Model of a domestic boiler

Question

I have to design a model of a domestic boiler. I represent the boiler as a heat exchanger (HEX1), as shown in the figure. In many articles I have found that it is schematized with this equation:

$$ M C \frac{\text{dT}_{10}}{\text{dt}}=\text{c m$\_$dot} \left(\text{Ti}-T_{10}\right)-Q $$

where $M C$ is the thermal capacity of the boiler, and $Q$ is the heat product by the burner.

My question is: why there is $\frac{\text{dT}_{10}}{dt}$ at the left hand side of the equation? Why not $\frac{d\frac{\text{T}_{10} + \text{T}_{i}}{2}}{dt}$ or $ \frac{\text{dT}_{i}}{dt}$?

Thank you!

Update: Connor thank you for the answer. However I don't understand:

And finally, the rate at which heat is stored within the control volume is proportional to the rate of change of temperature of the exchanger: $$Q_{stored} = M C \frac{dT_{10}}{dt}$$

Why the temperature is $T_{10}$? I have found that a possible answer is ralated to the so colled "well-mixed conditions", infact in a article:

"the time derivative of the total energy in the mth block is calculated on the basis of the well-mixed condition"

Is there a better explanation of the well-mixed condition?

Answer 1

The short answer is: it appears that the boiler is treated as a "lumped" mass; that is, that the temperature throughout the boiler is a uniform $T_{10}$, and that the working fluid exits the heat exchanger at the same temperature as the temperature of the heat exchanger itself.

The first law of thermodynamics essentially states: $$Q_{stored} = Q_{in} - Q_{out}$$ where $Q_{stored}$ is the heat stored within the control volume, $Q_{in}$ is the heat entering the control volume, $Q_{out}$ is the heat exiting the control volume. I assume a positive sign convention for each of these terms; that is, a positive value of $Q_{stored}$ indicates that the energy stored within the control volume is increasing, a positive value of $Q_{in}$ indicates energy transfer from outside of the control volume into the control volume, and a positive value of $Q_{out}$ indicates energy transfer from inside the control volume out of the control volume.

We see that the rate at which heat enters the heat exchanger equals the rate at which the burner supplies heat to the exhanger plus the rate at which heat is transferred into the exchanger by the incoming working fluid: $$Q_{in} = Q + \dot{m}c T_i$$ Furthermore, the rate at which heat leaves the control volume is equal to the rate at which heat is transferred by the exiting working fluid: $$Q_{out} = \dot{m}c T_{10}$$ And finally, the rate at which heat is stored within the control volume is proportional to the rate of change of temperature of the exchanger: $$Q_{stored} = M C \frac{dT_{10}}{dt}$$ Putting it all together, we have: $$M C \frac{dT_{10}}{dt} = Q + \dot{m}c T_i - \dot{m}c T_{10}$$ $$M C \frac{dT_{10}}{dt} = \dot{m}c (T_i - T_{10}) + Q$$ I grew up in the "HIP to WIN" (Heat In Positive, Work In Negative) school of thought, and in my answer I have taken $Q$ from the perspective of the heat exchanger; thus $Q$ is positive when there is a net transfer into the heat exchanger. However, the paper you read may have defined the $Q$ value from the perspective of the burner, in which case a net transfer into the heat exchanger (i.e. net transfer out of the burner) would give a negative $Q$. That would explain the negative sign ($-Q$) in the equation you referenced.