Introduction

Buildings are typically heated by the combustion of a fuel (wood, fossil). In modern systems the heat released is absorbed by a heat transfer fluid, which is often water. The hot water is circulated through the building where it releases heat, for example via radiators, and is subseqently pumped back to the boiler. There it comes in at a lower temperature and the process is repeated until the desired temperature in the building is reached.

In case of complete combustion of hydrocarbons this would yield CO\(_2\) and water. During combustion, we could argue that it is pretty damn hot. As such, water leaves the combustion zone in gaseous state. By condensing this water, additional energy could be recovered, namely the enthalpy of vaporization (2.2 MJ / kg). For the combustion of any fueltype that contains hydrogen atoms water forms and this strategy works. In this example I focus on natural gas (i.e. methane). Every mole of methane yields two moles of water:

  • CH\(_4\) + 2O\(_2\) -> CO\(_2\) + 2 H\(_2\)O

Let’s calculate

The lower heating value (i.e. when water is not condensed) of methane is 50 MJ / kg. 1 kg of methane (16 g/mol) yields 2.25 kg of water (18 g/mol). Condensing this water returns 2.25 \(\cdot\) 2.2 = 5 MJ of energy. Thus methane has a higher heating value (HHV) of 55 MJ / kg. Importantly, the theoretical energy efficiency difference between (full) condensing and non condensing is 10%!

Above 100$^\circ$C water is a gas, so we need to cool down the outlet stream of the boiler at least below 100C to achieve condensation. However, the outlet gas has a water vapour “carrying” capacity often referred to as humidity. As most people know, higher temperatures allow higher level of absolute humidity.

Now the water carrying capacitiy depends on the temperature of “air” and abviously on the amount of “air” available. Typically the oxygen fed to a gas boiler is run at 10 to 20% compared to methane. The excess of oxygen ensures that no carbon-monoxide (CO) is formed, which a toxic gas. For a 10% excess of oxygen the in and outlet gas composition is displayed in Table 1, see below. Note that the majority of the gas is composed of nitrogen as air contains nearly 4 times as much nitrogen than oxygen.

Table 1: In and outlet gas composition, in mol

Component Inlet Outlet
CH\(_4\) 1 (8.7%) 0
O\(_2\) 2 \(\cdot\) 1.1 = 2.2 (19.1%) 2 \(\cdot\) 0.1 = 0. (1.7%)
CO\(_2\) 0 1 (8.7%)
H\(_2\)O 0 2 (17.4%)
N\(_2\) 2 \(\cdot\) 1.1 / 0.21 \(\cdot\) 0.79 = 8.3 (72%) 8.3 (72%)
Sum 11.5 11.5

The total pressure of the system (1 atmosphere) is equal to the sum of the partial pressures (Dalton’s law) Wherein the partial pressure is equal to the molar fraction of a component times the total pressure. For water the partial pressure in the flue gas would be 2/11.5 = 0.17 atmosphere. However, water will only be in the gaseous state when its vapour pressure is higher than the aforementioned 0.17 atmosphere. “The saturation vapour pressure is the pressure at which water vapour is in thermodynamic equilibrium with its condensed state. At pressures higher than vapour pressure, water would condense, whilst at lower pressures it would evaporate or sublimate.” Alternatively stated: air is saturated with moisture when the following holds:

  • \[P_{vap} = x \cdot P_{Total}\]

In which x is the molar fraction of water, P\(_{Total}\) is the total system pressure (1 atm) and P\(_{vap}\) is the vapour pressure of water.

The vapour pressure of water can be described by the Antoine equation: Where P is in atmosphere an T in degree C.

  • \[P_{vap} = 0.00132 \cdot 10^{\left(A - \frac{B}{C+T}\right)} = 0.00132 \cdot 10^{\left(8.07131 - \frac{1730.63}{233.426+T}\right)}\]

The normal boiling point (at 1 atm) is found for the temperature that results in a pressure of 1 atmosphere. Which, obviously, is 100$^\circ$C for water. Indeed, filling in T = 100$^\circ$C delivers a pressure of 1 atmosphere. So when the pressure of a system is lower than the vapour pressure of a component the component will be in its gaseous state (Pvap > Psystem). In our case we have a mixed system, i.e. there are “permanent” gas molecules such as nitrogen, oxygen, etc and water. As previously discussed, to achieve condensation of water the vapour pressure must be lower than 0.17 atmosphere. This point is indicated in Figure 1 below.


Figure 1: Water vapour pressure as function of temperature

0.17 atmosphere corresponds with 57$^\circ$C, which means that no water will condense for temperatures greater than 57$^\circ$C. Thus, recover energy by the condensation of water the boiler needs to operate at a temperature below 57$^\circ$C. When the operating temperature is 50$^\circ$C, the flue gas will become supersaturated with air and water will condense, until the flue gas reaches a new equilbrium (saturation). For 50C this saturation pressure is 0.12 atmosphere. The differential is 0.05 atmosphere. The energy corresponding to this pressure decrease is recovered. I have plotted this for a temperature range of 0 to 100C, see Figure 2 below.


Figure 2: Water pressure difference between vapour pressure and water vapour concentration in fluegas

At 0C, dP is 0.17 and all water vapour can be condensed. We also see the non-linear characteristic as function of temperature. Reduction of the boiler temperature from 60 to 40 results in relatively more water condensation than from 40 to 20. Nevertheless, I am not so much interested in in \(\Delta\)P. Therefore, I have translated this chart to the amount of energy that is recovered by water condensation from the flue gas stream as function of the boiler operating temperature, see Figure 3 below.


Figure 3: Energy recovered from fluegas by water condensation as function of boiler operating temperature

We can see that at 0$^\circ$C approximately 5 MJ is recovered per kg of Methane combusted. This corresponds to the difference between the LHV and HHV mentioned in the beginning of this piece of text. Typically we like to heat up a room to at least 20$^\circ$C. To avoid the need for an infinite surface area, some delta T is required, for example \(\Delta\)T = 10$^\circ$C. This results in a minimum operating temperatur of 30$^\circ$C. For such temperature we are able to condens roughly 80% of the water and retrieve its corresponding condensation enthalpy. Runnig at 40$^\circ$C still recovers 60% (3MJ) of energy which is 6% of the LHV (50 MJ / kg). Running a boiler in the condensing regime, preferably towards 30$^\circ$C is highly recommended. You will get energy for “free”, if you know what I mean.

The non-condensing part

Obviously, further cooling the flue gas allows more energy recovery. I was lazy and assumed that the thermodynamic properties of air would resemble flue gas representatively and were constant. I considered 100$^\circ$C as the reference state. Operating at 30$^\circ$C would gain 1.5 MJ per kg of methane combusted compared to the reference state. Alomst 0.5 MJ more than an operating temperature of 50$^\circ$C, see Figure 4 below. Although less, this contribution is signficant. Running at 30$^\circ$C compared to a reference state of 100$^\circ$C gains 4 (from water condensation) + 1.5 = 5.5 MJ per kg of methane (LHV = 50 MJ), which 11%!


Figure 4: Additional energy recovered from the non-condensable fraction as function of boiler operating temperature

Summary

Almost 5 MJ (10%) more energy is reclaimed when running a boiler at 30$^\circ$C instead of 60$^\circ$C.

  • Operate your (condensing boiler) at the lowest temperature possible to maximize the amount of water condensed from the flue gas and thereby increasing the energy efficiency. This can be as high as 4 MJ when condensing to 30$^\circ$C for every kg of methane combusted (LHV = 50MJ).
  • This strategy is relevant for any fuel type that contains hydrogen atoms.
  • Besides recovery of the condensation energy, cooling down the flue gas allows more energy recovery. This can be up to 1.5 MJ for an outlet temperature of 30$^\circ$C compared to a reference state of 100$^\circ$C.