Apply the 0.6 power law to chemical plant construction

0.6 power law
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I will explain the 0.6 power law , which is often used in chemical plant construction projects .

Combined with the Lang coefficient, it is very useful for rough estimation of capital investment.

If you know this content, it is possible to super-speed up the in-house estimate.

I want a quick estimate! Even for a request from the manufacturing department, you will be able to reply on the same day and your trust will rise at once.

You will be able to work profitably within the company.

What is the 0.6 power law?

The 0.6th power law is the law that the capital investment amount is proportional to the 0.6th power of the plant capacity .

You can’t tell just by looking at it.

0.6 power (Plant construction)

This law states that when the capital investment amount for plant capacity X is Y, the capital investment for plant capacity 2X is not 2Y, which is a simple proportion , but 1.52Y, which is proportional to 2 raised to the power of 0.6 (≒1.52) .

Generalizing, we get

$$ Y_2 = (\frac{X_2}{X_1})^{0.6}Y_1 $$


This seems to be an empirical rule of chemical plant construction, a law that was discovered after collecting a huge amount of data.

Especially often used in continuous plants.

This is because the production method for a continuous plant is almost fixed to make a certain product, and the configuration of the plant is also almost fixed.

It seems that it will be a similar configuration anywhere in the world.

Not so with a batch plant.

The products to be made are different, and the number of factories that make the same product is small in the first place.

However, if you want to put aside such details and make a rough estimate, you can use the 0.6 power law even in a batch plant.

If anything, it’s a drastic idea that can’t be helped because there’s no other way …

I would like a rough estimate by tomorrow!

Eh… I don’t have time… That’s right! Let’s finally put it out with the 0.6 power law. you won’t get angry

I use it when I make an estimate like this.

The 0.6th power law is literally 0.6th power, but depending on the case, it seems that the numerical value is somewhat manipulated.

There may be companies that are swinging the value around 0.6 to 0.8 .

Application to Plant construction

Consider the case of using the 0.6 power law when building a batch plant.

It is basically impossible to discuss on a production capacity basis like a continuous plant.

In this world, the amount of investment is determined by what kind of equipment is installed in the plant.

In this case, the Lang coefficient comes into play.

FacilityLang coefficientunit pricequantitySubtotal (10,000 yen)
Reactor A2,000Five10,00020200,000
Reactor BXFive5X15X
Heat exchanger500Five2,50040100,000
Unique equipment10,000Five50,000150,000

When constructing a plant, the Lang coefficient is easy to use because a complete set of equipment must be purchased.

Since the plant size is mostly determined by the number of reactors, we apply the Lang factor on the basis of the facility.

If you know that reactor A is 10m 3 and costs 20,000,000 yen, just apply the Lang coefficient.

If the Lang coefficient is 5, then

5 x 2,000 = 100 million yen (100 million yen)

and the result will come out.

Complementation for unknown installations

Reactor B is 20m3 but I don’t know the cost! In that case, the 0.6 power law comes into play.

Just apply the formula as is.

(20/10) 0.6 x 2,000 = 1.52 x 2,000 = 31.4 million yen

Multiply this by the Lang factor

5×3,140=157 million yen


A summary is as follows.

FacilityLang coefficientunit pricequantitySubtotal (10,000 yen)
Reactor A2,000Five10,00020200,000
Reactor B3,140Five15,700115,700
Heat exchanger500Five2,50040100,000
Unique equipment10,000Five50,000150,000
  • I know the price of most of the equipment, but I don’t know the price of a few special equipment
  • Not enough to ask the manufacturer for a detailed estimate

In that case, we will use it to complement with the 0.6 power law .

Apply 0.6 to the whole?

Here, one might think that the production capacity is proportional to the number of reactors and use the power of 0.6 .

In this example, it is 365,000 yen for 20 reactors, so in the case of 21 reactors with the addition of reactor B,

(21/20) 0.6 x 365,000 = 1.03 x 365,000 = 3,759,500,000 yen


quantitySubtotal (10,000 yen)
Reactor A series20365,000
Reactor B1X
totaltwenty one375,950

There is not much difference in the result in this order.

Since the unique equipment should be considered outside, 315,000,000 yen for 20 reactors, in the case of 21 reactors

21/20) 0.6 x 315,000 = 1.03 x 315,000 = 3,244,500,000 yen

Add 50,000,000 yen to the outside equipment

324,450+50,000= 374,450,000 yen

quantitySubtotal (10,000 yen)
Reactor A series20315,000
Reactor B1X
Unique equipment150,000
totaltwenty two374,450

The results are similar.

It makes me feel like I’m playing with numbers.

A characteristic of batch plants is that the standard investment amount is not well known, so I think that applying the 0.6 power law to it will result in a rough estimate.

There was a premise that the 0.6th power law was aggregated data of similar plants, wasn’t it?

It would be nice if there was enough data to be aggregated in a batch plant, but it is not.

It would be safer to apply the Lang factor to individual installations.

Application example for facility renewal

At the batch plant level, the 0.6 power law flies around at the detailed capital investment level.

This is the case when remodeling or updating a single piece of equipment in a plant.

Plant construction

The concept is the same as for the entire plant construction.

Apply Lang coefficients to individual equipment.

For example , when estimating the cost of installing 20m 3 of reactor B where 10m 3 of reactor A is installed ,

(20/10) 0.6 x 2,000 = 1.52 x 2,000 = 31.4 million yen

5×3,140=157 million yen

I introduced that it will be the calculation result.

Be skeptical when using the 0.6 power law for individual equipment upgrades.

Just because it’s calculated according to the general rule of thumb doesn’t mean it’s absolutely correct…

If the power of 0.6 is applied to the extent that the effect on the whole is small, the error will not be noticeable, but if you try to apply the power of 0.6 to a single facility, the error will become noticeable.

equipment amount

We have to decide whether to apply the 0.6 power law as the price difference between 10m 3 and 20m 3, but basically there is no problem .

As long as there are no factors that increase the unit price, such as soaring material costs or soaring labor costs, it generally fits.

This is because equipment manufacturers also think in the same way.

However, please be aware that can manufacturing companies are relatively water products.

Stainless steel is a material that tends to rise in price relatively easily, so there are cases where the price rises so much that the 0.6 power law does not work.

construction cost

When the Lang coefficient is fixed, it is assumed that the 0.6 power law is also applied to construction costs.

This is actually pretty weird.

Construction cost (Plant construction)

In this way, the coefficients are applied differently for plumbing work, building work, and electrical work.

The 0.6th power law assumes that this area is somehow summarized.

For example, at the level of updating one reactor,

  • How often does the interfering object detach?
  • foundations and beams need to be updated (will construction work occur)?
  • the motor increase in size
  • instruments be considered?

It makes a big difference at that level of detail.

It also leads to the question of whether the Lang factor should be applied at such a granular level .

Of course, we use the 0.6th power rule because we don’t have time to do such a detailed investigation.

But if the numerical value is abnormally high when using the 0.6th power rule, we should question this area and correct it as necessary. I think it’s a good feeling to keep going.

It is doubtful whether the equipment itself is effective at 0.6 power, and it is also doubtful whether the construction work is effective at 0.6 power, so I think the recent trend is to put out all at 0.6 power and there is no stopping even if the amount increases.

Even if the value of the 0.6th power law is apparently high, even if we fine-tune the number and lower it, we are unable to absorb the soaring material and labor costs, resulting in an increase in budget overruns.

Either way, if you go over budget, you can use the 0.6 power law as an excuse.

Even if it depends on the 0.6 power law as a result, it is introduced as an example that it is better to keep in mind the possibility of lowering the numerical value and the division of costs into elements.


Capital investment is an important element in which a company’s strategy is questioned.

There is not much information available to the public, but it would be one way to look at the following books.

created by Rinker
¥1,650 (2024/02/29 07:27:41時点 楽天市場調べ-詳細)

Related information

Related information

DB example

DB basic

Lang factor


I introduced an example of applying the 0.6 power law to capital investment in a batch chemical plant.

The 0.6 power law shows the relationship between plant production capacity and investment amount.

It is especially effective in continuous plants, closely related to the Lang coefficient.

It would be safer to use the 0.6 power law in a complementary sense.

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

*We will read all the comments and reply seriously.