# Why enthalpy decreases with pressure increase? and how we can understand this relation with the fact which we get from steam power plant operation?

why enthalpy decreases with pressure increase? and how we can understand this relation with the fact which we get from steam power plant operation, when we reduce the boiler pressure from 1800 psi to 1500 psi, turbine increase its valve opening value more to admit more steam to meet the power requirement at the same produced power. also in it is recommended in ASME when doing capacity test of turbine to insure the turbine is VWO (valve wide open) and we get this status by reducing the boiler pressure as per ASME. so, that means when pressure decrease, the enthalpy decrease which conflict with the phenomena above. thanks Ahmad Rabie

• For an ideal gas, enthalpy is independent of pressure. Jan 14, 2017 at 12:51

Enthalpy(H) is defined as H=U + PV. U is internal energy and P and V are pressure and volume respectively.

I do not know much about steam power plants, so I looked up steam boilers on the net. Since they are closed vessels, if the pressure falls, the temperature also falls. The internal energy is directly related to temperature, so it decreases. Moreover, the PV term also decreases with temperature ( for both ideal and real gases). Hence, equating with U and PV, H(enthalpy) should also decrease with pressure in this particular situation.

• Like I said, enthalpy of an ideal gas is a function only of temperature. Jan 14, 2017 at 16:08
• Steam at 1800 psi is not an ideal gas. Jan 26, 2017 at 6:57

I think no one is answering because your question is made complicated by the fact that you are dealing with a non-ideal compressible flow mixtures. Most people who went to school barely got through this stuff.

I deal particularly with air flows in underground mines and I noticed similar behavior as your question (this is where google sent me). When I have a given fixed mass flow of moist air at a constant wet-bulb and dry-bulb, the internal energy of the flow itself has to remain constant.

So what is happening? What happens when pressure on this fixed flow increases? The vapor pressure increases and water begins to condense out of the flow itself. It falls out of the flow, and the enthalpy decreases by an amount equal to the latent heat of condensation. The same thing should happen in reverse, when pressure is decreased, more water can join the airflow and this causes its combined total internal energy to increase.

I suggest reading McPherson's psychrometry chapter, specifically when he talks about constant sigma heat (section 14.5.2), see figure 14.2. He talks about omitting the sensible heat of liquid water term.

While this applies to regular water vapour and not steam, I imagine there are parallels. I imagine the useful part of steam is the vapour part, and not the liquid part, and in this way our problems are common. I am only interested in the water vapor.

Note in equation 14.52 enthalpy is defined as the sigma heat + some constant that depends on the moisture fraction (eq 14.48). Sigma heat itself (eq 14.46) depends on a saturated vapour fraction (eq 14.44). If the wet-bulb temperature and dry-bulb temperature are constant, and the pressure increases, the vapour pressure increase must result in a lower moisture fraction (inversely proportional to pressure in equation 14.44).

I hope this answer spurs some discussion as I am simply philosophizing. Happy to provide clarification if the above is not clear. Must insist on further reading in McPherson and parsing of the above in hand with section 14.6.

The thermodynamic chart shows that specific enthalpy of superheated steam isotherms shows a very small decrease in specific enthalpy with pressure, and rapid decrease in specific enthalpy at pressures that are closer to critical point.

But this is enthalpy per kg of steam. At lower pressure, the density of steam is much lower (PV being nearly constant). So the valve needs to be opened to allow the same quantity of steam into the turbine, and get the required enthalpy flow into the turbine.

The specific enthalpy goes down as pressure goes up because the specific volume will decrease more than the pressure increases. Mathematically, h = u + P(↑)*v(↓↓) so overall h(↓).

As far as your boiler example: Lowering the boiler pressure will lower the driving force that causes steam flow and the steam gets less dense, so the mass flowrate of steam entering the turbine goes down (remember we're talking about enthalpy per pound), so the overall rate of work that the steam is able to do on the turbine goes down. The turbine throttle valve will open more to bring this mass flowrate back up until you have enough mass of steam entering the turbine to perform the required work. (In actuality the valve is sensing turbine speed, but the above explanation covers what is happening thermodynamically).

Side note: 1800 pound steam is not an ideal gas. At higher pressures, the PV term will actually do the opposite of what temperature does (above about 450 psi). The steam from the boiler is probably saturated (it's expensive and unnecessary to superheat it in most powerplants), but that does not matter. The enthalpy of saturated steam at 1500 psi is ~18 btu/lbm higher than saturated steam at 1800 psi even though the 1500 psi steam would be about 25 °F colder.