In the expander, (engine). Heat is converted into work. Here we will be taking a look at various engines and the process of producing work from steam under pressure. Steam engines fall into two major classes: Positive displacement and other.
Positive displacement engines produce work in mechanical cycles. This type engine is made up of one or more expanders operating together producing a common output. Each expander increases and decreases volume in continuos alternating cycle. At the start of increasing volume steam is admitted to the chamber producing a force that is transmitted to the output through mechanical linkages. At some point the steam is cutoff and left to expand (exception noted for nonexpanding engines) still producing force that is transmitted to the output but to a lesser and lesser degree as the pressure drops during expansion. At the end of expansion the expander is at it's max volume (exception noted for uniflow engines). At this point the exhaust valve or port is open and the steam is allowed to escape the expander. The expander continues it's cycle decreasing its volume forcing steam out to the exhaust. At some point the exhaust is closed and the residual trapped steam is compressed into the clearance space of the expander.
The expansion and compression are isentropic processes. That is that they occur along constant entropy lines. To better understand this concept I refer you to the isentropic plots on the pressure temperature chart.
In this chart you see lines of constant entropy and the saturation line in red. As you can see pressure and temperature are dependent on the entropy line. Along an entropy line volume decreases as pressure and temperature increase and viceversa. In other words expansion always results in a decrease in temperature and an increase in specific volume. Compression is just the opposite. Note that as you increase pressure during compression a rise in temperature results. From these plots it can be seen that you cannot compress steam into a liquid. To further illustrate this we refer to the pressure enthalpy chart.
This is a plot of pressure verses enthalpy (BTU/lb) steam. The lines of constant entropy are in black. For reference the saturation line is shown in red. Points below the saturation line are a mixture of steam and water. The pressure is a logrithmatic scale so low pressures references show in more detail. On this chart I have drawn in a Rankine compression cycle. The expansion and compression take place along constant entropy lines. Note that the exhaust takes place along a constant enthalpy resulting in higher entropy. Also note that as entropy increases so does the enthalpy to pressure ratio. What this means is that compression back to the inlet pressure will result in a higher quality residual steam than the original inlet steam. (It will have a higher specific enthalpy and be at a higher temperature). The mixture resulting form this residual compressed steam and the admitted steam will be at a higher entropy and do more work during it's fixed expansion then steam at the lower admitted steam entropy. The result is a higher engine efficiency but of a lower output for a given displacement.
From the two charts above you can see those compression results in increases of enthalpy (heat) and temperature. Unless this heat is dissipated the steam will not condense. Further compression from a saturated mixture can result in superheated steam. No condensation! The heat generated during compression must be dissipated in order for any steam to condense. Condensation by compression can only work if the heat can be dissipated. Compression by increasing the temperature can increase heat transfer and in the proper vessel could assist condensation. Within the engine and at speed the heat would not be dissipated fast enough for condensation to occur. Just the opposite. It would become superheated as in the example above.
To figure the thermal efficiency of the engine we calculate the work produced by one cycle of the engine and divide by the heat put into the admitted steam from the boiler. In thermodynamic cycle calculations it is common to figure the cycle based on a unit mass to simplify the calculations. Using a unit mass allows the use of specific volume in place of actual volume. The calculations produce values per unit mass. For instance we may figure the cycle as if 1 pound of steam is cycled through the expander on each cycle. The calculations then produce work per pound of steam, ft-lbs. or BTUs per pound of steam. When we have an engine with clearance, part of the mass expanded by the expander is the residual or compressed steam and part freshly admitted steam from the boiler. At cutoff we have a mixture of the residual steam that was compressed into the clearance space and the admitted fresh steam. To deal with this I selected the cutoff point, as having the unit mass of steam so the total mass expanded would be 1 unit mass. Let mc be the mass of the compressed steam in that mixture. The amount of admitted steam would then be 1 - mc to keep in line with a unit mass expanded by the cycle. We will call the admitted steam mb. So:
m = 1 = m
c + mb
In the diagram above the work is the sum of the admission and expansion work plus the additional work done by the difference of the ending expansion pressure and exhaust pressure through the entire stroke. From this we must deduct the compression and pump work. Taking Qin as the heat from the boiler. Our expression for efficiency is:
Eff = (W
admission + Wexpansion + Wx - Wcompression - Wpump) / Qin
In calculating the thermal efficiency of the "ideal" engine cycle we only consider the energy conversion and ignore all mechanical and heat transfer losses. So the steam is heated in the boiler and transferred to the engine with no heat loss. Qin from the boiler is the difference in enthalpy of the steam and the feed water enthalpy at boiler pressure times the mass mb of the admitted boiler steam.
Q
in = mb (Hf - Hl)
The admission work is seen to be the product of the inlet steam pressure minus the expansion pressure and the displacement at the cutoff point. The total volume at cutoff is the cutoff displacement plus the clearance. As no work is done by filling the clearance space with steam only that part of the steam that fills the displacement space must figured in the work calculations. That part can be figured as a ratio. We are given the clearance and cutoff as a percentage of the total displacement. So the total cutoff volume as a percentage of the displacement is cutoff + clearance. And the ratio of cutoff volume to cutoff and clearance volume is that part of the volume doing work.
Work_Volume = V
cutoff (cutoff / (cutoff + clearance))
As we are using specific volumes in out calculations. The specific volume at cutoff must be adjusted by the above ratio to calculate the admission work.
W
admission = (pcutoff - pexpansion) vcutoff (cutoff) / (clearance + cutoff)
The expansion of steam in the engine is an adiabatic process. Remember from the general energy equation discussion that in an adiabatic process no heat is transferred into or out of the substance. Q = 0. In an adiabatic process work is equal to the change in internal energy during the process.
W
expansion = Ucutoff - Uexpansion
Ucutoff is the internal energy of the steam at the cutoff point and Uexpansion is the internal energy at the end of expansion before the exhaust port opens.
U
cutoff = Hcutoff - Pcutoff vcutoff / J Uexpansion = Hexpansion - Pexpansion vexpansion / J J = 778.17 a conversion factor of BTU to ft lb.
This adiabatic expansion is a special type of expansion, an isentropic expansion. A property of a substance is its entropy. I am not going to try and explain entropy here. All that is needed here is to know that during an isentropic expansion the entropy of the substance does not change. Isentropic means constant entropy. With this fact we can determine the end point of the isentropic expansion. The end point of the expansion is determined knowing the expansion ratio, specific volume and entropy of the cutoff steam. Take the specific volume and multiply it by the expansion ratio.
expansion = (1 + clearance) / (cutoff + clearance)
When the exhaust opens it is common in a positive displacement engine that the steam is not fully expanded to the exhaust pressure. The shaded rectangular area along the bottom of the diagram under the admission and expansion work areas is work done by this difference in pressure and is not accounted for in the internal energy difference above. The work done by this pressure difference, adjusting for clearance volume, is:
W
x = (Pexpansion - Pexhaust) vexpansion) / (1 + clearance))
Compression is the process of reducing the volume of some amount of the residual steam into the clearance space. The point of exhaust valve/port closure determines the amount of residual steam left. At that point the remaining steam is compressed into the clearance. In order to figure the amount of residual steam we must first figure the amount of steam at the start of the expander's volume reduction process. Sense we are using specific volumes and a unit mass in our calculations. The volume of the expander at max volume is the specific volume at the end of expansion. When the exhaust port opens the steam is allow to freely expand to the exhaust pressure. At that point we only have a partial volume remaining in the expander. That part is a ratio of the expansion volume to that of exhaust volume. If we have a pound of steam in the expander at the end of expansion and the volume of the steam doubles when released to exhaust pressure. We have a half pound of steam remaining in the expander. The other half has exited the expander as exhaust. So the amount of the unit mass in the expander at max volume after the exhaust opens is
V
expansion / Vexhaust
But that is not the amount of residual steam just yet. As the expander decreases in volume still steam is forced out the exhaust. When the exhaust valve/port closes. This remaining amount is left to mix with the admitted steam. At the point of exhaust close the percentage of remaining steam, using exhaust_close as a percentage of displacement, is:
(exhaust_close + clearance) / (1 + clearance)
To get the mass of the residual steam is now an easy step. It is the product of the above two ratios:
m
c = ((exhaust_close + clearance) / (1 + clearance)) (Vexpansion / Vexhaust)
This residual steam may be compressed by further reduction in volume of the expander. The compression ratio is calculated as:
compression = (clearance) / (exhaust_close + clearance)
If the expander is at the lowest point in volume reduction when the exhaust closes there is no further reduction in volume and the above ratio would be 1/1 as exhaust_close is 0 at min volume of the expander.
The compression process is an isentropic adiabatic process and the work is again the difference in internal energies of the ending and beginning state.
W
compression = (Ucompression - Uexhaus) mc
Only in this case we are doing work to the substance. The ending state is found in the same way the expansion end point was found for expansion. Being compression the ending volume is computed by dividing the specific volume of the exhaust steam by the compression ratio and locating the steam point with this specific volume and entropy.
Now remember the masses mb and mb. The real trick is to calculate the relative masses as the state of the steam at cutoff is circular interdependent. The mass of the compressed steam is dependent on the specific volume of the exhaust steam which is dependent on the state of the steam at the end of expansion which is dependent on the state of the steam at cutoff which is a mixture of the compressed steam and admitted steam and it's state is dependent on the states of both the admitted steam and compressed steam which is dependent on the state of the exhaust steam, which is dependent etc. etc. Well you get the idea. In the text books this problem is never addressed and the cycle is simplified to exclude the compression or any residual steam.
As an example: If clearance space is 3% of the engine displacement and cutoff is at 10% of the displacement volume Inlet pressure of 1000 pounds per square inch at 800 degrees. Below is the point properties calculated by the Steam Cycle Calculator. I have set the compression so as to just, bring the cutoff point up to the admittance state.
W
admission = (pcutoff - pexpansion) vcutoff (cutoff) / (clearance + cutoff) = (1000-71.7)(PSI) 144(in^2/foot^2) 0.751704(foot^3/lb) 0.10 / (0.03 + 0.10) = 77295.5 foot lb / lb
Or
99.33 BTU/lb = 77295.36 / 778.17
The above calculation (with necessary conversion factors introduced) is the work done by the constant pressure during the admission of steam to the cutoff point. As we are calculating the cycle for a pound of steam at the cutoff point the above work is that done per pound of steam through the cycle.
From the steam properties at cutoff take the specific volume 0.751704 and multiply it by the expansion ratio 7.92
0.751704 (1 + 0.03)/(0.10 + 0.03) = 5.9558
Using entropy at cutoff 1.60745. Locate the steam point with that specific volume and entropy in a steam table or use my Steam Cycle Calculator to get the ending point of the expansion. 71.7 PSI at 304.5 F. It is left up to the reader to get solve for the work 217.342
We determine mc = 0.163 making mb = .837 And compute the compression work
W
compression = 69.04 BTU = .163 (1706.9 - 1000 144 1.0616/J) - (1164.0 - 14.696 144 27.9909/ J)
The work done by the expansion to exhaust pressure difference is the only work left to figure? The reader should check the value to be 47431.2 ft LB or 60.95 BTU The pump work is 2007.86 ft lb or 2.58 BTU. The heat added per cycle is (1389.6 - 183.3) * 0.837 = 1009.76 The efficiency is
30% = (99.33 + 217.34 + 60.95 - 69.04 - 2.58) / 1009.76
Our same engine running with no compression calculating residual steam would comes out to 29% and the text book Rankine calculates out to 25.9%
Realize that a pound of steam at the cutoff point is not a pound of admission steam. There was residual steam left in the clearance volume from the previous cycle. The steam at the cutoff point is a mixture of that residual steam and the admitted steam. The state of the residual can greatly effect the state of the mixture as the cutoff decreases. Low or no compression leaves cool steam in the clearance spacing reducing the energy of the mixture. High compression can increase the energy of the mixture. This is why it is important to analyze an engine including the compression.
In the chart below we see two efficiency lines charted for the same engine. The lower line is starting efficiency, based on a two cylinder double acting engine at 75% cutoff for accelerating smoothly from zero RPM. As RPM increases cutoff can be cut back and efficiency increased. The upper line is the efficiency limit for a given output. The Max output is full throttle at 75% cutoff. It is unknown the RPM for which we can be operating at max efficiency but it is not all that high. If we were to design for 10% output to maintain 60 MPH on the level we would be well above the I.C. engine efficiency on an average driving cycle. If we increase the number of cylinders we can increase the starting efficiency as the starting cutoff can be reduced.
