7 Economic efficiency analysis of heat waste utilization

Heat recovery systems should have a direct payback in fuel savings no more than 8 years to be economically feasible. Factors influencing payback period are:

•Distance between Power plant and User.

•Amount of available recovered heat.

•Condition of power plant and user’s system.

•Quality of recovered heat needed.

The Need for Economic Analysis

Economics play a dominant role in the decision whether the management/owner and the financing institutions will invest in an energy savings project or not. The communication of energy managers with the decision makers is very important in investment decisions. The energy manager must learn to speak management’s language. The energy manager must present projects in economic terms in order to help the decision makers to make their decisions.

Project Profitability Analysis

Profitability analysis is a method used for assessing financial feasibility of a project. Life Cycle Cost Analysis (LCCA) is used for project profitability analysis because it takes into account the entire life of a project and time value of money.

Life cycle cost analysis

Life cycle cost (LCC) analysis

• considers all cost incurred from design to disposal phase;

• separates savings from investments;

• outputs both absolute (NPV) and relative (SIR) savings;

Cost of capital perspective

• The value of energy savings diminishes each year;

• High interest rate limits the scope of projects

Main definitions for economic analysis:

1. Future Value (FV) – shows how much money/project will worth at a specific time in the future.

, n is a life cycle period.

2. Present Value (PV) – shows how much future money/project worth today. We use discounting to calculate PV.

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3. Discount rate (i) - is a rate at which the future benefit (loss) can be compared to the present value. Discounting rate includes risk free rate, inflation and risk premium.

4. Net Present Value (NPV):

B – benefits;

C – Costs;

i – interest rate;

t – period.

The net present value (NPV) of a project is its life cycle net savings. It is the absolute monetary value of a project. NPV shows the total potential earnings of a project and considers the effect of interest on future net savings. NPV is a major decision making tool for project owners.

5. Internal Rate of Return (IRR) is the interest in percent that the project's investment will earn. IRR is a hypothetical discount rate that causes NPV = 0.

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Theoretically, any project with an IRR greater than the company's cost of capital is profitable (and will have positive NPV). Companies set "hurdle rates" for IRR (minimum IRR that the company will accept for implementing projects). Hurdle rates are normally higher than a company's cost of capital. So only higher profit projects are selected.

6. Payback period (PBP) – the length of time required to recover the cost of an investment in device for heat waste utilization.

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Simple payback period (SPB), expressed in years, does not discount its input or consider future re-investment costs. It is only useful for projects with very quick return. If a project can back pay in a year, there is a little need to calculate discounted future values. For longer paybacks, SPB becomes inaccurate.

7. Savings-to-investment ratio (SIR) is the same as a benefit / cost ratio and it shows a project’s relative feasibility in percent.

.If SIR > 1.0 a project is profitable (economically feasible).

Steps to determine Financial Feasibility of an Energy Project.

1. Determine old costs (existing baseline conditions).

2. Determine new costs (implementation and beyond).

3. Calculate differences.

4. Choose discount rate.

5. Choose analysis period.

6. Estimate residual value of equipment at end of service life.

7. Calculate present value of annual savings.

8. Calculate present value of investments (PV).

9. Calculate net present value (NPV).

10. Calculate savings-to-investment ratio and internal rate of return (IRR).

8. Annual cost savings (AS) = old energy cost – new energy cost + old operations and maintenance costs – new operations and maintenance costs + old other costs – new other costs. Factors in annual savings may be negative.

For determination of annual savings its necessary to find out fuel savings from development of heat waste recovery system.

With the thermal use of heat waste fuel savings is determined by the flow rate in the main or substitutable process for the production of the same amount and at the same parameters of the heat, which are obtained by the heat waste. In this case, the primary fuel savings B_h (tons/year) will be

.

_r is efficiency of the power plant, which is associated with indicators of the efficiency of waste heat recovery (coefficient of 0,0341 indicates that 1 GJ is equivalent to 0,0341 of tce, in case of Gcal this coefficient is 0,123);

Q is the heat content that can be produced through the heat waste.

In case of application of heat waste for cooling systems

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 is a coefficient of performance of cooling device.

In case of power generation with heat waste fuel savings are determined by its cost for production in power plant.

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b_p is fuel consumption for electricity generation in the power system or in substitutable device, which is compared with the power generation through heat waste, g/(kW h).

W is a production of electrical or mechanical energy from heat waste in the utilization plants, million kW h/year.

When heat waste are used as combustible resource the fuel savings can be determined:

,

B is the use of combustible waste, tce/year.

₁ is a coefficient of utilization of fuel heat in device (furnace, boiler, etc) that uses combustible waste.

₂ is a coefficient of utilization of fuel heat in the same device (furnace, boiler, etc) that uses primary energy.

Example of definition of energy saving effect in case of heat waste utilization in heat pump.

For any energy savings project, the basic goal in estimating savings in operating costs is to establish the difference in costs between current and future case-base operation.

For passive-heat recovery projects, establishing this cost difference is relatively simple, because the value of steam or fuel saved is readily calculated.

In the case of heat pumps, the energy savings is equal to the value of steam or fuel saved, minus the cost of operating the heat pump. The quantity of energy saved and the cost of operating the heat pump depend on the application and the heat-pump characteristics.

The example that follows illustrates the steps involved to estimate the savings that a mechanical heat-pump application would generate.

To estimate savings for a mechanical heat pump, we need to know how much energy we will save, along with its value. We can determine the relationships between work input, temperature lift, and heat output using a parameter known as the heat pump Coefficient of Performance (COP_HP).

COP_HP is defined as: ,

where Q_out is the heat delivered by the heat pump and W_in is the energy or “work” supplied to the driver. For an ideal, Carnot-cycle heat pump, the COP is related to the heat delivery temperature and the temperature lift:

,

where T_in and T_out are the temperatures, in degrees Rankine, at which the heat pump receives and delivers heat respectively (for example, the evaporator and condenser). Note that these temperatures are not the process operating temperatures, but the heat-pump operating temperatures. Because a temperature difference between the process streams and the heat-pump working fluid must exist, the actual temperature lift internal to the heat pump is greater than the temperature lift applied to the process streams.

The COP_HP of an actual machine will be 65 to 75% of that for an ideal machine. For the purpose of estimating heat-pump work requirements, and operating cost, it is only necessary to know the expected operating temperatures. (Note that COP_HP is not the same as the more common term, ‘refrigeration COP (COP_REF),’ which is defined as

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Consider the example in which a process stream cooling from 77° C to 60° С is a 2,5 Gcal/h heat source, and a process stream heating from 82° C to 100° C is the heat sink.

To calculate the work input, we need to determine the T_in and T_out for the heat pump.

T_in is determined by the temperature approach between the heat-pump working fluid and the lowest temperature of the heat source. In this case, the heat source is cooling to 60° C and there is a 18° C approach temperature in the evaporator, so T_in = 48° C.

T_out is determined by the temperature approach of the heat-pump working fluid to the hottest temperature that the heat sink achieves. In this case, the heat source is heating to 100° C and there is 10° C approach temperature in the condenser, so T_out = 110° C.

The ideal COP_HP for this example is:

Assuming that the actual COP HP is 70% of ideal, the estimated operating COP_HP will be:

COP = 6,2  0,7 = 4,3.

The energy balances for the heat pump are as follows:

Some mathematical rearranging gives the relationship between Q_in and Win as:

For the example, the work required to deliver the waste heat at the higher temperature is:

kW.

And

Gcal/h.

Therefore, our heat pump is going to save 3,3 Gcal/h of heat input to the process, at the cost 881 kilowatts (kW) of electrical power.

If we assume that, in the base case, the cold process stream is heated with steam costing 15$/Gcal, and that power costs 4,4 cents per kilowatt-hour (kWh), then we can calculate estimated savings as:

If the process runs 8 500 hours per year the project could be justified based on obtaining 90 950$/yr in annual savings.

The fuel savings associated with obtaining this heat in heat pump are (in case of substituted boiler efficiency of 92%):

Note that if the driver were not an electric motor, the cost of shaft work for the particular device would be substituted for power cost. In instances where turbines or combustion engines are drivers, considering the overall site energy balance is necessary to correctly determine the cost of work.

The method described above helps determine if there are sufficient savings and incentive to pursue a heat-pump application. Although simplified, this method gives a reasonable idea of the energy-cost economics of installing a mechanical heat pump.

When evaluating economics, also consider non-energy cost benefits:

• Product quality: In lumber-drying applications, the gentler heating resulting from use of heat pumps results in better quality dried lumber and higher yields.

• Offset capital costs: In evaporation applications, using a heat pump means that boiler steam load and cooling water duty are avoided. This leads to reduced capital for boilers and/or cooling towers, together with lower NO_x emissions.

In addition, other costs may exist:

• Operating costs for auxiliary pumps and fans associated with the heat pumps

• Maintenance costs for the heat-pump equipment.

In conclusion, try to include all relevant costs and benefits in the detailed economic calculations.