6.1.03 Informative

2024-02-01

Insulation of a system implies capital expenditure. In any insulation system it is important to analyse the thermal insulation with respect to cost. There is a definite economic limit to the amount of insulation, which is justified.

The simplest method of analysing the extent of insulation is by comparing the cost of energy losses with the cost of insulating the pipe or equipment. The insulation thickness for which the total cost is the minimum is termed as economic thickness.

For understanding the preferred method of calculating the economic insulation thickness there are two costs associated with the insulation type chosen. For any given thickness, there is:

• The cost of the insulation itself; and
• The cost of the energy lost through the insulation.

The bases is already introduced in 2.1

Equation 1:
(Capital Cost × Capital Service Factor) + Annual Operating Cost = Total Cost
Remark: This is the overall equation.

In figure 1, the total cost line is drawn for several different thicknesses of insulating material.

The economic insulation thickness will be the thickness corresponding to the minimum point on the total cost line. When two of more layers of insulation are used, each layer may have its own minimum.
The Minimum Total Cost Method is particularly useful when more than one layer of insulating material is expected, since the results can be presented graphically, and the minimum or minima are immediately obvious. Three values need to be calculated:

• Capital Cost.
• Capital Service Factor; and
• Annual Operating Cost.

Calculation of Capital Cost

The Capital Cost can be calculated according:

Equation 2:

Capital Cost = (Ip × Fr × Fc + s × Fa) × Fi in EUR/j × m¹ of m² (2)

Where:

I_p is the insulation price per unit of flat surface (m²) or unit of pipe length (m¹). The value given should include:

• The current price of insulation;
• Fastening and weather-proofing materials;
• Installation;
• Packing;
• Transport and tax costs (variable costs);
• The values should exclude scaffolding and metal surface preparation costs, for example rust removal and painting (fixed costs).

F_r is the currency rate required to express costs in Euro (EUR), or the appropriate local currency (for this guide the ¿ (Euro) is used in the examples);

F_c is the complexity factor. Normally prices are quoted for straight line piping. The complexity factor allows for the extra cost incurred due to fittings. (See Table 1);

s is the insulation thickness in meters;

F_a is an additional cost which is added to the insulation price in certain cases to cover the costs of building pipe-racks, increased civil work, bigger platforms, and so on. An indication of the additional costs are:

• 700 EUR/m of insulation thickness for diameters ≤ 20 inches, for example piping;
• 350 EUR/m of insulation thickness for diameters > 20 inches, for example vessels/columns; and;
• Nothing to be added for very large diameters, for example tanks.

F_i is the investment escalation factor which escalates the present insulation costs to the actual costs at the time of installation.

Piping Complexity Code and Factors

The complexities of table 1 are defined as:

• None: No fittings or prices of fittings are already included.
• Simple: 10 fittings every 30 m (100 feet) of pipe.
• Average: 15 to 20 fittings every 30 m (100 feet) of pipe.
• Complex: 30 to 40 fittings every 30 m (100 feet) of pipe.

Notes:

Tabel 1: Factoren voor de complexiteit van leidingen

• The above factors apply only to welded piping with flanged valves and line pipe flanges. For welded piping with no flanges use an average piping complexity factor of 1.05. These factors are not to be used when valves and flanges are not to be insulated.
• Fittings are denoted as:
  • Flange pair: 1 fitting.
  • Valve flanges: 2 fittings.
  • Valve body: 1 fitting.
  • Elbows, tees, reducers, and so on: 1 fitting.

Calculation of Capital Service Factor

The Capital Service Factor is calculated according to the VDI 2055. As there are three different methods explained in the VDI 2055, one of the methods should be known at the beginning of a project.

The methods are explained hereof.

The capital service factor b contains depreciation or repayment  (n expected service life in years), interest rate z, maintenance r and general cost g percent are defined as equation (3):

b = (anual fixed cost)/(investment for the insulation) (3)

For the determination of the capital service factor different approaches are used. The values are inserted into these formulas in  (per annual year) and they relate to the investment for the insulation.

a) Addition (in 1/a):
b1 = 1/n+1/100×(z+r+g) in 1/a
Where:
n expected service life, in a (”years”)
z interest rate, in %
r maintenance, in %/a
g overhead, in %/a

b) as a, however, instead of the full interest rate only: in %/a is inserted to take the repayment over the service life into account:
b2 = 1/n+1/100×((z+1)/2+r+g) in 1/a

c) Annuity:
b3 = (z/100)/(1-(1/(1+z/100))^n )+(r+g)/100 in 1/a

The interest rate z may be either a company-internal interest rate or an external interest rate. Often r+g (maintenance and overhead) is taken 2%.

In general, the last method should be used or as otherwise determined by the financial department. It is always recommended to involve financial experts on this subject.

For insulation projects, the economic lifetime of the insulation is usually taken to be 10 years, although this lifetime can vary depending on the inspection requirements of local authorities.

Other factors affecting the Capital Service Factor (b) are CAPEX phasing, depreciation, tax rate at margin, inflation and increases in heat cost.

The Capital Service Factor (b) should be defined for each project.

The maintenance percentage (r), which is as default 1% may be changed into higher percentages depending on the complexity of the process facility.

For overhead percentages (g) 1% should be maintained or as recommended by the financial department.

Calculation of Annual Operating Cost

The Annual Operating Cost can be calculated according to equation (4):

Annual Operating Cost = (Ft×Hc×Q×N)×f in €/y×m1 or m2 (4)

Where:

F_t is a conversion factor, which is required to convert the units J/s.m2 or J/s.m1 into GJ/day.m2 or GJ/day.m1 since the heat cost is given in /€GJ. The conversion factor (F_t) has the value: 3600×24/1,000,000,000

H_c is the heat cost which varies according to the heating medium used. Piping and equipment which can be related to a certain heating medium shall be insulated with the economic insulation thickness calculated for that heating medium. The following heating media can be used (prices are indicative and should be checked).

• Refinery fuel;
• LP steam, condensing temperature between 130 °C – 180 °C;
• MP steam, condensing temperature between 180 °C – 250 °C;
• HP steam, condensing temperature above 250 °C;
• Heat Transfer Fluid (HTF) system;
• Emission CO2 op top of above mentioned media (if applicable);
• The heat cost of refinery fuel can be calculated as follows:
  • Refinery fuel price, for example, 150 €/ton = 0.150 €/kg;
  • Lower heating value 0.0402 GJ/kg, Efficiency 0.88;
  • Heat cost = 0.150/ (0.0402 * 0.88) = 4.24 €/GJ;
  • The heat cost of LP, MP and HP steam are related to the cost of refinery fuel and should be higher than the heat cost for refinery fuel.
  • The emission cost of CO2 should be added to the heat cost price.

A standard point of discussion is the value of heat cost related to surplus steam, associated gas, heavier oil fractions, and so on. Before deciding to take the advantage of the apparently low heat cost value of these alternatives, the following points shall be considered: auditing of utilities, future extensions, town heating, more effective use of gas, environmental conditions, and so on.

When the heat cost is genuinely very low, then only insulation for personnel protection, process or acoustic reasons may be required.

Since the economic insulation thickness is required to be specified at a very early stage of the project (to be able to fix the width of pipe bridges and tracks), the assumed heat cost shall be correct or even on the conservative side. This is because changes later on which result in an increased insulation thickness, are almost impossible to achieve.

Q is the heat loss per unit of flat surface (m²) or unit of pipe length (m¹). The calculation of the heat loss will be explained in the next section.

N is the number of working days per year. Scheduled shutdowns should be taken into account. For batch processes it is important to know whether equipment is allowed to cool down between batches or whether the temperature has to be maintained.

F is the price variation factor that will be determined with the following equation:

F = S1/S2 (5)

S1 = (1-((1+p/100)/(1+z/100))^n)/(1-(1+p/100)/(1+z/100)) (6)
Note: For p=z follows a limit S1 = n

S2 = (1-(1/(1+z/100))^n)/(1-1/(1+z/100)) (7)
Note: For p=z follows a limit S1 = n

Where: n expected service life, in a (”years”) z interest rate, in % p price change factor, in percent

Heat Loss

This section contains an example of the minimum total cost method of calculating the economic insulation thickness.

A limited set of input data is specified, which is the minimum required. In the case of indirect input data the resulting value is given, for example, Capital Service Factor. For other data, more information can be found in tables 1, 2, 3 and 4.

The example is not fully worked out in detail, but is shown to give an indication of the results that can be obtained. Al input should be checked by the financial, process and engineers.

For the example, the following input data is used:

Table 5: Data Economic Insulation Calculation Example

The economic insulation thickness can be calculated with the mentioned equations hereof. In the first instance, however, it should be checked whether insulation is actually required.

Therefore, the first step is to calculate the heat loss, and hence the cost of heat loss, of the bare pipe using the applicable equations.

Remark: Normally for simplification of the calculation, the resistances over the boundary layer on product side and the pipe wall may be ignored.

The temperature at the outer wall surface is assumed to be equal to the product temperature and may be used to calculate the heat transfer coefficient due to radiation. The emissivity for cladding can be found in table 4 (more can be found in the CINI Manual or VDI 2055). For the bar surface of the pipe ε = 0.8 is adopted.

Air side coefficient (convection + radiation) bar surface:

h = h_r + h_r = 10.30 + 44.10 = 54.40 W/m² ×K

Heat loss bare surface: 7111 W/m¹

Cost of energy (for 330 days/year):

Annual operation cost (heat loss): 2027.49e/m¹ ×y

The capital cost and the annual operating cost (energy cost) together with data required to follow the calculation sequence are given in table 5. Again, not all intermediate results are shown.

The piping complexity factor can be found in table 1 and the allowance for type of material and application with respect to the thermal conductivity in table 2.

The resistance over the boundary layer on air side is once again ignored. This means that in part of internal film coefficient and pipe wall are neglected in equation 14 , only the resistance over the insulation layer for a curved wall and external film coefficient is used.

Table 6: Data Minimum Cost Method

Above is plotted into a curve (see figure 3 )

The discounted payback method takes into account the time value of money and is therefore an upgraded version of the simple payback period method. Companies use this method to assess the potential benefit of undertaking a particular business project. In this case the project cost of insulation and related fixed heat cost should be taken into account, against the heat loss cost of the bar surface (”non-insulated piping/equipment”).

For input the same information of the minimum total cost method is used and as example the economical thickness of 80 mm is taken, along with all related calculations.

Table 7: Data Discounted Payback Period

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What is the Discounted Payback Period (DPP)?

The discounted payback period is a measure of how long it takes until the cumulated discounted net cash flows offset the initial investment in an asset or a project. In other words, DPP is used to calculate the period in which the initial investment is paid back.

In project management, this measure is often used as a part of a cost-benefit analysis, supplementing other profitability-focused indicators such as internal rate of return or return on investment. It can however also be leveraged to measure the success of an investment or project in hindsight and determine the point at which an initial investment has actually paid back.

The discounted payback period is calculated by discounting the net cash flows of each and every period and cumulating the discounted cash flows until the amount of the initial investment is met. You will find the formula in reference: 17. This requires the use of a discount rate which can be either a market interest rate or an expected return. Some organizations may also choose to apply an accounting interest rate or their weighted average cost of capital. While the discounted payback period is still a comparatively handy indicator, it is more precise than the generic payback period as it considers the time of the occurrence of cash flows. Discounting the cash inflows of the project for each period at a suitable discount rate.

How is the Discounted Payback Period Calculated (DPP)?

The formula to calculate the discounted payback period is:

DPP = y + ABS(n)/p (17)

Where: y = the period preceding the period in which the cumulative cash flow turns

positive,

p = discounted value of the cash flow of the period in which the cumulative cash flow is => 0 abs(n) = absolute value of the cumulative discounted cash flow in period y.

The equation is worked out in a spreadsheet, the outcome is shown in table 8

Table 8: Discounted Payback Period (DPP) for piping

DPP = 0+146/1568 = 0.10 year ∼ 38 days

The payback period of 38 days is not a realistic scenario and therefore is this method not suitable for the calculation of economical insulation calculated with this kind of input. More interesting is if in cases like additional heat losses the capacity of e.g. gas turbines or boil off compressors need to be increased or by thicker insulation it could be decreased. The Discounted Payback Period is often mentioned and used as a measure for the viability of projects. In this case the capacity of turbines or compressors.

Therefore the use of the Discounted Payback Period should be discouraged when calculating an economic insulation thickness. This is because it does not involve all factors (as mentioned above), used in the ”Minimum Total Cost” method. Therefore, it does not take into account the changing value of money over time. In short, the Discounted Payback Period should only be used for very broad-brush screening and not for the final choice of insulation thickness. Therefore a more better example is taken. In this example the Discounted Payback Period is a valid instrument.

DPP Tank farm Roof Insulation

Given: A tank farm with 8 oil tanks (fixed roof tanks) of which is investigated if insulation of the tank roofs is a savings-potential.

Table 9: Information Tanks for determining DPP

Question: Determine discounted payback period (DPP), based on a service life of 10 years.

Solution:

Table 10: Discounted Payback Period (DPP) Tank

DPP = y + abs(n)/p = 2 + 446.901/488.355 = 02.93 years

A discounted payback period of less than 3 years is very attractive for this project and justified to continue with this project.