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BOILER EFFICIENCY CALCULATIONS 

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boiler-efficiency

BOILER EFFICIENCY CALCULATIONS 

BOILER EFFICIENCY CALCULATIONS

BOILER EFFICIENCY CALCULATIONS 
    Efficiency is a level of work ability of a tool. While the efficiency of the boiler is the work performance or the level of performance of the boiler or steam boiler obtained from a comparison between the energy transferred to or absorbed by the working fluid in the kettle with the input of chemical energy from the fuel. For the level of efficiency in boilers or steam boilers, the efficiency ranges from 70% to 90%. (Agung.N, 2007) 
There are two methods of evaluating boiler efficiency: 
– Direct Method: energy obtained from working fluids (water and steam) compared to energy contained in boiler fuel.
– Indirect method: efficiency is the difference between loss and energy that enters 

Combustion 
    Combustion occurs in a chemical process between combustible materials and oxygen from the air to produce heat energy which can be used for other purposes. The main components of combustible materials are carbon, hydrogen, and other mixtures. In the combustion process this component burns into carbon dioxide and water vapor. A number of sulfur is also found in most fuels (Singer, 1991).    
    In the combustion process the amount of oxygen used can affect the quality of combustion. Oxygen is one of the air elements which amounts to 20.9% of all airborne components. The fuel will burn in normal conditions if there is enough air. 

Heat Balance
    Boiler performance parameters, such as efficiency and evaporation ratio, decrease with time due to poor combustion, dirty surface heat exchanger and poor operation and maintenance. Even for new boilers, reasons such as poor fuel quality and water quality can result in poor boiler performance. A heat balance can help in identifying heat losses that can or cannot be avoided. Boiler efficiency tests can help in finding boiler efficiency deviations from the best efficiency and target problem areas for corrective actions.

    The combustion process in the boiler can be described in the form of an energy flow diagram. This diagram graphically illustrates how the incoming energy from fuel is converted into energy flows of various uses and becomes a flow of heat and energy loss. Thick arrows indicate the amount of energy contained in each stream. (UNEP, 2008). 
. Boiler energy balance diagram. 

    The heat balance is the total energy balance that enters the boiler against the one leaving the boiler in a different form. The following figure illustrates the various losses that occur for steam generation. 

Picture. Heat loss in the Heat Exchanger coal fired boiler    Heat exchanger is a tool to move heat energy from a fluid to another fluid. Hot fluids give heat to cold fluid through a medium or directly so that changes will occur according to what is desired, both decreases and increases in temperature. In general, this heat transfer occurs in combination between conduction and convection. 

Logarithmic average temperature difference (ÄTlmtd) 
The amount of ÄTlmtd can be calculated based on the type of flow arrangement applied in a heat exchanger. The equation for calculating tdTlmtd in parallel, opposite and cross flow is explained in the following explanation.

    The logarithmic average temperature difference method is used as a first step in the analysis of heat exchangers, when the fluid in and out temperature of the heat exchanger is known, both for hot and cold fluids, so it can determine the logarithmic average temperature difference. If the condition of the hot fluid is identified in condition 1 and the fluid out state is identified 2, while the cold fluid state is identified according to those points. 
For the opposite direction 

flow for unidirectional flow of 

Boiler Classification of the 1X25 MW Lombok Jeruk PLTU
    Based on the purpose and construction of the boiler in the Jeruk PLTU, including the industrial boiler category because it has specifications, among others, the boiler is used to drive the turbine, can use fuel in the form of oil, has a steam capacity of 130 tons per hour (210,000 kg / hour), pressurized design on 9.8 MPa superheater (pressurized design 104 kg / cm2, superheater outlet design pressure 91 kg / cm2), superheater outlet design temperature 540 ° C (513 0C), temperature at feedwater 215 ° C, temperature at primary air heater 150 ° C, temperature on the secondary air heater 150 ° C, the exhaust gas temperature is 150 ° C, with the assembly done at the PLTU Jeranjang where the boiler will be used.

    Based on the area that experienced heating, boilers in the PLTU Jeranjang include boiler water tube types. There it is clear that the circulating water in the boiler enters through pipes and the combustion heat is passed through the outer surface of the pipe.
    Based on the type of PLTU Jeruk boiler including FBC (Fluidized Bed Combustion) boiler. The boiler. evenly distributed air or gas is passed up through the bed of solid particles such as sand in a fluidized state heated to the flame temperature of coal and coal is injected continuously into the bed, the coal will burn quickly and the bed reaches a uniform temperature and is supported by a fine filter, particles will not be disturbed at low speeds. Once the air velocity gradually rises, a state is formed where the particles suspended in the air stream – the bed is called “fluidized”. With the subsequent increase in air velocity, bubble formation, strong turbulence, rapid mixing and the formation of a tight bed surface occur.

Analysis of Condition of Commissioning of 
    PLTU Jeranjang in the normal operation plan using coal fuel. To do further analysis, it is necessary to calculate to obtain the empirical formula and molecular formula of the fuel. Data obtained from the results of fuel analysis used during commissioning are: 
– Carbon, C = 85.46 wt% 
– Hydrogen, H = 11.68 wt% 
– Sulfur, S = 2.24 wt% 
– Nitrogen, N = 0.07 wt% 
– Oxygen, O = 0.20 wt% 
– Ash, a = 0.04 wt% 
– Moisture = 0.31 wt% 
– HHV = 10473 kcal / kg 
– LHV = 9840 kcal / kg 
– Specific gravity = 0.9475 

Load 25 MW 

Boiler 
Input:
~ Mass flow rate from feed water 
    = (139580 kg / h) 
    = 38.77 kg / s 
~ Enthalpy from feed water 
    = 183.19 kcal / kg 
    = 766315.04 joules / kg 
Output: 
~ Mass flow rate from steam outlet boiler 
    = 139760 kg / h 
    = 38.82 kg / s 
~ Enthalpy from the steam outlet boiler 
    = 817.9 kcal / kg 
    = 3421337.86 joule / kg 
~ The heat energy received by water 
    = 38.82 kg / sx 3421337.86 joules / kg – 38.77 kg / sx 
       766315.04 joules / kg 
    = 103112201.69 joules / s 

Furnace 
Input: 
~ Fuel mass flow rate 
    = 9246 kg / h 
    = 2.57 kg / s 
~ Heating value of fuel
    = 9840 kcal / kg 
    = 41161467.84 joule / kg 
~ Combustion air mass flow rate 
    = 136836 kg / h 
    = 38.01 kg / s 
~ Enthalpy from combustion air 
    = 324426.50 joules / kg 
Output: 
* Flue gas 
~ Flue gas mass flow rate 
    = 139912.2 kg / h 
    = 38.86 kg / s 
~ Entalphi from exhaust gas 
    = 341819.64 joules / kg 
~ Heat energy from combustion 
    = (2.57 kg / sx 41161467.84 joules / kg + 38.01 kg / sx 324426.50 joules / kg) – (38.86 kg / sx 341819.64 joules / kg) 
    = 104763171.77 joules / s 
~ Energy losses not transferred to water 
    = 104763171.77 joule / s – 103112201.69 joule / s 
    = 1650970.08 joule / s

Analysis of Current Conditions 
Data obtained from fuel 
analysis results are: 
    – Carbon, C = 85.12 wt% 
    – Hydrogen, H = 12.09 wt% 
    – Sulfur, S = 2.04 wt% 
    – Nitrogen, N = 0.65 wt% 
    – Oxygen, O = 2.13 wt% 
    – Ash, a = 0.01 wt% 
    – Moisture = 0.5 wt% 
    – HHV = 18497 Btu / Lb 
    – LHV = 17619 Btu / Lb 
    – Specific gravity = 0.9816 

Load 25 MW 
Boiler 
Input: 
~ Mass flow rate from feed water 
    = 155667 kg / h 
    = 43.24 kg / s 
~ Enthalpy from feed water 
    = 245.98 kcal / kg 
    = 1028941.92 joules / kg 
Output: 
~ Mass flow rate of the steam outlet boiler
    = 155000 kg / h 
    = 43.06 kg / s 
~ Enthalpy from steam outlet boiler 
    = 813.63 kcal / kg 
    = 3403476.13 joule / kg 
~ Heat energy received by water 
    = 43.06 kg / sx 3403476.13 joule / kg – 43.24 kg / sx 1028941.92 joule / kg 
    = 102046249.35 joule / s 

Furnace 
Input: 
~ Fuel mass flow rate 
    = 9634 kg / h 
    = 2.68 kg / s 
~ Heating value fuel 
    = 9826.54 kcal / kg 
    = 41105163.64 joule / kg 
~ Mass air combustion rate 
    = 143640 kg / h 
    = 39.90 kg / s 
~ Enthalpy from combustion air 
    = 176.14 kcal / kg 
    = 736816.18 joules / kg 
Output:
* Flue gas 
~ Flue gas mass flow rate 
    = 153090 kg / h 
    = 42.53 kg / s 
~ Entalphi from exhaust gas 
    = 486932.23 joule / kg 
~ Combustion heat energy 
    = (2.68 kg / sx 41105163.64 joules / kg + 39.90 kg / sx 736816.18 joule / kg) – (42.53 kg / sx 486932.23 joules / kg) 
    = 118694157.63 joules / s 
~ Energy losses not transferred to water 
    = 118694157.63 joule / s – 102046249.35 joule / s 
    = 16647908.29 joules / s 

Calculation of Heat Loss 
Calculation of commissioning boiler efficiency
    This efficiency calculation uses a heat loss method. From this calculation we can get a decrease in boiler efficiency caused by heat loss from the system in the boiler. The heat losses include: 
Loss of acid price correction 
    = acHac / HHV / (1 + a) x 
    = 0.11% 
Dry exhaust gas loss 
    = Dry gas at AH inlet x 0.24 x (Uncorrected gas temp at AH outlet – Temp dry bulb) x / (1 + a) 
    = 11.25% 
Loss of water in fuel 
    = Moisture in fuel x Entalphy diff x / (1 + a) 
    = 0.003% 
Loss for burning hydrogen in fuel 
    = Moisture from hydrogen x Entalphy diff x / (1+ a) 
    = 1.14%
Disadvantages CO 
    = Dry gas at AH inlet x CO wt% at AH inlet x 24.2 x / (1 + a) 
    = 0% 
Loss of unburned carbon 
    = 81 x Solid combustible loss / HHV / (1 + a) x 
    = 0 
Loss vapor for sample 
    = Total sampling losses / Fuel heat input / (1 + a) x 
    = 0.02% 
Heat loss of water content in combustion air 
    = Moisture in water x 0.46 x Uncorrected gas temp at AH outlet – Temp dry bulb 
       x / (1 + a) 
    = 0.03% 
Steam losses for fuel oil 
    removal = Moisture from steam x atomizing burner (Saturated steam at drum) – Feed water enthaly at drum inlet) x / (1 + a) 
    = 0.02%
Heat transfer and flow losses 
    = 0.07% (American Boiler Manufacturer Association chart) 
    = Dry gas at AH inlet x CO wt% at AH inlet x 24.2 x / (1 + a) 
    = 0% 
Loss of unburned carbon 
    = 81 x Solid combustible loss / HHV / (1 + a) x 
    = 0% 
Steam losses for samples 
    = Total sampling losses / Fuel heat input / (1 + a) x 
    = 0.11% 
Heat loss for combustion air 
    = Moisture in air x 0.46 x Uncorrected gas temp at AH outlet – Temp dry bulb x / (1 + a) 
    = 0.35% 
Steam losses for fuel oil removal
    = Moisture from the burner atomizing steam x (Saturated steam at drum) – Feed water depth at drum inlet) x / (1 + a) 
    = 0.21% 
Heat transfer and flow losses 
    = 1.14% (American Boiler Manufacturer Association chart) 

Boiler heat balance diagram for the conditions of commissioning. 

Calculation of boiler efficiency Current conditions The 
following is the percentage of each loss. 
Loss of acid price correction 
    = acHac / HHV / (1 + a) x 
    = 0.81% 
Dry exhaust gas loss 
    = Dry gas at AH inlet x 0.24 x (Uncorrected gas temp at AH outlet – Temp dry bulb) x / (1 + a) 
    = 14.85% 
Loss of water in fuel
    = Moisture in fuel x Entalphy diff x / (1 + a) 
    = 0.001% 
Loss for burning hydrogen in fuel 
    = Moisture from hydrogen x Entalphy diff x / (1 + a) 
    = 9.33% 
Loss CO. 
    Dry gas at AH inlet x CO wt% at AH inlet x 24.2 x / (1 + a) 
    = 0% 
Loss of unburned carbon 
    = 81 x Solid combustible loss / HHV / (1 + a) x 
    = 0% 
Steam losses for sample 
    = Total sampling losses / Fuel heat input / (1 + a) x 
    = 0.11% 
The heat loss for combustion air 
    = Moisture in water x 0.46 x Uncorrected gas temp at AH outlet – Temp dry bulb 
       x / (1 + a) 
    = 0.35%
Steam losses for fuel oil 
    removal = Moisture from burner atomizing steam x (Saturated steam at drum) – Feed water depth at drum inlet x / (1 + a) 
    = 0.21% 
Heat transfer and flow losses 
    = 1.14% (American Boiler Manufacturer Association chart) 

Pictures. Graph of heat loss difference between 
 commissioning conditions and current conditions 
    From the above calculations, it is found that there is a decrease in boiler efficiency caused by heat loss with the greatest heat loss occurring in the exhaust gas. The higher the exhaust gas means the higher the heat released from the boiler. This means that there is a waste of heat which results in a decrease in the efficiency of the boiler.
    The causes of rising boiler exhaust temperature include a lack of heat transfer surface and fouling that occurs in heat transfer tubing. Which is caused by crust, both from fuel and feed water. The increase in exhaust temperature is also caused by a decrease in the performance of the air heater. 

Discussion of boilers The 
    efficiency of the FBC boiler has decreased by 4.17%. This decrease can be caused by many things, here are the factors that affect the efficiency of the boiler.
    The rate of clean air supplied through the water heater. Boilers must be operated with an air flow rate that is more than the theoretical air requirement calculated based on the smoke gas analysis. But too much excess air will also cause losses due to the taking of the heat itself by excess air to be carried along with the exhaust gas, for this reason the gas gas analysis is carried out to determine the actual air requirement (actual air W).
    The burner factor, the function of the burner is to mix fuel and air with proportions that are suitable for the occurrence of ignition of the fire and to maintain the condition of combustion that continues to run well. Burners that are not properly tuned will result in mixing air and fuel incorrectly and at each loading rate will increase excess air demand and waste fuel so that the boiler efficiency will decrease. 
    Combustion air temperature is also a factor that affects boiler efficiency, combustion air temperatures can be increased by utilizing high exhaust gas temperatures through the air heater, boiler efficiency can be increased by 1% at each increase in combustion air temperature by 25 C.  
    Fouling is the main factor that affects boiler efficiency. Fouling is the occurrence of deposits or crust on the heat transfer surface which can result in inefficient combustion results so that the exhaust gas temperature will be high.
    Blowdown also affects boiler efficiency. Deposits formed on the tube walls on the water side can reduce efficiency and even crust can damage the body due to over heating. These deposits are caused by high concentrations of suspended solids and dissolved solids, which can also cause foam (foam) to cause carry over. Therefore the concentration of solids must be maintained under certain conditions, and this is done by a blowdown process, where water is discharged out and immediately replaced by boiler feed water. Because blowdown is water released in high temperature conditions, this is a heat dissipator which results in a decrease in efficiency.
    The fourth factor is the use of condensate, if water vapor provides thermal energy into a process system, the heat absorbed by the process generally is latent heat, while the condensate which still carries sensible heat will leave the process system at high temperatures. Given that condensate is basically high-temperature pure water, so far not contaminated with condensate can be used as ideal boiler feed water. Condensate contamination is one of the things that does not allow condensate to be used as feed water, as an alternative, the heat present in condensate can be utilized through a heat exchanger.
    Another factor that also influences efficiency is the ultimate residual data, as we can compare that the high heating value of the residue used at commissioning is greater than 113 kcal / kg than that used in the current condition. This obviously affects the combustion process that occurs. The combustion process will produce higher heat energy if the fuel used has a high HHV, and vice versa. The content of carbon, hydrogen, nitrogen, oxygen, sulfur, moisture, ashcontent in residual fuel also affects the results of the combustion process. Carbon, hydrogen and sulfur are burning constituents. It is different with nitrogen which will burn only with a high combustion temperature,

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