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ENGINEERING ARTICLES
Easy Formula
Estimate Strength of Cable
Rule:
1. Change line diameter to eigths2. Square the numerator
3. Divide by the denominator
4. Read the answer in tons
Example:
Estimate the strength of 1/2 inch steel cableDiameter = 1/2=4/8
42/8 = 16/8 = 2
The approximate strength of 1/2 inch steel cable is 2 tons
Find the working strength of manila rope
The working strength of Manila rope is approximately 900*(diameter)2. orW=900d2
where
d is expressed in inches.
W is given in pounds.
Example:
What is the working strength of a 3/4 inch Manila rope ?The maximum recommended pull is W=900*(3/4)2 = 506 lbs.
For rope diameter greater than 2 inches , a factor lower than 900 should be use : in working with
heavier rigging it is well to refer to accepted handbooks to find safe working strength .
How to estimate length of material contained in roll
Where material of uniform thickness, like belting ,is in a roll , the total length may be obtained by the following rule:Measure the diameter of the hole in the center , and of the outside of the roll , both measurements in inches ;
count the number ot turns ; multiply the sum of the two measured diameter by the number of turns ,
and multiply this product by 0.13 ; the result is the total length of the material in feet.
Example:
A roll of belting contains 24 turns . The diameter of the hole is two inches, and of the outsideof the roll is 13 inches:
(2+13)*24*0.13 = 46.8 : the roll contains 46.8 feet of belting
Note:
The rule can be even be applied to materials as thin as pipe line felt ; counting the turns isnot as difficult as might appear with a trial.
How to determine the crop acreage included in a right-of-way strip
Multiple the width of the strip in feet by the length in rods ; divide this by 2640 to obtain the acreage .If the ends of the strip are not parallel , use the length of the center line of the right-of-way.
Example:
A right-of-way 35 feet wide crosses a cultivated field for a length of 14 rods ; how manyacres of crop were destroyed ?
35*14/2640 = 0.18 acres , or almost one-fifth of an acre.
The rule is exact,not to be approximation.
Estimate weight of pipe in metric tons per kilometer
To estimate the weight of pipe in metric tons per kilometer , multiply the nominal diameter by the number ofsixteenths of an inch in wall thickness.
Example:
Find the weight of pipe in metric tons per kilometer for 20 inch diameter pipe , wallthickness 1/4 inch.
16*(1/4)*20 = 80 tons (metric ) per kilometer
Actual answer is 79 metric tons per kilometer
This rule of thumb is based on a density of 490 pounds per cubic foot for steel .For lager diameter thin wall pipe,
this approximation gives an answer usually about one percent low.
How to find pipe weight from outside diameter and wall thickness
Weight (pounds per foot) =(Dt-t2)*10.68where
D = outside diameter in inches
t = wall thickness in inches .
Example:
Outside diameter = 4.500 inch , thickness = 0.25 inchW = (4.5*0.25 - 0.252)*10.68
= (1.125 - 0.0625 )*10.68 = 11.35 pounds per foot .
The above formula is based on a density of 490 pounds per cubic foot for the steel . High-yield-point thin wall
pipe may run slightly heavier than indicated
When should steel be pre-heated before welding
From the chemistry of the steel determine the carbon equivalent which isCarbon Equivalent = C + Mn/4
If it exceed 0.58 the steel may be crack sensitive and should be preheated before welding in ambient temperature
Example:
If steel pipe having a carbon content of 0.25 and manganese content of 0.70 is to be weldedin spring time temperature , ranging from 40oF to 80oF , is preheat necessary?
Carbon Equivalent = 0.25 + 0.70/4 =0.25+0.175 = 0.425
It is not necessary to preheat this particular steel before welding it.
But for another example:
Carbon = 0.20Manganese =1.60
Carbon Equivalent = 0.20 + 1.60/4 = 0.60
This steel should be preheated , particularly for early morning welding.
Why does preheating prevent cracking ? It slows the cooling rate and reduces the amount of austenite retained
as the weld cools . This prevents microcracking.Other alloying elements and pipe wall thickness may also
influence , when joints of high strength pipe should be preheated.
How many pounds of electrodes are required per weld on line pipe
Divide the norminal pipe size by two and multiply the result by one-fourth pound.Pounds of electrode =(N/2)*2.5
Example:
How many pounds of electrode will be used per weld on 10 inch line pipe?Pounds of electrode =(10/2)*2.5 =1.25 pounds per weld.
Calculate V-belt length using simple equation
Rule:
L = 1.57 (D+d)+2CWhere:
L=Belt length , inch
D = Diameter of larger sheave , inche
d = Diameter of smaller sheave , inche
C = Distance between sheave center , inche
Estimate the horsepower that can be transmitted by a shaft
1. Where there are no stresses due to bending , weight of the shaft , pulleys , gear or sprockets,use :HP = (D3N) / 50
Where
D = diameter of shaft in inches , and
N = revolutions per minute.
2. For heavy duty sevice use :
HP = (D3N) / 125
Example:
What horsepower can be transmitted to an atmospheric cooling coil by a two-inch shaft turning at 1,800 RPM ?HP = (23*1,800) / 125
= 115.2 horsepower
Find the stress in pipe wall due to internal pressure
Multiple 1/2 by diameter by pressure and divide by wall thickness to get stress in psiExample:
Find the wall stress due to internal pressure in a 24 inch pipe line having a 1/4 inch wall thickness if the pressuregauge reads 800 psi : (1/2 * 24 * 800 ) / 1/4 = 38,400 psi .
What wall thickness should be specified for a 30 inch discharge line where pressure runs 1,000 psi and wall
stress is to be limited to 30,000 psi ? : ( 1/2 * 30 * 1,000 ) / wall thickness = 30,000 psi
then Wall thickness = 1/2 inch.
This method of calculation transverse tensile stress due to internal pressure on the pipe is the Barlow formula.
Actually , there are two accepted variations , and some designers use inside diameter of the pipe in calculating
this stress. This gives a slightly lower figure for the transverse tensile stress and consequently gives less safety
factor . The method used here is based on the assumption that maximum stress occurs at outside diameter of the
pipe . The results give a conservative value for safety factor calculations.
Estimate the amount of gas blown off thru a line puncture
To calculate the volume of gas lost from a puncture or blowoff , use the formula :Q = D2P
Where
Q = volume of gas in Mcf / hour at a pressure of 14.9 psi , 60oF.and with a specific gravity of 0.60 :
D = diameter of the nipple or orifice in inches ; and
P = absolute pressure in pounds per square inch at some nearby point upstream from the opening.
Example:
How much gas will be lost during a five minute blowoff through a two inch nipple if the upstream pressure is1,000 psi absolutely ?
Q = D2P
Q = ( 2 )2 * 1,000
Q = 4,000 Mcf/hour
Q5 minute = 4,000 * 5/60 = 333 Mcf
Sizing pipe lines for water flow
This form gives a reasonable degree of accuracy.d5 = (GPM2D) / ( 1,000*H )
d = inside diameter of pipe in inches
GPM = gallons per minute
D = equivalent distance ( in feet ) from stand pipe ( or pump ) to outlet
H = total head lost due to friction in feet
Example :
What size line should be run from a water tower to a cooling tower for a minimum of 160 gpm if the equivalentdistance ( actual distance in feet plus allowances for valves and fittings ) is 1,000 feet and the minimum head
supplied by the tower will always be atleast 40 feet ?
Solution:
d5=( 1602*1,000) / ( 1,000*40 ) = 640 inchesd = 3.64 inches
Volume of liquid in vertical cylindrical tanks
Measure the depth of the liquid and either the diameter of the tank ; then the volume inGallons = 0.0034d2h =5.88D2H
Barrels = 0.000081d2h =0.140D2H
Where
d = diameter in inches
h = depth in inches
D = diameter in feet
H = depth in feet
Example :
How many gallons will a tank 12 feet in diameter and 16 feet high hold when full ?Gallons =5.88D2H =5.88(144)(16)
Gallons =13,548 gallons
Estimate the discharge of a centrifugal pump at various speeds
Within the design limits of a pump , the discharge will be approximately proportional to the speed.Q2=Q1(N2 / N1 )
Where Q1 is the gallons per minute at a particular speed N1 and
Q2is the gallons per minute speed N2
Example:
If a pump delivers 350 gpm at 1,200 rpm what will it deliver at 1,800 rpm ?Q2 =350*(1,800 / 1,200 ) = 525 gpm
How to estimate the rate of liquid discharge from a pipe
A quick and surprisingly accurate method of determining liquid discharge from an open ended horizontalpipe makes use of the formula :
gpm = ( 2.56Xd2 ) / y1/2
Where
gpm = gallons per minute
d = diameter of pipe ; inches
X = inches
y = inches
Example:
From the drawing :
X = 24 inches
y = 25 inches
d = 2 inches
gpm = ( 2.56 * 24 * 22 ) / 251/2
=245.76 / 5 = 49.15
How to estimate hp required to pump at a given rate at a desired discharged pressure
Ignoring effects of temperature , viscosity etc .HP = 0.022 * psi * ( barrels per day / 1,000 )
The factor 0.022 is arrived at by using a pump efficiency of 80 percent and a gear efficiency of 97 percent .
Example:
What horsepower would be required to pump 15,000 barrels per day with an 800 psi discharge ?HP = 0.022*800*15,000/1,000 = 264 horsepower
The factor 0.022 was derived from the formula :
HP = (psi)(BPD) / (58,800* pump efficiency * gear efficiency )
How to select motors for field gathering pumps
The following simple formulas will provide a reasonably close approximation of the motor hp necessary todrive duplex double acting pumps normally used in the field.
Rule:
1. hp = (bpd*psi) / 48,0002. hp = ( bph * psi ) / 2,000
3. hp = ( gpm * psi ) / 1,400
Motors selected with these formulas will fit delivery rates used in the formula if they are within capacity of
the pump. If anticipated rate is below pump capacity , a saving may be made in first cost and power bills
by selecting a motor corresponding to the actual rate rather than to full pump capacity.
It should be noted that net bpd values are to be used in formula 1. That is , if 600 barrels are to be pumped
each day , but it is desired to accomplish this in only 8 hours ,the rate to be used in the formular is 1,800 bpd.
How to estimate the head for an average centrifugal pump
By slightly modifying the basic pump formula , H = ( DN / 1,840Ku )2 it is possible to find a means ofestimating the head a pump will develop for each impeller at various speeds.The terms are defind as follow:
H = Head in feet
D = diameter of the impeller in inches
N = RPM of the impeller
Ku = Pump design coefficient. The values of this expression may vary from 0.95 to 1.15 . For estimating purposes
substitute 1,900 for 1,840 Ku. The formula becomes , H = (DN / 1900 )2*( the number of impeller )
Example:
What hrad will be developed by a well designed centrifugal pump with four 11.75 inch inpellers turning at 1,200 rpm ?H = 4*((11.75*1,200) / 1,900 )2 = 4*7.42 =220 feet of head.
Find the capacity of reciprocating pump
Multiply the diameter of the plunger by itself , multiply this by the stroke , then by the number of strokes per minute , and finally by 0.11;the result is the capacity in barrels per day.
C = B2*S*N*0.11
Example:
four inch bore , six inch strike , 80 strokes per minute.4*4*6*80*0.11 = 845 barrels per day.
Note:
The rule is approximate only;it allows about 5 percent for "slip" which is actually somewhat variable.Also note thatit is applicable only to single-acting pumps ; for a double-acting piston type pump , N must be twice the rpm , and a reduction must
be made for the area of the piston rod on the crank end of the pump.
Determining horsepower requirement of pumps
in the selection of pumps it is necessary to calculate the horsepower requirement to handle a specified delivery rate of fluid , againstthe specified head.The formula for the horsepower required by a pump is as follow : HP=( G*H*SpGr )/(3,960*e )
Where:
HP = pump horsepower
G = flow rate , gpm
H = total dynamic head ( change in head ) , feet
SpGr = specific gravity of fluid
e = pump efficiency