Cross-sectional area of a rectangle formula. How to Calculate the Sectional Area of a Pipe

The parallelogram is quadrangular figure, whose opposite sides are pairwise parallel and pairwise equal. Its opposite angles are also equal, and the intersection point of the diagonals of the parallelogram divides them in half, while being the center of symmetry of the figure. Special cases of a parallelogram are such geometric shapes as a square, a rectangle and a rhombus. The area of a parallelogram can be found different ways, depending on what initial data is accompanied by the problem statement.

The key characteristic of a parallelogram, which is very often used in finding its area, is the height. The height of a parallelogram is called the perpendicular dropped from arbitrary point opposite side to the line segment that forms the given side.

In the very simple case The area of a parallelogram is defined as the product of its base times its height.
S = DC ∙ h

where S is the area of the parallelogram;
a - base;
h is the height drawn to the given base.
This formula is very easy to understand and remember if you look at the following figure.
As seen from given image, if we cut off an imaginary triangle to the left of the parallelogram and attach it to the right, then as a result we get a rectangle. And as you know, the area of a rectangle is found by multiplying its length by its height. Only in the case of a parallelogram, the length will be the base, and the height of the rectangle will be the height of the parallelogram lowered to this side.
The area of a parallelogram can also be found by multiplying the lengths of two adjacent bases and the sine of the angle between them:
S = AD∙AB∙sinα

where AD, AB are adjacent bases that form the intersection point and the angle a between them;
α is the angle between the bases AD and AB.
Also, the area of a parallelogram can be found by dividing in half the product of the lengths of the diagonals of the parallelogram by the sine of the angle between them.
S = ½∙AC∙BD∙sinβ

where AC, BD are the diagonals of the parallelogram;
β is the angle between the diagonals.
There is also a formula for finding the area of a parallelogram in terms of the radius of a circle inscribed in it. It is written as follows:

The cross section is formed at a right angle with respect to the longitudinal axis. Moreover, the cross section of different geometric shapes can be represented various forms. For example, a parallelogram has a cross section along appearance resembles a rectangle or square, a cylinder has a rectangle or circle, etc.

You will need

- calculator;
- initial data.

Instruction

To find the sections of a parallelogram, you need to know the value of its base and height. If, for example, only the length and width of the base are known, then find the diagonal using the Pythagorean theorem for this (the square of the length of the hypotenuse in a right triangle is equal to the sum squares of legs: a2 + b2 = c2). In view of this, c = sqrt (a2 + b2).

Having found the value of the diagonal, substitute it into the formula S \u003d c * h, where h is the height of the parallelogram. The result obtained will be the cross-sectional area of the parallelogram.

If the section passes along two bases, then calculate its area using the formula: S \u003d a * b.

To calculate the area of the axial section of the cylinder passing perpendicular to the bases (provided that one side of this rectangle is equal to the radius of the base, and the second is equal to the height of the cylinder), use the formula S = 2R * h, in which R is the value of the radius of the circle (base), S is the cross-sectional area, and h is the height of the cylinder.

If, according to the conditions of the problem, the section does not pass through the axis of rotation of the cylinder, but is parallel to its bases, then the side of the rectangle will not be equal to the diameter of the base circle.

Calculate by yourself unknown side by constructing the circle of the base of the cylinder, drawing perpendiculars from the side of the rectangle (sectional plane) to the circle and calculating the size of the chord (according to the Pythagorean theorem). After that, substitute in S \u003d 2a * h the resulting value (2a is the value of the chord) and calculate the cross-sectional area.

The cross-sectional area of a sphere is determined by the formula S = R2. Please note that if the distance from the center of the geometric figure to the plane coincides with the plane, then the cross-sectional area will be zero, because the ball touches the plane at only one point.

note

Recalculate the result twice: this way you will not make mistakes in the calculations.

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Pipe parameters are determined according to calculations made using special formulas. Today, most calculations are done using online services, but in most cases it is required individual approach to the question, therefore it is important to understand how the pipe cross-sectional area is calculated.

How are the calculations done?

As you know, a pipe is a cylinder. Therefore, its cross-sectional area is calculated from simple formulas, known to us from the geometry course. The main task is to calculate the area of a circle whose diameter is equal to the outer diameter of the product. In this case, the wall thickness is subtracted to obtain the true value.

As we know from the course secondary school, the area of a circle is equal to the product of the number π and the square of the radius:

R is the radius of the calculated circle. It is equal to half its diameter;
Π is a constant equal to 3.14;
S is the calculated cross-sectional area of the pipe.

Let's start the calculation

Since the task is to find the true area, it is necessary to subtract the value of the wall thickness from the obtained value. Therefore, the formula takes the form:

S \u003d π (D / 2 - N) 2;
In this entry, D is the outer diameter of the circle;
N is the pipe wall thickness.

To make the calculations as accurate as possible, you should enter more decimal places in the number π (pi).

D = 1 m; N = 0.01 m.

To simplify, let's take π = 3.14. Substitute the values in the formula:

S \u003d π (D / 2 - N) 2 \u003d 3.14 (1/2 - 0.01) 2 \u003d 0.754 m 2.

Some physical features

The cross-sectional area of the pipe determines the speed of movement of liquids and gases that are transported through it. It is necessary to choose the optimal diameter. No less important is the internal pressure. It is on its value that the expediency of choosing a section depends.

The calculation takes into account not only pressure, but also the temperature of the medium, its nature and properties. Knowledge of the formulas does not exempt from the need to study the theory. The calculation of sewer pipes, water supply, gas supply and heating is based on information from reference books. It is important that all the necessary conditions when choosing a section. Its value also depends on the characteristics of the material used.

What is worth remembering?

The cross-sectional area of the pipe is one of the important parameters that should be taken into account when calculating the system. But along with that, the strength parameters are calculated, it is determined which material to choose, the properties of the system as a whole are studied, etc.

It is quite simple to calculate the pipe section, because there is a series for this standard formulas, as well as numerous calculators and services on the Internet that can perform a number of simple actions. AT this material we will talk about how to calculate the cross-sectional area of \u200b\u200bthe pipe yourself, because in some cases it is necessary to take into account a number of structural features of the pipeline.

Calculation Formulas

When making calculations, it must be taken into account that the pipes are essentially in the form of a cylinder. Therefore, to find the area of their cross section, you can use geometric formula circle area. Knowing the outer diameter of the pipe and the value of the thickness of its walls, you can find the indicator inner diameter needed for calculations.

The standard formula for the area of a circle is:

S=π×R 2 , where

π – constant number, equal to 3.14;

R is the radius value;

S is the cross-sectional area of the pipe, calculated for the inner diameter.

Calculation procedure

Insofar as the main task is to find the area of the flow section of the pipe, the basic formula will be somewhat modified.

As a result, the calculations are made as follows:

S=π×(D/2-N) 2 , where

D is the value of the outer section of the pipe;

N is the wall thickness.

Keep in mind that the more digits in pi you plug into your calculations, the more accurate they will be.

Let's bring numerical example finding the cross section of a pipe, with an outer diameter of 1 meter (N). In this case, the walls have a thickness of 10 mm (D). Without going into subtleties, let's take the number π equal to 3.14.

So the calculations look like this:

S=π×(D/2-N) 2 = 3.14×(1/2-0.01) 2 = 0.754 m 2 .

Physical characteristics of pipes

Also, when designing pipelines, it is worth considering Chemical properties working environment, as well as its temperature indicators. Even if you are familiar with the formulas for how to find the cross-sectional area of a pipe, it is worth studying an additional theoretical material. So, information regarding the requirements for the diameters of pipelines for hot and cold water supply, heating communications or gas transportation is contained in special reference literature. The material from which the pipes are made is also important.

findings

Thus, the determination of the pipe cross-sectional area is very important, however, in the design process, it is necessary to pay attention to the characteristics and features of the system, the materials of the pipe products and their strength characteristics.

Instruction

Strip the cable cores. Using a caliper, or rather a micrometer (this will allow for a more accurate measurement), find the diameter of the core. Get the value in millimeters. Then calculate the cross-sectional area. To do this, multiply the coefficient 0.25 by the number π≈3.14 and the value of the diameter d squared S=0.25∙π∙d². Multiply this value by the number of cable cores. Knowing the length of the wire, its cross section and the material from which it is made, calculate its resistance.

For example, if you need to find the cross-section of a copper cable of 4 cores, and the measurement of the diameter of the core gave a value of 2 mm, find its cross-sectional area. To do this, calculate the cross-sectional area of one core. It will be equal to S=0.25∙3.14∙2²=3.14 mm². Then determine the cross section of the entire cable for this, multiply the cross section of one core by their number in our example, this is 3.14 ∙ 4 \u003d 12.56 mm².

Now you can find out the maximum current that can flow through it, or its resistance, if the length is known. Calculate the maximum current for a copper cable from the ratio of 8 A per 1 mm². Then the limit value of the current that can pass through the cable taken in the example is 8∙12.56=100.5 A. Keep in mind that for this ratio is 5 A per 1 mm².

For example, a cable is 200 m long. To find its resistance, multiply resistivity copper ρ in Ohm ∙ mm² / m, by the length of the cable l and divide by its cross-sectional area S (R = ρ ∙ l / S). Having made a substitution, you will get R=0.0175∙200/12.56≈0.279 Ohm, which will lead to very small losses of electricity during its transmission through such a cable.

Sources:

how to find out the cable size

If a variable, sequence or function has an infinite number of values that change according to some law, it can tend to o limited number, which is the limit sequences. Limits can be calculated in various ways.

You will need

- concept number sequence and functions;
- the ability to take derivatives;
- the ability to transform and reduce expressions;
- calculator.

Instruction

To calculate the limit, substitute the limit value of the argument into its expression. Try to do the calculation. If it is possible, then the value with the substituted value is the desired one. Example: Find the values of the limit with the common term (3 x?-2)/(2 x?+7) if x > 3. Substitute the limit into the expression sequences (3 3?-2)/(2 3?+7)=(27-2)/(18+7)=1.

If there is ambiguity in the substitution attempt, choose a way to resolve it. This can be done by transforming the expressions in which . By making cuts, get the result. Example: Sequence (x+vx)/(x-vx) when x > 0. Direct substitution results in uncertainty 0/0. Get rid of it by removing the common factor from the numerator and denominator. AT this case it will be vx. Get (vx (vx+1))/(vx (vx-1))= (vx+1)/(vx-1). Now the lookup field will get 1/(-1)=-1.

When, under uncertainty, it is impossible to reduce (especially if the sequence contains irrational expressions) multiply its numerator and denominator by the conjugate to remove from the denominator. Example: Sequence x/(v(x+1)-1). The value of the variable x > 0. Multiply the numerator and denominator by the conjugate expression (v(x+1)+1). Get (x (v(x+1)+1))/((v(x+1)-1) (v(x+1)+1))=(x (v(x+1)+1) )/(x+1-1)= (x (v(x+1)+1))/x=v(x+1)+1. After substitution, get =v(0+1)+1=1+1=2.

With uncertainties like 0/0 or?/? use L'Hopital's rule. For this, the numerator and denominator sequences imagine as functions, take from them . The limit of their ratios will be equal to the limit of the ratios of the functions themselves. Example: Find the limit sequences ln(x)/vx, for x > ?. Direct substitution gives the ambiguity?/?. Take the derivatives of the numerator and denominator and get (1/x)/(1/2 vx)=2/vx=0.

For disclosure of uncertainties, use the first wonderful sin(x)/x=1 for x>0, or the second wonderful limit (1+1/x)^x=exp for x>?. Example: Find the limit sequences sin(5x)/(3x) for x>0. Convert the expression sin(5 x)/(3/5 5 x) factor out the denominator 5/3 (sin(5 x)/(5 x)) using the first limit you get 5/3 1=5/3.

Example: Find the limit (1+1/(5 x))^(6 x) for x>?. Multiply and divide powers by 5x. Get the expression ((1+1/(5 x))^(5 x)) ^(6 x)/(5 x). Applying the rule of the second wonderful limit, get exp^(6 x)/(5 x)=exp.

Tip 9: How to find the area of the axial section of a truncated cone

To solve this task, it is necessary to remember what a truncated cone is and what properties it has. Be sure to draw. This will determine which geometric figure is a section. It is quite possible that after this the solution of the problem will no longer be difficult for you.

Instruction

A round cone is a body obtained by rotating a triangle around one of its legs. Straight lines coming from the top cones and intersecting its base are called generators. If all generators are equal, then the cone is straight. At the base of the round cones lies a circle. The perpendicular dropped to the base from the top is the height cones. At the round straight cones height coincides with its axis. The axis is a straight line connecting to the center of the base. If the horizontal cutting plane of the circular cones, then its upper base is a circle.

Since it is not specified in the condition of the problem, it is the cone that is given in this case, we can conclude that this is a straight truncated cone, the horizontal section of which is parallel to the base. Its axial section, i.e. vertical plane, which through the axis of a circular cones, is an isosceles trapezoid. All axial sections round straight cones are equal to each other. Therefore, to find square axial sections, it is required to find square trapezoid, the bases of which are the diameters of the bases of the truncated cones, and the sides are its generators. Truncated Height cones is also the height of the trapezoid.

The area of a trapezoid is determined by the formula: S = ½(a+b) h, where S is square trapezoid; a - the value of the lower base of the trapezoid; b - the value of its upper base; h - the height of the trapezoid.

Since the condition does not specify which ones are given, it is possible that the diameters of both bases of the truncated cones known: AD = d1 is the diameter of the lower base of the truncated cones;BC = d2 is the diameter of its upper base; EH = h1 - height cones.Thus, square axial sections truncated cones defined: S1 = ½ (d1+d2) h1

Sources:

truncated cone area

In the regulatory documents for the design of electrical networks, the cross-sections of the wires are indicated, and only the cores can be measured with a caliper. These values are interrelated and can be translated one into another.

Instruction

To translate the specified normative document section single-core wire to its diameter, use the following formula: D=2sqrt(S/π), where D is the diameter, mm; S - conductor cross-section, mm2 (it is electricians who call "squares").

A flexible stranded wire consists of many thin strands twisted together and placed in a common insulating sheath. This allows him not to break with frequent movements, which is connected with his help to the source. To find the diameter of one core of such a conductor (it can be measured with a caliper), first find the cross section of this core: s \u003d S / n, where s is the cross section of one core, mm2; S is the total cross section of the wire (indicated in the regulatory); n is the number of wires. Then convert the cross section of the wire to the diameter, as indicated above.

Flat conductors are used on printed circuit boards. Instead of a diameter, they have thickness and width. The first value is in advance from the technical data of the foil material. Knowing it, you can find the width by . To do this, use the following formula: W=S/h, where W - conductor, mm; S - conductor cross section, mm2; h - conductor thickness, mm.

Square conductors are relatively rare. Its cross section must be converted either to the side or to the diagonal of the square (both can be measured with a caliper). side is calculated as follows: L=sqrt(S), where L - side length, mm; S - conductor cross-section, mm2. To find out the diagonal by the side length, perform the following calculations: d=sqrt(2(L^2)), where d - square diagonal, mm; L - side length, mm.

If there is no conductor whose cross section exactly matches the required one, use another one that has a larger one, but in no case smaller section. Select the type of conductor and the type of its insulation depending on the application.

note

Before measuring the conductor with a caliper, remove the supply voltage and verify that it is absent with a voltmeter.

Sources:

diameter translation

For example, the diameter of the base of a straight cylinder is 8 cm, and its equal to 10 cm. Determine square its lateral surface. Calculate Radius cylinder. It is equal to R=8/2=4 cm. cylinder is equal to its height, that is, L = 10 cm. For calculations, use a single formula, it is more convenient. Then S=2∙π∙R∙(R+L), substitute the corresponding numerical values S=2∙3.14∙4∙(4+10)=351.68 cm².