Determine centrifugal pump specifications

We will explain the operating conditions required to determine the design conditions for centrifugal pump.

Designing centrifugal pumps is an essential skill for mechanical and electrical engineers in chemical plants.

In most cases, analogy from existing equipment is sufficient, but it is unexpectedly difficult to seriously determine design conditions.

After reading this article, you will be able to determine specification conditions based on operating conditions, which will lead to pump purchase and piping design.

It is also useful for trouble analysis during operation.

flow rate

There are two main reasons for determining the flow rate.

1. Flow rate that must be strictly followed during reaction
2. Approximate flow rate that should be secured

In either case, if you carefully calculate the required flow rate, you will end up with a wide variety of flow values.

In such a case, it is a good idea to select several patterns of standard flow rate values.

The patterns are divided into 100L/min, 200L/min, etc.

If the calculation result is 148L/min, set the specified flow rate to 200L/min with some margin.

It is an idea that emphasizes operation management and maintenance management by standardizing the model.

In particular , it is better to follow this concept for process equipment in plants.

Consider large and small exceptional pumps such as utility pumps individually.

Flow rate design

Select from several patterns of flow rate values ​​based on calculation results

lifting height

When designing the head , the first step is to calculate the pressure loss.

In order to standardize the model, it is often divided into 5m or 10m units.

I think 10m increments are just right for batch plants.

Lifting height design

Select from several patterns of head values ​​based on calculation results.

Design example of unification of centrifugal pump specifications

We will divide the steps to standardize pump specifications into three stages.

Calculate pressure loss [Individual optimal design]

First of all, we will steadily calculate the pressure loss.

Calculate using the following method:

• Piping friction loss
• System of pressure loss
• Elements of pressure loss

For example, let’s assume that we calculated the pressure loss of a pump of 0.1m3/min and obtained a result of 22m.

The problem is how to determine specifications based on these results.

First of all, we do not decide on 0.1m3/min x 22m.

At least the lifting height is rounded in units of about 5m.

Compare similar pumps [Overall]

Find similar pumps in step 2

Typically you list pumps from the same plant.

Recently, equipment databases have been developed, so it would be a good idea to take advantage of them.

Let’s say the following is written in the database:

There are only two pumps of 0.1m3/min x 25m.

Looking at this, should I choose a 0.1m3/min x 25m pump?

The second step is to focus on this.

Final decision [Overall optimal design]

Make a final decision in step 3.

It is common to raise the lift

In this example, the company I work for often makes decisions as follows.

The lift height will be increased to 30m, which is the most common type.

This has the following advantages:

• The pump itself can be reused
• It is also possible to reuse spare parts.
• Electric circuits can also be standardized.

It is important to develop a medium- to long-term plan and not just focus on when you purchase equipment.

This idea rarely comes to mind when purchasing equipment. Is difficult.

Do not increase flow rate

This is off topic, but there is also a direction to increase the flow rate to 0.2m3/min.

This is not normally done.

• It is necessary to increase the pipe diameter
• Large power loss
• Unable to ensure minimum flow

The flow rate is directly related to the performance of the pump.

Even if the lifting head is a little higher, it will not be a problem in a batch type chemical plant.

This is because the destination is atmospheric pressure.

This is because there is no need to consider the pressure balance between high and low pressure at the destination.

Physical properties

It is better to give some leeway to the physical properties.

For pumps, density and viscosity are important physical properties.

There are relatively common rules regarding the characteristics of liquids used in batch chemical plants.

If the density is an organic solvent, select 1000 kg/m3 according to water, and if it is other than water, select its physical properties.

It can be used almost unconditionally up to a viscosity of about 10mPa・s.

Physical property design

• If the density is an organic solvent, select 1000 kg/m3. If it is heavier than water, select its specific gravity.
• The viscosity is 10 mPa・s or more, so it is OK to just come up with something.

Operating conditions of centrifugal pumps and differences in pump performance

Now, when you calculate the flow rate and head and place an order with a pump manufacturer, you will notice that there is a slight difference in the operating point and pump performance.

This is a design margin determined by the pump manufacturer.

Although it does not dramatically increase the margin, there is a certain value.

As a user, it is best to assume that the manufacturer selects a pump with the maximum capacity within the range of minimum motor power and no impeller cut.

Tuning of centrifugal pump

We have summarized what happens in actual operation when there is a difference between the design specification point and the pump capacity.

If nothing is done, it will flow at a flow rate greater than the design point.

This is generally OK for batch operation.

Large and small.

From the operation side, the transfer time is shortened by several minutes, which is a positive direction.

Valve opening adjustment

A typical example of operational adjustment is adjusting the valve opening degree.

Reducing the valve opening moves in the direction of making the piping resistance curve steeper.

The intersection of the sloped piping resistance curve and the pump performance curve is shifted to the ” low flow/high head ” side,

This is a method to obtain the desired flow rate.

The cost is zero yen.

In some cases, control (FIC) is performed using a flow meter and regulating valve.

More than 99% of the flow rate control is controlled by this manual valve or adjustment valve.

Features of valve adjustment

• Zero cost yen
• In some cases, it is controlled by a flow meter and regulating valve.

inverter

One method is to use an inverter to control the speed.

When the pump rotation speed is lowered, the flow rate is proportional to the rotation speed, the head is proportional to the square of the rotation speed, and the power is proportional to the cube of the rotation speed.

The idea is that by lowering the pump performance curve but not changing the piping resistance curve, the specified flow rate can be obtained at some point.

If you use an inverter, the power reduction effect is high and it saves energy! That’s a strong opinion.

I haven’t had any success trying this.

The reason is that the performance curve changes rapidly relative to the adjustment range of the inverter’s rotation speed.

Since the head is proportional to the square of the rotational speed, changing just one inverter frequency can significantly change the performance curve.

In reality, even when adjusting the valve opening manually, there are fluctuations within the error range of turning the handle, but with an inverter it is more extreme.

Especially in batch-type chemical plants, large-capacity pumps are limited to utility equipment,

It is human nature to not want to interfere in the direction of lowering the utility ability poorly.

You don’t know when the inverter will break, and at that time it will be in commercial operation.

If only energy saving is the target, it would be better to reconsider the pump selection or challenge the impeller cut.

Number control

Pump number control is not used in batch chemical plants.

Although we may have spare units installed in parallel, we do not operate multiple units.

Number control is really difficult.

Parallel operation

Let’s consider a case where two pumps with the same specifications are operated in parallel.

Since there are two pumps, the performance is doubled!

It’s not going to be that easy…

Operating two pumps in parallel means that on the pump performance curve,

• double the flow rate
• Lifting height does not change

This is the relationship.

The flow rate is doubled compared to the capacity of the pump itself.

However, it must be noted that the operating point is the point where the pump performance curve and piping resistance curve match.

Even if two pumps are arranged in parallel, if the piping size is not changed.

The flow rate will be lower than twice due to the pipe resistance curve.

Series operation

Consider the case where two pumps are arranged in series.

This is used as a booster pump.

Operating two pumps in series means that on the pump performance curve,

• No change in flow rate
• Double the lift

This is the relationship.

When operating in series, there is a strong demand for a booster to increase the necessary head, so in many cases, there is no interest in increasing the flow rate.

The idea is to simply compensate for the lack of lift.

Piping combined resistance

As a supplement to pump performance curves, we will introduce the concept of composite resistance.

When the pipe diameter changes midway through a single liquid destination

Let’s consider a case where the pipe diameter changes midway through a single liquid destination line.

This means considering two pipe resistance curves.

The image above shows a large aperture at first and then narrowed down to a small aperture.

The slope of the piping resistance curve increases from the point where the diameter changes.

The idea is the same as when tightening a valve.

The difference is that the resistance curve changes depending on where the diameter changes along the pipe path.

The point where the slope of the resistance curve bends indicates exactly where the caliber changes .

The bending position will change depending on whether the distance is longer for large-diameter piping or small-diameter piping.

Calculation example: One liquid destination

We will introduce the simplest example of “one liquid delivery destination”.

Look at the diagram below.

There is a pump that sends liquid to 40A piping.

We designed the pump’s capacity and created equipment that can pump liquid at 0.1m3/min.

This is the result derived from pressure drop calculation.

Calculation example There are multiple liquid transfer destinations, but no simultaneous liquid transfer

Let’s look at a case where there are multiple liquid delivery destinations.

This is more complicated than the case where there is only one liquid destination.

The condition is that there is no simultaneous liquid transfer.

Please see the diagram below

The conditions on the 0.1m3/min side sent to tank A are the same as the above case.

Here is the case where the pipe that sends liquid to another tank B, which is a little far away, is extended.

The pipe diameter is 40A, the same as in the example above.

Consider sending only to tank A and only to tank B in this example.

In a batch chemical plant, this is the case when liquid separation is used to separate liquid destinations.

For simplicity, let’s assume that liquids with the same physical properties are sent to tanks A and B.

Tank B has a longer piping distance, so friction loss is greater and the flow rate is lower.

For example, 0.09m3/min is slightly lower than 0.1m³/min.

Once you understand the concept of pressure loss calculation, you can easily visualize it.

When sending liquid to multiple destinations at the same time

Let’s consider a case where one pump is used to simultaneously deliver to multiple locations.

There are not many examples in batch plants.

For simplicity, let’s consider a case where the pipe diameters of multiple destinations are the same and the diameter does not change from the pump outlet to the destination.

In this case, the slope of the piping performance curve becomes gentler after the bifurcation point.

This is because it can be treated almost the same as if the number of pipes were reduced to two at a branch point.

On a cross-sectional area basis, you can think of the diameter as having increased by 1.4 (√2).

The piping resistance curve becomes gentler and the flow rate increases.

Consult with the pump performance curve to determine whether this flow rate can be doubled.

If the flow rate in the branch case is 1.6 times that of when the water is sent to one location without branching, then the flow rate will be approximately 0.8 times the flow rate in each of the two branch pipes.

Strictly speaking, there must be differences in details such as the pressure loss of the branch T-pipe and the shape of the pipe after branching, but this tends to be an academic world.

Although it is possible to use dedicated software to perform calculations, there is no need for such calculations in batch plants.

As a concept, it’s OK if you can predict what the outcome will be.

Calculation example

Finally, let’s consider the case where liquid is sent to multiple tanks at the same time in the above example.

Look at the diagram below.

This flow rate calculation is rather troublesome.

This is the case for utility pumps in batch chemical plants.

Exact calculations will result in repeated calculations.

But I won’t do such difficult calculations.

The reason is “increase the diameter of collecting pipe”.

Check the diagram below.

It is difficult to calculate the pressure loss in the collecting pipe when transferring liquids at the same time.

Calculation of pressure loss on tank A side and calculation of pressure loss on tank B side should be performed first.

This is because you have to calculate by adding the two calculation results.

The usual idea is to increase the diameter of this collecting pipe and omit the pressure loss calculation itself.

Then, the flow rate of liquid sent to tank A and tank B during simultaneous liquid feeding can be calculated as follows.

• Tank A : 0.10/2= 0.050
• Tank B : 0.09/2 = 0.045

This is an approximate calculation.

It simply divides the result of a single calculation by 2. 2 because there are two destinations for liquid delivery.

Since 0.05+0.045=0.095, it seems that there is more margin than the pump capacity of 0.10m3/min.

If we calculate this a little more precisely, we can get the following calculation.

• Tank A : 0.050*(1.0/(0.050+0.045)) = 0.053
• Tank B : 0.045*(1.0/(0.050+0.045)) = 0.047

The error is about 5%, so it is almost negligible.

If it’s this much calculation, you can do additional calculations, but it’s probably not worth the effort.

reference

Excelで解く配管設計法 [ 斉藤義巳 ]

created by Rinker

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Related article

lastly

This article explained how to consider operating conditions in order to determine design conditions for centrifugal pumps used in batch chemical plants.

While looking for leeway in terms of flow rate, head, and physical properties, pump manufacturers are also looking for leeway.

This page summarizes design concepts using pump performance curves, such as valve control and inverter control for flow rate control, and parallel and series operation of two pumps.

Please feel free to post any concerns, questions, or concerns you may have regarding the design, maintenance, and operation of chemical plants in the comments section. (The comment section is at the bottom of this article.)

*We will read all comments received and respond seriously.