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\begin{document}


\section{abstract}

In this paper, a loss method is proposed to calculate the energy auditing and exergy auditing of domestic hot water boilers along with the Condensing heat exchanger.

In this method, the heat losses of the hot water boiler are eliminated by losses from ash, unburned carbon,radiation and etc. Also, exergy losses are broken down into unburnt hydrocarbon, flue gas, boiler wall.
By calculating these losses, the efficiency of the first law and the exergy efficiency of the hot water boiler are obtained.

To demonstrate the efficiency of the method, a laboratory set up is constructed and measured first rule efficiency and exergy efficiency by measuring parameters such as temperature and pressure,gas analysis.

Then, using this set up, each of the parameters has been experimentally modified and its effect on the efficiency of the first law and exergy efficiency has been investigated.

Finally, it has been shown that the efficiency of the first law is 53 and the exergy efficiency is 45 and the heat exchanger has increased the efficiency of 5.

\section{introduction}

Increasing demand for fossil fuels, rapid evacuation of energy resources and environmental problems, such as air pollution and global warming, have made it important to use energy efficiently.

One of the ways to improve the efficiency of boilers that are in operation is an energy audit that involves energy auditing and exergy auditing.

The goal is to measure the boiler efficiency using a series of parameters.

There is two methods for energy auditing of boilers. In the direct method, efficiency is estimated
directly by estimating input and output terms. This method can not be used to improve boiler performance.
In indirect or loss method, each of the thermal losses are calculated separately and
are used to estimate the efficiency. In this method, as the heat loss is calculated, it can be used to improve the boiler performance. One of the most well-known versions of this method, is the standard ASME ptc4.1 which is developed for energy audition of steam boilers. Other organizations
have published similar standards such as BS845 (British standard).

In connection with the energy auditing of hot water boilers standard BS-EN303-3 is presented. 

%maghale marbot be talafat hararati dig ab garm ezafe shavad


In connection with the exergy auditing of boilers, the only paper available is Dr. Bebehbahaninia's article.

For hot water boilers, Kondensing, an economizer is installed to cool the flue gas so that it changes to the liquid phase.
This condensing fluid is one of the energy losses and exergy losses.
%Which did not consider the loss methods in this term.

%tarikhchee

In this paper, an experimental setup is considered and for the first time, the terms of energy losses and exergy losses are considered in unburnd carbon and % other terms.
Each of the parameters is measured and the exergy efficiency is 53.
Also, the effect of the parameters on the efficiency is specified.
Exergy development is based on Dr. Behbahaninia's article.


\section{methods}

\subsection{energy analysis}

The heat loss method is used to calculate the energy efficiency.
%ezafe shavad
first the flow rate of the dry flue gas per kg of the fuel must be evaluated. The mass of flue gas per mass of fuel is calculated as follows. 

\begin{equation}
W_{dg}=\frac {11CO_{2}+8O_{2}+7(CO+N_{2})}{3(CO_{2}+CO)} \times [C_{ab}+ \frac{S}{267}]+ \frac{S}{160}
\label{b1} 
\end{equation}

In which $C_{ab}$ is carbon actually burned per kg of fuel.

%ravabete Cab chetori byarim??????

$L_{2}$ is energy loss due to dry flue gas and is calculated using equation (\ref{b2}).

\begin{equation}
L_{2}=\frac{0.24W_{dg}(T_{g}-T_{o})}{HHV}
\label{b2}
\end{equation}

In which $T_{g}$ is temprature of the gas leaving the boiler



$L_{3}$ is energy loss due to evaporating and superheating moisture in fuel and is calculated using equation (\ref{b3}) .

\begin{equation}
\begin{cases}
L_{3}=\frac{M_{f}(1089+0.46T_{g}-T_{f})}{HHV} \hspace{1cm} T_{g}< 575 F
\\
 L_{3}=\frac{M_{f}(1066+0.5T_{g}-T_{f})}{HHV} \hspace{1cm} T_{g}>575 F
\end{cases}
\label{b3}
\end{equation}
In which $M_{f}$ is moisture in fuel and $T{f}$ is temprature of fuel.


$L_{4}$ is energy loss due to evaporating and superheating moisture formed by combustion of hydrogen and is calculated using equation (\ref{b4}).


\begin{equation}
\begin{cases}
L_{4}=\frac{9H_{2}(1089+0.46T_{g}-T_{f})}{HHV} \hspace{1cm} T_{g}< 575 F
\\
 L_{4}=\frac{9H_{2}(1066+0.5T_{g}-T_{f})}{HHV} \hspace{1cm} T_{g}>575 F
\end{cases}
\label{b4}
\end{equation}


$L_{5}$ is energy loss due to incomplete combustion and is calculated using equation (\ref{b5}).

\begin{equation}
L_{5}=10160\hspace{0.1cm} C_{ab} \frac{CO}{CO_{2}+CO} \times \frac{1}{HHV}
\label{b5}
\end{equation}


$L_{6}$ is energy loss due to unconsumed carbon and is calculated using equation (\ref{b6}).

\begin{equation}
L_{5}=\frac{14600(C-C_{ab})}{HHV}
\label{b6}
\end{equation}

$L_{6}$ is energy loss due to radiation loss on boiler wall and is calculated using equation (\ref{b6}).
%rabete L6 gharar dade shavad.


\subsection {exergy analysis}

In order to obtain the exergy loss, mass, energy and exergy
balance must be established. Combining the energy equation and
the second law of thermodynamics at the steady state condition the
following equation is obtained in which potential and kinetic
exergies are neglected.

\begin{equation}
\dot E_{F}=\dot E_{p}+(\dot E_{L1}+\dot E_{L2}+\dot E_{L3}+\dot E_{L4})+(\dot E_{D1}+\dot E_{D2})
\label{a}
\end{equation}


\subsubsection{fuel exergy}

Fuel exergy contains 3 components of chemical exergy of
consumed fuel and physical exergy of atomizing steam and inlet air.

\begin{equation}
\dot E_{F}=\dot E_{f}+\dot E_{As}+\dot E_{a,11}
\end{equation}

In which $\dot E_{f}$ is chemical exergy of the consumed fuel, $\dot E_{AS}$ is exergy of atomizing steam, and $\dot E_{a;11}$ is physical exergy of the air.
Chemical exergy of consumed fuel is calculated using the following equation%(manbaa).

\begin{equation}
\dot E_{f}=\dot m_{f} \times \varepsilon^\circ
\end{equation}

In which $ \varepsilon^\circ$ is chemical exergy of consumed fuel.

The physical exergy of a single-component system can be
calculated by employing the following equation:

\begin{equation}
\varepsilon_{ph}=[(h-h_{0})-T_{0}(s-s_{0})]
\label{a2}
\end{equation}

In case of perfect gas assumption equation
%manbaa
may be written as follows

\begin{equation}
\varepsilon_{ph}=C_{p} \times (T-T_{0}-T_{0}\ln(\frac{T}{T_{0}}))+R\ln(\frac{P}{P_{0}})
\end{equation}

The exergy value of the atomizing stream is zero.
\begin{equation}
\dot E_{AS}=0
\end{equation}



\subsubsection{Product exergy}

Product exergy can be calculated as shown in equation (\ref{a}) and
based on %Fig. 1

\begin{equation}
\dot E_{P}=\dot m_{3}\varepsilon_{3}-\dot m_{1}\varepsilon_{1}
\end{equation}

The amount of exergy per kg of eachflow is calculated using
equation (\ref{a2}).


\subsubsection{exegy loss}


$E_{L1}$that is the exergy loss of the flue gas involves physical and
chemical components. 

\begin{equation}
 \tilde{\varepsilon_{ch}}=\sum x_{k}\varepsilon_{k}+\bar R T_{0}\sum x_{k} \ln x_{k}
\end{equation}


\begin{equation}
 \tilde{\varepsilon_{ph}}=(T-T_{0}) \sum n_{i} \tilde {C_{P}}\varepsilon +\bar R T_{0} \ln \frac{p}{p_{0}}
\end{equation}


Therefore total exergy loss of the flue gas is calculated as follows

\begin{equation}
\dot E_{L1}=\frac{\dot m_{g,7}}{M_{g,7}} \times \tilde{\varepsilon_{ch,7}}+\dot m_{g,7} \times {\varepsilon_{ph,7}}
\end{equation}

$M_{G}$ is the equivalent molecular mass of the flue gas which is
calculated as follows

\begin{equation}
M_{G}=\sum X_{i} \times M_{i}
\end{equation}
$X_{i}$ is mole fraction of the flue gas components.


$E_{L2}$ is the exergy loss due to emission of unburnt hydrocarbons
and is calculated as follows

\begin{equation}
\dot E_{L2}=\frac{\dot m_{g,7}}{M_{g,7}} \times (x_{k} \tilde{\varepsilon_{CH}}+\bar R T_{0} \ln x_{CH}
\end{equation}

$\tilde{\varepsilon_{CH}}$ denotes the molar exergy of unburned hydrocarbon and assumes equal to exergy of $C_{2}H_{4}$ namely $331.2868163 \frac {Kj}{Kmol}$ %manbaa zekr shavad


$E_{L3}$ is the exergy loss due to emission of non-fullybunburnt
hydrocarbons which leads to formation of CO in the flue gas and
is calculated as follows

\begin{equation}
\dot E_{L3}=\frac{CO}{CO+CO_{2}}(\varepsilon CO -\varepsilon CO_{2})C_{b} \times \dot m_{fuel}
\end{equation}

% wdg dar ghesmate ghabl bayad mohasebe shavad fk nakonam nyazi shavad


In order to calculate the exergy dissipation through the boiler walls, first the total heat loss of the walls must be calculated. According to ASME, this parameter is calculated via the following
equation % equation

\begin{equation}
Q_{radiation \hspace{0.1cm}   loss}=L \times A
\end{equation}

in which thermal loss that is due to radiation per area unit can be
obtained using the following equation

\begin{equation}
L= 0.548 \times [(\frac {T_{s}}{55.55})^4-(\frac {T_{o}}{55.55})^4]+1.957 \times ({T_{s}}-{T_{o}})^(1.25)  \times  \sqrt {(\frac{196.85 \times V_{m}+68.9}{68.9})}
\end{equation}

Therefor $\dot E_{L4}$ is calculated by using the following equation.

\begin{equation}
\dot E_{L4}=Q_{radiation \hspace{0.1cm}  loss} \times (1-\frac{T_{o}}{T_{s}})
\end{equation}


$E_{L5}$ is the exergy loss due to economizr wall and is calculated as $E_{L4}$

\begin{equation}
Q_{radiation \hspace{0.1cm}   loss}=L \times A
\end{equation}

\begin{equation}
L= 0.548 \times [(\frac {T_{s}}{55.55})^4-(\frac {T_{o}}{55.55})^4]+1.957 \times ({T_{s}}-{T_{o}})^(1.25)  \times  \sqrt {(\frac{196.85 \times V_{m}+68.9}{68.9})}
\end{equation} 

\begin{equation}
\dot E_{L5}=Q_{radiation \hspace{0.1cm}  loss} \times (1-\frac{T_{o}}{T_{s}})
\end{equation}


$E_{L6}$ that is the exergy loss of the flue gas condens and calculate as follow.
%mohasebe badn vared shavad



\end{document}