Mechanical energy added to a fluid - A pump

 An important component of the general mechanical energy equation

$\dfrac{p_1}{\gamma}+z_1+h_A-h_R-h_L+\dfrac{v_1^2}{2g}=\dfrac{p_2}{\gamma}+z_2+\dfrac{v_2^2}{2g}$        Eq. (01)

is $h_A$ since it represents the mechanical energy added to the fluid so that it can continue its trip through the pipe. At this point, two cases are visualized,

• the energy $h_A$ is one the system requires in order to the pipe system to do its work, and
• the energy $h_A$ is brought from the technical features of a real life pump, for example, so that the conditions of the pipe system are changed in turn.

Of course, in most technical problems both cases presented above are part of the solution in an iterative procedure.

What is the relationsip between $h_A$ and a pump

Since the pump takes electrical energy and transforms it into mechanical energy to be later transferred into the fluid, these  two parameters should be related. This is done through the effeciency of the motor,

$\mathbf{e_M}=\dfrac{Power\, output\, from \, the\, motor}{Power\, delivered\, by\, fluid}=\dfrac{P_O}{P_R}$    Eq. (02)

where $P_O$ is the mechanical power the motor so that the impeller may turn at certain speed with certain force and $P_R$ is the power the fluid actually receives. Since part of the power $P_O$ is wasted as heat or lost due to wearing of mechanical parts, $P_O>P_R$ and as a consequence $\mathbf{e_M}<1$. If $\mathbf{e_M}=1$ you would have an impossible thermodynamical machine. $P_O$ is a parameter measured and supplied, in the name plate, by pump manufacturers so that this data is easy to get.

Notice that Eq. (02) is the same if instead of a pump the case were that of a turbine (actioned by a mechanical energy of the fluid). We would be talking about $h_R$ instead of $h_A$ too.

For a centrifugal pump the power transferred into the fluid and the energy added $h_A$ are related as follows,

$P_R=h_A\gamma Q$        Eq. (03)

where $Q$ is the volumetric flow rate and $\gamma$ a property of the fluid. Also, from Eq. (03) it is obvious that $P_R$ would be very hard to measured. Therefore, it is more common to speak of the efficiency of a pump which we know is smaller than 1 but with present technological advances could be in the range of 0.8 - 0.9 for new equipments. Then, the usual case would be that $P_R$ is unknown, so that,

$P_R=\mathbf{e_M}P_O$        Eq. (04)

and consequently, Eq. (01) becomes,

$\mathbf{e_M}P_O=h_A\gamma Q$        Eq (05)

From Eq. (05), $h_A$ is isolated to be,

$h_A=\dfrac{\mathbf{e_M}P_O}{\gamma Q}$        Eq. (06)

In this way $h_A$ can only be estimated if the efficiency of the pump $\mathbf{e_M}$ and power of the motor of the pump $P_O$ are known.

About the contactor

 For short, a contactor is an encapsulated relay on steroids.

Just as the encapsulated relay, a contactor has:

• a coil,
• a set of contacts and
• a test button.

However, a contactor is built for heavy duty applications.

You can read about the encapsulated relay here: About the encapsulated relay.

Knowing the contactor

A contactor should be easy to identify. You should see coil terminals, auxiliary contacts (NO or NC or a combination of these), power contacts and a small level working as a test button. Depending on the manufacturer and model, these features can be arranged in different ways.

Most models are design for rail mounting inside a cabinet. However, more expensive models may have other features such as small pilot lights or self-coupling to other components such as overload relays, to say an example.

Images of sample contactors

The coil

The contactor coil can be easily identified by the A1 and A2 terminals. You should  remember that depending on the model, the coil may be for DC or AC and design for a given voltage too.

In the contactor, the coil does the same work as the encapsulated relay: when it is energized a set of contacts is open or closed.

The contacts

In the contactor, two types of contacts appear,

• auxiliary contacts and
• power contacts.

Auxiliary contacts are meant for turning on/of devices of low voltage such as: pilot lights or alarms, for example. On the other hand, power contacts are used for powering equipment requiring high voltaje and representing a potential danger for an operator, such as: motors or electrical resistances, to name some.

In a contactor, the number of contacts are,obviously, specified according to the application of the component. It is common practice that auxiliary contacts are NO but one NO and one NC or only NC contacts can be looked for. You can identify the auxiliary contacts in the contactor by NO or NC legend.

Contacts in a contactor. Manufacturer: WEG, Model: CWB9

On the other hand, power contacts are always NO since these are meant for energizing an equipment. You can identify the power contacts by the letters L1, L2, etc. and T1, T2, etc.. The letters L and T stand for: line and terminal; while the numbers indicate correspondence between each terminal, this is: L1 and T1 are a NO contact, and so on. Also, all NO power contacts are independent so that L1 and T2 do not form a contact.

Specification of a contactor

Specifying a contactor is usually reduced to determine the number of NO/NC auxiliary and NO power contacts. Therefore, you should, first, determine what is to be turn on/off by the auxiliary contacts so that a ladder diagram needs to be well established before anything else.

The power contacts are related to the power diagram or circuit. You need to know if the equipment is single or triple phase or something else so that the number of poles (NO contacts) can be determined.

Rating

Contactor rating usually refers to the NO power contacts since these are the ones to be subjected to high voltage and current. These contacts need to be such that can withstand a given maximum voltage and current intensity.

Again, you need to know the electrical features of the equipment to plug into the contactor so that you can safely use the proper option.

Sometimes, rating is also related to the selection of the coil. However, the coil is selected according to ladder diagram and features of other components in the same circuit.

Any question? Write in the comments and I shall try to help.

Other stuff of interest

• Understanding the law of Boyle
• Measuring atmospheric pressure
• Some examples of temperature instruments
• Minor losses  - Formulas
• What is a process variable?
• What are the most important process variables?
• Time dependence of process variables
• A list of process variables
• Composition variables for mixtures

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Ildebrando.

How important is the estimation of minor head loss in transport of fluids

 It depends on several factors. In some cases minor head losses may be too small to waste time in its calculations but in some other it may not.

A discussion and some guidlines can be given about this matter.

Two pipe design cases

Head loss $h_L$, head added $h_A$ and head removed $h_R$ are adjustments to the Bernoulli equation so that these only make sense when the piping components are such that need to be taken into account. For example,

• when control valves are present in a pipe or branch,
• when the piping design is a network,
• when pumps or turbines are part of the pipe,
• among other cases,

heads need to be calculated for a proper balance. Head losses could be of the order of two digits.

However, if the pipe is not too long or intricated then heads could be expected to be small. In this case, head loss could be of the order of one digit.

Factors incresing/decreasing the head loss $h_L$

Head losses may be increased by some factors. Here are some:

• large pipe sections,
• pipe length to diameter ratio larger than three digits,
• pipe accesories such as: elbows, high pressure drop valves, invasive flow meters (orifice plate, venturi meter, etc.),
• pipe elevations.

Warning

You should always estimate head losses to give a reliable result and reasonable design. However, rough estimations discarding some head losses could only be possible in early design stages to have a quick idea of the overall piping.

Any question? Write in the comments and I shall try to help.

Other stuff of interest

• LE01 - AC and DC voltage measurement and continuity test
• LE 02 - Start and stop push button installation 24V DC
• LE 03 - Turn on/off an 24V DC pilot light with a push button
• Minor losses  - Formulas
• What is a process variable?
• What are the most important process variables?
• Time dependence of process variables
• A list of process variables

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Ildebrando.

Does it matter to know how an instrument works?

 The short anwer is YES, it does.

However, one can always make a further question.

Choosing among different instrument techonologies

Why does it?

This question is anwered by recalling about two important activities in engineering:

• instrument selection and
• instrument specification.

You cannot select/specify the most convinient instrument for your control loop or monitoring or recording task unless you know the process requirements such as

• precision,
• installation location,
• available power,
• available budget,
• control mode,
• compatibility with already installed hardware/software, etc.

Once you know the above you need to find the instrument technology that matches this requirements. The only way of doing this is by also knowing the working principles of a number of instruments or technologies available.

You should remember, as well, that selection/specification may become an iterative procedure. This is, a choice for instrument with certain capabilities may be changed in further stages of the design of a control loop so that models, power input, communication, etc. can change a number of times.

Knowing the working principles of the instruments installed in the process may also help in troubleshooting assesment. This not only will save you time but money.

Any question? Write in the comments and I shall try to help.

Other stuff of interest

• LE01 - AC and DC voltage measurement and continuity test
• LE 02 - Start and stop push button installation 24V DC
• LE 03 - Turn on/off an 24V DC pilot light with a push button
• LE 04 - Latch contact with encapsulated relay for turning on/off an AC bulb light
• What is a process variable?
• What are the most important process variables?
• Time dependence of process variables
• A list of process variables

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Ildebrando.

Gross estimation of pipe size (no head calculations)

 Can pipe size be determined without engineering calculations involving the Bernoulli or the energy balance equations?

The short answer is yes; but some experience may be needed and the result may only be point of start or guess for formal calculations.

Here is how to do it, pros and cons.

Some restraints to this approach

Before using this approach you better know the limitations of it:

1. since no pressure gradient nor head losses are being considered the estimated pipe size may not be in agreement with the hydraulic load balance,
2. the pipe size may be larger or smaller than required so that experience is to be used to reduce the margin of error,
3. you will need data sheets for different pipe materials,
4. this approach works better for pipe sections without fittings,
5. most data sheets only have data for water as working fluid (air is sometimes used as well).

Some advantages of this approach

Despite the errors, possibly involved by this method, there are some good things about it:

1. only real pipe sizes are used so that the result is better adapted to already sized equipment characteristics (pumps, heat exchangers, etc.),
2. you can get a good approach to the pipe size in a shorter time than by performing large engineering calculations,
3. pipe material, ranges for flow rate and fluid velocity so that head loss may be balanced are only required.

Gross estimation of the pipe size (no calculations)

The key is in a table having the following data:

• pipe nominal size,
• flow rate,
• fluid velocity,
• working fluid,
• pressure drop (a head loss indication).

A good example for commercial steel pipe Sch. 40 is given in the book Flow of fluids through valves, fittings and pipe. Crane Co. Look in its engineering data appendix. We shall not reproduce the table here since you may find the bok over the internet.

An example of pipe size gross estimation

Consider the a situation in which about $2.5\times 10^{-2}$ m$^3$/s of water are to be transported across 200 m. The pipe to be installed will need a number of bendings and little elevations.

What would be an estimated pipe size for this case?

Solution.

Let us first take a flow rate margin of 5% for the flow rate. Then, we would be looking for pipes capable of transporting about:

$Q=1,500 \pm 75$ L/min of water

Now, if you take a look to the data table from the book Flow of fluids through valves, fittings and pipe. Crane Co several pipe candidates, capable of carrying this amount of fluid, appear!

In fact, pipes of nominal diameters: 3 1/2 inch, 4 inch, 5 inch, 6 inch and 8 inch fit our needs.

What is the best option?

Here is where your engineering experience is important and integration of other information (fittings, elevations, pumps, etc.) help to make a decision.

You may choose one of the pipe sizes by answering the following questions with approximate values:

• How fast will the fluid travel through the pipe?

You should recall that as the internal diameter increases the fluid velocity reduces and vice versa. You may compare different pipe sizes for the same flow rate and see how the fluid velocity dicreases with the size.

• Do you expect the fluid to be highly pressurizsed?

Two options may arise here: to increase the pipe size or to increase the wall thickness (schedule). One way of going around this issue is lby chosing a pipe size whose average capacity matches your needs.

• Economic factors

The larger the pipe size the larger the economic bill. If the extension of the pipe line is short then perhaps economicas may not be a trouble.

• Do you expect head losses to be important?

The parameter $h_L$ is becomes important with the number of fittings, pipe elevation and when pipe line is too long. If you forsee a large $h_L$ in the order of 10 m, for example, then perhaps a larger pipe size would work better. However, the fluid velocity will be reduced too.

Now, back to the question on the most suitable pipe size.

For this case the pipe of ND 4 inch has been selected. The main reasons are:

1. it is large enough to allow a maximum flow rate of 2,200 L/min which is 147 % more than required,
2. being only the second pipe in size it may give a reasonable economic advantage,
3. if fittings are installed this pipe may still give a good margin to keep $h_L$ at reasonable values.

Note: Remember that this is just a gross estimation before more detailed and formal calculations but it may be a good guess.

Any question? Write in the comments and I shall try to help.

Other stuff of interest

• Estimation of internal pipe diameter
• Who was Clausius?
• Units conversion
• LE 04 - Latch contact with encapsulated relay for turning on/off an AC bulb light
• About PID controllers
• Ways to control a process
• About pilot lights
• Solving the Colebrook equation

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Ildebrando.