How to simplify pump pressure drop

Pump pressure drop calculations can be found in chemical engineering textbooks, but the calculations are somewhat complicated.

At least, it is impossible to calculate by hand, and it is often calculated with Excel or special software.

However, it cannot be used when making prompt decisions at the site level.

Therefore, I tried to simplify the calculations to a mental arithmetic level that can be used in the field.

If you can use this content freely, the trust from the manufacturing department will suddenly soar.

Flow of pressure loss calculation

Introduce the flow of pressure loss calculation .

Mechanical engineers sometimes calculate pressure loss when installing mechanical equipment or performing performance verification in a chemical plant.

Calculation of pressure loss is systematized from a chemical engineering point of view, and there are many materials in textbooks and on the Internet.

I myself have written an article, but in practice, only simple calculations are performed.

The following conditions are generally considered for pressure loss in batch chemical plants.

• Liquid below 500kPa
• gas below atmospheric pressure

Liquids often have a density of 1000 kg/m ^{3 and a viscosity of around 10 cP.}

check the flow

Before calculating the pressure loss, first check the flow.

Check the flow to determine whether to calculate pressure loss.

The model is as follows.

Grasping flow velocity

First, find out the velocity .

In some cases, calculations are made, but most companies design standard flow velocities and standard diameters.

Isn’t the idea of ​​standard diameter limited to piping that sends liquids?

This is because the specific volume of a gas varies depending on the temperature and pressure, so it is difficult to grasp the flow velocity.

Whether or not to grasp the flow velocity. The steps would be:

• Check the pipe diameter　　　　　　from the relationship between the liquid standard flow velocity and the standard diameter
• Check the pipe diameter　　　from the relationship between the normal pressure gas standard flow velocity and the standard diameter
• 　　Calculate the specific volume by examining the gas temperature pressure under reduced pressure and calculate the flow velocity

Even if you say to check the flow velocity, first check the pipe diameter.

Most pipe sizes should be sufficient.

The problem is when the diameter goes up and down in one piping line.

If the pipe diameter is changed without knowing the reason, it is often the case that the standard flow velocity is not considered.

In a batch-type chemical plant, the pipe diameter at the gas line heat exchanger outlet is smaller than that at the inlet .

It is necessary to check the part where there is such a change in the pipe diameter.

Piping friction loss calculation

Now, let’s finally introduce the actual calculation of pipe friction loss in a batch-type chemical plant.

Actually, it is done quite roughly .

It is roughly as follows.

Mostly this idea is OK.

What I want to say here is that there is no need to use academic formulas .

For beginners and when doing precision calculations, we do serious calculations.

It is not enough to focus only on the calculation.

Versatility is more important

Even if you don’t like it, you will acquire it if you experience the operation manager / maintenance staff.

As a result of earnest calculation, how much trouble will the site be troubled by not being able to share spare parts?

Once you realize this, the method of equipment design will change.

1. Gather information on similar equipment
2. View the operating conditions, layout, and flow of the equipment to be installed
3. Determine the specifications of the equipment to be installed

This is OK. I don’t count (laughs).

I will do the calculation in the sense of unifying the ideas of the equipment in the plant ^^

Piping height

The pipe height is directly linked to the pump lift.

If anything, pipe friction loss is a minor existence, and pipe height is a major pump head factor.

See the diagram below. This is an image diagram of the plant viewed from the side.

Even if the position of the source tank does not change, there are many candidates for the height of the destination tank.

Send to the 2nd floor, send to the 3rd floor, send to the 4th floor…

The height of the piping is almost automatically determined for the height of the destination tank.

In the factory where I work, one floor is determined by 5m, so the piping height can be easily determined as follows.

Plumbing height is viewed on the safe side of “height to the ceiling of each floor” .

If you have a 5m high floor and you want to lay pipes in a tank on the 2nd floor, the maximum would be 7-8m.

If the pipe height is 10m and applied to the pump lift calculation , there is a margin of 2 to 3m on the pump side.

piping resistance

In calculating the pipe friction loss, the pipe resistance must be calculated.

In reality, this is a 5-10m world .

This is because it limits the standard flow velocity to 1-2m/s.

Although it is said to be 5 to 10m, there are actually two options: 5m or 10m .

See the diagram below. A top view of the plant.

Plants are usually rectangular in shape when viewed from above.

Considering the pump as the center, let’s consider the length of the piping for the source and destination of the liquid.

There are two options: when the pipe length is short and when it is long .

A rough idea like this is enough.

plumbing parts

Plumbing parts must be taken seriously as resistance .

It’s much more important than doing poorly mathematical calculations of friction loss.

Here is what you should think of as resistance.

I arbitrarily call them the three major resistances.

These are “things that are likely to clog” the piping flow .

1. strainer
2. Flowmeter
3. spray nozzle

Strainers and flow meters are often regarded as 5m for the time being .

There are times when I take things a little seriously, but not often.

Spray nozzles have to be taken pretty seriously.

You should check the spray nozzle specifications with the manufacturer.

Considering this as 5m appropriately, buy a pump,

By the time we found out that the spray nozzle had a specification of 20m, it was too late.

Normally, there is a tendency to go from pump design to piping design (spray nozzle design) , but this will fail.

We have to change the order of priority from spray nozzle design to pump design .

In addition, the specifications of “things that easily clog” must be decided before designing the pump.

Formula for calculating pressure loss of a pump

There is a formula for calculating the pressure loss of a pump.

See the figure below first.

The formula for calculating the pressure loss of a pump is generally written as:

$$P_1+ρgH_1+\frac{1}{2}ρ{v_1}^2+W=P_2+ρgH_2+\frac{1}{2}ρ{v_2}^2+ΔP_2$$

This is a little complicated, isn’t it?

Verbalizing it might be easier.

(energy at source) + (energy applied by pump to fluid) = (energy at destination)

This is the basic relationship.

The source energy and the destination energy consist of the following three components.

• pressure energy
• potential energy
• velocity energy

The energy that the pump adds to the fluid is now

$$W$$

It is simplified and expressed as

Consider pipe friction loss here. this

$$ΔP$$

It is simplified and expressed as

Calculation formula for piping friction loss

The calculation formula for pipe friction loss is shown.

$$ ΔP = 4F\frac{1}{2}ρv^2\frac{L}{D} $$

The notation for this varies from textbook to textbook, but the idea is of course the same.

Piping friction loss is proportional to kinetic energy

Pressure loss is proportional to kinetic energy.

For fluid mechanics, the kinetic energy is

$$\frac{1}{2}ρv2$$

is defined by

Pressure loss is friction loss itself.

Friction loss is usually defined as the square of the velocity .

There are some fields that are defined by the absolute value of velocity…

You can intuitively understand that the higher the density, the higher the friction loss.

Which has greater pressure loss, water or air? It’s water.

Putting them all together, we consider the pressure drop to be proportional to the kinetic energy .

In the figure below, the pressure loss is higher in the lower stage where the velocity is faster for the objects flowing in the same pipe.

Proportional to piping route

Pressure loss is proportional to the piping route.

Even though it is a piping route, a simple expression is used here.

$$\frac{L}{D}$$

L is the pipe length and D is the pipe diameter.

The longer and thinner the pipe, the greater the resistance.

The longer the pipe length L, the greater the resistance.

The smaller the pipe diameter, the greater the resistance.

This is also intuitive and easy to understand.

When it comes to piping shapes, we have to consider fittings such as elbows, tee, and reducers.

Replace the friction loss on the fitting with the friction loss for the length of the piping of 〇m .

This is called the equivalent length .

This means that the more complicated the piping, the larger the L.

Please check the figure below for an image.

It is an image of replacing bends such as elbows with straight piping.

Proportional to the roughness of the piping

Pipe friction loss is proportional to the surface roughness of the pipe .

It’s a hard part to see, so it’s hard to be conscious of it.

It will be easier to understand if you imagine the ground of soil and the ground of ice.

Even inside the pipe, the roughness differs depending on whether it is steel, stainless steel, or fluororesin lined pipe.

Calculation example of piping friction loss

Here is an example of calculation of pipe friction loss in a batch chemical plant.

Remove unnecessary terms

It starts with organizing the above formula.

• \(P_1=P_2=0\)
• \(H_1=0\)
• \(\frac{1}{2}ρv^2=0\)
• \(ΔP_1=0\)

This is organized and explained below.

\(P_1=P_2=0\)

In a batch-type chemical plant, the pressure inside the tank before and after liquid transfer is considered to be zero .

• All tanks are open to the atmosphere.
• Batch operation is based on total liquid transfer.
• Pump at the beginning or end of the process.

Based on this idea, the pressure during liquid transfer is assumed to be zero.

\(H_1=0\)

The height of the source tank is considered zero .

Strictly speaking, it is very rare to consider the height from the bottom of the tank to the pump.

It is safer to set the height to zero without considering its height.

In reality, the liquid height in the tank is the available energy.

Expecting this, I mean “better not calculate the head required for the pump”.

\(\frac{1}{2}ρv^2=0\)

In the first place, the proportion of kinetic energy to the whole is very low .

It is common to think of v = 1 to 2m/s. Even if v = 2, when ρ = 1000 (water),

$$\frac{1}{2}pv^2=\frac{1}{2}*1000*2^2=2$$

When considering H=10m as potential energy

$$ρgH=1000*9.8*10=98$$

becomes.

Overwhelming!

A negligible amount.

\(ΔP_1=0\)

This comes from the relationship “v ₁ < v _{2 “.}

If the diameter of the pipe at the source > the diameter of the pipe at the destination

$$v_1<v_2$$

becomes.

The reason why the diameter of the pipe to which the liquid is sent > the diameter of the pipe to which the liquid is sent is to prevent cavitation on the suction side of the pump.

If the pipe diameter changes by one size, the diameter will change by about 25%.

$$\frac{v_1}{v_2}=(\frac{1}{1.25})^2=0.64$$

Since the pressure loss is the square of the flow velocity,

$$(\frac{v_1}{v_2})^2=0.64^2=0.4$$

At ΔP ₁ (suction side), you can see that the kinetic energy, which is important in calculating pressure loss, is quite small.

Considering the pipe length L and pipe diameter D, we can see that ΔP ₁ can be ignored compared to ΔP _{2 .}

Formula after tidying up

Let’s take a look at the cleaned-up formula above, with unnecessary terms removed.

$$W=ρgH_2+ΔP_2$$

Only this.

It’s that easy!

The model considered in the actual calculation can be simplified so far.

This means that the calculation process becomes very simple.

• Determine pipe height
• Calculate Pipe Friction Loss
• The head of the pump is determined

This principle is very important in pump pressure drop calculations for batch chemical plants.

Pipe friction coefficient 4f

The pipe friction coefficient is 4f or λ, and the notation is slightly different.

A common method is to use the moody diagram.

In a batch chemical plant

$$4f = 0.02$$

About is enough.

This is because the density and viscosity of the liquid are almost the same, and the piping diameter design is based on the standard flow velocity.

This alone makes the Reynolds number Re almost constant.

The shape of the piping is also fixed, such as iron or stainless steel.

The factor of ε/d determined by the surface shape of the piping is also fixed.

From here, the pipe friction coefficient of 4f is often within the range of 0.01 to 0.03.

Roughly assuming the most troublesome pipe friction loss calculation of pipe friction loss calculation,

It can greatly simplify the calculation steps.

kinetic energy of fluid

The kinetic energy of the fluid is the following fraction.

$$\frac{1}{2}ρv^2$$

Considering ρ = 1000 for water and a velocity of 1m/s,

$$\frac{1}{2}ρv^2 = \frac{1}{2}×1000×1^2 = 500$$

This is also easy, but even if you do detailed calculations, the results will not change the digits significantly, so it is OK.

Piping shape

The kinetic energy of the fluid is the following fraction.

$$\frac{L}{D}$$

This is an influential factor here.

Let’s proceed with L=50 and D=0.05.

50m is quite high for the size of a batch chemical plant.

0.05mm, that is, 50A is also standard for batch chemical plants.

$$\frac{L}{D} = \frac{50}{0.05} = 1,000$$

Pipe friction loss results

Let’s summarize the calculation results.

$$ΔP =4f \frac{1}{2}ρv^2 \frac{L}{D} = 0.02×500×1,000 = 10,000 (J)$$

Since the unit of the calculation result is J, to replace it with the unit of m

$$ρg$$

to rebate.

Since it is water, ρ=1000 and the gravitational acceleration g is about 9.8, roughly 10.

$$ρg = 1000×10 = 10,000$$

Therefore,

$$ΔP = 1 (m)$$

This is the result.

Since the pipe height and valve loss are considered in units of 5m, it is easy to see that the pipe friction loss of 1m can be ignored!

Easy validation of results

Even if the pipe friction loss of 4f, which had been roughly assumed, doubled, the pipe friction loss would only be 2m.

Even if the pipe diameter is 50A to 25A, if the flow velocity does not change, the pipe friction loss will only be 2m.

If the flow velocity changes, the effect is large, but the flow rate is decreased accordingly.

As a result, it can be seen that there are not many factors that increase pipe friction loss.

Exceptions exist, but if you know the formula for piping friction loss and its results,

You can handle exceptions as an application.

Simple calculation of piping route and pressure loss

The results of a simple calculation of pressure loss are shown using piping often seen in batch chemical plants as an example.

For pump up

First, we will introduce the most common situations in calculating pressure loss in batch chemical plants.

In the case of pump-up as shown in the figure below.

Here we introduce the elements that require pressure loss calculation and their numerical values.

Ignore suction conditions

In a batch-type chemical plant, when liquid is sent from tank A to tank B, the suction side is free.

The condition of tank A is not a constraint like it is in a continuous plant.

Tank A pressure is 0, strainer pressure loss is 0, pump suction pressure loss is 0

It is normal to think in this condition.

You can consider the pressure loss of the strainer, but there is no limit.

If tank A has a certain height, it is normal that the strainer pressure loss can be covered by the head pressure.

Discharge conditions are piping, height, and CV

There are three elements to be considered for discharge conditions: friction loss of piping, piping height, and CV .

This is because in a batch system, it is common for tank B to be free, just like tank A.

In some cases, the pump may run while tank B is pressurized, but this is extremely limited.

Friction loss and height of piping are items that must be considered when calculating the pump head.

CV calculation assumes that there is a control valve in the line .

It is not so common to adjust the flow rate while pumping up in a batch system.

It’s not that there aren’t any, but many people don’t use pumps to limit flow.

This is because even though the pump can increase the pressure, fluctuations due to deterioration, etc. can occur.

The reason for restricting the flow rate is that there is a constraint to ensure the flow rate necessary for operation.

The pump, which is an uncertain factor, is rarely used there .

This is the concept of the batch system.

In addition, there are the following special conditions.

Simple pressure loss

• Strainer 5m
• Outlet nozzle is spray type 5-10m

for the head

In a conflicting relationship with Pump Up is Head.

This is the following scene.

As with pump-up, list the elements required for pressure loss calculation.

Tank height and flow meter pressure drop

In the case of the head, it is the same as the pressure loss of the pump, and the internal pressure of tank A, the strainer, and the pressure loss of tank B should be 0.

In addition to this, piping pressure loss is also 0

This is because the flow velocity at the head is very small.

It does not produce high pressure like a pump, and if the flow velocity is slow, piping friction loss can be almost ignored.

So, the difference in height between tank A and tank B and the CV value of the flow meter are almost settled.

The height of tanks A to B is assumed to be 5m, but this will change depending on the size and layout of the factory.

If the difference in height is less than 1m, be careful!

The CV calculation may also give unsatisfactory results.

In this case, we will do the following:

• Drop with the head while pressurizing tank A (increase the internal pressure of tank A)
• Change the operation method so that a certain amount is always stored in tank A (take the height of tank A to tank B)

Basic interpretation of performance curves

The performance curve of a pump is an important curve for knowing the capacity of the pump.

It generally looks like the figure below.

This figure draws a single curve, but this alone is almost meaningless.

Several plots are superimposed to form an important plot showing various capabilities of the pump.

I will explain the basic curve of the performance curve.

Bernoulli’s law

A law called Bernoulli’s law appears in fluid mechanics.

Bernoulli’s law is the application of the law of conservation of energy in mechanics to fluids.

The law of conservation of energy in mechanics is a law that shows the relationship that potential energy + driving energy = constant .

Bernoulli’s law generalizes this as the law of conservation of energy for fluids.

Potential energy + kinetic energy + pressure energy = constant

It can be said that it is a law to which the pressure energy is added.

The potential energy is

$$ρgh$$

the kinetic energy is

$$\frac{1}{2}ρv^2$$

pressure energy is

$$P$$

define each.

Bernoulli’s law is sometimes introduced by converting units such as the unit of pressure and the unit of head.

I find it easy to understand the method this time, which is aligned with the unit of pressure.

Because it can be related just by replacing the mass m of the dynamic potential energy and kinetic energy with the density ρ.

Performance curve and Bernoulli’s law

Performance curves are related to Bernoulli’s law.

The performance curves shown above are reproduced.

Let’s start by explaining what this curve means.

Horizontal axis is flow rate

The horizontal axis indicates flow rate.

It varies depending on the size of the pump, such as m3/h or L/min.

Vertical axis is lift

The vertical axis indicates the lifting height.

This is because pump performance is determined by the relationship between flow rate and head.

By arranging various parameters on the vertical axis, many curves can be superimposed.

Interpretation of performance curves

The performance curve of the pump is

• Almost maximum lift at 0 flow rate
• Lift decreases as flow increases

There is a relationship.

This is actually related to Bernoulli’s law.

The pump moves → Energy is added to the fluid → Dispersed into potential energy and kinetic energy

The relationship between the flow rate and the head is the relationship that shows the relationship.

In formula we write:

$$H=H_{0}+\frac{1}{2}ρ(Q/d)^2$$

H and H0 are lifting heights respectively.

ρ is the density, Q is the flow rate, and d is the pipe diameter.

Q/d indicates flow velocity.

Let v be the flow velocity

$$\frac{1}{2}ρ(Q/d)^2=\frac{1}{2}ρv^2$$

You can express it.

The pressure energy is not expressed in Bernoulli’s law because

This is because the definition of lift is “the height of the liquid level at which the pressure is 0”.

Relationship between pressure drop curve and performance curve

This may sound abrupt, but pressure loss is determined by the relationship between flow rate and pressure.

Details are in the article below.

Let’s write down the formula for calculating pressure loss again.

$$ΔP=4f\frac{1}{2}ρv^2\frac{L}{D}$$

When discussing pump pressure loss here, the following values ​​are fixed:

$$f,ρ,L,D$$

f is the frictional resistance of the pipe, and is automatically determined when the pipe material and construction method are determined.

When designing a pump, the piping specifications are fixed, so the idea of ​​changing f is usually not considered.

L is the pipe length and D is the pipe diameter, which are determined at the pump design stage.

Both L and D have a low degree of freedom in terms of design elements.

When the destination is decided, L is almost decided.

When the liquid transfer volume is determined, Q is almost determined as well.

There is no degree of freedom for the contents.

Once these values ​​are determined

$$ΔP∝Q^2$$

The pressure loss is proportional to the flow rate (proportional to the square of the flow rate).

Let’s take a look at the relationship between the pressure loss curve and the performance curve.

The operating point is the intersection of the “pressure loss” curve and the performance curve.

Once the piping specification is established and the process specification is determined, one pressure drop curve is obtained.

In contrast, let’s line up the performance curves of a single pump.

This is how much flow rate can be secured when designing a new pump for the existing piping.

This is a common example of such a case.

The intersection of these two curves is the operating point .

The pipe pressure loss curve shows that the more energy is required to increase the flow rate, the more the flow rate is fixed where the shape of the pipe is fixed.

The pump performance curve shows how the pumping height changes as the flow rate of a single pump increases.

Where the piping and pump conditions match

That’s the driving point. It’s a chance encounter.

nAh, H1 is the highest point of the pipe shape.

Piping pressure loss is almost determined by pipe height + pipe friction loss .

You can see the operating point by squeezing the valve

The operating point of the pump can be determined from the pipe geometry and pump capacity.

However, in the field, we see the phenomenon that “the flow rate drops when the valve is throttled.”

Rather, it is the same with domestic water supply ^ ^

This can also be discussed with performance curves.

In conclusion, squeezing the valve should look like the picture below.

The angle of the pipe pressure loss curve becomes steeper, and the intersection with the pump performance curve shifts to the left.

Shifting to the left means that the flow rate decreases and the lift increases.

It’s a good idea to thin the tip of the hose with your finger when you want the water to fly far away .

That’s the same thing that happens when you throttle a valve.

Although the resistance has increased and the flow rate has decreased, the pump capacity is the same, so the lift increases.

There is such a relationship.

Throttling the valve is an operation to change the pipe length L in terms of friction loss calculation for each pipe.

This is because we are discussing the valve resistance as the equivalent length of a straight pipe.

Don’t be fooled by constant mass flow

Some people who studied fluid dynamics in college fall into the trap of the law of constant mass flow.

$$ρAv=const.$$

As A decreases, v increases.

Even if you put a throttle, the mass flow rate should not change.

I tend to think like this.

This only works if we assume that Q is fixed.

An academic story that does not assume that the liquid transfer capability will change.

In reality, there are problems of supply capacity and pressure loss. Be careful!

Relationship between power curve and performance curve

There are relationships between motor power, shaft power, and water power as parameters related to pump power.

These parameters are a bit of a confusing factor.

Let’s organize.

　There is a flow of motor power → shaft power → water power .

Take a look at the diagram below.

Motor power is the power actually input to the motor.

Shaft power is the force that converts the electrical power of the motor to the motor and applies it to the fluid in the pump as a mechanical force.

Hydraulic power is the force itself that is purely applied as fluid energy.

The higher the shaft power/motor power value, the better the energy efficiency in the motor.

The higher the water power/shaft power value, the higher the efficiency of the pump.

The magnitude relationship between motor power, shaft power, and water power is shown below.

Hydraulic power is well defined physically.

If the hydraulic power is P, the following relationship exists.

$$ P=0.163ρQH $$

Let’s look at the relationship between the power curve and the performance curve.

Power curve diagram

As a pump power curve, there are many cases where the shaft power and efficiency curves are superimposed on the performance curve.

As shown in the figure below.

Hydraulic power is Q cubed

Before shaft power and efficiency, let’s look at water power.

Hydraulic power is proportional to the cube of Q and inversely proportional to Q.

$$ H∝Q^2$$

$$ P∝QH=Q^3$$

The relationship that hydraulic power is proportional to the cube of the flow rate is a story that often comes up in the topic of motor inverters.

Shaft power is Q to the first power?

Shaft power appears to be proportional to Q raised to the first power.

In reality, it is close to the power of 2 or 3, but when superimposed on the performance curve, it looks like the power of 1.

Here, it is simply assumed to be the first power.

This is because even if Q changes in the pump performance curve, the amount of change in H is extremely small.

$$H∝Q^2 → const.$$

Considering the relationship

$$P∝Q$$

It will be a relationship.

Shaft power Q=0, that is, it takes a constant value even in cut-off operation.

This is because a certain amount of energy is required to raise the fluid in the pump to the cut-off pressure.

Efficiency is Q squared?

The pump efficiency appears to be a double curve with a peak at a constant flow rate.

Efficiency seems to be effective at about the square of Q.

Efficiency = water power/shaft power

Hydraulic power: 3rd power of Q, Shaft power: 1st power of Q,

Efficiency is calculated as Q squared.

This is the physical interpretation of having a peak with efficiency.

As I wrote in the section on shaft power, since the shaft power is neither Q to the 1st power nor the 3rd power, it is not possible to make an accurate discussion.

However, I would say that it is a trend.

Let’s also explain why we have peaks for efficiency.

Q=0

In Q=0, shut-off operation, water power=0 and shaft power is a constant value.

This is a relationship of efficiency = water power / shaft power = 0.

0<Q<max

As the flow rate increases from Q=0, the pump efficiency gradually increases.

At low flow rate, there is a relationship of “Increase in water power>Increase in shaft power”.

The pump efficiency starts at 0 and increases steadily.

When the flow rate becomes high, the relationship of “Increase in hydraulic power < Increase in shaft power” will appear.

The pump efficiency will gradually decrease.

Efficiency peaks when the relationship of this “increase in water power < increase in shaft power” changes.

Q=max

At the maximum value of Q, the pump efficiency remains constant.

This is because both hydraulic power and shaft power have constant values.

reference

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lastly

I explained the pressure loss calculation of a chemical plant.

Height, piping friction loss, pressure.

Explains the concept of Bernoulli’s law, pump pressure loss curve, and pipe resistance curve.

In practice, a simple calculation that considers only the height and the pressure loss in areas where clogging is likely to occur is sufficient.

Please feel free to post your worries, questions, and questions about the design, maintenance, and operation of chemical plants in the comments section. (The comment section is at the bottom of this article.)

*I will read all the comments and answer them seriously.