The technology of electric boilers and their availability have allowed pathways for decarbonization in industrial heating processes. Several research studies have used analytical and computational methods to model the energy usage of fossil-fuel-fired boilers and their efficiency. However, an investigation into electrical boilers has yet to be made available.
The system requires some input to generate a pattern using simulation. In the model, the user has the flexibility to choose a location or user-defined degree days, number of boilers, number of end users, boiler capacity, flow rate, set temperature and pressure, combustion air temperature, makeup water amount and temperature, blowdown amount and temperature, boiler surface area and temperature, flue gas information, fuel type, steam type, auxiliary unit presence, control type, and facility operational hour and capacity probability. All these input factors enable the model to be robust. The simulation program is interactive and detail-oriented, which allows the user to change the values of every variable. All the input variables are explained in the next chapter with their functionality. The simulation is developed in Microsoft Excel® using the Visual Basic Application (VBA)® based on discrete-event Monte Carlo Simulation (MCS). The uniformly distributed random number is generated in Excel, depending on the user input. A heuristic approach is used to determine the temperature based on the generated random number concerning heating and cooling degree days. Similarly, the heuristic rule is used to calculate the boiler system losses and finally translate the energy usage by the system.
2.2. Case Development
For this study, five cases will be considered to increase the versatility of the fuel usage by boiler and equivalent electrical boiler.
Case I: Four boilers running in a facility that generates hot water to provide the energy required for space heating are considered. Since the requirement is for space heating, the demand for hot water from the boiler is continuous in wintertime; however, during summer, the boilers will barely run.
Case II: In this case, the four boiler settings remain the same as in Case I. The purpose of the boilers is similar to serving space heating needs. However, in this model, the location will be more temperate. All other operating parameters will remain the same as Case I.
Case III: This case will have one large-capacity boiler for process heating. Since process heating is required during the production hours only for this instance, weekdays, 8 AM to 4 PM, the simulation will imitate the boiler fuel consumption only during those hours.
Case IV: This case will be designed similarly to Case III for comparison. The only difference will be on the demand side, where space heating is needed alongside process heating. This will require the boiler to work more, resulting in more fuel consumption.
Case V: This last case will be again designed similarly to Case III, with the difference that improved boiler operating parameters will be used. Using an improved boiler will reduce the fossil-fuel-fired boiler’s energy losses, reducing CO2 emissions. The demand side energy usage will remain the same. Therefore, the electrical boiler should require a slightly lower capacity than Case III.
After providing all the required information, the user will click the “Click to Run Simulation” button at the top of the input sheet to start the simulation to generate temperature variation every hour for a year. The VBA program will generate a random range for temperature for each day. With the Monte Carlo Simulation and heuristic rules, the daily temperature will vary on the randomly generated range for 24 h. The simulation will calculate the heat energy of steam, condensate return, blowdown, flue gas losses, shell losses, and the end-user need at a 1 h interval for 365 days. Finally, an output spreadsheet will be created to summarize the results. Current energy and fuel usage for each boiler will be represented graphically and numerically. The result will include the required electrical boiler sizing and energy usage for a year. The current cost of operation and proposed electrical boiler operating cost will be included for comparison. Lastly, the CO2 emissions for all possible scenarios will be included to justify the decarbonization effort through electrical boilers.
2.3. Heat Balance
Heat balance is the underlying tool for calculating the energy usage of a boiler and steam system. Heat balance comes from the fundamental concept of the first law of thermodynamics. The first law is a statement of the principle of conservation of energy. In a steam system, the heat is generated by the combustion of fossil fuels, and it is transferred to the steam. The steam travels inside the system, finally transferring the heat energy to the end-user, and returns to the boiler after it has given away its heat content. In heat transfer, it is common practice to refer to the first law as the energy conservation principle or simply as an energy or heat balance. A heat balance determines the heat coming into a system from all sources and the heat leaving the system. The input and output are then balanced to account for all the heat. Therefore, the calculation can be divided into two groups. The first group includes all the heat inputs or gains to the boiler system, and the second group is all the heat outputs or losses from the system.
Heat in Fuel: The amount of heat energy available from the fuel is the fuel heating value multiplied by the fuel used per hour. This can be calculated as:
where
QF = Heat Content in Fuel (Btu/h) (1 Btu/h = 0.293071 W)
ṁF = Amount of Fuel (lb/h) (1 lb/h = 0.000126 kg/s)
HHV = Higher Heating Value of Fuel (Btu/lb) (1 Btu/lb = 2326 J/kg)
In addition to the previous formula, the firing factor and capacity of the boiler can be utilized to calculate the energy in the fuel. It will be easier for the user to obtain the capacity of the boiler from the name specification rather than from the mass flow rate of the fuel. Therefore, the model will integrate the latter concept for convenience.
where
xff = Firing Factor of Boiler (no unit)
C = Capacity of Boiler (MMBtu/h) (1 MMBtu = 293 kWh)
CC1 = Conversion Constant (Btu/MMBtu)
Heat Gain from Combustion Air: Since the enthalpy of flue gas will be calculated relative to the temperature of the entering combustion air, the relative enthalpy of the combustion air is zero. This convention avoids counting the enthalpy of the combustion air in both the input and output computations.
Heat Gain from Makeup Feedwater: If the makeup water enters the boiler at ambient temperature, it will not represent a heat gain or a heat loss. However, if a feedwater economizer exists, the user must provide the feedwater temperature to calculate the energy content. The temperature will be used to find the enthalpy of feedwater from the steam tables. The energy content in feedwater can be calculated as:
where
QMW = Heat Content in Makeup Feedwater (Btu/h)
ṁMW = Amount of Feedwater (gal/min) (1 gal/min = 0.063090 L/s)
hMW = Enthalpy of Feedwater (Btu/lb) (1 Btu/lb = 2326 J/kg)
k1 = Constant 1 (min/h)
k2 = Constant 2 (lb/gal) (1 lb/gal = 0.119826 kg/L)
Heat Gain from Condensate Return: Condensate returned to the boiler accounts for a substantial amount of heat energy input. This input can be calculated as:
where
QC = Heat Content in Condensate (Btu/h) (1 Btu/h = 0.293071 W)
ṁC = Amount of Condensate Return (lb/h) (1 lb/h = 0.000126 kg/s)
hC = Enthalpy of Condensate (Btu/lb) (1 Btu/lb = 2326 J/kg)
Heat in Steam: Steam going out of the boiler contains most of the heat energy generated by the boiler. The heat energy contained in steam is called enthalpy. It is made up of latent heat or the heat to vaporize the water, plus sensible heat, the energy required to heat pure steam—this is generally proportional to the temperature difference through which the steam was heated. The following equation can calculate the energy contained in the steam:
where
QS = Heat Content in Steam (Btu/h) (1 Btu/h = 0.293071 W)
ṁS = Amount of Steam Flow (lb/h) (1 lb/h = 0.000126 kg/s)
hS = Enthalpy of Steam at Given Temperature and Pressure (Btu/lb) (1 Btu/lb = 2326 J/kg)
Heat Loss in Blowdown: The blowdown loss can be calculated by determining the mass of water and the energy content of the discharged water. The amount of energy lost through blowdown depends on factors such as the temperature and pressure of the water being discharged and the volume of water being blown down. The following equation presents the energy content in the blowdown water:
where
QBD = Heat Content in Blowdown (Btu/h) (1 Btu/h = 0.293071 W)
ṁBD = Amount of Blowdown (lb/h) (1 lb/h = 0.000126 kg/s)
hBD = Enthalpy of Blowdown at Given Temperature (Btu/lb) (1 Btu/lb = 2326 J/kg)
Heat Loss from Surface: Much of the energy lost from boilers is radiated to the environment or transferred to the environment via convection. The following equation is used to calculate the heat losses from the surface by radiation and convection:
where
QSLradiation = Radiative Heat Loss from Surface (Btu/h) (1 Btu/h = 0.293071 W)
ABS = Boiler Surface Area (ft2)
C1 = Stefan-Boltzmann Constant (0.1714 × 10−8 Btu/h-ft2-°R4) (1 Btu/h-ft-°F = 1.730735 W/m·K)
Ts = Boiler Surface Temperature (°R = °F + 460)
TA = Ambient Temperature (°R = °F + 460)
where
QSLconvection = Convective Heat Loss from Surface (Btu/h) (1 Btu/h = 0.293071 W)
ABS = Boiler Surface Area (ft2) (1 ft2 = 0.092903 m2)
C2 = Convection Coefficient (0.18 Btu/h-ft2-°F) (1 Btu/h-ft-°F = 1.730735 W/m·K)
Ts = Boiler Surface Temperature (°F) (°C = (°F − 32)/1.8)
TA = Ambient Temperature (°F)
Heat Loss in Flue Gas: There are several ways to calculate the heat loss from the boiler through flue gas. (i) Use the fossil-fuel composition and the chemical reactions in the combustion to estimate the amounts of each flue gas component. Then determine the enthalpy of each component and add them all up. This requires a good knowledge of chemistry and extensive calculation. (ii) Use the combustion efficiency curve, which requires measuring the percent of flue gas oxygen. (iii) Calculate the heat energy in the flue gas using mass flow rate, specific heat, and temperature difference. This third model option will be used as mass flow rate can be found from the combustion motor specification, and the standard value for the specific heat of flue gas can be utilized. The first option can be cumbersome to integrate into the model, and the second option requires the oxygen percentage in flue gas, which might need to be more readily available. To calculate the flue gas heat loss, the following equation is used:
where
QFG = Flue Gas Heat Loss (Btu/h)
νFG = Volumetric Flow Rate of Flue Gas (ft3/min) (1 ft3/min = 0.000472 m3/s
ρFG = Density of Flue Gas (lb/ft3) (1 lb/ft3 = 16.018463 kg/m3)
k1 = Constant 1 (min/h)
Cp = Specific Heat of Flue Gas (Btu/lb-°F) (1 Btu/lb-°F = 4.1868 J/kg·K)
TFG = Flue Gas Temperature (°F) (°C = (°F − 32)/1.8)
Heat Loss from Flue Surface: Heat loss from the flue surface can be similar to heat loss from the boiler surface. Therefore, a similar equation can calculate the heat loss from the flue surface.
End User Energy Usage: The steam generated by the boiler transfers its heat energy content to the end user. The heat transfer equation can be utilized on the end user side to determine heat energy gained from the steam based on the required temperature. After providing the heat energy to the end user, the steam returns to the system as condensation. The following heat transfer equation can be used if the end-user is solid/metal:
where
QEUsolid = Heat Energy Requirement of End User (Btu/h) (1 Btu/h = 0.293071 W)
k = thermal conductivity (Btu/h-ft-°F) (1 Btu/h-ft-°F = 1.730735 W/m·K)
AEU = Surface Area of Solid End User (ft2) (1 ft2 = 0.0929 m2)
t = Thickness of the Solid End User (ft) (1 ft = 0.3048 m)
TEU = End User Required Temperature (°F) (°C = (°F − 32)/1.8)
If the end user is fluid (liquid/gas), the following heat transfer equation can be used:
where
QEUfluid = Heat Energy Requirement of End User (Btu/h) (1 Btu/h = 0.293071 W)
νEU = Volumetric Flow Rate of End User (ft3/min) (1 ft3/min = 0.000472 m3/s)
ρEU = Density of End User (lb/ft3) (1 lb/ft3 = 16.018463 kg/m3)
k1 = Constant 1 (min/h)
CpEU = Specific Heat of End User (Btu/lb-°F) (1 Btu/lb-°F = 4186.8 J/kg·K)
TEU = End User Required Temperature (°F) (°C = (°F − 32)/1.8)
Figure 3 shows the heat energy input and output of the developed model.
A similar heat energy balance can be applied to the demand side to balance the energy in the end user.
Figure 4 depicts the energy input and output at the end user.
Based on the diagram, the energy balance formula for the model used is as follows:
The demand side balance can be expressed with the following formula:
The enthalpy of the flue gas will be calculated relative to the combustion air; the relative enthalpy of the combustion air is zero. The enthalpy of steam, condensate return, and blowdown is determined relative to makeup water enthalpy; therefore, the enthalpy of makeup water will not be considered in Equation (15). Moreover, the firing factor used in Equation (5) dictates the operating load of the boiler, which will be reflected in the flue gas loss. The energy balance’s outcome determines the boiler’s hourly firing factor. Using the firing factor, the hourly fuel usage amount can be determined, which can be used to calculate the CO
2 emissions. Using the information mentioned above, Equation (15) can be rewritten as follows,
The hourly firing factor will be used to determine the hourly fuel usage by combining Equations (4) and (5) as follows,
Table 1 shows the higher heating values used for different fuels in the model [
63].
Finally, the electrical boiler capacity and energy requirements are calculated using the following energy equivalency formula,