TTP Practical Wireline Electronics Turning Point Student Rev2

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


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


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


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



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



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



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



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


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



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



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

Documents

TTP Practical Wireline Electronics Turning Point Student Rev2