When a fluid moves from a tank or vessel into a pipe system or vice versa there are pressure losses. This article provides K-values for pipe entrances and exits of various geometries. These K-values may be used to determine the pressure loss from a fluid flowing through these entrances and exits. Listed below are K-values for several common entrance and exit geometries. These may be used in conjunction with the velocity of the fluid in the pipe to calculate the entrance and exit pressure losses.

Refer to the article on pressure loss calculation using the K-value or excess head method for the formula by which the pressure loss may be calculated from the K values below. When a fluid exits a pipe into a much large body of the same fluid the velocity is reduced to zero and all of the kinetic energy is dissipated, thus the losses in the system are one velocity head regardless of the exit geometry.

Email Name. Summary When a fluid moves from a tank or vessel into a pipe system or vice versa there are pressure losses. Definitions : Pipe Diameter : K-value or excess head : radius of rounded pipe inlet. Article Created: January 3, Excess Head. Pressure Drop. Pressure Loss. Tank Nozzle.Hot Threads. Featured Threads. Log in Register. Search titles only. Search Advanced search…. Log in. Forums Engineering Mechanical Engineering.

JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding. Head loss due to sudden expansion. Thread starter previah Start date Aug 28, The underlying assumption in deriving this equation is that the pressure just after the expansion say, p0 is equal to the pressure before the expansion p1. Now, I was wondering how accurate is this assumption?

And what is the underlying physics behind this assumption?

2 sqn no 1 elementary flying training schoolRelated Mechanical Engineering News on Phys. Gold Member. Well, the g should be there, otherwise the dimensions of your equation does not match: m n. About the pressure: yes the static pressure is not the same between a and b.

What I mean is the static pressure at border between small area and big area so at position before b but just after a is assumed to be equal to the static pressure at a.

## Head loss due to sudden expansion

This is the underlying assumption in the momentum balans equation to get to the head loss equation I post earlier it is far easier to explain if I can post a picture, but I do not know how.

Well, the g should be there, otherwise the dimensions of your equation does not match:. FredGarvin Science Advisor. FredGarvin said:. How accurate is this assumption then? I mean if the two pressures are not the same, the equation will have different form. Some colleague say the value of k is to be taken from practical values or practical charts.

Last edited: Mar 15, Long time passed since your question, Dear previah This is for the people who is concerned to head loss in sudden expansioning. I am sure you're misunderstanding the underlying pressure equality assumption. This assumption says "the pressure at the annular area, A2-A1,is equal to the pressure in A1" for the calculation of pressure difference perpendicular to control volume.

This annular area contacts turbulent eddies. So, as you say, " the pressure just after the expansion say, p0 is equal to the pressure before the expansion p1 " has completly different meaning.

You must log in or register to reply here. Related Threads on Head loss due to sudden expansion How to calculate head loss due to sudden enlargement in case of compressible fluid? Last Post Jan 3, Replies 2 Views 7K. Minor loss due to Sudden contraction help needed.

Last Post Mar 26, Documentation Help Center. The Gradual Area Change block represents a local hydraulic resistance, such as a gradual cross-sectional area change. The resistance represents a gradual enlargement diffuser if fluid flows from inlet to outlet, or a gradual contraction if fluid flows from outlet to inlet.

## Need a helping hand?

The block is based on the Local Resistance block. It determines the pressure loss coefficient and passes its value to the underlying Local Resistance block. The block offers two methods of parameterization: by applying semi-empirical formulas with a constant value of the pressure loss coefficient or by table lookup for the pressure loss coefficient based on the Reynolds number.

If you choose to apply the semi-empirical formulas, you provide geometric parameters of the resistance, and the pressure loss coefficient is determined according to the A.

Gibson equations see [1] and [2] :. If you choose to specify the pressure loss coefficient by a table, you have to provide a tabulated relationship between the loss coefficient and the Reynolds number.

In this case, the loss coefficient is determined by one-dimensional table lookup. You have a choice of two interpolation methods and two extrapolation methods.

The pressure loss coefficient, determined by either of the two methods, is then passed to the underlying Local Resistance block, which computes the pressure loss according to the formulas explained in the reference documentation for that block.

The flow regime is checked in the underlying Local Resistance block by comparing the Reynolds number to the specified critical Reynolds number value, and depending on the result, the appropriate formula for pressure loss computation is used.

The Gradual Area Change block is bidirectional and computes pressure loss for both the direct flow gradual enlargement and return flow gradual contraction.

If the loss coefficient is specified by a table, the table must cover both the positive and the negative flow regions. Connections A and B are conserving hydraulic ports associated with the block inlet and outlet, respectively.

N4 jace serial shellThe block positive direction is from port A to port B. By semi-empirical formulas — Provide geometrical parameters of the resistance. This is the default method. Tabulated data — loss coefficient vs. Reynolds number — Provide tabulated relationship between the loss coefficient and the Reynolds number. The loss coefficient is determined by one-dimensional table lookup.

**Fluid Mechanics: Minor Losses in Pipe Flow (18 of 34)**

The table must cover both the positive and the negative flow regions. Internal diameter of the small port, A. The default value is 0. Internal diameter of the large port, B.

This parameter is used if Model parameterization is set to By semi-empirical formulas. The enclosed angle.Documentation Help Center.

### Sudden expansion of a pipe (total head loss)

The Sudden Area Change TL block models the minor pressure losses due to a sudden change in flow cross-sectional area. The area change is a contraction from port A to port B and an expansion from port B to port A. This component is adiabatic. It does not exchange heat with its surroundings. Sudden Area Change Schematic. If the Loss coefficient specification parameter is set to Semi-empirical formulationthe loss coefficient for a sudden expansion is computed as. In the transition zone between sudden expansion and sudden contraction behavior, the loss coefficient is smoothed through a cubic polynomial function:.

If the Loss coefficient specification parameter is set to Tabulated data — Loss coefficient vs.

Reynolds numberthe block obtains the loss coefficient from tabular data provided as a function of the Reynolds number. Area normal to the direction of flow at inlet A. This value must be greater than the cross-sectional area at B.

Area normal to the direction of flow at inlet B. This value must be smaller than the cross-sectional area at A.

The default value is 1e Average distance traversed by the fluid from inlet A to inlet B. This value must be greater than zero. The default value is 0. Parameterization for calculating the loss coefficient due to the sudden area change. Select Semi-empirical formulation to automatically compute the loss coefficient from the cross-sectional areas at ports A and B.

Select Tabulated data — Loss coefficient vs. Reynolds number to specify a 1-D lookup table for the loss coefficient with respect to the flow Reynolds number. The default setting is Tabulated data — Loss coefficient vs.

Reynolds number. Scaling factor for adjusting the loss coefficient value in the contraction portion of the sudden area change. The block multiplies the loss coefficient factor calculated from the semi-empirical expression by this factor.

This parameter is visible only when the Loss coefficient specification parameter is set to Semi-empirical formulation. The default value is 1. Scaling factor for adjusting the loss coefficient value in the expansion portion of the sudden area change. Reynolds number at which flow transitions between laminar and turbulent regimes in the contraction portion of the sudden area change.

The default value is Vector of Reynolds numbers with which to build a loss coefficient lookup table. You specify the Contraction loss coefficient vector and Expansion loss coefficient vector parameters at these Reynolds numbers. This parameter is visible only when the Loss coefficient specification parameter is set to Tabulated data — Loss coefficient vs.

The default vector is a element array ranging from Vector of loss coefficients for the contraction portion of the area change.

Specify the loss coefficients at the Reynolds numbers in the Reynolds number vector parameter.Jump to content. Low Flow in Pipes - posted in Ankur's blog. Posted 30 June - PM. If a gas passing through a 10" pipe has a pressure of 15 psig enters a stack that is 36" in diameter. The stack is 80 ft long and is open to the atmosphere. Would the gas as it enters the stack still have a pressure?

The 15 psig was calculated right before the gas entered the stack. Also, can the transition from the pipe to the stack be treated as a sudden expansion from 10" to 36" or can it be assumed that the pressure will go to atm as it enters the stack?

Posted 01 July - AM. Posted 05 July - PM. Posted 07 July - AM. Posted 07 July - PM. Posted 08 July - AM.

### Borda–Carnot equation

Terry, Same question here Community Forum Software by IP. Featured Articles Check out the latest featured articles. File Library Check out the latest downloads available in the File Library. New Article Product Viscosity vs.

Featured File Vertical Tank Selection. This topic has been archived. This means that you cannot reply to this topic. Posted 30 June - PM If a gas passing through a 10" pipe has a pressure of 15 psig enters a stack that is 36" in diameter. So, at base of the stack I am getting 4. However, if i work back from the top of the stack it does not converge.

The pressure drop across the 36" stack is only 0. The pipe is made up of a combination of 8" and 10" pipe. The pipe starts as 8" turns into 10" and then ties into a 36" stack.

Spyro enter the dragonfly remakeA quick reply: I believe you can consider the transition from 10" to 36" as a sudden expansion using the appropriate K-value. But Bernoulli also tells you that, as the velocity decrease in the pipe when the diameter increase, the pressure will also increase accordingly.

So you will have to calculate a net pressure drop equal to the difference between the pressure loss sudden expansion and the pressure increase due to decrease in velocity.This article provides methods to calculate the K-value Resistance Coefficient for determining the pressure loss cause by changes in the area of a fluid flow path.

These types of pressure drops are highly dependent on the geometry and are not usually covered in simple pressure loss estimation schemes such as a single k-value, equivalent length etc. As the name suggests the permanent pressure loss is not recoverable, and like the pressure loss through ordinary fittings such as a pipe elbow, it is lost to friction, eddies, and noise.

Pressure changes due to acceleration are reversible, for example in a system where the pipe area decreases, some potential energy in the form of pressure is converted to kinetic energy as the fluid is accelerated in the smaller pipe, reducing the fluid pressure in this pipe. In a perfect system if the fluid then entered a larger pipe and decelerated the kinetic energy of the fluid would be converted back into pressure. The degree of permanent pressure lost through a pipe size change is dependent on the geometry of the size change.

Generally the more abrupt the change the higher the losses, while more gradual changes result in much lower pressure drops.

Section 3 presented the equations required to determine the pressure loss through various geometries in terms of K values.

The K value may be converted into head loss through multiplication with the fluid velocity head as shown in the equation below. Some methods require the friction factor to be known of the pipe. For full details of the method of calculating the friction factor see pressure loss from pipe.

Email Name. Summary This article provides methods to calculate the K-value Resistance Coefficient for determining the pressure loss cause by changes in the area of a fluid flow path. Definitions : Internal diameter of pipe : Diameter of the upstream pipe : Diameter of the downstream pipe : Darcy friction factor : Resistance Coefficient : Length as indicated : Reynolds number in upstream pipe : Average velocity : Angle indicate in degrees.

Article Created: November 2, Flow Rate. Fluid Flow. Pressure Drop. Pressure Loss. Sudden Contraction. Sudden Expansion.If you need a quick calculation, but you are not still familiar how to use the calculator, you can order calculation service from the calculator developer.

Order calculation service. Pressure drop or head loss is proportional to the velocity in valves or fittings. For the most engineering practices it can be assumed that pressure drop or head loss due to flow of fluids in turbulent range through valves and fittings is proportional to square of velocity. To avoid expensive testing of every valves and every fittings that are installed on pipeline, the experimental data are used. These values are available from different sources like tables and diagrams from different authors and from valves manufacturers as well.

Kinetic energy, which is represented as head due to velocity is generated from static head and increase or decrease in velocity directly is proportional with static head loss or gain. The number of velocity heads lost due to resistance of valves and fittings is:.

Resistance factor K calculation for elbows, bends, tees, pipe entrance, pipe exit, pipe reducers, pipe enlargement. The head loss due to resistance in valves and fittings are always associated with the diameter on which velocity occurs. The resistance coefficient K is considered to be constant for any defined valves or fittings in all flow conditions, as the head loss due to friction is minor compared to the head loss due to change in direction of flow, obstructions and sudden or gradual changes in cross section and shape of flow.

Head loss due to friction in straight pipe is expressed by the Darcy equation:. For geometrically similar valves and fittings, the resistance coefficient would be constant. Actually there are always smaller or bigger geometrical non similarity in valves and fittings of different nominal size, so the resistance coefficient is not constant.

The resistance coefficient K for a given type of valves or fittings, tends to vary with size as does friction factor for straight clean commercial steel pipe at the same flow conditions. Some resistances in piping like sudden or gradual contractions and enlargements, as well as pipe entrances or exists are geometrically similar. Using momentum, continuity and Bernoulli equation the resistance due to sudden enlargements may be expressed as:.

In order to express the resistance coefficient in terms of larger pipe diameter, following relation should be used:. If the enlargement is not sudden but gradual, or if angle of gradual enlargement is different from OGibson coefficient C e can be used for different angle of divergence as follows:. In other words, if angle of divergence is bigger than 45 Othe resistance coefficient is equal to one for sudden enlargement.

For gradual contraction the resistance coefficient on the same basis based on Crane test data, contraction coefficient C c can be used for different angles of convergence, as follows:. Using above expressions for enlargement and contraction coefficient, resistance coefficient can be calculated as:.

Equations for gradual enlargement and contraction can be used for resistance coefficient calculation for reduced bore straight-through valves like ball valves and gate valves. The total resistance coefficient for this type of ball and gate valves is the summation of resistance coefficient for gradual contraction and gradual enlargement.

Calculate flow coefficient Cv or Kv for known flow capacity and pressure difference. Calculate maximum flow capacity of control valve for known flow coefficient Cv or Kv and pressure difference. Calculate pressure difference for known flow coefficient Cv or Kv and flow capacity.

All the mods 3 java argumentsSelecting the correct valve size for a given application requires knowledge of process conditions that the valve will actually see in service. In the industry of control valves it is practice to use flow coefficient and flow characteristics.

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