4) =
1/ (1/2) + (1/2) + (1/2) + (1/2) = 0.5 Farads
2 F 2 F 2 F 2 F
What is the equivalent capacitance of the circuit below?
50 pF 50 pF 50 pF 50 pF
Series Capacitance Review
1. 0.005 pF
2. 0.08 pF
3. 12.5 pF
4. 200 pF
Equivalent Capacitance for Parallel Circuits
Equivalent Capacitance = C1 + C2 + C3 = 2 + 2 + 2 = 6 Farads
2 F 2 F 2 F
What is the equivalent capacitance of the circuit below?
50 pF 50 pF 50 pF
Parallel Capacitance Review
1. 0.007 pF
2. 0.06 pF
3. 16.7 pF
4. 150 pF
Equivalent Inductance for Series Circuits
Equivalent Inductance = L1 + L2 = 2 + 2 = 4 Henrys
2 H
2 H
Equivalent Inductance for Parallel Circuits
Equivalent Inductance =1/(1/L1 + 1/L2) = 1 / (1/2) + (1/2) = 1 Henry
2 H 2 H
Magnets
Bar magnet has a NORTH POLE and SOUTH POLE
LIKE POLES REPEL, UNLIKE POLES ATTRACT
It is surrounded by a MAGNETIC FIELD running north to south
This field passes through some materials (like iron) more easily.
Iron is said to have a low RELUCTANCE.
The lines of force prefer to pass through lower reluctance
materials. We use this principle in some of our tools
Electromagnetism
When current travels through a
wire, a magnetic field is formed
around the wire. If the wire is
coiled, the lines of force link
with each other. The result is a
field with the same shape as
that surrounding a bar magnet.
The field strength is determined
by the number of turns and
current through the coil. We
use this principle in some of our
tools
Phase shifts
Two sine waves(Voltage and Current) in a purely inductive circuit
which start at different times. This is called being out of phase.
The current is LAGGING voltage by 90 degrees. This happens
when magnetic fields induce currents in wire. We use this
difference in phase in some of our tools.
Voltage Current
Relays
Relays are electromagnetic switches which can be made to either SWITCH ON or OFF depending on the current in their coil
Transformers
The voltage induced in the secondary is determined by the TURNS RATIO.
Primary voltage Number of primary turns
--------------------- = ----------- ------------------------
Secondary voltage Number of secondary turns
Primary winding has 100 turns
Secondary winding has 50 turns
So is it a step up or step down? If our primary voltage is 220 volts
What is the secondary voltage? If our primary current is 10 amps
What is our secondary current?
What is our transformer rating in kVA?
Transformers Exercise
Primary winding has 100 turns
Secondary winding has 200 turns
If Primary voltage is 120 volts, what is the secondary voltage?
Transformer Review
1. 60 volts
2. 120 volts
3. 240 volts
4. 440 volts
Primary winding has 100 turns
Secondary winding has 125 turns
If Primary Current is 35 amps, what is the secondary current?
Transformer Review
1. 28 amps
2. 35 amps
3. 43.75 amps
4. 50 amps
Wirelines
Open Hole Wireline has a typical resistance of 10 ohms/1000’
Cased Hole Wireline (5/16”) has a typical resistance of 3 ohms/1000’
You have 20,000 feet of cable on your drum.
You are logging a tool which requires 120 volts and 1 amp to operate.
How much voltage must be placed at the surface to generate sufficient power at the
tool?
Equivalent resistance of cable =(20,000/1000) * 10 = 200 ohms
So how much voltage is required at the surface to power up a tool
requiring 120 volts at 1 amp at the end of 20,000 feet of open hole
cable?
Voltage drop in cable with 1 amp of current flowing= 200 ohms * 1
amp = 200 volts.
Therefore we will require 320 volts at surface to supply 120 volts
at 1 amp at tool
Wireline Power Example
So how much voltage is required at the surface to power up
a tool requiring 120 volts at 850 milli-amps at the end of
23,500 feet of open hole cable (9.8 ohms/1000’)?
Wireline Power Requirement Review
1. 120 volts
2. 230.3 volts
3. 316 volts
4. 340 volts
How Does Temperature Influence It?
• For 7J46 cable with typical resistance of 10 ohms/kft
• Rl = (8.52 + .0218 Ts)L + ((8.57 + .0109(Tb – Ts))D)
• Rl=Total Conductor Resistance (ohms)
• Ts = Surface Temperature (F)
• L = Total length of cable on winch (Kft)
• Tb = Bottom Hole Temperature
• D = Depth of tool (Kft)
• We are logging in a well with BHT of 400 F. Surface temperature is 68 F. Our tool is at a depth of 15,000’ and we have a total of 20,000’ of wireline on our drum.
• What is the resistance of the wireline?
• How much voltage is required at the surface to power a tool requiring 120 volts and 1 amp current?
Rl = (8.52 + .0218*68)5 + ((8.57 + .0109(400-68))15) = 233 ohms
Voltage = 120 volts + 233 volts lost in wireline = 353 volts
A Meg-ohmeter determines quality of the insulation in our wireline
By using high injection
voltage (1000 volts),
resistance values in
excess of 2 Giga-ohms
can be measured.
We consider the leakage
to be acceptable when the
resistance is in excess of
50 Mohms
First determine if the leakage is in the cable head or
rotary connector. These are the most common areas
of leakage.
Cable Head is Clean and Rotary Connector is clean.
Now What??
Gather up a digital voltmeter (Fluke or equivalent)
and a 9 volt transistor battery and some clip leads
and let’s get to work
You Find Leakage on Your Wireline. Now What??
Connect 9 volt battery between the ends of the leaking conductor
Voltage
between
conductor at
cable head and
armor (V1)
Voltage
between
conductor at
Drum End and
armor (V2)
Voltage
between
conductor end
to end (V3)
Location of Leakage
Distance of leakage from Cable head = L * (V1/V3)
Distance of leakage from Drum end = L * (V2/V3)
Cable has same unit resistance along its length
Treat the line on either side of leakage as a resistor
9 volts
Measure voltage drop across line
Wireline Leakage Exercise
Given: Wireline 20,000’ long and the measurements below.
Where is the leakage located?
Cable Leak Review
1. 3667’ from
cable head
(whip)
2. 3667’ from
reel end
3. 16,500’ from
cable head
(whip)
4. 16,500’ from
reel end
Wireline Cable
Cable can experience Capacitance, Inductance, and Resistance
This can generate Attenuation and Distortion
Diodes
+
Diode is a electronic device that allows
current to flow in one direction ONLY
The Diode Effect
can be checked by
reversing the
polarity on the meter
while checking
resistance. The
circuit should show
some resistance in
one polarity and an
open circuit in the
other
The resistance of a
diode while
conducting is
dependant on the
meter used so is
only qualitative.
AC to AC transmission & DC conversion
DC
AC AC
AC
System – Communications
• Data
• Control
• PTC/DIMP
• CHIP
System Tool
Telemetry
Block Diagrams
Good for Troubleshooting
We speak into the microphone but no sound comes out of
the loudspeaker. We are able to measure a signal at the
output of the audio mixer. What can we conclude?
Problem is either in amplifier or loudspeaker
Tool won’t work, there’s no output from the analog board nor the
digital board, but the output from the AC/DC switch board seems
ok – where to start?
Troubleshooting Review
1. Analog Board
2. Digital Board
3. AC/DC Board
4. Power Board
Resistivity & Conductivity
• Resistivity (ρ); Resistance (R) = ρ*L/A
• So the resistance of a wire is related to its length, its
diameter, and the material it is made from.
– Solids ρ increases with temperature
– Liquids ρ decreases with temperature
• Conductivity (σ); σ = 1/ ρ; Conductance = 1/R
– We normally report conductivity in millimhos/m which
would be 1000/ ρ with ρ given in ohm-meters (I Mho/m =
1 Seimen/m)
What will the resistance in ohms of a wire which is 0.5 millimeter in diameter and 10 meter long composed of
copper wire which has a resistivity of 0.0000000168 ohm-m
Resistance Review
1. 0.0000856 Ohms
2. 0.214 Ohms
3. 0.672 Ohms
4. 0.856 Ohms
What will the conductance in mmhos of a wire which is 0.5 millimeter in diameter and 10 meter long composed of copper
wire which has a resistivity of 0.0000000168 ohm-m
Conductance Review
1. 0.000856 mmhos
2. 0.001168 mmhos
3. 1.168 mmhos
4. 1168 mmhos
Review
• Perform calculations using Ohm’s Law
• Perform calculations for equivalent resistance, capacitance,
and inductance
• Describe the basics of electromagnetism and its relationship to
relays, transformers, and phase shifting.
• Perform basic transformer calculations
• Identify problems faced in basic wireline transmission
• Define the relationship between resistance,resistivity, and
conductivity
• Identify the purpose of transformers, diodes, rectifiers,
capacitors, resistors, and inductors in electronic circuits