WER015 Final Report Assignment 1

Akademin för hållbar samhälls- och teknikutveckling

REPORTVästerås, den 5 Oktober. 2010

Thermal Engineering

Projektarbete utfört av group 4: Teacher: Jan SandbergPhotchara, WER015, [email protected], WER015, [email protected]

AbstractIn this study we are going to design how many boiler we have to install and how to manage them during the year to reach the optimum cost. To solve a problem we have developed combinations of boiler capacity installation. In addition we also stimulate the running period of each boiler to find the best combination. The optimization is completed by a simulation based optimization procedure, Microsoft Excel, which calculate various costs factor and energy consumption in each period. The optimum results are confirmed the using combination.

Table of contents

Table of figures...........................................................................................................................3Introduction.................................................................................................................................4

Background.............................................................................................................................4Problems.................................................................................................................................4Limitation of study..................................................................................................................4

Methodology...............................................................................................................................5Optimization Method..............................................................................................................5Efficiency Calculation..........................................................................................................18Total Cost Calculation..........................................................................................................19

Main text...................................................................................................................................22Under titles in different levels...............................................................................................22

Result........................................................................................................................................23Conclusion................................................................................................................................25

Table of figures

Figure 1 Power production for the first case...............................................................................5Figure 2 Power production for the second case..........................................................................8Figure 3 Power production for the third case............................................................................11Figure 4 Zoom-in of the third case...........................................................................................11Figure 5 Power production for the forth case...........................................................................15Figure 6 Zoom-in of the forth case...........................................................................................15Figure 7 Total cost of first case.................................................................................................21Figure 8 Setting up solver parameters......................................................................................21Figure 9 Optimization of Case 2...............................................................................................23Figure 10 Optimization of case 2..............................................................................................23Figure 11 Optimization of case 3..............................................................................................23Figure 12 Optimization of case 4..............................................................................................24Figure 13 Construction cost sensitivity of case 4.....................................................................24Figure 14 Fuel cost sensitivity of case 4...................................................................................25

Table 1 Cost Description............................................................................................................5Table 2 Fuel Consumption........................................................................................................19Table 3 Case comparison..........................................................................................................24

3

Introduction

BackgroundThere are always more than one production unit in a district heating system. The load duration curve is very pointed (rather high load for a short time) and it is unlikely to find one unit that can manage the peak load (often 90 % of the installed load) and that is possible to regulate down to the minimum load (7-8 % of installed load). More production units increase the availability and one may choose a cheap boiler with rather expensive fuel for the peak load and a more expensive boiler with a cheaper fuel for the base load. To achieve an acceptable reliability one often installs a stand by capacity corresponding to the largest boiler in the system.

In this exercise we will examine a district heating system with 100 MW installed load with hot water boiler for the maximum load calculated by the load curve.

Problems Find the best performance of boiler Find the number and capacity of boilers after optimization Compare the total energy production Find heat energy production cost per MWh Make a simple sensitivity analysis if;

o Fixed costs varies between 70-150%

o Fuel cost varies between 70–200%

Draw a total energy ratio diagram

Limitation of studyAn experimental study has been performed on the 8 953 MW steam boilers that burn 3 types of fuel compose of oil, wood, and coal to optimize boiler efficiency. Cost calculation is consider economic criteria includes construction cost, staff cost, fuel cost, auxiliary power cost, and interest rate as shown in table 1.

Fixed cost Variable cost

Tax and Insurance

Oil 1,5% of construction cost Wood and Coal 2,0% of construction

cost

Fuel cost

Wood - 145 kr/MWh Coal - 450 kr/MWh Oil - 550 kr/MWh

Auxiliary power cost

380 kr /MWh

Construction cost

Interest rate 5% for 20 years

Staff cost for 6 persons

3 600 000 kr /yearTable 1 Cost Description

The variable cost for a boiler is SEK per kW and varies with the size.

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Methodology

Optimization MethodFrom the data given in excel Oil has lowest construction cost while has highest fuel cost. In another Wood has high construction cost while has lowest fuel cost. So both Oil and Wood as boilers have been choose for developed optimization methods by using Wood as much as possible while Oil used as small as possible to produce the energies.

First Case For this case we use 2 boilers, one for Oil as peak load and another for Wood as base load. According to Figure 1,

A

B

C

Figure 1 Power production for the first case

1. Point A to BAt zero hour we run Wood Boiler at maximum load all the time and make up other load with Oil Boiler until Oil cannot be operated at lower than minimum capacity, 10% of full capacity at point A. At this point to save fuel cost, we reduce running load of Wood Boiler following the Oil Boiler run at the minimum load, point A to B. Now we can shut down Oil Boiler and start up Wood Boiler only to produce the load until it reach at minimum load of 25%,

2. Point B to CWe shut down Wood Boiler and start up Oil Boiler again, run it until end of year, 8760th hour. We have also a Standby Boiler in case of the largest boiler cannot work. We have chosen Oil Boiler, because it has lowest cost compared with others two, wood and coal.

Limitation of Boiler Capacity

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To set up the Limitation of Boiler Capacity, first condition we assume that the maximum capacity of Oil Boiler has to be greater than or equal to the minimum capacity of Wood Boiler (0.25) at point C from figure 1.

Pmax.oil-cap. >= min.wood-load * Pmax.wood-cap. ---- Eq. a_1

The second condition is the summation of both main boilers has to be equal to the maximum load curve.

Pmax.oil-cap. + Pmax.wood-cap. = 89.53 MW ---- Eq. a_2

From solving Eq. a_1 and a_2 we get possible maximum capacity of Oil Boiler and minimum capacity of Wood Boiler.

Pmax.oil-cap. >= 17.906 MWPmax.wood-cap. <= 71.624 MW

For the third condition Oil Boiler must be able to run at minimum load curve, 6.303 MW.

min.oil-load * Pmax.oil-cap. <= 6.303 MW ---- Eq. b_1

Then we can get possible minimum capacity of Oil Boiler and maximum capacity of Wood Boiler, with min.oil-load = 0.1.

Pmax.oil-cap. <= 63.03 MWPmax.wood-cap. >= 36.97 MW

Load Calculation and Condition in Excel for the First CaseWood Boiler

0th to point A,

Ppart = Pmax.wood-cap.

Point A to B,

Ppart = (Load_Curve) – (min.oil-load * Pmax.oil-cap.)

IF condition in Excel: Load_Curve – Pmax.wood-cap. <= (min.oil-load * Pmax.oil-cap.)

Point B to C,

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Ppart = Load_CurveIF condition in Excel: Load_Curve <= Pmax.wood-cap.

Point C to 8760th hour,

Ppart = 0

IF condition in Excel: Load_Curve <= (min.wood-load * Pmax.wood-cap.)

Oil Boiler 0th to point A,

Ppart = Load_Curve - Pmax.wood-cap.

Point A to B,

Ppart = (min.oil-load * Pmax.oil-cap.)

IF condition in Excel: (Load_Curve) – Pmax.wood-cap. <= (min.oil-load * Pmax.oil-cap.)

Point B to C,

Ppart = 0IF condition in Excel: Load_Curve <= Pmax.wood-cap.

Point C to 8760th hour,

Ppart = Load_CurveIF condition in Excel: Load_Curve <= (min.wood-load * Pmax.wood-cap.)

Second CaseThe second case has been applied from the first case to reduce energy production from oil and replace it by using the energy production from wood between point B and C. And another

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thing of this case is instead of using only one Oil Boiler we use 2 Oil Boilers as peak load and run them as the same ratio from each boiler. But disadvantage of this case is we have to install 2 of Oil Boiler so the construction cost will be more expensive than the big 1 Oil Boiler as shown in Figure 2.

A

B

C D

Figure 2 Power production for the second case

Limitation of Boiler CapacityFrom solving the optimization cost for the first case we got the combination of each boiler capacity, there are 17.906 MW for Oil Boiler and 71.622 MW for Wood Boiler. So this second case we just want to separate Oil Boiler from the first case to 2 smaller Oil Boiler. Then we assume the total of both small boilers must be equal to 17.906 MW.

Pmax.oil 1-cap. + Pmax.oil 2-cap. = 17.906 MW

Load Calculation and Condition in the Second CaseWood Boiler

0th to point A,


Point A to B,

Ppart = (Load_Curve) – (min.oil-load * Pmax.oil 1-cap.) - (min.oil-load * Pmax.oil 2-cap.)IF condition in Excel: Load_Curve – Pmax.wood-cap. – (min.oil-load * Pmax.oil 1-cap.) <= min.oil-load * Pmax.oil 2-cap.

Point B to C,

Ppart = (Load_Curve) – (min.oil-load * Pmax.oil 2-cap.)

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IF condition in Excel: (Load_Curve) – (min.oil-load * Pmax.oil 2-cap.) <= Pmax.wood-cap.

Point C to D,


Point D to 8760th hour,

Ppart = 0IF condition in Excel: Load_Curve <= (min.wood-load * Pmax.wood-cap.)

Oil_1 Boiler 0th to point A,

Ppart = (Load_Curve - Pmax.wood-cap.) * ((Pmax.oil 1-cap.)/( Pmax.oil 1-cap. + Pmax.oil 2-cap.))

Point A to B,

Ppart = min.oil-load * Pmax.oil 1-cap.

IF condition in Excel: Load_Curve – Pmax.wood-cap. - (min.oil-load * Pmax.oil 2-cap.) <= (min.oil 1-load * max.oil-cap.)

Point B to D,

Ppart = 0IF condition in Excel: Load_Curve - (min.oil-load * Pmax.oil 2-cap.) <= Pmax.wood-cap.


Ppart = Load_Curve * ((Pmax.oil 1-cap.)/( Pmax.oil 1-cap. + Pmax.oil 2-cap.))IF condition in Excel: Load_Curve <= (min.wood-load * Pmax.wood-cap.)


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Ppart = (Load_Curve - Pmax.wood-cap.) * ((Pmax.oil 2-cap.)/( Pmax.oil 1-cap. + Pmax.oil 2-cap.))

Point A to C,

Ppart = min.oil-load * Pmax.oil 2-cap.

IF condition in Excel: Load_Curve – Pmax.wood-cap. - (min.oil-load * Pmax.oil 1-cap.) <= (min.oil 2-load * Pmax.oil-cap.)

Point C to D,

Ppart = 0IF condition in Excel: Load_Curve <= Pmax.wood-cap.


Ppart = Load_Curve * ((max.oil 2-cap.)/( max.oil 1-cap. + max.oil 2-cap.))

IF condition in Excel: Load_Curve <= (min.wood-load * Pmax.wood-cap.)

Third caseIn this case we change how to run both oil boilers. Instead of running both oil boilers as the same load ratio we run Oil_1 Boiler at full load at the beginning and run Oil_2 Boiler to produce other load. When Oil_2 Boiler is run at its minimum load we just shutdown it then run another oil boiler. The reason is to use Oil_1 Boiler as much efficient as possible as shown in Figure 3 and Figure 4.

E F

Figure 3 Power production for the third case

A C

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B D

Figure 4 Zoom-in of the third case

Limitation of Boiler CapacityFor this case we use the same condition with the second one. We also do the optimization cost and the combination of boiler capacity is 71.622 MW for Wood boiler, 7.913 MW for Oil_1 Boiler, and 9.992 MW for Oil_2 Boiler. And the optimization cost for this is cheaper than the second one only 1 Kr/MW-h.

Load Calculation and Condition in the Third caseWood Boiler

0th to point A,


Point A to B,


IF condition in Excel: (Load_Curve) – Pmax.wood-cap. – Pmax.oil_1-cap. <= (min.oil-load * Pmax.oil_2-cap.)

Point B to C,


IF condition in Excel: (Load_Curve) – Pmax.wood-cap. <= Pmax.oil_1-cap.

Point C to D,

Ppart = Load_Curve - min.oil-load * Pmax.oil_2-cap

IF condition in Excel: (Load_Curve) – Pmax.wood-cap. <= min.oil-load * Pmax.oil_1-cap.

Point D to E,

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Point E to F,

Ppart = 0IF condition in Excel:Load_Curve <= min.wood-load * Pmax.wood-cap.

Point E to 8760th hour,

Ppart = 0IF condition in Excel:(Load_Curve) - Pmax.oil_1-cap. <= min.oil-load * Pmax.oil_2-cap.


Ppart = Pmax.oil_1-cap.

Point A to B,

Ppart = Pmax.oil_1-cap. - Pmax.wood-cap. - (min.oil-load * Pmax.oil_2-cap.)IF condition in Excel: (Load_Curve) – Pmax.wood-cap. – Pmax.oil_1-cap. <= (min.oil-load * Pmax.oil_2-cap.)

Point B to C,

Ppart = Load_Curve - Pmax.wood-cap.

IF condition in Excel: (Load_Curve) – Pmax.wood-cap. <= Pmax.oil_1-cap.

Point C to D,

Ppart = min.oil-load * Pmax.oil_1-cap

IF condition in Excel: (Load_Curve) – Pmax.wood-cap. <= min.oil-load * Pmax.oil_1-cap.

Point D to E,

Ppart = 0

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IF condition in Excel: Load_Curve <= Pmax.wood-cap.

Point E to F,

Ppart = Pmax.oil_1-cap.

IF condition in Excel:Load_Curve <= min.wood-load * Pmax.wood-cap.

Point E to 8760th hour, reduce,

Ppart = Pmax.oil_1-cap. – (min.oil-load * Pmax.oil_2-cap.)IF condition in Excel:(Load_Curve) - Pmax.oil_1-cap. <= min.oil-load * Pmax.oil_2-cap.


Ppart = (Load_Curve) - Pmax.wood-cap. - Pmax.oil_1-cap.

Point A to B,

Ppart = (min.oil-load * Pmax.oil_2-cap.)IF condition in Excel: (Load_Curve) – Pmax.wood-cap. – Pmax.oil_1-cap. <= (min.oil-load * Pmax.oil_2-cap.)

Point B to C,

Ppart = 0IF condition in Excel: (Load_Curve) – Pmax.wood-cap. <= Pmax.oil_1-cap.

Point C to D,

Ppart = 0IF condition in Excel: (Load_Curve) – Pmax.wood-cap. <= min.oil-load * Pmax.oil_1-cap.

Point D to E,

Ppart = 0

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IF condition in Excel: Load_Curve <= Pmax.wood-cap.

Point E to F, run following load curve.

Ppart = Load_Curve - Pmax.oil_1-cap.

IF condition in Excel:Load_Curve <= min.wood-load * Pmax.wood-cap.

Point E to 8760th hour, run at minimum load.

Ppart = (min.oil-load * Pmax.oil_2-cap.)IF condition in Excel:(Load_Curve) - Pmax.oil_1-cap. <= min.oil-load * Pmax.oil_2-cap.

Forth CaseWe develop this case from the third one. Instead of choosing Oil_1 in third case we just choose wood as Wood_2 Boiler. And at ending of year we run only Wood_2 Boiler again. For the capacity of Oil Boiler we don’t want to use it so much. The only thing choosing oil and locate it as peak load is to make up this peak in short time. Because oil has lowest construction cost but in otherwise it has highest of fuel cost. This case is our cheapest cost per MW-h as shown in Figure 5 and Figure 6.

C D E

Figure 5 Power production for the forth case

A

14

B

Figure 6 Zoom-in of the forth case

Limitation of Boiler CapacityThe condition that we use for this case is only letting capacity of Oil Boiler equal to minimum load of Wood_1 Boiler.

Pmax.oil-cap. = min.wood-load * Pmax.wood_1-cap.

Pmax.wood_1-cap. + Pmax.wood_2-cap. + Pmax.oil-cap. = 89.53 MW

Load Calculation and Condition in the Forth CaseWood_1 Boiler

0th to point A,

Ppart = Pmax.wood_1-cap.

Point A to B,


IF condition in Excel: (Load_Curve) – Pmax.wood_1-cap. – Pmax.wood_2-cap. <= (min.oil-load * Pmax.oil-cap.)

Point B to C,


IF condition in Excel: (Load_Curve) – Pmax.wood_1-cap. <= Pmax.wood_2-cap.

Point C to D,

Ppart = (Load_Curve) - min.wood-load * Pmax.wood-cap

IF condition in Excel: (Load_Curve) – Pmax.wood_1-cap. <= min.wood-load * Pmax.wood_2-cap.

Point D to E,

Ppart = Load_CurveIF condition in Excel:

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Load_Curve <= Pmax.wood_1-cap.


Ppart = 0IF condition in Excel:Load_Curve <= min.wood-load * Pmax.wood_2-cap.

Wood_2 Boiler 0th to point A,


Point A to B,

Ppart = (Load_Curve) - Pmax.wood_1-cap. – (min.oil-load * Pmax.oil-cap.)IF condition in Excel: (Load_Curve) – Pmax.wood_1-cap. – Pmax.wood_2-cap. <= (min.oil-load * Pmax.oil-cap.)

Point B to C,

Ppart = (Load_Curve) - Pmax.wood_1-cap.

IF condition in Excel: (Load_Curve) – Pmax.wood_1-cap. <= Pmax.wood_2-cap.

Point C to D,

Ppart = min.wood-load * Pmax.wood_2-cap.

IF condition in Excel: (Load_Curve) – Pmax.wood_1-cap. <= min.wood-load * Pmax.wood_2-cap.

Point D to E,

Ppart = 0IF condition in Excel: Load_Curve <= Pmax.wood_1-cap.


Ppart = Load_CurveIF condition in Excel:

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Load_Curve <= min.wood-load * Pmax.wood_2-cap.

Oil Boiler 0th to point A,

Ppart = (Load_Curve) - Pmax.wood_1-cap. - Pmax.wood_2-cap.

Point A to B,

Ppart = min.oil-load * Pmax.oil-cap.

IF condition in Excel: (Load_Curve) – Pmax.wood_1-cap. – Pmax.wood_2-cap. <= (min.oil-load * Pmax.oil-cap.)

Point B to C,

Ppart = 0IF condition in Excel: (Load_Curve) – Pmax.wood_1-cap. <= Pmax.wood_2-cap.

Point C to D,

Ppart = 0IF condition in Excel: (Load_Curve) – Pmax.wood_1-cap. <= min.wood-load * Pmax.wood_2-cap.

Point D to E,

Ppart = 0IF condition in Excel: Load_Curve <= Pmax.wood_1-cap.


Ppart = 0IF condition in Excel:Load_Curve <= min.wood-load * Pmax.wood_2-cap

Efficiency CalculationWhen we run boiler the efficiency is not constant, it very with Part Load (Ppart). And the formula of Part Load Efficiency is.

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η partload=η full∗[0.6+0.8∗(P part /P full)– 0.4∗(P part /P full)2] --- Eq. 1

Example 1 We are running 30 MW Wood Boiler which has maximum capacity of 50 MW.

η full=0.87 , P part=30MW ,P full=50MW

η partload=0.87∗[0.6+0.8∗(30MW /50MW )−0.4∗(30MW /50MW )2]

η partload=0.81432 Ans.

Energy Consumption Calculation in 1 hour (MW-h)In this assignment we calculate Power Input in every 1 hour during a year. In otherwise we could say calculated Power Input is MW-h in each hour.

Power Input in each hour (MW) * 1 hour = Energy Consumption in each hour (MW-h)

We can find Power Input from the Efficiency Equation.

η partload=(PowerOutput (P part ))

(Powe r Input)

Power Input=(P part )

(η partload ) ---- Eq. 2

Substitute Eq. 2 in Eq. 1

Power Input=(P part )/(η partload=η full∗[0.6+0.8∗(P part /P full )– 0.4∗(P part /P full)2]) ---- Eq. 3

Example 2 From example 1, we will get Power Input by using Eq. 3.

Power Input=(30MW )(0.81432 )

Power Input=36.84MW

And from this Power Input we will get Energy Consumption in 1 hour.

36.84MW∗1hour=36.84 (MW−h )Fuel Consumption Calculation in period of time (MW-hfuel)To calculate Fuel Consumption in period of time we use “SUM ()” function in Excel to sum up all of Fuel Consumption in a year. According to the first case we will get the Fuel Consumption in a year as shown in Table 2.

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hourOil input

(MW)Oil_Safe input

(MW)Wood input

(MW)

1 19.46304 82.3241

2 19.40851 82.3241

3 19.35415 82.3241

4 19.29997 82.3241

5 19.24596 82.3241

. . . .

. . . .

. . . .

8756 8.232142 0

8757 8.230652 0

8758 8.229157 0

8759 8.227659 0

8760 8.226156 0

oil sum input

oil_safe sum input

wood sum input

35638.9 0 331451

MWh-fuel MWh-fuel MWh-fuel

Table 2 Fuel Consumption

Highlights are the yearly fuel consumption, for oil is 35638.9 MW-h fuel and for wood is 331451 MW-hfuel.

Total Cost CalculationIn this assignment there are 5 costs that we have to pay in a year. There are interest rate, fuel, construction tax, auxiliary power cost and the last one is staff cost.1. Interest Rate

This cost come from loaning money from bank and we have to pay back equally money to the bank in each year.

2. Fuel CostThis cost has been calculated from total energy produced of each boiler per year.

3. Tax and InsuranceThis cost has been calculated from energy capacity of each boiler.

4. Auxiliary PowerThis cost come from electricity that we use to operate the boiler, such as to convey coal to the boiler or to pump oil fluid in the system.

5. Staff CostWe use only 6 persons to operate our plant and the cost for them is 600 000 Kr per year. So for 6 persons, Staff Cost is 3 600 000 Kr per year.

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Finding Optimization CostIt is impossible to find Optimization Cost by using mathematical method, so in this assignment we use “Data Solve” function to solve the Optimization Cost. And we have example what parameters we have to use. Figure 7 is our result for the first case.

Figure 7 Total cost of first case

Now we want to find the optimization of total cost per MW-h (cell U9145 in figure 7). By changing oil capacity (cell U121). And we also set up the constraints, 17.906 <= oil capacity <= 63.03. Then we put all of this data in Data Solver (figure 8), we get the minimum cost per MW-h equal to 282.98 Kr/MW-h from figure 7.

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Figure 8 Setting up solver parameters

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ResultWe have done the optimization from each case as showing.

Case 1 - 1 oil boiler 1 wood boiler

0 1 2 3 4 5 6 7 8 9 100

102030405060708090

100

Oil_1Wood_1

Figure 9 Optimization of Case 2


0 1 2 3 4 5 6 7 8 9 100

102030405060708090

100

Oil_2Oil_1Wood_1

Figure 10 Optimization of case 2


0 1 2 3 4 5 6 7 8 9 100

102030405060708090

100

Oil_2Oil_1Wood


22


0 1 2 3 4 5 6 7 8 9 100

102030405060708090

100

Oil_1Wood_2Wood_1


We also make the compariton table to find which is the cheapest case.

CASE 1 CASE 2 CASE 3 CASE 4Wood 1 17,91 17,91 71,62 52,10Wood 2 - - - 24,41Oil 1 71,62 71,62 17,79 13,02Oil 2 - - 10,11 -Oil 3 (Standby) 71,62 71,62 71,62 52,10Cost (kr/MWh) 283,97 283,08 281,44 239,24

Table 3 Case comparison

Finally we found that the forth case is the cheapest one. So we have also analyse sensitivity curve with variation of both Construction Cost and Fuel Cost as shown in figure 13 and 14 restpectively. Base load ratio in both figures is the ratio of the biggest load to total load.

0.6 0.8 1 1.2 1.4 1.6 1.80.5810

0.5815

0.5820

0.5825

0.5830

0.5835

0.5840

Sensitivity of Construction Cost Variation

Base Load RatioPower (Base Load Ratio)

Figure 13 Construction cost sensitivity of case 4

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0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.20.5780

0.5790

0.5800

0.5810

0.5820

0.5830

0.5840

0.5850

0.5860

0.5870

Sensitivity of Fuel Cost Variation

Base Load RatioPower (Base Load Ratio)

Figure 14 Fuel cost sensitivity of case 4

ConclusionIn this study, boiler size, fuel type and fuel consumption in period of time are optimized using Microsoft Excel based on an hourly operating cost. This method is suitable for complicate problem analysis because errors are less than integration method.

From construction cost sensitivity curves in figure 13 it’s very less variation of base load ratio. But for fuel cost sensitivity curve in figure 14 we can see bigger change of base load ratio about 3 times comparing with another one. Because the construction cost is very small comparing with fuel cost. But this changing is also very small comparing with boiler size changing. From this fact we can conclude that our case has very low size variation of boiler when either construction cost or fuel cost.

The methodology described in this study provides an important and systematic approach for design and analysis combination of boiler. The analysis shows the lowest energy production cost is expected for the boiler. In addition the optimization gives the overview of results obtained by the different combinations of parameters. This is useful in choosing the solution to a given problem. In contrast analyzing all possible size by stimulation energy consumption of each boiler is time consuming. Thus, by using stimulation based approach, we obtained the optimum point at a lowest energy production cost.

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WER015 Final Report Assignment 1

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