Southern Marine Engineering Desk Reference: Second Edition Volume I

By Rolf N. Ekenes

The information contained within this reference compilation is intended to be a helpful guide for the marine engineer in solving problems or answering questions that he or she may encounter daily, as well as problems or questions that may be encountered on a much less common basis. A good deal of this information is also necessary knowledge for any tests or examinations that may be required for the advancement of his or her career in the marine industry. The source primarily used for the direction of this compilation has been the USCG merchant marine engineering question bank for motor-propelled vessels, accessible on the internet at www.uscg.mil/stcw/. Another source is experience. All units of measurement are in imperial/standard units unless otherwise noted. SI/metric units have been used where appropriate.

• Language: English
• Publisher: Xlibris US
• Release date: Feb 23, 2022
• ISBN: 9781664191495

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Copyright © 2022 by Rolf N. Ekenes.

All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without permission in writing from the copyright owner.

Any people depicted in stock imagery provided by Getty Images are models, and such images are being used for illustrative purposes only.

Certain stock imagery © Getty Images.

Rev. date: 02/17/2022

Xlibris

844-714-8691

www.Xlibris.com

773015

To my daughters—Jenifer, Kristine, and Rachel—

for making the world a brighter place.

CONTENTS

Foreword

Chapter 1     General Mathematical Formulae/Terms and Definitions

Air Volume

Apparent Slip

Apparent Slip

Density

Draft

Force

Gear

Horsepower

Hydraulics/Pneumatics

Pipe/Tube/Hose

Pressure Relief Valve

Pumps

Shafts

Specific Gravity

Tapers

Torque

Mensuration

Barrel

Circle

Cone

Cube

Cylinder

Helix

Parallelogram

Polygon

Pyramid

Rectangle

Rectangular Prism

Square

Trapezoid

Triangle

Wedge

Torque

Air

Circular Earth

Line Measuring

Oil

Pipe

Steel Weights

Water

Weight

Glossary

Safety

Chapter 2     Electricity

Ohm’s Law

Variations of Ohm’s Law

Capacitive reactance

Capacitive time constant

Capacitors in parallel

Capacitors in series

Impedance of a parallel circuit

Impedance of a series circuit

Impedance in an R/C circuit (series)

Impedance of an R/L circuit (series)

Impedance with resistance, capacitance, and inductance in series

Inductance

Inductive energy

Inductive reactance

Inductive time constant

Inductors in parallel

Inductors in series

Q of a coil

Quantity of charge

Resistors in parallel

Resistors in series

Resonance of a circuit

Rules for DC Series Circuits

Kirchhoff’s Current and Voltage Laws

Rules for DC Parallel Circuits

Rules for DC Series Parallel Combination Circuits

Henry’s Law

Farad’s Law

L/R Time Constant

R/C Time Constant

Rules for AC Circuits

Transformer

Amperes

Horsepower

Kilowatts

Kilovolt Amperes

Load

Power Factor

Reactive Power

Three-Phase Voltage

Notes

Capacitance, Inductance, and Impedance

Current

DC

Dielectric Strength

Generator

Horsepower

Lead Acid Battery

Motors

Phase

Power Factor, Power Loss, Reactive Power, True Power, and Voltage Drop

Resistance

Watt

Wire Measuring

Glossary

Appendix

Three 29246.png Generator Amperage Chart

Imperial / Standard US conversion tables

Standard US conversion tables

Imperial / Standard US to Metric / SI conversion tables

Potable Water Chlorine Conversion Information

Appoximate Bulk Cargo Weights & Load Factors

Wind and Sea Scale For Fully Arisen Sea

Wind and Sea Scale For Fully Arisen Sea cont.

Decimal and SI Equivalents of Fractional parts of the Inch

Millimeter Equivalents of the Inch

Inch² Equivalents of the Centimeter²

Centimeter² Equivalents of the Inch²

Feet³ Equivalents of the U.S. Gallon

Feet³ Equivalents of the U.S. Gallon

U.S. Gallon Equivalents of the Foot³

U.S. Gallon Equivalents of the Foot³

U.S. Gallon Equivalents of the Barrel

Barrel Equivalents of the U.S. Gallon

U.S Gallon Equivalents of the Liter

U.S Gallon Equivalents of the Liter

Liter Equivalents of the U.S. Gallon

Inch³ Equivalents of the Liter

Equivalents of Mercury (Hg)

Pound per Inch² (psi) Equivalents of Kilogram per Centimeter² (kg / cm²)

Kilogram per Centimeter²(kg / cm²) Equivalents of the Pound per Inch² (psi)

SI & Standard Pressure conversions

Horsepower Equivalents of the Kilowatt

Horsepower Equivalents of the Metric Horsepower

Fractional, Letter, Wire gauge & Metric drill sizes and decimal equivalants

Fractional drill bit speeds for various metals

Tap Drill Sizes for National Coarse & National Fine Threads

Tap Drill Sizes for Taper & Straight Pipe Threads

Metric tap drill sizes

Standard Hex Head Bolt Torque Values, (non lubricated)

ISO Metric Hex Head Bolt Torque Values

ISO Metric Hex Head Bolt Torque Values

Common Hex Head Bolt Data

Common Hex Head Bolt Data

Common Metric Hex Head Bolt Data

Hardness Conversion Table

Machine Screw Torque Values (approximate)

Standard Machine Screw Head Diameter

Standard Flat-Head Socket Head Cap Screw Head Height & Diameter

Standard Socket Head Cap Screw Head Diameter / Height & Hex Key Size

Standard Button-Head Socket Head Cap Screw Head Height & Diameter

Metric Machine Screw Head Diameter

Metric Socket Head Cap Screw Head Diameter / Height & Hex Key Size

Metric Flat-Head Socket Head Cap Screw Head Height & Diameter

Metric Low-Head Socket Head Cap Screw Data

Self Drilling Screw Data

Standard Tapping Screw Head Data

Wood Screw Data

Standard Hydraulic Fitting Wrench & Thread Sizes

Standard Hydraulic Fitting Torque Values

SAE J512 45° Swivel Nut

ISO Metric Hydraulic Fitting Torque Values

Cirumferences, Radii and Areas of Circles

General Drawing List

Electrical Drawing List

General Table List

Electrical Table List

Abbreviations and Acronyms

References

Foreword

The information contained within this reference compilation is intended to be a helpful guide for the marine engineer in solving problems or answering questions that he or she may encounter daily, as well as on a much less common basis. A good deal of this information is also necessary knowledge for any tests or examinations that may be required for the advancement of his or her career in the marine industry. The source primarily used for the direction of this compilation has been the USCG merchant marine engineering question bank for motor-propelled vessels, accessible on the internet at www.uscg.mil/stcw/. All units of measurement are in imperial/standard units unless otherwise noted. SI/metric units have been used where appropriate.

Chapter 1

General Mathematical Formulae/

Terms and Definitions

The information contained within this chapter deals with basic mathematical formulae that are encountered daily and some less common but useful to an engineer while in the performance of his or her duties, accompanied by examples, illustrations, useful notes, and a glossary.

Air Volume

To find the cubic foot per minute of air traveling through a duct (6 × 12) at 100 ft/min

Examples:

6 × 12 = 72 in²

72 ÷ 12 = 6 ft²

6 ft² × 100 = 600 ft²

600 ft² ÷ 12 = 50 ft³

or

6 × 12 = 72

72 × 100 = 7,200

7,200 ÷ 144 (square inch in one square foot) = 50 ft³

To find the cubic feet per minute of air traveling through a duct (5 × 12) at 100 ft/min

Examples:

5 × 12 = 60 in²

60 ÷ 12 = 5

5 × 100 = 500

500 ÷ 12 = 41.6 ft³

or

5 × 12 = 60

60 × 100 = 6,000

6,000 ÷ 144 = 41.6 ft³

To find the difference in construction material per unit length between a round duct and a rectangular duct of the same capacity

A rectangular duct of equal capacity has a perimeter twice the width and depth as compared with a round duct.

Example:

20 in round duct

19 in × 18 in rectangular duct

20 × 3.1416 = 62.83 in

19 + 19 + 18 + 18 = 74 in

74 - 62.83 = 11.17 in/unit length

To find the equivalent circular duct as compared with that of a rectangular duct

A = length of one side of rectangular duct in inches

B = length of adjacent side of rectangular duct

cd = circular equivalent of rectangular duct for friction and capacity in inches

cd = 1.30 × (A × B) × 0.625 ÷ (A + B) × 0.250

To find the horsepower required to drive a fan or blower

Multiply 5.2 × ft³/min × water gauge pressure in inches ÷ (33,000 × efficiency of the fan).

Water gauge in inches = 1.728 oz/in²

To determine the horsepower requirements needed to increase the cubic feet per minute of an existing ventilation fan

1. Calculate the ratio of the new cubic feet per minute to the existing one.

Ratio = new cubic feet per minute ÷ existing cubic feet per minute

2. Calculate the new revolutions per minute required.

Revolutions per minute = new cubic feet per minute ÷ existing cubic feet per minute × existing revolutions per minute

3. Calculate the new static pressure (sp) value.

Static pressure = (new cubic feet per minute ÷ existing cubic feet per minute)² × existing static press

4. Calculate the new horsepower requirements.

Horsepower = (new cubic feet per minute ÷ existing cubic feet per minute)³ × existing horsepower

Example:

Existing ventilation: rpm = 1,725; cfm = 4,000; sp = 0.6; hp = ¾

New ventilation requirement: 8,000 cfm

8,000 ÷ 4,000 = 2 (ratio)

1,725 × 2 = 3,450 (new revolutions per minute required)

0.6 × 4 = 2.4 (static pressure)

0.75 × 8 = 6 (new horsepower requirement)

To determine the velocity of air in a length of pipe

Find the ID of the pipe and the length of the pipe.

V = air velocity

P = pressure loss due to friction

D = the inside diameter of the pipe

L = the length of the pipe

Example:

√(25,000 × D × P) ÷ L = V

√(25,000 × 0.5 × 0.02) ÷ 10 = 176.77662

To determine the volume (cubic feet per minute) of air discharged from a pipe

V = air velocity

A = cross-sectional area of the pipe measured in feet

Example:

Volume (cfm) = 60 × V × A

60 × 176.77662 × 3.1416 = 33,321.685 cfm

Apparent Slip

To calculate the apparent slip of a propeller

A ship travels 250 nautical miles in a 24-hour period with a propeller speed of 60 rpm and having a propeller with a pitch of 20 ft.

Example:

20 ft (propeller pitch) × 60 (revolutions per minute) = 1,200 ft/min

1,200 ft/min × 60 (minutes in an hour) = 72,000 ft/hr

72,000 ft/hr × 24 (hours in a day) = 1,728,000 ft/day

1,728,000 ft/day ÷ 6,076 (feet in a nautical mile) = 284.39763 nautical miles

284.39763 nautical miles - 250 (actual distance traveled) = 34.39763 nautical miles

34.39763 nautical miles ÷ 284.39763 nautical miles = 0.120949074

0.120949074 × 100 = 12.0949074% slip

To calculate the apparent slip of a propeller

A ship left port A at 12:06 with a counter-reading of 616,729 and arrived at port B at 11:48 the next day with a counter-reading of 731,929. The propeller had a pitch of 20 ft, and the observed distance traveled was 404.18 miles at a speed of 16.85 knots.

Example:

Mile = 6,076.12 ft

Pitch = 20 ft

731,929 - 616,729 = 115,200

(115,200 × 20) ÷ 6,076.12 = 379.189

(379.189 - 404.18) ÷ (379.189 × 100) = rate of slip

-24.991 ÷ 37,918.9 = 6.59% slip

Density

To find the density of a liquid

Divide the volume of liquid into its gross weight.

Example:

Gross weight of liquid filling a container = 1,497.6 lbs

Container dimensions = 4 × 3 × 2 = 24 ft³

1,497.6 lbs ÷ 24 ft³ = 62.4 lbs/ft³

Draft

To find the weight, in long tons, of an empty barge with one foot of draft in seawater

Example:

A barge measures 40 ft wide × 20 ft long × 10 ft high.

40 × 20 × 10 = 8,000 ft³

40 × 20 × 1 = 800 ft³ displacement

800 ft³ × 62.425 lbs (density of the seawater at 39.2°F) = 49,940 lbs

49,940 ÷ 2,240 = 22.2946 long tons

To find the weight added, in long tons, to an empty barge with one foot of draft in seawater

Example:

A barge measures 40 ft long × 20 ft wide × 10 ft high. After being loaded, the barge has a draft of five feet.

40 × 20 × 10 = 8,000 ft³

40 × 20 × 1 = 800 ft³

800 ft³ × 62.425 lbs (density of the seawater at 39.2°F) = 49,940

49,940 ÷ 62.425 = 22.2946 long tons empty

40 × 20 × 5 = 4,000 ft³

4,000 ft³ × 62.425 lb (density of the seawater at 39.2° F) = 249,700 lbs

249,700 ÷ 2,240 = 111.47321 long tons

111.47321 - 22.2946 = 89.17861 long tons added

To find the maximum weight load, in long tons, of a barge with a maximum draft of eight feet

Example:

A barge measures 40 ft long × 20 ft wide × 10 ft high.

40 × 20 × 8 = 6,400 ft³

6,400 ft³ × 62.425 lbs (density of the seawater at 39.2°F) = 399,520 lbs

399,520 ÷ 2,240 = 178.35714 long tons maximum load

To find the mean final draft of a vessel with a mean draft of 22.5 ft and a tons per inch immersion (TPI) of 42, after loading 1,175 tons of product or ballast

TPI = 42

Mean draft = 22.5 ft (22 ft, 6 in)

Forward draft = 21.3 ft (21 ft, 4 in)

Aft draft = 23.2 ft. (23 ft, 2 in)

Tons product loaded = 1,175

Mean final draft = tons product loaded ÷ tons per inch immersion

Mean final draft = 1,175 ÷ 42

Mean final draft = 27.976

Force

To find the force required to hold a weight with a block and tackle

Divide the number of times the line passes through the block into the object weight.

Example:

Object weight = 398 lbs divided by 4 passes through a block and tackle

398 ÷ 4 = 99.5 lbs

Gear

Rules for Calculating Gears or Sheaves

(When calculating for gears, use pitch diameter.)

To find the revolutions per minute of a driven wheel or gear, the driver revolutions per minute, diameter, and the diameter of the driven wheel or gear being known

Revolutions per minute = diameter of the driver wheel or gear × revolutions per minute of the driver ÷ diameter of the driven wheel or gear

Example:

4 (diameter) × 1,700 (revolutions per minute) ÷ 12 (diameter) = 566.6 rpm

A 30 hp motor rotating a gear with 24 teeth at 300 rpm driving a gear with 8 teeth at 67% efficiency. The horsepower and revolutions per minute at the driven gear can be calculated:

Example:

24 ÷ 8 = 3

The turns ratio is 1 turn of the drive gear to 3 turns of the driven gear = ratio of 1:3

100% efficiency of the driven gear = 30 hp and 900 rpm (3 × 300)

67% efficiency of the driven gear = 30 hp × 0.67 = 20.1 hp

67% efficiency = 20.1 hp and 900 rpm

A motor with a revolution per minute of 1,750 will be used to drive an air compressor with a 12-inch flywheel. The compressor needs to rotate at 510 rpm. The motor pulley size can be calculated:

1,750 ÷ 510 = 3.43 inches

To find the turns ratio and Rpm of a multiple gear train

Divide the driven gear into the drive gear to find the ratio and multiply the driven gear ratio by the drive gear revolutions per minute to obtain driven gear revolutions per minute. To find drive gear revolutions per minute with a known driven gear revolutions per minute, divide the gear ratio by revolutions per minute.

Gear A = 90 teeth at 150 rpm

Gear B = 75 teeth

Gear C = 22 teeth

Gear D = 30 teeth

1.jpg

Example:

90 ÷ 22 = 4.09 a ratio of 1:4.09

4.09 × 150 = 613.63 rpm

75 ÷ 30 = 2.5 a ratio of 1:2.5

2.5 × 613.63 = 1,532.82 rpm

A 5 hp motor rotating an 8-inch diameter wheel at 300 rpm, driving a 20-inch diameter wheel with an efficiency of transmission of 80%. The revolutions per minute and torque delivered to the driven wheel can be calculated:

Example:

20 ÷ 8 = 2.5

The turn’s ratio is 2.5 turns of the drive wheel to 1 turn of the driven wheel, a ratio of 2.5:1.

100% efficiency of the driven wheel = 5 hp and 120 rpm (300 ÷ 2.5)

80% efficiency of transmission = 5 hp × 0.80 = 4

Rpm = 120

Torque = 4 hp

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Horsepower

To find the horsepower required to drive a pump

Gallons per minute × the total head (including frictional losses for pipe, fittings, and valves) ÷ (33,000 × rate of efficiency of the pump)

or

Gallons per minute × total head ÷ (3,690 × rate of efficiency of the pump)

or

Gallons per minute × pounds per gallon ÷ (33,000 × rate of efficiency of the pump)

or

Liters per minute × kilogram per liter × meters of head ÷ (6,120 × rate of efficiency of the pump)

To calculate the horsepower generated by a hydraulic fluid

Torque × revolutions per minute ÷ 63,025

or

Torque × revolutions per minute ÷ 9,543 = kilowatt

To calculate the horsepower generated by a hydraulic pump

Pounds per square inch gauge × gallons per minute ÷ 1,714

or

Bar (14.50 psi) × cubic decimeter per minute ÷ 600

or

Theoretical horsepower × (100 ÷ rate of efficiency of hydraulic pump and system)

To find the theoretical horsepower generated by a hydraulic fluid

0.000583 × system operating pressure (psi) × pump gallons per minute

To find the theoretical horsepower required to raise water

Gallons per minute × total head in feet (including frictional losses due to pipe, valves, and fittings) ÷ 3,960

or

Gallons per minute × total head in pounds ÷ 1,714

For liquids other than water, multiply the gallons per minute by the specific gravity of the liquid being pumped.

19578.png

To determine water horsepower

The potential power of a stream is the product of feet of head and the weight of the water flowing per second.

Example:

H = total head (including frictional losses due to pipe, valves, and fittings)

C = cubic feet per second or minute

P = pound per cubic feet

Whp = horsepower

C × P × H = work per second, in foot-pound

Whp = work per second, in foot-pound, ÷ 550

or

Whp = 0.144 × C × H

To determine water horsepower when the flow is expressed in gallons per minute

Whp = (gallons per minute × foot of head × 8.33) ÷ 33,000

or

Whp = gallons per minute × foot of head ÷ 3,960

To find the horsepower required to drive a fan or blower

5.2 × cubic feet per minute × water pounds per square inch gauge ÷ (33,000 × rate of efficiency of the fan)

Water pounds per square inch gauge = 1.728 oz/in²

To determine the horsepower requirements needed to increase the cubic feet per minute of an existing ventilation fan

1. Calculate the ratio of the new cubic feet per minute to the existing one.

Ratio = new cubic feet per minute ÷ existing cubic feet per minute

2. Calculate the new revolutions per minute required.

Revolutions per minute = new cubic feet per minute ÷ existing cubic feet per minute × existing revolutions per minute

3. Calculate the new static pressure (sp) value.

Static pressure = (new cubic feet per minute ÷ existing cubic feet per minute)² × existing static press

4. Calculate the new horsepower requirements.

Horsepower = (new cubic feet per minute divided by existing cubic feet per minute)³ × existing horsepower

Example:

Existing ventilation: revolutions per minute = 1,725 cfm = 4,000; sp = 0.6; horsepower = ¾

New ventilation requirement: 8,000 cfm

8,000 ÷ 4,000 = 2 (ratio)

1,725 × 2 = 3,450 (new revolutions per minute required)

0.6 × 4 = 2.4 (static pressure)

0.75 × 8 = 6 (new horsepower requirement)

To find the theoretical horsepower to compress air, at sea level and dry air

Determine the system airflow rate in cubic feet per minute.

Q = the system cubic feet per minute

Hp = 0.2267 Q [([pounds per square inch ÷ 14.696] + 1)⁰.²⁸³ - 1]

Hydraulics/Pneumatics

To determine the correct air compressor cubic feet per minute for a new or existing system

Total all the cubic feet per minute requirements of the system, desired pounds per square inch, air receiver size, system piping size, power source, and the compressor duty cycle. The duty cycle should not exceed 80%. Divide the system cfm requirements by the duty cycle to obtain the operating cubic feet per minute for the required compressor.

Example:

Required cubic feet per minute = 50

Required pounds per square inch = 120

Duty cycle = 80%

50 ÷ 0.80 = 62.5 cfm at 120 psi

Pipe Size for Compressed Air Systems

To calculate the required cubic feet per minute to operate an air cylinder

Piston area × piston stroke × pressure drop constant × compression factor ÷ time (in seconds) × 29

Example:

2-inch piston (piston area = 3.1416)

24-inch stroke

0.064 pressure drop constant

3.7 compression factor

10-second travel time

3.1416 × 24 × 0.064 × 3.7 ÷ (10 × 29) = 0.06156 cfm

To find the required pressure to hold a static hydraulic load

Divide piston area into the load.

Example:

Load = 500 lbs

Piston area = 0.63 in²

500 ÷ 0.63 = 793.6 gauge pressure

To find the force applied to a hydraulic cylinder

Multiply the oil pressure (pounds per square inch) to the area of the cylinder in square inches.

Hydraulic Output Forces in Pounds for Various-Sized Cylinders

Displacement = bore square diameter × 0.7854

Cylinder force = pounds per square inch × piston displacement

To find the pounds per square inch applied to a hydraulic cylinder

Divide the force by the area of the cylinder in square inches.

To find the area of a hydraulic cylinder

Multiply 3.1416 to the square radius of the cylinder in square inches.

or

Square diameter × 0.7854

To calculate the flow and pressure performance of a hydraulic pump

Flow (gallons per minute) = displacement (cubic inches per revolution) × motor revolutions per minute × volumetric efficiency ÷ 231

or

[Rod speed (in inches per minute) × area (in square inches)] ÷ 231

Pressure = motor horsepower × volumetric efficiency × 1,714 ÷ flow (gallons per minute)

To calculate the torque generated by hydraulic oil

(Horsepower × 5,252) ÷ revolutions per minute

or

(Kilowatt × 9,543) ÷ revolutions per minute

To calculate the motor torque generated by hydraulic oil

(Pounds per square inch × displacement in cubic inches) ÷ 62,822

To calculate hydraulic oil viscosity

[20 × system operating pressure (pounds per square inch)] ÷ revolutions per minute = SUS viscosity

or

(300 × bar) ÷ revolutions per minute = SUS viscosity

To calculate the revolutions per minute generated by hydraulic oil

(Horsepower × 5,252) ÷ torque

or

(Kilowatt × 9,543) ÷ torque

or

(231 × gallons per minute) ÷ displacement in cubic inches

To calculate the horsepower generated by hydraulic oil

(Torque × revolutions per minute) ÷ 63,025

or

(Torque × revolutions per minute) ÷ 9,543 = kilowatt

To find the theoretical horsepower generated by hydraulic oil

0.000583 × system operating pressure (pounds per square inch) × pump gallons per minute

To calculate the horsepower generated by a hydraulic pump

(Psig × gallons per minute) ÷ 1,714

or

Bar (14.50 pounds per square inch) × cubic diameter per minute ÷ 600

or

Theoretical horsepower × (100 ÷ rate of efficiency of hydraulic pump and system)

To calculate the required volume of oil to operate a hydraulic cylinder

Area of the piston × piston stroke × 60 ÷ time (in seconds) × 231

Example:

4-inch piston (piston area = 12.5564 in²)

12-inch stroke

10-second travel time

12.5564 × 12 × 60 ÷ (10 × 231) = 3.9168 gpm

To find the speed of a hydraulic cylinder in feet per second

(231 × gallons per minute) ÷ (12 × 60 × cylinder area in square inches)

To find the velocity of hydraulic oil flow in pipe or tubing

(0.3208 × pump gallons per minute) ÷ the internal area of the pipe or tubing in square inches

or

Using the chart, velocity × the internal area of the pipe or tubing in square inches ÷ 0.3208

Single- and Double-Acting Cylinder Power Factors

(in pounds)

Hydraulic Oil Flow Capacity of Tubing

19710.png19716.png

Hydraulic Oil Flow and Pressure Drops per Ten Feet of Pipe

Hydraulic Oil Flow and Pressure Drops in Fittings (psi)

Pipe/Tube/Hose

Standard Pipe Schedules and Capacities

Unthreaded Pipe Dimensions

To determine the pound per foot weight of a pipe

Multiply the wall thickness by 10.6802 and then by the outside diameter minus the wall thickness.

Wall thickness × 10.6802 × (OD - wall thickness)

To determine the ID, in square inches, area of a pipe

0.785 × ID²

To find the pressure loss of a liquid per foot of pipe

Velocity of liquid × gallons per minute ÷ (18,300 × diameter⁴)

To determine the volume, in gallons per minute, of a fluid through a pipe

Multiply the square diameter by the length of the pipe, in feet, and then by 0.048 divided by the fluid velocity per minute through the pipe.

0.0408 × diameter² × feet ÷ water velocity per minute = gallons per minute

or

3.61 × the inside area in square inches × diameter × the rate of vertical volume of fluid in the pipe

or

square diameter × 2.448 × the fluid velocity in foot per minute through the pipe

To determine the velocity of flow in cubic feet per minute of a fluid through a pipe

Multiply the gallons per minute by 0.408 divided by the inside diameter of the pipe.

Gallons per minute × 0.408 ÷ ID = velocity

or

Gallons per minute × 449 = cubic feet per second

To determine the velocity required in cubic feet per minute to discharge a given volume of water in a specific amount of time

Cubic feet of water × 144 × the inside area of the pipe in square inches

To determine the weight of water in a pipe

Multiply the length of the pipe by the square diameter and then by 0.34.

Length × square diameter × 0.34 = weight of water

To determine the maximum pressure change in a 1-inch Sch 40 pipe during a water/fluid hammer

For water/liquids, c = [(E × g) ÷ p]¹/²

Flow rate = 10 gpm

Ambient temperature = 70°F

∆P = change in pressure from the fluid hammer in pounds per square inch

p = fluid density

c = speed of sound in feet per second

v = fluid velocity of 10 gpm or 3.71 ft/sec

Q = fluid flow rate

A = internal area of pipe, 0.00600 ft²

∆V = change in velocity of fluid

g = 32.2 feet per second per second, a gravitational constant

E = bulk modulus of the fluid media, converted from pounds per square inch to pounds per square foot

c = [(E × g) ÷ p]¹/² = [(320 × 10³ psi) × (144 in²/ft²) × (32.2 ft/sec²) ÷ 62.3 lbs/ft³]¹/² = 4,880 ft/sec

∆P = (p × c × ∆V) ÷ g

= (62.3 lbs/ft³ × 4,880 ft/sec × 3.71 ft/sec) ÷ 32.2 ft/sec² = 3502g lbs/ft² = 243 psi

For gases, c = (K × g × Rx T)¹/²

K = ratio of specific heats (1.4 for air)

R = specific gas constant (foot pounds per pound mass per Rankine degree)

T = absolute temperature in Rankine degree

To calculate the rupture pressure of pipe or tubing in pounds per square inch

Multiply twice the wall thickness by the tensile strength of the pipe or tubing material being used divided by the outside diameter. Refer to ASME B31, formally known as ANSI B31, and 46 CFR § 56.07–10(C).

To determine the length of pipe needed for a 90° bend

3.1416² × radius ÷ 4.

Subtract the bend radius from each leg of the 90° angle and add the sum of all three equations.

3.jpg

Example:

3.1416² = 9.8696505

9.8696505 × 18 = 177.6537

177.6537 ÷ 4 = 44.413425

68 - 18 = 50

38 - 18 = 20

50 + 20 + 44.413425 = 114.41342 in

To find the length of the long radius (distance between points A and B) of a 3″ standard seamless pipe bent to 90°

4.jpg

Outside diameter of pipe = 3.5 in

Radius of bend = 5 × diameter of pipe

Outside diameter × 5 = radius + ½ diameter = outside dimension

Outside dimension² + outside dimension² = number AB

√number AB = distance between points A and B

Example:

3.5 × 5 =17.5

3.5 ÷ 2 = 1.75

17.5 + 1.75 = 19.25

19.25² + 19.25² = 741.125

√741.125 = 27.223611 inches (distance between points A and B)

17.5 - 1.75 = 15.75

15.75² + 15.75² = 496.125

√496.125 = 22.273863 inches (distance between points C and D)

or

(2 × 3.1416 × pipe outside diameter × 90° bend) ÷ 360 = greater length of A - B

To determine the minimum bend length of a hose

When determining the bend radius, the hose should be measured to the innermost surface of the curved portion. The bend should take place over the entire minimum length of the hose to avoid damage or kinking.

Angle of bend ÷ 360° × 2 × 3.1416 × radius = minimum hose length required for bend

Example:

To find the minimum length of a 2-inch hose required to make a 90° bend, the hose having a 4.5-inch bend radius:

(90° ÷ 360°) × 2 × 3.1416 × 4.5

0.25 × 2 × 3.1416 × 4.5 = 7.0686 inches

Pressure Relief Valve

To find the pressure at which a relief valve disk reseats

A relief valve disk having an area of 0.85 when seated lifts at 250 psi. The area of the disk increases by 20% when lifted.

Example:

250 psi = 100% closed

250 psi = 100% + 20% open

250 ÷ 120% (1.2) = 208 psi

Pumps

To find the volumetric capacity of a duplex single-acting reciprocating pump

6″ diameter, 14″ stroke, 150 strokes per minute, and pump at 95% capacity, duplex single acting

Examples:

Square diameter × 0.7854 × stroke × strokes per minute ÷ 231 (cubic inches in one gallon of water) × capacity or rate of efficiency

6² × 0.7854 × 14 × 150 ÷ 231 = 257.04 × 0.95 = 244.18 gpm

or

3.1416 × square radius × stroke × strokes per minute ÷ 231 (cubic inches in one gallon of water) × capacity or rate of efficiency

3.1416 × 3² × 14 × 150 ÷ 231 = 257.04 × 0.95 = 244.18 gpm

or

G = discharge in gallons per minute

L = length of stroke in inches

A = area of cylinder in square inches

N = number of strokes (150)

E = efficiency of the pump (95%)

14 × (6² × 0.7854) × 150 × 0.95 ÷ 231

14 × 28.2744 × 150 × 0.95 ÷ 231 = 244.18 gpm

To find the volumetric capacity of a duplex double-acting reciprocating pump

6″ diameter, 14″ stroke, 150 strokes per min, and pump at 95% capacity, duplex double acting

Examples:

Dia² × 0.7854 × stroke × strokes/min ÷ 231 (in³ in one gallon of water) × 2 × capacity or % efficiency

6² × 0.7854 × 14 × 150 ÷ 231 × 2 = 514 × 0.95 = 488.37 gpm

or

3.1416 × square radius × stroke × strokes per minute ÷ 231 (cubic inches in one gallon of water) × 2 × capacity or rate of efficiency

3.1416 × 3² × 14 × 150 ÷ 231 × 2 = 514 × 0.95 = 488.37 gpm

or

GLANE

14 × (6² × 0.7854) × 150 × 2 × 0.95 ÷ 231

14 × 28.2744 × 150 × 2 × 0.95 ÷ 231 = 488.37 gpm

To find the volumetric capacity of a simplex single-acting reciprocating pump

6″ diameter, 14″ stroke, 150 strokes per minute, and pump at 95% capacity, simplex single acting

Examples:

Square diameter × 0.7854 × stroke × strokes per minute ÷ 231 (cubic inches in one gallon of water) ÷ 2 × capacity or rate of efficiency

6² × 0.7854 × 14 × 150 ÷ 231 ÷ 2 = 128.52 × 0.95 = 122.09 gpm

or

3.1416 × square radius × stroke × strokes per minute ÷ 231 (cubic inches in one gallon of water) ÷ 2 × capacity or rate efficiency

3.1416 × 3² × 14 × 150 ÷ 231 ÷ 2 = 128.52 × 0.95 = 122.09 gpm

or

GLANE

14 × (6² × 0.7854) × 150 ÷ 2 × 0.95 ÷ 231

14 × 28.2744 × 150 ÷ 2 × 0.95 ÷ 231 = 122.09 gpm

To find the volumetric capacity of a simplex double acting reciprocating pump:

6″ diameter, 14″ stroke, 150 strokes per minute, and pump at 95% capacity, simplex single acting

Examples:

Square diameter × 0.7854 × stroke × strokes per min ÷ 231 (cubic inches in one gallon of water) × capacity or rate of efficiency

6² × 0.7854 × 14 × 150 ÷ 231 = 257.04 × 0.95 = 244.18 gpm

or

3.1416 × radius² × stroke × strokes per minute ÷ 231 (cubic inches in one gallon of water) × capacity or rate of efficiency

3.1416 × 3² × 14 × 150 ÷ 231 = 257.04 × 0.95 = 244.18 gpm