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Power Plant Efficiency
John Denker

## 1  Introduction

Let’s see if we can understand how the efficiency of an ordinary engine varies as a function of how hard we drive the engine.

## *   Contents

### 1.1  Two-Parameter Model

We imagine that the engine is operating at constant cycle-time, i.e. constant rotational rate, so the amount of power output is proportional to the torque. For details on how this might be arranged, see section 1.3.

Figure 1 shows some (but not all) of what happens when the engine is driven hard. The inner yellow-shaded region tells us roughly how much useful energy the engine puts out per cycle. (If this were a PV diagram instead of a TV diagram, the area would be exactly the energy.) The outer unshaded loop shows what the engine would do if there were no losses of the kind considered in reference 1, i.e. no series dissipation. Figure 1: Engine, Driven Hard

The white lines in these diagrams are contours of constant pressure. They are not particularly important for the current discussion. They stand in contrast to the contours of constant V (which are vertical) and the contours of constant T (which are horizontal). The contours of constant entropy are curves, trending roughly from the upper left to the lower right; the blue and black curves are examples.

In operation, the engine moves clockwise around the path shown in these figures. The red and black portions involve expansion of the working fluid, while the magenta and blue portions involve compression.

Figure 2 shows some of what happens when the engine is driven less hard. Remember, this is a constant-speed engine. Because we are transferring less entropy – or, loosely speaking, less “heat” – into the engine per cycle, the series temperature drop is less. Therefore the unshaded areas at the top and bottom of the cycle are smaller in the dT-direction. They are also smaller in the dV direction, just because the whole cycle is smaller in that direction. Putting the two factors together, we see that the series dissipation depends more-or-less quadratically on how hard we drive the engine. Figure 2: Engine, Not Driven Hard

Beware that these diagrams show only the series dissipation, not the parallel losses. The importance of this will be discussed in section 1.2.

The spreadsheet for making these Carnot plots is cited in reference 2.

### 1.2  Power and Efficiency versus Drive Figure 3: Power Plant Contributions (and Efficiency) versus Drive

In figure 3, the horizontal axis represents how hard we are driving the engine. We quantify the drive in terms of S2, the entropy the flows through the subengine.

Terminology: In the spreadsheet (reference 3) and in this document, the term “subengine” refers to the inner engine that operates reversibly between temperatures T2 and T1 ... in contrast to the overall engine that operates irreversibly between temperatures T3 and T0. The overall engine is irreversible because of the parallel losses and series dissipation. The definition of E3, T2, et cetera can be inferred from figure 6.

Since E3 is a nice monotonic one-to-one function of S2, we can replot the data using E3 as the abscissa. This is shown in figure 4. Note that E3 is the total energy used by the overall system. Figure 3 has the advantage of more closely reflecting the details of the model, while figure 4 makes some of the final results easier to interpret. Figure 4: Power Plant Contributions (and Efficiency) versus Total Energy

In both figures:

• The horizontal red band represents zeroth-order losses, i.e. losses that are independent of how hard we drive the engine.
1. An example of such a loss is a heat leak in parallel with the main energy path through the system, such as a leak R3x out of the high-temperature reservoir. (There could also be a leak R0x into the low-temperature reservoir, but the spreadsheet does not account for this.)
2. Another example of such a loss, approximately speaking, is a fluid leak past the piston-rings in a piston engine, or past the blades in a turbine. Such a leak is not exactly zeroth order, but it has a similar effect, insofar as it forces the engine efficiency to zero at low speeds.
• Next above that, the orange band represents first-order losses, i.e. losses that are simply proportional to how hard we drive the engine. Specifically, this is the waste heat that would be generated by an engine operating reversibly between the highest temperature (T3) and the lowest temperature (T0), ignoring the effect of the series dissipation.
• Next above that, the yellow band represents various nonlinear losses. It includes the second-order and higher-order contributions. In other words, it includes all the energy not counted as useful work, zeroth-order losses, or first-order losses.
1. One type of second-order loss arises if there is a thermal resistance somewhere in series with the main energy path through system. The more we pull (or push) energy through that resistance, the greater the temperature drop across it. Strictly speaking, no energy loss occurs in the resistive element, because we imagine that the element is thermally isolated from the environment. (This makes it conceptually different from an electrical resistor, which does dump energy into the environment.) To say the same thing another way: The series dissipation is not so much a decrease in energy as a gain in entropy. In any case, the series resistance will provoke an energy loss later in the process, in the form of increased waste heat from the subengine.
2. Another type of second-order loss arises from plain old mechanical friction. A frictional force that is proportional to velocity makes a contribution to the power that is second order in the velocity (yellow band). This is relevant to some models of sliding friction, including viscous lubricants at very low Reynolds number. (The spreadsheet does not model any contributions of this form.)
A frictional force proportional the square of velocity would make a contribution to the power that is third-order in the velocity. This would apply to fluids at high Reynolds number. I’m not sure how relevant this is to typical engines. This would require a new band to represent the third-order contributions.
• The blue band represents the energy available to do useful work. It is calculated by finding T2 and T1 and then computing the Carnot efficiency of the subengine.
• At the top of the blue band, the blue line represents the total amount of energy taken in by the engine.

Note the contrast: The zeroth-order term represents thermal losses in parallel, while the second-order term represents series dissipation.

The model we have just set forth can be described via an equation:

 E3 total power in = β zeroth-order loss + (T2−T1) S2 subengine work + T1 S1 subengine waste heat
(1)

where S1 is equal to S2 and represents the entropy flowing through the subengine. The RHS of this equation is considered a function of S2. It is nonlinear, because T2 and T1 are functions of S2. This stands in contrast to T3 and T0, which are considered fixed, independent of S2.

We can expand the RHS in a Taylor series. This gives us a phenomenological two-parameter model:

 E3 total power in = β zeroth-order loss + (T3−T0) S2 linear term + γ (S2)2 second-order term + ⋯ higher-order terms
(2)

The energy here represents the energy per cycle. Ditto for the entropy. Because this engine is running at a constant rate, the distinction between power and energy per cycle is not very important. As a related point, the “time” variable does not appear in equation 1 or in the spreadsheet in reference 3.

The constant-rate simplification does not result in any loss of generality. You could perfectly well run the model engine at constant torque and variable rate. There would still be zeroth-order, first-order, and second-order losses. The calculations would follow the same pattern and the conclusions would be the same. There would be a slight change in terminology, but that’s all.

Normally a second-order polynomial is a three-parameter model, but in equation 2 there are only two adjustable parameters. That’s because the coefficient of the linear term is determined for us in terms of the specified parameters T3 and T0, determined by conservation of energy in the limit of small S2.

In the legend of figure 3, the word “cume” stands for cumulative. It reminds us that the brown line at the top of the nonlinear band represents the cumulative losses, including zeroth-order, first-order, and nonlinear losses (not just nonlinear).

1. If we doubly-degrade this model by throwing away the zeroth-order term and the nonlinear terms, we are left with the Carnot model. The simplified Carnot efficiency is:

η c =
 T3 − T0 T3
(3)

and is independent of how hard we drive the engine. In the example shown in figure 3 and figure 5, the Carnot efficiency is just over 0.8. This is shown in figure 5 by the horizontal dashed black line.

This Carnot efficiency could be obtained by using a high-side temperature of T3 = 1400 C and a low-side temperature of 27 C, which are reasonable numbers for a real-world power plant. However, in the spreadsheet in reference 3, we normalize the temperatures so that T3 = 1. In normalized units, our example uses β = 0.05 and γ = 1.0.

2. If we degrade the original model by throwing away only the second-order term, so that we keep the zeroth-order and first-order terms, the model tells us that we should run the engine as hard as possible, right up to the point where it explodes due to centrifugal forces et cetera. This gives us the maximum power output as well as the maximum efficiency. This is grossly unrealistic. This hypothesis is dead on arrival.
3. If we degrade the original model by throwing away only the zeroth-order term, so that we keep only the first-order and second-order terms, the model tells us we should run the engine as slowly as possible. This means we would need a huge number of engines. This is grossly unrealistic, for numerous reasons discussed in section 3, yet it is the procedure followed by reference 1 and reference 4. This is another dead-on-arrival hypothesis. We need a zeroth-order term to account for the inevitable heat leaks. Furthermore, it could be argued that engine hardware is not free, so we need a zeroth-order term to account for amortization of the capital cost of the engines, which applies even if the engine is completely powered down. This corresponds to defining the ordinate in figure 3 to represent economic value out (not merely energy out).

Figure 5 shows the same situation. The power out curve is shown by itself, riding on a zero baseline, rather than riding on top of the cumulative losses. This makes it easier to recognize the max-power point as being the max-power point. Figure 5: Power Plant Output (and Efficiency) versus Drive

Let’s do a little bit of engineering analysis. Let’s suppose, temporarily and hypothetically, that we start out operating at the max-power point, at about 0.85 on the “S1” axis as shown in the figure. Nobody in his right mind would continue to operate at this point. That’s because throttling back a little saves fuel to first order, but leaves the power output unchanged to first order. If we throttle back all the way to 0.6, there is only a tiny decrease in power, but there is a substantial increase in efficiency ... as you can see by comparing the blue and black curves in the figure.

Because nobody ever operates near the max-power point, engines are not even designed to permit that. The redline operating limit of the engine will be set much lower, so that the internal parts do not need to be nearly as strong.

The peak efficiency comes out to be approximately 1−√(T0/T3), as shown by the short horizontal dotted line. This is a coincidence, devoid of any deep physical meaning.

### 1.3  Modeling a Constant-Speed Engine

The power produced at the drive shaft of any engine is given by the rotational rate multiplied by the torque.

We now restrict attention to the special case where the rotational rate is constant. This simplifies the math. It is not unduly unrealistic, because:

• In the real world, each engine in the power plant of an electrical utility runs at near-constant speed. It is phase-locked to the regional AC power grid. Throttling the engine changes the amount of torque it produces.
• A wide class of aviation engines run at near-constant speed in normal cruising flight.

We can build a simplified pedagogical model of an engine that works this way by slightly modifying the usual Carnot-type engine. The key modification is to provide for an adjustable compression ratio. The general idea is shown in figure 6. Figure 6: Piston with Variable Compression Ratio

We can adjust the compression ratio by moving the cylinder head, shown in tan, at the top of the cylinder. Any such adjustment happens very slowly relative to the cycle time of the engine.

The compression ratio in figure 1 is about 12.05, while the compression ratio in figure 2 is about 6.57. You can read the graphs to get the max volume (at the black/magenta corner) and the min volume (at the blue/red corner) and take the ratio.

As previously mentioned, the red and black parts of the path involve expansion of the working fluid, while the magenta and blue parts involve compression. There is some timing involved. That is, we need to engineer the engine so that the proper fraction of the expansion is isothermal and the proper fraction is isentropic. Ditto for the contraction. The mechanism for doing this is not shown in figure 6. The timing changes quite a bit, depending on how hard we are driving the engine.

The spreadsheet (reference 3) needs to calculate the temperature at the low side of the subengine. This involves solving two simultaneous equations

 series temperature drop = T1 − T0 = R10 E1
(4)

entropy through subengine =
S2 =
 E2 T2

=
S1 =
 E1 T1

(5)

Plugging one into the other we find

T1 =
 T0 1 − S2 R10
(6)

Also note: in the spreadsheet, the resistor from T0 to ground is ignored. All of the parallel losses are attributed to the resistor from T3 to ground.

## 2  Some Real-World Data

### 2.1  Background: Specific Fuel Consumption

The inverse of efficiency is specific fuel consumption. Efficiency is elegant, and is usually expressed in dimensionless units. Specific fuel consumption is practical, and is usually expressed in practical units, such as pounds of fuel per horsepower-hour.

In a real power plant, vehicle, or aircraft, there are various places to measure the power. In an electric power plant, we could measure the mechanical power at the output shaft of the prime mover, or we could measure the electrical power at the output of the dynamo.

The term Brake Specific Fuel Consumption (BSFC) refers to mechanical power at the output shaft, as measured by a device known as a prony brake.

### 2.2  Familiar Qualitative Result : Idling

Let’s consider the energy budget of a car engine when it is idling in neutral. The output shaft is delivering zero power, yet the fuel consumption is nonzero. Therefore the BSFC is infinite. The efficiency is zero. At any speed less than the idle speed, we are faced with some difficult concepts, because in some sense the BSFC is “more than infinite” and the efficiency is “less than zero”.

### 2.3  Efficiency versus Rate : Waukesha VGF F18

Reference 1 sets forth an oversimplified one-parameter model and compares it to “some” power-plant data. It checks the predicted efficiency against observed efficiency. Alas, this is not a very incisive check.

It is vastly better to check the BSFC as a function of operating speed. Figure 7 shows the data for a Waukesha VGF F18, which is a stationary natural-gas-fired 6-cylinder engine, used for electrical power generation. It is rated for 400 hp at 1800 RPM. The data is from reference 5. In the figure, the various colored curves correspond to various experimental lubricants. Figure 7: Waukesha VGF F18 : Brake Specific Fuel Consumption

The data in figure 7 shows that for this engine, normal operation is actually slower than the max-efficiency point. It is far, far slower than the max-power point. Also, it completely disproves one of the central claims made by reference 1, namely that the engine should become more efficient at lower speeds. I’ve seen lots of BSFC maps, and in every case I can think of, the BSFC gets worse (not better) as the speed goes down, in the lower part of the operating range.

### 2.4  Power and Efficiency versus Rate : Lycoming O-320

Logic suggests that the Curzon/Ahlborn argument should apply even more strongly to trucks and aircraft, where one is interested not only in power, but also in power-to-weight ratio. There is considerable incentive to install an undersized engine and run it extra-hard, even if this entails some loss in thermodynamic efficiency.

However, the Curzon/Ahlborn argument fails miserably when applied to real-world aircraft engines. Figure 8 shows the power versus rotation rate for a Lycoming O-320 engine, a widely-used aircraft engine. The circles are taken directly from the O-320 Operator’s Manual. The green line comes from a numerical model of engine performance, as used in a flight simulator. Figure 8: Lycoming O-320 : Power Curve

The manual specifies that redline is 2700 RPM. That is to say, the engine is never operated at any higher rotation rate. This is marked by – you guessed it – a red radial line on the tachometer.

Typical cruising flight is conducted at 2400 RPM or thereabouts, depending slightly on altitude.

It should be obvious from the figure that power is an increasing function of rate, throughout the permissible operating range. This engine is never operated at the max-power point. Not even close.

The BSFC is a declining function of rate throughout the permissible operating range, as shown in figure 9. The corresponding efficiency (not plotted) is an increasing function of rate. The various colored curves correspond to various operating altitudes. Figure 9: Lycoming O-320 : Brake Specific Fuel Consumption

Operating the engine at higher rates would be be less efficient ... especially if we factor in the propeller, i.e. if we think in terms of effective thrust power (not just brake power), as we should. In response, the engine components were designed with only enough strength to handle redline speeds, where redline is far below the max-power point. Designing the engine to operate at the max-power point would have made it heavier and more expensive, for no good reason.

The data in figure 9 shows that for this engine, normal operations are conducted near maximum efficiency. Also, it completely disproves – again – one of the central claims made by reference 1, namely that the engine should become more efficient at lower speeds.

### 2.5  Power and Efficiency versus Rate : Alt-Fuel Engine

Figure 10 shows more of the same. The data is from reference 6. The normal operating point is near max efficiency. It is nowhere near max power. The BSFC gets worse, not better, at the lowest throttle settings. Figure 10: Alt-Fuel Engine : Brake Specific Fuel Consumption

### 2.6  Purchasing Decisions

Airline companies care a great deal about fuel efficiency. They sometimes decommission perfectly serviceable airliners and buy new ones – at \$200,000,000.00 apiece – in order to get improved fuel economy. See reference 7.

## 3  Discussion

Let’s compare the two-parameter model presented here to the one-parameter model presented in reference 1 and elaborated in reference 4.

 jsd two- C/A one- parameter parameter model model Operating point is at or no yes near the max-power point. Operating point is at or yes no near the max-efficiency point. Efficiency at operating point yes yes is on the order of 50% of the Carnot efficiency. Driving the engine less hard decrease increase would cause the efficiency to: At zero drive, the efficiency is less than zero at a maximum Zeroth-order losses are yes no essential to understanding the concepts and principles.

The two models disagree on five out of the six predictions... and even the one point of agreement is largely illusory. The operating points are wildly different because the abscissas are different, so the fact that the ordinates are similar means little, if anything. In other words: I consider the agreement as to efficiency at the operating point to be fortuitous and devoid of real significance.

Not only are the conclusions of the C/A model grossly wrong, even the premises of the model are absurd. The model is predicated on the assertion that efficiency doesn’t matter. This assertion is exceedingly implausible, and is supported by precisely zero evidence.

The model in reference 1 is sometimes summarized by saying

1. The cost of engines is more important than the cost of fuel.
2. Therefore it makes sense to operate at the max-power point.

which is a wrong explanation in support of an absurd conclusion.

## 4  References

1.
F. Curzon and B. Ahlborn,
“Efficiency of a Carnot engine at maximum power output”
Am. J. Phys. 43, 22 (1975)
http://ajp.aapt.org/resource/1/ajpias/v43/i1/p22_s1

2.
John Denker,
“spreadsheet for making Carnot plots”
./carnot-plot.xls

3.
John Denker,
“spreadsheet for computing engine power and efficiency”
./power-efficiency.xls

4.
Massimiliano Esposito, Ryoichi Kawai, Katja Lindenberg, and Christian Van den Broeck,
“Efficiency at Maximum Power of Low-Dissipation Carnot Engines”
Phys. Rev. Lett. 105, 150603 (2010)
http://prl.aps.org/abstract/PRL/v105/i15/e150603

5.
Kathleen H. Tellier et al.,
“What does it take to make a modern engine oil?”
Power Engineering International
http://www.powerengineeringint.com/articles/print/volume-18/issue-6/features/what-does-it-take-to-make-a-modern-engine-oil.html

6.
Ioannis Gravalos, Dimitrios Moshou, Theodoros Gialamas, Panagiotis Xyradakis, Dimitrios Kateris and Zisis Tsiropoulos,
“Performance and Emission Characteristics of Spark Ignition Engine Fuelled with Ethanol and Methanol Gasoline Blended Fuels”
http://www.intechopen.com/books/alternative-fuel/performance-and-emission-characteristics-of-spark-ignition-engine-fuelled-with-ethanol-and-methanol-

7.
Wikipedia article, “Boeing 767”
http://en.wikipedia.org/wiki/Boeing_767
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