Figure 3.15 One possible implementation of a Stirling engine.

Real efficiency departs from from the theoretical because:

1. The efficiency of mechanical heat engines increases with increasing compression ratio. The Stirling engine is no exception (see Problem 3.1). The internal volume of the heat exchangers and the regenerator cause the compression ratio to be smaller than if heat could be applied directly to "Space #1."

2. The combustion gases leaving the input heat exchanger are still quite hot and carry away considerable energy. To recover part of the latter a preheater is used as explained previously.

3. The regenerator cannot operate ideally and return all the heat deposited in it during part of the cycle.

4. There are frictional losses when the working fluid is transferred from one space to the other through the heat regenerator. This is the main reason for using hydrogen as a working fluid, as it leads to minimum frictional losses.

5. Ideally, the motion of the pistons should be intermittent, but this is difficult to implement in machines operating at reasonably high rpm. Thus, one has to make a compromise in the piston programming.

6. The limited time for the heat exchanges in the working fluid lead to temperatures that never reach the desired steady state levels.


The power output of a Stirling engine can be controlled by either adding or removing working fluid. To reduce power, a compressor removes some working fluid from the engine and stores it at high pressure in a holding tank. To increase power, gas from this tank can be quickly delivered back to the engine. The input heat exchanger temperature is continuously monitored and this information is used to control the fuel flow.

3.10 Cryogenic Engines

As pointed out, a heat engine depends for its operation on a heat source and a sink which, by definition, must be at a lower temperature. Almost invariably, the sinks are at nearly the ambient temperature while the source is driven by combustion, nuclear reaction, solar energy, ans so forth, to a relatively elevated temperature. However, it is possible to use a source that is at ambient temperature, while the temperature of the sink is lowered by a supply of a cryogenic substance.

Ordonez and his collaborators at the University of North Texas have proposed such cryogenic engines. Their experimental engines operate on an open Rankine cycle and use liquid nitrogen for the sink. Figure 3.16 shows a schematic representation of the setup.

The liquid nitrogen in the cryogenic reservoir is under the (adjustable) pressure, pi. The liquid is fed to a heat exchanger to which enough heat is added (from the outside air) to produce gaseous N2 at the pressure, pi, and the temperature, Ti.

The exhaust valve is closed and the intake valve is opened so that y kilomoles of nitrogen enter the cylinder. This causes the piston to be pushed down until the volume reaches a value, v2. The process is isobaric (p2 = p1) and isothermic (T2 = Ti) because sufficient heat is added to the heat exchanger.


Cooling fins Gas

Heat exchanger


Cooling fins Gas

Heat exchanger

Exhaust /valve

Liquid nitrogen


Figure 3.16 A cryogenic heat engine using an open Rankine cycle.

Exhaust /valve

Liquid nitrogen


Figure 3.16 A cryogenic heat engine using an open Rankine cycle.


We have:

P2 Pi

The work produced by the piston is

Next, the intake valve is closed. The high pressure nitrogen continues to expand until the pressure inside the cylinder is p3 and the volume reaches v3. This process may be carried out isothermally (T3 = Ti) or adiabatically (T3 < Ti), depending on the type of hardware used. Consider the more favorable isothermal case:

The total work done by the piston during the 1-to-3 phase is

However, this is not the available work. Part of the total work is done by the piston on the atmosphere:


hence, assuming Ti = TatTO,

The net useful work done is

This means that the net work is exactly that owing to the isothermal expansion of the gas.


Choose pi = 1 MPa and assume that T1 = Tatm = 298 K and P3 = patm = 0.1 MPa. The expansion ratio, r = p1 /p3, is 10:1. The net work per kilomole of "fuel" is 5.7 MJ/ kmole of N2 or 204 kJ/ kg of N2.


It may be difficult to achieve isothermal expansion—it would require an additional heat exchanger to warm up the nitrogen as it forces the piston down. However, something similar is required in a Stirling engine, so it is possible.

Let us examine the case of adiabatic expansion that is easier to achieve. If no heat is added during the expansion, the temperature will fall:

Here, y = 1.4 for nitrogen. The work during the expansion is

For the example with p1 =1 MPa, we get T3 = 154 K and W23 = 3 MJ/kmole.

When the volume, v3, is reached, the gas will have cooled to T3 so that

consequently the work done on the atmosphere is

The net work is now,

Still for the example being considered, the specific net energy is 4.2 MJ/kmole or 150 kJ/kg. Compare with the 5.7 MJ/kmole or 204 kJ/kg for the isothermic case and with the 47,000 kJ/kg for the typical gasoline.

Clearly, the specific energy of the cryogen will increase with increasing operating pressure. The gain, however, is logarithmic. Thus, by raising the pressure to 10 MPa (a factor of 10) the specific energy rises to 11.4 MJ/kmole, a gain of 2. Observe that 10 MPa correspond to approximately 100 atmosphere and would lead to a rather heavy (and expensive) engine.

Gasoline has to be used in an internal combustion engine with some 20% efficiency, while the pneumatic motor used in the cryogenic engine can have very high component efficiency. This would reduce the practical specific energy advantage of gasoline to a factor of 40 over the nitrogen.


In practice, the Ordonez engine has yielded around 19 kJ/kg thus far. A demonstration car using the engine does 0.3 miles to the gallon, which is not practical. The efficiency will probably be improved.


Vehicles equipped with the cryogenic engine achieve only very modest mileage. Demonstrated mileage is 0.3 miles/gallon. Present day cost of liquid nitrogen is $0.50/kg or $1.52/gallon, essentially the same as that of gasoline. This means that fuel cost would be about 100 times larger than that of a gas driven car.

Assume a 10-fold improvement in performance for the cryogenic car— fuel cost would still be one order of magnitude larger than for the gasoline one. Add to this the need to carry huge amounts of cryogenic material, which would constitute a considerable hazard in case of accident.

The advantage of the system is its low pollution (but not necessarily zero pollution because there is need of energy to produce the liquid nitrogen). The question is whether these advantages compensate for the serious disadvantages that we discussed?


Heywood, John B., Internal Combustion Engine Fundamentals, McGraw-Hill, Inc., 1988.

Ordonez, C. A., and M. C. Plummer, Cold thermal storage and cryogenic heat engine for energy storage applications, Energy Sources, 19:389396, 1997.

Ordonez, C. A., Cryogenic heat engine, Am. J. Phys. 64 (4), April 1996.

Plummer, M. C., C. P. Koehler, D. R. Flanders, R. F. Reidy, and C. A. Ordonez, Cryogenic heat engine experiment, 1997 Cryogenic Engineering Conference/International Cryogenic Materials Conference, Portland, Oregon. July 1997 (submitted).



1. Demonstrate that the theoretical efficiency of a Stirling engine without regenerator is -1

where nCARNOT is the Carnot efficiency associated with the engine temperature differential, v is the number of degrees of freedom of the working gas and r is the compression ratio.

2. What gas would you suggest as a working fluid? Why?

3. In the example in the text, a compression ratio of 10 was used. What would the efficiency of that engine be if the ratio were raised to 20? What are the disadvantages of using this higher compression ratio? Is it worth the effort?

3.2 Plot a pressure versus volume diagram and a temperature versus entropy diagram for the Stirling engine in the example given in the text. What do the areas under the pV and the TS lines represent?

3.3 Consider two cylinders, A and B, equipped with pistons so that their internal volume can be changed independently. The maximum volume of either cylinder is 10 m3 and the minimum is zero. The cylinders are interconnected so that the gas is at the same pressure anywhere in the system. Initially, the volume of A is 10 m3 and of B is 0. In other words, piston A is all the way up and B is all the way down. The system contains a gas with 7 = 1.4.

1. If this is a perfect gas, what is the number of degrees of freedom of its molecules and what is its specific heat at constant volume?

2. The pressure is 0.1 MPa and the temperature is 400 K. How many kilomoles of gas are in the system?

3. Now, push piston A down reducing the volume to 1 m3, but do not change the volume of cylinder B. What are the temperature and pressure of the gas assuming adiabatic conditions? What energy was expended in the compression?

4. Next, press A all the way down and, simultaneously, let B go up so that the volume in cylinder A is zero and that in cylinder B is 1 m3. What are the pressure and temperature of the gas in B?

5. The next step is to add heat to the gas in B so that it expands to 10 m3 at constant temperature. How much heat was added? How much work did Piston B deliver? What is the final pressure of the gas?

6. Now press B all the way down while pulling A up. This transfers gas from one cylinder to the other and (theoretically) requires no energy. Cylinder A rejects heat to the environment and the gas cools down to


400 K. The pistons are not allowed to move. The cycle is now complete. How much heat was rejected?

7. What is the efficiency of this machine—that is, what is the ratio of the net work produced to the heat taken in?

8. What is the corresponding Carnot efficiency?

9. Sketch a pressure versus volume and a temperature versus volume diagram for the process described.

10. Derive a formula for the efficiency as a function of the compression ratio, r. Plot a curve of efficiency vs r in the range 1 < r < 100.

11. If this ratio were to reach the (unrealistic) value of 10,000, what would the efficiency be? Does this exceed the Carnot efficiency? Explain.

3.4 A car is equipped with a spark ignition engine (Otto cycle). It uses gasoline (assume gasoline is pure pentane) as fuel, and, for this reason, its compression ratio is limited to 9. The highway mileage is 40 miles/gallon.

Since pure ethanol is available, the car owner had the engine modified to a compression ratio of 12 and is using this alcohol as fuel. Assuming that in either case the actual efficiency of the car itself is half of the theoretical efficiency, what is the mileage of the alcohol car?

The lower heats of combustion and the densities are Pentane: 28.16 MJ/liter, 0.626 kg/liter; Ethanol: 21.15 MJ/liter, 0.789 kg/liter. Do this problem twice, once using 7 = 1.67 and once using 7 = 1.4.

3.5 Consider a cylinder with a frictionless piston. At the beginning of the experiment it contains one liter of a gas (7 = 1.4, cv = 20 kJ K-1 kmole-1) at 400 K and 105 Pa.

1. How many kmoles of gas are in the cylinder?

2. What is the pV product of the gas?

Move the piston inwards reducing the volume of the gas to 0.1 liters. This compression is adiabatic.

3. What is the pressure of the gas after the above compression?

4. What is the temperature of the gas after the above compression?

5. How much work was done to compress the gas?

Now add 500 J of heat to the gas without changing the temperature.

6. After this heat addition phase what is the volume of the gas?

7. After this heat addition phase, what is the pressure?

8. Since the gas expanded (the piston moved) during the heat addition, how much work was done?

Let the gas expand adiabatically until the volume returns to 1 liter.

9. After the expansion, what is the pressure of the gas?


10. After the adiabatic expansion what is the temperature of the gas?

11. How much work was done by the expansion?

Now remove heat isometrically from the gas until the pressure reaches 105Pa. This will bring the system to its original State 1.

12. What is the net work done by the piston on an outside load?

13. What is the total heat input to the system (the rejected heat cannot be counted)?

14. What is the efficiency of this machine?

15. What is the Carnot efficiency of this machine?

16. Sketch a pressure versus volume diagram of the cycle described in this problem statement.

3.6 Assume that gasoline is pure pentane (actually, it is a complicated mixture of hydrocarbons best represented by heptane, not pentane). Consider a 1:4 ethanol-gasoline mixture (by volume). The gasoline has an 86 octane rating. The blending octane rating of ethanol is 160. Use 7 = 1.4.

1. What is the energy per liter of the mixture compared with that of the pure gasoline?

2. What is the octane rating of the mixture?

Assume that the maximum tolerable compression ratio is r = 0.093 x Or where Or is the octane rating.

3. What is the highest compression ratio of the gasoline motor? Of the mixture motor?

4. What is the relative efficiency of the two motors?

5. What is the relative kilometrage (or mileage) of two identical cars equipped one with the mixture motor and the other with the gasoline motor?

3.7 An open-cycle piston type engine operates by admitting 23 x 10-6 kmoles of air at 300 K and 105 Pa. It has a 5.74 compression ratio.

Compression and expansion are adiabatic. Heat is added isobarically and rejected isometrically. Heat addition is of 500 J.

Air has cv = 20,790 J K-1 kmole-1 and a 7 = 1.4. What is the theoretical efficiency of this engine? Compare with its Carnot efficiency.

Proceed as follows:

Calculate the initial volume of cylinder (at end of admission). Compress adiabatically and calculate the new V, p, T and the work required.

Add heat and calculate new state variables. Expand and calculate the work done.


3.8 A certain Stirling engine realizes one half of its theoretical efficiency and operates between 1000 K and 400 K. What is its efficiency with

1. A perfect heat regenerator, argon working fluid, and 10:1 compression?

2. The same as above but with a 20:1 compression ratio?

3. The same as in (1.) but without the heat regenerator?

4. The same as in (2.) but without the heat regenerator?

3.9 Rich mixtures reduce the efficiency of Otto engines, but mixtures that are too lean do not ignite reliably. The solution is the "stratified combustion engine."

Consider an engine with a 9:1 compression ratio. A rich mixture may have a gamma of 1.2 while, in a lean one it may be 1.6. Everything else being the same, what is the ratio of the efficiency with the lean mixture to that with the rich mixture?

3.10 Consider an Otto (spark-ignition) engine with the following specifications:

Maximum cylinder volume, V0: 1 liter (10-3 m3). Compression ratio, r: 9:1.

Pressure at the end of admission, p0: 5 x 104 Pa.

Mixture temperature at the end of admission, T0: 400 K.

Average ratio, 7, of specific heats of the mixture: 1.4.

Specific heat of the mixture (constant volume), cv,: 20 kJ K-1 kmole-1

The calculations are to be based on the ideal cycle—no component losses. At maximum compression, the mixture is ignited and delivers 461 J to the gas.

If the engine operates at 5,000 rpm, what is the power it delivers to the load?

3.11 If pentane burns stoichiometrically in air (say, 20% O2 by volume), what is the air-to-fuel mass ratio?

Atomic masses:

C: 12 daltons.

N: 14 daltons.

O: 16 daltons.

Neglect argon.

3.12 The higher heat of combustion of n-heptane (at 1 atmosphere, 20 C) is 48.11 MJ/kg. What is its lower heat of combustion?


3.13 One mole of a certain gas (7 = 1.6, cv = 13,860 J K-1 kmole-1) occupies 1 liter at 300 K. For each step, below, calculate the different variables of state, p, V and T.

Compress this gas adiabatically to a volume of 0.1 liters.

How much is the energy, W12, is used in this compression?

Add isothermally 10 kJ of heat to the gas.

What is the external work, if any?

Expand this gas adiabatically by a ratio of 10:1 (the same as the compression ratio of Step 1 ^ 2).

Reject heat isothermally so as to return to State 1. What is the energy involved?

What is the overall efficiency of the above cycle?

What is the Carnot efficiency of the cycle?

What power would an engine based on the above cycle deliver if it operated at 5000 rpm (5000 cycles per minute)?

3.14 The Stirling engine discussed in the example in the text uses an isothermal compression, followed by an isometric heat addition, then by an isothermal expansion, and, finally, by an isometric heat rejection.

Isothermal compression may be difficult to achieve in an engine operating at high rpm. Imagine that the engine actually operates with an adiabatic compression.

Assume that the other steps of the cycle are the same as before. Thus, the isometric heat addition still consists of absorption of 293 joules of heat. This means that the "hot" cylinder is no longer at 652 K, it is at whatever temperature will result from adding isometrically 293 J to the gas after the initial adiabatic compression.

Calculate the theoretical efficiency of the engine (no heat regeneration) and compare it with the Carnot efficiency.

Calculate the power produced by this single cylinder engine assuming that the real efficiency is exactly half of the ideal one and that the engine operates at 1800 rpm. Each full rotation of the output shaft corresponds to one full cycle of the engine. Use 7 = 1.4.

3.15 Assume an engine (Engine #1) that works between 1000 K and 500 K has an efficiency equal to the Carnot efficiency.

The heat source available delivers 100 kW at 1500 K to the above engine. A block of material is interposed between the source and the engine to lower the temperature from 1500 K to the 1000 K required. This block


of material is 100% efficient—the 100 kW that enter on one side are all delivered to the engine at the other side.

What is the Carnot efficiency of the system above? What is the power output of the system?

Now, replace the block by a second engine (Engine #2) having a 10% efficiency. The heat source still delivers 100 kW. What is the overall efficiency of the two engines working together?

3.16 A boiler for a steam engine operates with the inside of its wall (the one in contact with the steam) at a temperature of 500 K, while the outside (in contact with the flame) is at 1000 K.

1 kW of heat flows through the wall for each cm2 of wall surface. The metal of the wall has a temperature dependent heat conductivity, A, given by A = 355 - 0.111T in MKS units. T is in kelvins.

1. Determine the thickness of the wall,

2. Determine the temperature midway between the inner and outer surface of the wall.

3.17 A 4-cycle Otto (spark ignition) engine with a total displacement (maximum cylinder volume) of 2 liters is fueled by methane (the higher heat of combustion is 55.6 MJ/kg, however, in an IC engine, what counts is the lower heat of combustion). The compression ratio is 10:1. A fuel injection system insures that, under all operating conditions, the fuel-air mixture is stoichiometrically correct. The gamma of this mixture is 1.4.

Owing to the usual losses, the power delivered to the load is only 30% of the power output of the ideal cycle. Because of the substantial intake pumping losses, the pressure of the mixture at the beginning of the compression stroke is only 5 x 104 Pa. The temperature is 350 K.

If the engine operates at 5000 rpm, what is the power delivered to the external load?

3.18 Consider a spark-ignition engine with a a 9:1 compression ratio. The gas inside the cylinder has a 7 = 1.5.

At the beginning of the compression stroke, the conditions are:

At the end of the compression, 10 mg of gasoline are injected and ignited. Combustion is complete and essentially instantaneous.

Take gasoline as having a heat of combustion of 45 MJ/kg.

1. Calculate the ideal efficiency of this engine.

2. Calculate the Carnot efficiency of an engine working between the same temperatures as the spark-ignition engine above.


3. Prove that as the amount of fuel injected per cycle decreases, the efficiency of the Otto cycle approaches the Carnot efficiency.

3.19 In a diesel engine, the ignition is the result of the high temperature the air reaches after compression (it is a compression ignition engine). At a precisely controlled moment, fuel is sprayed into the hot compressed air inside the cylinder and ignition takes place. Fuel is prayed in relatively slowly so that the combustion takes place, roughly, at constant pressure. The compression ratio, r, used in most diesel engines is between 16:1 and 22:1. For diesel fuel to ignite reliably, the air must be at 800 K or more.

Consider air as having a ratio of specific heat at constant pressure to specific heat at constant volume of 1.4 (7 = 1.4). The intake air in a cold diesel engine may be at, say, 300 K. What is the minimum compression ratio required to start the engine?

3.20 We have a machine that causes air (7 = 1.4) to undergo a series of processes. At the end of each process, calculate the state of the gas (pressure, volume and temperature) and the energy involved in the process.

The initial state (State #1) is p1 = 105 Pa, V1 = 10-3m3, T1 = 300 K.

1. 1st Proc. (Step 1^2): Compress adiabatically, reducing the volume to 10-4 m3.

2. 2nd Proc. (Step 2^3): Add 200 J of heat isobarically.

3. 3rd Proc. (Step 3^4): Expand adiabatically until V4 = 10-3 m3.

4. List all the heat and mechanical inputs to the machine and all the mechanical outputs. From this, calculate the efficiency of the machine. (Hint: Don't forget to add all the processes that deliver energy to the output.)

3 21 A--Adiabatic B--Isothermal

A crazy inventor patente d the following (tot ally useless) device: Two geometrically identical cylinders (one adiabatic and the other isothermal) have rigidly interconnected pistons as shown in the figure.

The system is completely frictionless and at the start of the experiment (State #0), the pistons are held in place so that the gases in the cylinders are in the states described below:


Cylinder A Cylinder B

(Adiabatic) (Isothermal)

1. Now, the pistons are free to move. At equilibrium, what is the temperature of the gas in Cylinder A? The 7 of the gas is 1.5. An external device causes the pistons to oscillate back and forth 2500 times per minute. Each oscillation causes VB to go from 0.1 m3 to 1 m3 and back to 0.1 m3.

2. How much power is necessary to sustain these oscillations?

Consider the same oscillating system as above with the difference that in each compression and each expansion 1% of the energy is lost. This does not alter the temperature of the isothermal cylinder because it is assumed that it has perfect thermal contact with the environment at 300 K. It would heat up the gas in the adiabatic cylinder that has no means of shedding heat. However, to simplify the problem assume that a miraculous system allows this loss-associated heat to be removed but not the heat of compression (the heat that is developed by the adiabatic processes).

3. How much power is needed to operate the system? 3.22 In a diesel cycle one can distinguish the following different phases: Phase 1 —> 2 An adiabatic compression of pure air from Volume Vi to Volume V2.

Phase 2^3 Fuel combustion at constant pressure with an expansion from Volume V2 to Volume V3.

Phase 3^4 Adiabatic expansion from Volume V3 to Volume V4. Phase 4 —> 1 Isometric heat rejection causing the state of the gas to return to the initial conditions.

This cycle closely resembles the Otto cycle with a difference that in the Otto cycle the combustion is isometric while in the diesel it is isobaric.

Consider a cycle in which V1 = 10-3 m3, V2 =50 x 10-6 m3, V3 = 100 x 10-6 m3, p1 = 105 Pa, T1 = 300 K, and (for all phases), 7 = 1.4.

1. Calculate the theoretical efficiency of the cycle.

2. Calculate the efficiency by using the efficiency expression for the diesel cycle given in Chapter 4 of the text.

3. Calculate the efficiency by evaluating all the mechanical energy (compression and expansion) and all the heat inputs. Be specially careful with what happens during the combustion phase (2 ^ 3) when heat from the fuel is being used and, simultaneously, some mechanical energy is being produced.

You should, of course, get the same result from 2 and 3.


Chapter 4

Getting Started With Solar

Getting Started With Solar

Do we really want the one thing that gives us its resources unconditionally to suffer even more than it is suffering now? Nature, is a part of our being from the earliest human days. We respect Nature and it gives us its bounty, but in the recent past greedy money hungry corporations have made us all so destructive, so wasteful.

Get My Free Ebook

Post a comment