# Is high pressure combustion hotter than atmospheric pressure combustion

let's say i have the same quantity of coal and air. Will the combustion achieve higher temperature if the air is under pressure?

• this is an interesting question. First of all, it depends on the not only starting temperature but also starting pressure of air. Let's clarify by asking temp increase due to combustion, rather than just final temp. Starting T and P determine density of air. Assuming any starting temp change in the air is due to compression, I think it would depend on whether burning rich or lean. If burning lean, I think there would be less temperature increase (higher P-> similar combustion energy but higher air density -> less increase in gases temp at end of combustion). Less sure about burning rich. Mar 12 at 20:51
• Yes, higher partial pressure of oxygen ; same reason fire is more dangerous and energetic with oxygen enrichment at one atmosphere. Mar 12 at 21:35

The maximum temperature is the adiabatic flame (combustion) temperature. All else being equal, the combustion temperature will approach this to the extent that using higher pressure air puts the reaction process closer to being truly adiabatic.

The adiabatic flame temperature is determined by assuming stoichiometric feed to full reaction with all combustion enthalpy going entirely to heat the products. Assuming constant molar specific heats $$\bar{C}_{p,j}$$ (J/mol K) for each product, the simple energy balance becomes

$$\Delta_{rxn}\bar{H} = \left(\sum\ \nu_j\ \bar{C}_{p,j}\right) \left(T_{ad} - T_o\right)$$

where $$\Delta_{rxn}\bar{H}$$ is the molar reaction enthalpy (J/mol) and $$\nu_j$$ are the stoichiometric coefficients on the products. Essentially, you start with the reactions and do the combustion in a closed system (no mass flow in/out) under adiabatic conditions (no heat flow in/out).

Two factors play a role in defining the extent to which the real combustion process is adiabatic at any instant in time.

One is the instantaneous rate of heat release. This is controlled by the reaction rate. When the overall empirical rate law for the combustion reaction has a positive reaction order in oxygen, higher air pressure will increase combustion rate. When the combustion is zero order in oxygen, air pressure will have no effect on the combustion rate. When the combustion reaction is overall negative order, increasing pressure will decrease combustion rate. The same is true by extension to keeping the total pressure constant but increasing or decreasing the mole fraction of oxygen in the air.

The second factor is the instantaneous rate of heat transfer away during the combustion. Pressure can change the conduction and convection coefficients in heat transfer. It has less (if not negligible) effect on radiation. Correlations and first principles formulations can be investigated elsewhere to address all of the competing effects in detail.

Putting this all into a summary, we can say that, as pressure increases:

• If the reaction rate increases and the heat convection rate decreases, the combustion temperature will increase. Inverting the two rates (reaction and convective heat transfer) will also invert the direction of combustion temperature change.
• If the reaction rate increases but the heat convection rate increases, the combustion temperature may go up, stay the same, or even decrease. The same is true that when the two rates (reaction and convection) are inverted. The direction of the change in temperature is unknown.
• If the reaction rate is unaffected by pressure, the prediction will depend on whether heat convection rate increases or decreases with pressure.

Real combustion reactions are never perfectly adiabatic. Heat flows from the system to the surroundings. We need only also to realize that radiation will occur from the hot combustion zone to anything colder around it. The most basic correction could be to subtract the heat loss per mole of reaction $$\bar{q}_{loss}$$ as a "lumped" term.

$$\Delta_{rxn}\bar{H} - \bar{q}_{loss} = \left(\sum\ \nu_j\ \bar{C}_{p,j}\right) \left(T_{real} - T_o\right)$$

Real combustion reactions are also never closed systems. In the case of coal and oxygen, gas phase oxygen flows in to the solid coal and gas phase products, carbon dioxide, carbon monoxide, and water, flow out from the combustion zone. Each gas species carries its own enthalpy. The net enthalpy difference will change the real case versus what is determined by the adiabatic flame calculation. The most basic correction might be to subtract an enthalpy difference for gas flows in and out $$\Delta_{flow}\bar{H}$$.

$$\Delta_{rxn}\bar{H} - \bar{q}_{loss} - \Delta_{flow}\bar{H} = \left(\sum\ \nu_j\ \bar{C}_{p,j}\right) \left(T_{real} - T_o\right)$$

By example, take complete combustion of carbon C(s) + O$$_2$$(g) $$\rightarrow$$ CO$$_2$$(g) with only radiation to the surroundings. We could write a starting point as

$$\Delta_{comb}\bar{H}(C(s)) - \sigma \epsilon A (T^4 - T_o^4) - \left(\Delta_f\bar{H}(T,p)(CO_2(g)) - \Delta_f\bar{H}^o(O_2(g)) \right)\\ = \bar{C}_{p}(CO_2(g)) \left(T - T_o\right)$$

Lets start by making a few assumptions about the combustion environment.

• The combustion is occurring at constant pressure. The relevant energy-temperature term for changes in temperature at constant pressure is enthalpy.
• The solid fuel isn't affected by the different pressures.
• The oxidizer is an ideal gas starting out at some pressure P. Its enthalpy is pressure dependent.
• The combustions product mix is identical for the two different pressures. And let's say it's all ideal gases at these pressures.
• The kinetic energy terms of the combustion produces are negligible. This may be a bad assumption because exhaust velocities of hundreds of miles per hour are common.
• Unlike in Jeffrey J Weimer's answer, assume there is no heat loss to the environment. This is normally a really bad assumption.

Begin by looking up the formation enthalpy of the fuel and oxidizer. Combine using molar weighting. Do the same for the combustion products by weighting them with respect to one mole of fuel and oxidizer.

Next, calculate the enthalpy change to get the fuel and oxidizer to the initial conditions. There will be a small difference with the oxidizer enthalpies due to pressure.

Now subtract the combustion-product enthalpy (per mole of fuel and oxidizer) from the fuel+ox enthalpy to get the heat evolved during combustion. Then use the molar weighted specific heat of the products to compute the change in temperature of the combustion gasses from the standard temps in the heat of formation tables.

Lastly, get the combustion products to the proper pressure adiabatically, and see what the temps are. Note this is the only part that changes with pressure.

The only terms that change with pressure are the P-V work terms used in adiabatically moving the system from standard pressure in the Formation tables to the actual pressures. The higher the pressure, the more P-V work and the higher the temps.

If you combust in a fixed volume, the temps won't change unless the chemistry of the combustion products change, which normally does happen at different pressures.

High pressure combustion occurs faster than atmospheric combustion. If the energy release is the same, the faster combustion means it is more powerful than slower combustion. Greater power (shorter time) means less time for the heat released in the reaction to diffuse away, and the combustion products get hotter (but for a brief time).

Q. let's say i have the same quantity of coal and air. Will the combustion achieve higher temperature if the air is under pressure?

Yes

To put all the technical jargon to one side and answer this in a very simplistic way.

Yes that it why a Blacksmith fans the flames in his forge.

However you are making a mistake in your assumption, you are assuming that the amount of air does not increase and this is the "pivotal point". In fact you are using the same amount of air but rate of use increases. In general it takes x amount of air to burn x amount of coal. If you deliver the air to the coal quicker, the coal will burn faster and more heat will be generated.