Reference: Daniel V. Schroeder, *An Introduction to Thermal Physics*, (Addison-Wesley, 2000) – Problems 4.22 – 4.24.

A steam engine differs from the Otto and Diesel engines that we’ve looked at so far in that the working substance is not an ideal gas. A steam engine follows the Rankine cycle on a diagram, which looks like this:

The working substance in a steam engine is, not surprisingly, steam, which is condensed to liquid water for part of the cycle. Starting at point 1, the water is in liquid form and is compressed at constant volume to a pressure at point 2. Liquid water doesn’t change its volume much when compressed, so the constant volume is a good approximation.

Along edge 2 to 3, the water is boiled at constant pressure, producing steam which, by the time it reaches point 3, is superheated to a temperature much greater than its boiling point. This is the stage in which heat is absorbed from the hot reservoir, which is typically produced by burning fuels such as coal or gas, or from nuclear power.

The edge 3 to 4 is the stage where the steam does work by turning a turbine (usually to produce electricity, although in the 19th century, steam engines were used in many other places, most nostalgically in steam trains). This stage is an adiabatic expansion to point 4.

Finally, the stage 4 to 1 condenses the steam back to liquid water at constant pressure, expelling waste heat .

The efficiency of the steam engine is defined in the usual way as

Although we have a nice, simple diagram for the Rankine cycle, we can’t use it to calculate the efficiency directly, since we can’t use any of the equations for an ideal gas. However, we can make a start by observing that the two edges along which heat is exchanged both occur at constant pressure, so the heats and are equal to the corresponding enthalpy changes along these edges:

The enthalpies can be read off from steam tables. Schroeder gives a couple of brief tables in his Tables 4.1 and 4.2; more complete tables can be found on the web. The two tables correspond to different parts of the diagram above. One table gives enthalpy and entropy values for saturated water and steam, which, for a given pressure, occurs at the boiling point of water (the boiling point decreases with decreasing pressure). Thus in this case, the temperature is determined by the pressure (or vice versa). This corresponds to points 1 and 4 since this is condensing phase, where the steam is being converted back into liquid water.

The other table gives values of entropy and enthalpy for superheated steam, which for a given pressure, is at a temperature above the boiling point. We can find the enthalpy for point 3 from here, given the pressure and temperature.

We can typically approximate the enthalpy at point 2 to be the same as that for point 1, so an approximate efficiency is

Example 1We have a steam engine operating between temperatures of and with a maximum pressure of . To calculate the efficiency using the approximate formula, we need 3 enthalpies. can be read off Schroeder’s Table 4.1, since here we have water at at its boiling point (implying a pressure of ), so the enthalpy isPoint 3 is in the superheated steam regime, so we can use Table 4.2 to find

To get , we use the fact that curve is an adiabat, so the entropy of the steam is the same at points 3 and 4. From Table 4.2, we have

We need to find a mixture of water and steam at in Table 4.1 that has the same entropy. That is, the fraction of water is given by

Also from Table 4.1, we can now find the enthalpy of this mixture

The efficiency is therefore

Example 2How good is the approximation we used above? We can get an estimate as follows. Enthalpy is defined asso an infinitesimal change in enthalpy is

The change in energy is given by the thermodynamic identity (for constant particle number) as

so

Along edge , the volume is (approximately; water does compress a little when the pressure is increased) constant, but what about the entropy change? If we assume that this compression is approximately adiabatic (I’m not 100% sure we can do this, but it seems like we’re merely increasing the pressure and keeping the temperature and volume constant, so there doesn’t appear to be much heat flow involved here), then the entropy is also roughly constant, so

The values given in Schroeder’s tables are all for 1 kg of water, which occupies so for the numerical example given in Schroeder’s book, where the pressure increases from 0.023 bar to 300 bars, the enthalpy change from 1 to 2 is

Schroeder’s example uses (giving ), and . This changes the efficiency given by Schroeder to

Which is slightly less than the value of 0.48 when we assume .

Example 3We can see the effect of changing the key values in a steam engine by starting with the values given in the previous example. We’ll use the approximation .First, we reduce the maximum temperature to . This changes to (using Table 4.2):

To find , we need to redo the interpolation from Example 1. The entropy at point 3 is

We want a mixture of water (at a fraction ) and steam at that has the same entropy, so

The resulting enthalpy is

with a modified efficiency of

Next, we reduce the maximum pressure to 100 bars (with the high temperature back to ). Again, this affects

To find , we need to redo the interpolation from Example 1. The entropy at point 3 is

We want a mixture of water (at a fraction ) and steam at that has the same entropy, so

The resulting enthalpy is

with a modified efficiency of

Finally, with the original high temperature and pressure, we reduce the minimum temperature to . This gives

To find , we need to redo the interpolation from Example 1. The entropy at point 3 is

We want a mixture of water (at a fraction ) and steam at that has the same entropy, so

The resulting enthalpy is

The efficiency is

The first two changes produce an efficiency lower than the original of 0.48, but increasing the temperature range increases the efficiency slightly.

Pingback: Entropy of water and steam | Physics pages

Pingback: Steam engines in the real world | Physics pages

Pingback: Refrigerators in the real world | Physics pages

HassanIn the last calculation, when we change the minimal temperature to 10, you calculated x=0.305, but used x=0.184 to find the enthalpy H4. The result should have been 0.494, which is higher than the initial efficiency. So lowering the minimal temperature increases the efficiency.

Aside from that, your website makes me smarter everyday, thank you! Keep it up!

gwrowePost authorFixed now. Thanks.