Lesson 1

THIS LESSON IS DESIGNED TO BE USED IN CONJUNCTION WITH LESSON 1 ON THIS INTRODUCTORY VIDEO 1

NATIONAL CURRICULUM REFERENCE

At KS4 it is recognised that

Mike Mackay’s case study involves mathematics from the higher tier

CASE STUDY 1 Mike Mackay, Chemical Engineer KS4 HIGHER TIER

Background Information

The chemical engineer featured in the video is

Name: Mike Mackay

Profession: Chemical Engineer

Everyone is familiar with the image of a scientist working in a laboratory with different chemicals or raw materials to discover wonderful innovative products from new drugs, plastics, and solvents to new hair care products or foodstuffs. This image involves the scientist working with test tubes or beakers, essentially low volumes of materials.

A chemical engineer’s role is to take processes from small-scale laboratory experiments and to design, operate and optimise large-scale production plants to supply the desired products to satisfy the demands of consumers. One example of this type of work is the role we perform at Esso and Exxon, taking crude oil as our raw material and refining it, or converting it, into useful products which we use in our every day lives such as petrol for our cars or rubber for car tyres.

Mathematics is a critical tool used in engineering of all descriptions and chemical engineering is no different.

We continue by considering two areas of mathematics used daily in refineries.

The first case will consider direct proportionality and the second simultaneous equations.

Each case will set a real life problem and demonstrate through a worked example how to solve the problem. Then will follow a series of questions which pupils can work through to practise their mathematical skills in a real life setting. Answers are provided.

Case 1 Flow rates and Pressure Drop

Have you ever thought how it is possible to know how much fluid is passing through a steel pipe without ever being able to see it? We calculate it by measuring the drop in pressure of the fluid as it passes though a small opening or orifice.

Pipe

Flow

P1 = Pressure upstream of the orifice

P2 = Pressure immediately after the orifice

The flowrate is directly proportional to the square root of the drop in pressure

Written as

We can assume P1 is fixed at 100 kPa (pressure is measured in kilo Pascals)

When P2 = 50 kPa we know that the flowrate = 100 m³/hr

From this information we calculate the constant of proportionality, C

C = 14.14

What is the flowrate travelling through the pipe when P2 is 75 kPa?

(kPa : Kilo Pascals)

Flow when P2 = 75 KPa

F = 70.7 m³/hr

QUESTIONS FOR PUPILS

A) Given the formula

a) P1 = 250 kPa P2 = 40 kPa

b) P1 = 132 kPa P2 = 71 kPa

c) P1 = 86 kPa P2 = 81 kPa

In the oil refinery the bounds for flowrate are 25 < F < 300 m³/hr

2. Calculate the constant of proportionality, C given the following information:

Case 2 Optimising Production of Products from Crude Oil

A refinery takes crude oil extracted from wells bored into the earth. Crude oil is a natural product formed over thousands of years by a transformation of plant life and organic materials which have been buried and exposed to high temperatures and pressures. At Fawley Refinery we take crude from all over the world including wells in the North Sea and Saudi Arabia.

The composition of crude oil is not fixed or constant, it is not a scientific compound, but a mixture of many components that we separate in our processes and refine into useful products. Some examples of these many components are petrol, aviation fuel, liquefied petroleum gas and naphtha that are sold for various industrial and commercial uses.

Each component has different properties and we separate them using distillation columns, which split the components according to their boiling point. Essentially, we heat up the crude oil and the light products, such as LPG, come off the top of the distillation column, streams such as petrol are taken out of the middle of the column and heavier products, such as the components used in motor oils or bitumen, come off the bottom of the tower.

It is important for us to be able to calculate the proportion of these various components in our feed crude oil as each product is worth a different amount of money to the refinery and has an effect on the way we run our equipment.

Chemical models of the various crudes are used to predict the overall refinery production and matched with the markets or demands for the various products. Below is a simple example of the type of calculation we use to determine what types of crude we use and in what amounts.

Worked Example

We have two crude oil feed streams that are blended to make up a single feed to a distillation column. For simplicity we will assume that the crude oil is being split into 3 components: Liquefied Petroleum Gas, Light Virgin Naphtha and Petrol. One crude comes from the North Sea oilrigs and one crude comes from Saudi Arabia. Each crude is composed of a different percentage of each product.

We need to be able to calculate the percentage of LPG, LVN and petrol in each of the two feed crudes.

LPG

Crude 1 LPG A%

LVN B%

Crude 2 Petrol C% LVN

LPG X%

LVN Y%

Petrol Z% Petrol

The table below shows the input flowrates of crude 1 and 2 and the resulting amounts of LPG, LVN and Petrol that come out of the distillation column.

This information can be shown diagrammatically

CASE 1 100

CASE 2 130

All amounts are in m³/hr

So in case 1, 100 m³/hr of crude 1 is mixed with 100m³/hr of crude 2 and 25, 45, 130 m³/hr respectively of LPG, LVN and Petrol are produced. Similarly for case 2.

What are the percentages of each component in each crude?

ANSWERS

From the two cases above we can produce two simultaneous equations for each product.

Consider LPG 100A + 100X = 25

Consider LVN 100B + 100Y = 45

Consider Petrol 100C + 100Z = 130

So we have been able to determine the composition of the two feed streams of crude oil.

In summary

Altering the flowrate of each crude will alter the mix of products. This in turn will have profit implications. We will consider these now.

Financial Implications

If the products are worth LPG $10 /m³

LVN $20 /m³

Petrol $30 /m³

Crude 1 costs $15 /m³

Crude 2 costs $25 /m³

How much per day will we make in each case?

Let us look at what we calculated the amounts of each of the three products to be in case 1

From these figures we can calculate the money made and the costs

(In each case we multiply by 24 to calculate for a day rather than an hour)

Money Made

For LPG 25 x 10 x 24 = $ 6 000

For LVN 45 x 20 x 24 = $ 21 600

For Petrol 130 x 30 x 24 = $ 93 600

Total $121 200

Costs of Crude

Crude 1 100 x 15 x 24 = $ 36 000

Crude 2 100 x 25 x 24 = $ 60 000

Total $ 96 000

PROFIT (assuming no other costs) $ 25 200 per 24 hours

QUESTIONS FOR PUPILS

LPG

Crude 1 LPG A%

LVN B%

Crude 2 Petrol C% LVN

LPG X%

LVN Y%

Petrol Z% Petrol

A)

As you can see from the diagram two feed streams of crude oil 1 and 2 are flowing into the distillation column. By heating the mixture the distillation column will separate the streams into three components, LPG, LVN and Petrol.

Crude oils vary enormously and we do not know the percentage of each of the three components in each input stream, but we can measure how much of each component comes out of the distillation column.

We need to work out the input percentages A, B, C and X, Y, Z so that we can ensure that the correct mix of crude1 : crude2 is being fed in to the column.

We can alter the rate at which each of the crude oils flow into the column.

We then measure how much of LPG, LVN and Petrol comes out for that particular case.

The information below shows two cases. The first case has crude 1 and 2 both flowing in at 100 m³/hr and the second has crude 1 at 200 m³/hr and crude 2 at 150 m³/hr.

Using the information below produce three sets of simultaneous equations and hence calculates the percentages A, B, C, X, Y and Z.

B)

Following the worked example now look at the financial implications for running the distillation column using case2 i.e. 130 m³/hr of crude 1 and 150 m³/hr of crude 2.

The values and costs remain the same

If the products are worth LPG $10 /m³

LVN $20 /m³

Petrol $30 /m³

Crude 1 costs $15 /m³

Crude 2 costs $25 /m³

How much profit will you make? Which way would you run the process?

ANSWERS

A)

100A + 100X = 15

200A + 150X = 25

SOLVING X = 10% , A = 5%

100B + 100Y = 50

200B + 150Y = 87.5

SOLVING Y = 25% , B = 25%

100C + 100Z = 135

200C + 150Z = 237.5

SOLVING Z = 65% , C = 70%

B)

(In each case we multiply by 24 to calculate for a day rather than an hour)

Money Made

For LPG 34.5 x 10 x 24 = $ 8 280

For LVN 62.5 x 20 x 24 = $ 30 000

For Petrol 183 x 30 x 24 = $131 760

Total $170 040

Costs of Crude

Crude 1 130 x 15 x 24 = $ 46 800

Crude 2 150 x 25 x 24 = $ 90 000

Total $ 136 800

PROFIT (assuming no other costs) $ 33 240 per 24 hours

Case 2 gives the most profit and so all other things being equal choose case 2.

Worked Example

Consider the following extensions to this initial problem

If you need to supply 300 m³/hr to the distillation column what is the most profitable split of crude.

This type of calculation is done by computer as it involves an enormous number of calculations done very quickly.

However we can mimic the computer by choosing just two particular ways of making 300 m³/hr and calculating and comparing the end profits.

Let us take the following two cases

For each example we can calculate the costs and money made for the two different input flowrates. In each case a total of 300 m³/hr is being fed in but because the streams are in different proportions different amount of LPG, LVN and Petrol are produced.

If the products are worth LPG $10 /m³

LVN $20 /m³

Petrol $30 /m³

Crude 1 costs $15 /m³

Crude 2 costs $25 /m³

How much per day will we make in each case?

ANSWERS

CASE 1

Money made 24 ((37.5 x 10 ) + ( 67.5 x 20 ) + ( 195 x 30 )) = $181 800

Costs 24 (( 150 x 15 ) + ( 150 x 25)) = $144 000

PROFIT $ 37 800

CASE 2

Money Made 24 (( 40 x 10 ) + ( 70 x 20 ) + ( 190 x 30 )) = $180 000

Costs 24 (( 200 x 15 ) + ( 100 x 25 )) = $132 000

PROFIT $ 48 000

All of these types of issues are worked by refinery staff using powerful computer tools to come up with an optimised daily run plan for the entire refinery. No small task considering that the refinery is the size of a village and has many distillation columns as well as reactors and blending technology which can convert one type of product into another!