I am currently studying a boiler without many technical data.

Below a short description:

Fuel is put inside the boiler furnace. Primary (F1) and secondary air (F2) fans allow fresh air to go through the boiler. At the outlet of the gas circuit, there is also another fan (F3) at constant speed that is here to maintain a -300 Pa depressure inside the furnace. The fresh air temperature is increased up to 650°C thanks to combustion.

After combustion, hot air calories are exchanged with oil circulating along a coil inside the boiler. This oil is at 100°C before exchange, and at the end of the exchange, it is around 200°C and air temperature is around 250°C. The whole purpose of this boiler is to maintain this 200°C temperature of oil.

For security measures, a controller is here to ensure that the depressure on air side, inside the furnace remains constant by adjusting the primary and secondary air fans speed. Also, if oil temperature goes beyond 200°C, no more fuel will be added inside the boiler. A pressure decrease will increase these fans speed, and a pressure increase will decrease it.

So, for the heat exchange we have these values:

  • Air flowrate: $\dot{m}_{air} = 100.000 kg/h$
  • Air temp in: $T_{air1} = 650°C$
  • Air temp out: $T_{air2} = 250°C$
  • Specific heat of air: $C_{p_{air}} = 1.063 kJ/kg.K$
  • Oil flowrate: $\dot{m}_{oil} = 212.600 kg/h$
  • Oil temp in: $T_{oil1} = 100°C$
  • Oil temp out: $T_{oil2} = 200°C$
  • Specific heat of oil: $C_{p_{oil}} = 2 kJ/kg.K$

Thanks to this formula, we can estimate the heat exchange power (11,8 MW):

$$\dot{Q} = \dot{m}_{air}\cdot C_{p_{air}}\cdot (T_{air1}- T_{air2}) = \dot{m}_{oil}\cdot C_{p_{oil}}\cdot (T_{oil2} - T_{oil1}) $$

Now, I would like to predict the boiler bevahiour if I change 1 parameter: inlet oil temperature. As stated above, it is currently at 100°C before exchange. Now, let's say it is lower: 50°C.

What would happen?

Knowing from experience that we have a little margin regarding the heat exchange power capacity, my guess is that more fuel will be consumed to increase gas temperature and allow the oil outlet temperature to remain at 200°C. Though, pressure inside the boiler will most likely increase because of hotter air and the fans F1 and F2 will consequently end up running slower to maintain the set depressure. This would mean that we would get hotter air, but at a lesser quantity.

Though, if we manage to get as much air as before with a hotter air temperature, I would then think that oil will absorb the extra calories, leaving the outlet air temperature as it is.

Please let me know if you think that I am on the right path. I am pretty uncertain of this because I do not know the limitations of the boiler.

So, I would like to check all of this with calculation. I thought of following these steps:

  1. Find the surface area thanks to LMTD method
  2. Trying to use the NTU method to find the outlet temperatures
  3. Predicting the resulting pressure with perfect gas law inside the boiler to understand how the controller will manage fans F1 and F2

I am really uncertain on how to go through these steps though. LMTD method is ok, I found 72 m² of surface area, but after this, I am stuck.

I am not sure how to use NTU method and perfect gas law for this. For the volume part, should I consider the volume of the boiler, of the air volume going through it?

  • $\begingroup$ Looks like you have confused your temperatures oil is air… $\endgroup$
    – Solar Mike
    Feb 23, 2022 at 19:57
  • $\begingroup$ And why more oil than air? About 14: 1 is stoichiometric … $\endgroup$
    – Solar Mike
    Feb 23, 2022 at 19:59
  • $\begingroup$ Sorry, bad copy paste... I have edited my message. For the quantity of oil and air, I cannot give your more details as this goes beyond my knowledge. These are the values I have $\endgroup$
    – Snite
    Feb 23, 2022 at 21:32

1 Answer 1

  1. Pressure inside the furnace won't change because it is controlled independently. You will need more forced draft (inlet) air to burn the extra fuel, but the induced draft fan (outlet) will speed up to compensate.

  2. Outlet gas temperature will likely increase a bit: you are increasing the heat transfer, so the outlet gas temp will probably be a bit higher.

  3. Note that this isn't a boiler because you're heating oil, not boiling water, but the principles are the same as any two-fan industrial boiler.

  • $\begingroup$ Thank you for your answer. Though, please note that fan on the outlet is not controlled by pressure. It is running at 100% at all times. So I think inlet fans will see their speed reduced because of the increase of gas pressure. For point 2. considering that flowrate remains the same (which should not) I am not sure to understand why outlet gas temp will be hotter since extra calories provided by combustion should be absorbed by the oil, no? At the end of the heat exchange, shouldn't oulet gas temp be the same? Thank you for clarification on point 3. $\endgroup$
    – Snite
    Feb 24, 2022 at 18:59
  • $\begingroup$ Now, I would like to know if there is any way to predict inlet fans speed, since outlet fan speed will not be reduced. What is the calculation method? What about gas temp? $\endgroup$
    – Snite
    Feb 24, 2022 at 18:59
  • 1
    $\begingroup$ @Snite, I've never seen an induced draft fan that wasn't controlled, this system may only work for a single set of conditions and may not respond well to drastic changes. It must be running at way over stochiometric if pressure is only controlled by inlet fans. $\endgroup$
    – Tiger Guy
    Feb 24, 2022 at 20:48
  • $\begingroup$ Sorry, I should also have mentionned that the fuel quantity is also dependant of the controller handling the pressure. But you are correct, this heater does not respond well to drastic changes $\endgroup$
    – Snite
    Feb 25, 2022 at 6:06

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