Heat Engines

One of the most important applications of thermodynamics is in heat engines. A heat engine converts heat energy into mechanical energy. This chapter will cover the most common types.

Figure 1 is a graphical way to depict the three main elements of a heat engine. These elements include:

1. A source of high temperature heat input, usually obtained by combustion of a fuel.

2. A cyclic process that acts on a working fluid, producing work.

3. A low temperature heat sink, to which heat is rejected.

The heat engine may operate in reverse, so it becomes a refrigeration device, air conditioner or heat pump. These applications are covered in other chapters.

A Simple Heat Engine

Figure 2 shows the main components of a simple heat engine. ‘C’ is the cylinder, open at one end and closed at the other, in which a gas-tight piston, ‘B’, is fitted. The piston moves freely within the cylinder. A gas is trapped in the cylinder space, ‘G’. The rod, ‘R’, fixed to the piston, is attached to the rack (or ratchet), ‘N’, at the joint, ‘J’. The rack contacts the toothed wheel, ‘M’, which is prevented from turning in an anti-clockwise direction by the pawl, ‘Q’.

The toothed wheel is attached to the drum, ‘D’, around which is wrapped a cord, ‘A’, carrying a mass, ‘W’. When the piston moves to the right, the ratchet wheel, ‘M’, will be turned clockwise by the rack, but if the piston moves to the left the wheel will remain stationary, held by the pawl, ‘Q’. The motion of the piston to the right will be resisted by the action of the mass, ‘W’ (through the ratchet wheel and rack) and by the pressure of the atmosphere on the outer face of the piston. The motion of the piston will cause work to be done, raising the mass to a height of ‘h’.

If the gas in the cylinder is heated, its volume will increase, due to thermal expansion, and the piston will move to the right, performing work. However, the amount of work that may be done in this way is limited by the extent of the travel of the piston in the cylinder. If the gas can be cooled, the piston can be made to move to the left, decreasing the volume of the gas (the piston being moved to the left by virtue of the pressure differential between the atmosphere and the gas in the cylinder). In this way, the condition of the gas might be restored to its initial state, or condition, such that subsequent heating would produce further work.

It is seen, therefore, that by alternately adding and removing heat, the engine can be made to raise the mass, in stages, through any desired height.

This simple example allows for the following important observations to be made and some conclusions drawn which apply equally well to all heat engines:

• A conversion of energy from heat to mechanical work has taken place. This has been done through the alternate supply and rejection of heat to and from the working fluid in the engine cylinder.

• It is apparent that these operations must be repeated to maintain the work output.

• There must be a relationship between the quantities of heat involved and the amount of work produced.

These observations, supported by experiments and calculations lead to the following statements:

• All heat engines must have a source of heat, a working fluid and a sink.

• In order to produce continuous work, the processes of heat supply and heat rejection must be carried out cyclically.

• The working fluid must be capable of returning to its original state at the end of each cycle.

• During the cycle of operations, the work done will be equivalent to the difference between the heat supplied and the heat rejected.

Types of Heat Engines

Heat engines may be described by the different types of thermodynamic cycle they utilize. The cycles explained in this chapter include:

1. The Carnot cycle: a theoretical cycle with the highest possible efficiency.

2. The Rankine cycle: applicable to steam plants.

3. The Otto cycle: applicable to spark ignition engines.

4. The Diesel cycle: applicable to compression ignition (Diesel) engines.

5. The Brayton cycle: applicable to gas turbines.

Heat engine cycles may also be classified into two types, based on the nature of the working fluid. These two types are:

• Gas cycle

• Vapour cycle

Gas Cycle

The gas cycles use air as the main working fluid. They take a fuel, combust it with atmospheric air, and use the heat of combustion to heat and expand the air. The available energy in the air is used to perform work, then the air and combustion products are released to atmosphere. The Otto, Diesel and Brayton cycles are all gas cycles. The closed cycle gas turbine (Brayton Cycle) is the only gas cycle engine that is not restricted to using only air as the working fluid.

Vapour Cycle

The vapour cycle uses a condensable vapour as the working fluid. The fluid is in the liquid phase during part of the cycle. The most common working fluid is steam. The Rankine cycle is the classic vapour cycle. A refrigeration cycle is also a vapour cycle.

The Carnot Cycle

In 1824, Sadi Carnot proposed an engine cycle that remains the model for the ideal upper limit of efficiency for heat engines of all types. In brief, he stated that this engine would operate in such a way that all heat supplied must be at the highest temperature of the cycle and all heat rejected must be at the lowest temperature of the cycle. Further, the expansion of the working fluid must take place without any heat transfer to or from the surroundings and without any internal loss of energy.

Such an engine cycle is capable of producing the maximum work output of any cycle working between the same temperature limits. The Carnot cycle of operations, taking place on the working fluid in a heat engine, consists of an isothermal heat addition followed by an adiabatic expansion, an isothermal heat rejection and finally an adiabatic re-compression, as shown in Figure 3.

The work done during the Carnot cycle is equal to the area (shaded) within the boundaries of the PV-diagram. The area within the temperature-entropy diagram represents the enthalpy.

Let’s consider an ideal Carnot engine, as shown in Figure 4, and imagine the cylinder and piston as perfect heat insulators (ie. will not conduct heat). Imagine the cylinder cover to be a perfect conductor so that all heat addition or rejection occurs only through the cover.

At point 1, let the hot body, A, at temperature, T1, and having infinite capacity for supplying heat, be placed in contact with the cylinder cover. The cylinder contains 1 kilogram of a perfect gas at temperature, T1, volume V1 and pressure P1. The piston is allowed to move slowly so that isothermal expansion takes place at temperature, T1. The gas does work as it expands from point 1 (P1, V1) to point 2 (P2, V2) and must take in an amount of heat equivalent to the work done. The work done during this expansion is:

Now, remove the hot body, A, and replace it with non-conductor B, which is a perfect thermal insulator. Allow the piston to continue moving to the right, powered only by the expansion of the gas in the cylinder. Since no heat can be taken in or rejected under these conditions, the gas expands adiabatically (ie. with constant entropy), doing work at the expense of its own internal energy. The loss of internal energy causes the gas temperature to fall. This adiabatic expansion continues until the temperature is T2, the pressure is P3, and the volume is V3.

Now, remove the insulating cover, B, and install cover C, which is capable of absorbing an infinite amount of heat. Cover C will remain at a constant, low temperature, T2, while absorbing heat from the gas in the cylinder. As the piston is forced back to the left, it is compressed and all of the heat of compression will be removed by cover C, so that the compression is isothermal. The isothermal work done on the gas is rejected as heat to cover C. This isothermal compression continues until point 4 is reached. Here, the pressure and volume are such that if further adiabatic compression occurs, the original conditions of the gas (P1, V1, T1) will be restored. In fact, this is what happens in the final step of the cycle.

At point 4, remove cover C and reinstall the perfectly insulating cover, B. Now, since no heat can be transferred to or from the gas, the continued compression will be adiabatic. The pressure and temperature will rise to P1 and T1, while the volume decreases to V1.

This ideal engine suffers no losses due to radiation, conduction or friction. In other words, it is an engine in which the total supplied heat is equal to the sum of the heat that is converted to work and the heat that is rejected (Figure 1). Further, it is an engine in which all the supplied heat is supplied at a constant temperature (the temperature of the hot body) and all the rejected heat is rejected at a constant temperature (the temperature of the cold body).

Definition of Efficiency

The efficiency of any thermodynamic cycle is represented by the ratio of the net mechanical work extracted from the cycle and the heat supplied. The heat converted into work is equal to the heat supplied minus the heat rejected to the heat sink. This gives the formula

For cycles that use air (Otto, Diesel and Brayton), the efficiency is often called the air standard efficiency although the term thermal efficiency is equally applicable.

An important note is that the efficiency equations derived relate to ideal efficiency and that actual values are always somewhat lower. Friction and other heat losses will result in an increase in entropy so that compression and expansion processes are not necessarily isentropic.

Carnot Cycle Efficiency

Even though it is impractical to implement, the Carnot Cycle has the highest possible efficiency, so it is valuable as a means of comparison.

The general definition of efficiency for any process is given by:

In the Carnot cycle, the heat supplied occurs during isothermal expansion. For an ideal gas, this heat can be expressed in terms of the properties of a gas using the equation:

In the same cycle, the heat rejected occurs during the isothermal compression, and is found using the equation:

These expressions for heat supplied and heat rejected can be substituted into the efficiency equation above as follows:

But this equation can be simplified further by cancelling like terms. The two volume ratios (V3/V4 and V2/V1) are equal, so can be cancelled, as can ‘m’ and ‘R’. The result is an equation that expresses the Carnot efficiency (denoted by the Greek letter, η) in terms of the absolute temperatures, T1 and T2. That is:

Example 1

A steam plant produces steam at 520°C. The steam is condensed back to water at 30°C. What is the maximum theoretical efficiency, based on the Carnot cycle?

Solution 1

Apply the Carnot Cycle efficiency equation:

The Rankine Cycle

The Rankine Cycle is the most fundamental vapour cycle and is in wide use in steam power plants. Figure 5 illustrates a basic steam plant cycle. Heat is supplied to a boiler to produce either saturated or superheated steam. This steam is then expanded through a turbine to extract mechanical work. The steam is then converted back to water by cooling it in a condenser. This condensate is re-pressurized to the required boiler feed pressure with a boiler feed pump and then returned to the boiler.

Whereas the Carnot Cycle added and rejected heat at constant temperatures, the Rankine Cycle specifies that the addition of heat to the working fluid (in the boiler) occurs at a constant pressure and the rejection of heat (in the condenser) also occurs at a constant pressure.

The Rankine Cycle can be illustrated on a pressure-volume (PV) diagram, as shown in Figure 6.

Temperature-Entropy Diagram

Figure 7 illustrates the heat process that occurs when feedwater, at 40°C, is heated in a boiler and converted to steam at 200°C, and then gives up heat by performing work. In the boiler, the feedwater must first be raised from 40°C to 200°C before any vaporization begins. The heat added to the water (in kJ/kg) is indicated by the area MNCD. This area represents approximately the difference between the enthalpies of the water. Using the steam tables, this is shown to be 852.45 – 167.57 = 684.88 kJ. The horizontal, entropy scale shows the difference in entropies of the water to be (2.33 – 0.57) or about 1.76 kJ/kgK.

The curve, NC, (the liquid line) is constructed by plotting, from the steam tables, the values of the entropy of the water for a number of different temperatures between 40°C and 200°C.

When water at 200°C is converted into steam at that temperature, the curve representing the change is a horizontal, constant temperature line, CE. The heat added in this process is the latent heat of evaporation at 200°C, which is 1940.7 kJ/kg. On the temperature-entropy diagram, this latent heat is represented by the area, DCEF. The change in entropy during evaporation is found to be 4.10 kJ/(kgK).

There are many variations on this basic Rankine Cycle. These include reheating after partial expansion and then a second expansion through another turbine as well as regeneration, which preheats the feedwater before it enters the boiler.

Rankine Cycle Efficiency

The efficiency of a steam plant on the Rankine Cycle is determined from the heat supplied to the water/steam in the boiler and the work extracted as the steam expands in the steam turbine. The work done by the boiler feed pump can be ignored, since it is relatively small.

The heat supplied to the water/steam in the boiler is the difference between the enthalpy of the feedwater as it enters the boiler (h1) and the enthalpy of the steam as it leaves the boiler (h2). That is, heat supplied = h2 – h1. With reference to steam tables, h1 is equivalent to hf. Therefore, the heat supplied = h2 – hf. The value of h2 will depend on the exit condition of the steam (saturated or superheated).

The work done in the turbine is the difference between the enthalpy of the steam at the boiler exit (h2) and the enthalpy of the steam at the exit of the turbine (h3). Therefore, the work done = h2 – h3. The value of h3 will depend on the condition of the steam leaving the turbine (ie. entering the condenser).

Applying the above to the general efficiency equation, will produce an equation for the efficiency (η) of a Rankine Cycle. That is:

Example 2

Steam conditions at the exit of a boiler are 4000 kPa abs at 500°C. Condenser operates at a pressure of 4 kPa abs and dryness fraction of steam is 0.88. If enthalpy of the feedwater in this case equals the enthalpy of the condensate, what is the efficiency of this system?

Solution 2

First, recognize that this is a steam cycle, so the Rankine efficiency should be applied.

At the exit of the boiler, the enthalpy (h2) of 4000 kPa steam, superheated to 500°C is found in the superheat steam tables to be 3445.3 kJ/kg.

At the entrance to the condenser, enthalpy (h3) can be calculated using the steam condition and values from the steam tables:

From steam tables, at 4 kPa: hf = 121.46 kJ/kg and hfg = 2432.9 kJ/kg

Since dryness fraction = 0.88, enthalpy of the steam is:

Enthalpy of the condensate at 4 kPa = hf = 121.46 kJ/kg. As stated in the question, this is also the enthalpy of the feedwater.

Efficiency (η) can now be found using the Rankine Cycle equation:

The Otto Cycle

The Otto cycle is applied to the spark-ignition that is used in internal combustion engines. The defining characteristic of the Otto cycle is that the addition and rejection of heat are done at (almost) constant volume. The main working fluid is air, which is taken in from the atmosphere, heated by combustion, and then exhausted along with the products of combustion.

Combustion occurs within the cylinder, unlike the Rankine cycle steam plant where combustion is external. Because the main part of the Otto cycle occurs inside the cylinder, it is called a “non-flow process”. A non-flow process is one in which the working fluid is confined and does not flow across its confining boundaries during the process.

Pressure-Volume Diagram

Figure 8 shows the pressure-volume diagram for a typical Otto Cycle. The intake and exhaust strokes are not shown on this diagram because thermodynamically, they cancel each other out.

The intake stroke ends at point 1 and compression begins. From point 1 to point 2, adiabatic compression occurs; the pressure and temperature both increase as the volume decreases.

From point 2 to point 3, heat is added as internal combustion occurs, when the piston is at the end of its stroke. This causes an increase in pressure, which occurs at approximately constant volume.

From point 3 to point 4, work is performed. The combustion gases expand adiabatic to atmospheric pressure, pushing the piston to the right.

From point 4 to point 1, heat is rejected as the gas exhausts from the cylinder at constant volume.

The temperature-entropy diagram for the Otto cycle is shown in Figure 9. Compression and expansion are both isentropic (constant entropy; reversible and adiabatic). The other two parts of the cycle cause a change in both temperature and entropy as heat is added and rejected.

Otto Cycle Efficiency

In the Otto cycle heat is supplied at constant volume and rejected at constant volume. The amount of heat in each case can be determined using the specific heat at constant volume (Cv) times the mass times the temperature difference. That is:

Heat supplied at constant volume = m Cv (T3 – T2)

Heat rejected at constant volume = m Cv (T4 – T1)

Substituting these into the basic definition of efficiency, an equation for the efficiency of an Otto Cycle, in terms of the temperatures, can be derived, as follows:

Another equation for Otto Cycle efficiency may be developed, based on volume rather than temperature. This uses the ratio of volumes (rv), which can be obtained from the physical dimensions of the piston stroke and clearance. The equation can be derived as follows.

First, referring to Figure 8, realize that the volume ratios for compression and expansion are equal. That is:

Now, already established equations for adiabatic compression and expansion for an ideal gas are:

The compression ratio (rv) can be substituted for V1/V2 and V4/V3 in these equations, giving:

These two equations can be rearranged as follows:

Now, subtract T2 from T3, then simplify, which gives:

Finally, substitute this expression for T3 – T2 into the standard Otto Cycle efficiency equation:

Internal combustion engines operating on the Otto cycle operate with a maximum compression ratio of about 10:1 to prevent pre-ignition of the air and fuel mixture. The corresponding ideal efficiency is 60.19% but the realistic value is somewhat less.

Example 3

Calculate the air standard efficiencies for a range of compression ratios (5, 10, 15 and 20) and describe the manner in which the efficiency increases with compression ratio. Use an isentropic exponent of 1.4.

Solution 3

Using the equation, η = 1 – $\frac{1}{r_v^{\gamma –1}}$ , the efficiencies are:

From these results, it can be seen that as the compression ratio increases, the efficiency also increases, but at a decreasing rate.

The Diesel Cycle

The Diesel cycle applies to compression-ignition, internal combustion engines. The important aspect of the Diesel cycle is that the addition of heat occurs at almost constant pressure. Air is the main working fluid. The air is compressed to a higher pressure inside the cylinder, which increases the temperature, and the fuel ignites spontaneously when it enters the cylinder.

Pressure-Volume Diagram

The variations in pressure and volume during the Diesel Cycle are shown in Figure 10. Intake and exhaust strokes are not shown on this diagram since they cancel each other out, thermodynamically.

The intake stroke ends at point 1. From point 1 to point 2, adiabatic compression occurs. Pressure and temperature both increase, while volume decreases.

From point 2 to point 3, internal combustion occurs as the fuel is injected. Pressure remains almost constant, since the piston is forced to the right (thus increasing the volume) for the first part of the power stroke.

From point 3 to point 4, during the second part of the power stroke, combustion gases expand adiabatically, producing work as the pressure decreases.

From point 4 to point 1, combustion gases are exhausted from the cylinder. Pressure is reduced back to starting pressure while volume remains constant.

Temperature-Entropy Diagram

The temperature-entropy diagram for the Diesel cycle is shown in Figure 11.

Compression and expansion are both isentropic, that is, reversible and adiabatic. The other two parts of the cycle cause a change in both temperature and entropy as heat is added and rejected. It is similar to the Otto cycle except that the addition of heat is at constant pressure which changes the nature of the curve from points 2 to 3.

Diesel Cycle Efficiency

In the Diesel cycle, heat is supplied at a constant pressure, but rejected at a constant volume. The efficiency equation can be developed much like the Otto efficiency was developed, except that the specific heats at constant volume and constant pressure must be used. This results in:

This equation can be expressed in terms of the isentropic exponent, γ, which is equal to the ratio of specific heats (ie. Cp/Cv.). Substituting this in the above equation produces the general equation for air standard efficiency, which is:

Another useful way to express the efficiency is in terms the compression ratio (r), and the cut-off ratio (R). Referring back to Figure 10, we see that expansion occurs during the portion of the power stroke that follows the cut-off point (point 3 on the P-V diagram). The cut-off volume, V3, is the volume at which constant pressure combustion ceases. The clearance volume, V2, is the volume when the piston is at the end of its stroke. Cut-off Ratio (R) is defined as the ratio between the cut-off volume and the clearance volume. That is:

An expression for R in terms of temperatures can also be derived. Since the combustion between points 2 and 3 is at constant pressure, the temperatures and volumes are related as:

The compression ratio (rv), as defined earlier, is: rv = $\frac{V_4}{V_2}$

The mathematical derivation of the efficiency equation in terms of R and rv is long and complex, so will not be covered here. Simply, the equation for air standard efficiency of a diesel engine is:

Since compression for a diesel engine is of air only, there is not the same restriction on compression ratio as the Otto engine. Although the Otto cycle is more efficient for the same compression ratio, the diesel engine can operate at substantially higher compression ratios, which make it more efficient. Mechanical considerations do limit diesel engines to a maximum compression ratio of approximately 25:1.

Example 4

Calculate the air standard efficiency of a diesel engine with a compression ratio of 13:1. Temperature at the start of compression is 60°C. After combustion, temperature is 1400°C. Isentropic exponent = 1.4.

Solution 4

First, determine the temperature (T2) after compression, so that the cut-off ratio can be established:

Now, cut-off ratio, R, can be found, using the ratio of temperatures, T3 and T2:

Then apply the air standard efficiency equation, given R and r.

Brayton Cycle

The Brayton cycle, also known as the Joule cycle, applies to gas turbines. Air is compressed in a compressor and then fuel is added and burned in a combustor. The hot gas leaving the combustor mixes with secondary compressed air to reduce it to a temperature that the turbine can safely handle. The hot gas/air mixture then expands through a turbine, producing work. A portion of this work powers the compressor, while the remainder is available to drive an electrical generator or a process compressor.

Most gas turbines are of the open cycle design, in which the hot gases are exhausted to atmosphere from the power turbine. Some gas turbines operate on a closed cycle, in which the exhaust gases are recirculated back to the compressor intake. This allows the use of a thermodynamically superior working fluid (such as helium). However, the efficiency of the open cycle design has progressed sufficiently that a closed cycle offers little advantage.

Figure 12 shows an open cycle gas turbine. It is single shaft design, in which the compressor and power turbine are on the same shaft. Dual and even triple shaft configurations are also quite common.

Pressure-Volume Diagram

Figure 13 shows the pressure-volume diagram for the Brayton Cycle. It is a steady-flow process, which means the working fluid flows steadily through the gas turbine (it is not “trapped” within a cylinder, like the previous processes). Compression (in the compressor section) and expansion (in the power turbine section) are both adiabatic. Combustion (in the combustor) is continuous and occurs at constant pressure. Heat rejection occurs in the external atmosphere, when the hot exhaust gases cool to ambient temperature.

Temperature-Entropy Diagram

Figure 14 shows the temperature-entropy diagram for the Brayton Cycle. The compression and expansion stages are close to isentropic and the heat addition and rejection are along the lines of constant pressure.

Several cycle modifications are possible with the gas turbine to make better use of the heat added. Though less common now, a regenerator was used to transfer heat from the power turbine exhaust gases to preheat the compressed air, prior to combustion. Intercooling may be used in the compressor to increase efficiency and power output. A few gas turbines reheat the combustion gases in a second turbine. A more common modification, called combined cycle, is the addition of a heat exchanger at the turbine exhaust to produce steam, which is then expanded through a steam turbine to power an electrical generator. A variation on this uses the steam for industrial or commercial heating purposes.

Brayton Cycle Efficiency

The Brayton Cycle features adiabatic compression and expansion with combustion at constant pressure. Using the specific heat at constant pressure (Cp), the efficiency is obtained from the basic equation for thermal efficiency:

Note that all temperatures must be absolute.

We can now derive an efficiency formula that is based on the pressure ratio, rp.

First, from Figure 13, we know that the pressure ratio is defined as:

Also, temperatures are related to pressures as follows:

Now, subtract eq. 1 from eq. 2 and simplify:

Now, let’s substitute this expression for “T3 – T2” into the denominator of the general efficiency equation, then simplify to obtain the ideal thermal efficiency of the Brayton Cycle in terms of the pressure ratio (rp).

In practice, compression ratios for gas turbines are usually between 10:1 and 15:1, although some engines exceed 20:1. High compression ratios are difficult to achieve efficiently, due to increased compressor temperatures. There is also a metallurgical limit to combustion and turbine components. The majority of open cycle gas turbines have an efficiency of 30-35%.

Example 5

A simple gas turbine operates with a pressure ratio of 12.5:1. Using an intake temperature of 20°C and a turbine inlet temperature of 1150°C, determine:

a) the temperature at exit of the compressor,

b) temperature at exit of the turbine, and

c) the thermal efficiency.

Use isentropic exponent = 1.4.

Solution 5

a) Temperature (T2) at the exit of the compressor:

b) Temperature (T4) at the exit of the turbine:

c) Efficiency is calculated using the equation:

This is substantially higher than the practical values achieved. Values of 35% are common for gas turbines with this pressure ratio.

Heat Balance Analysis

The term “heat balance” is applied to a schematic flow diagram for a thermal power cycle on which thermodynamic properties and mass flow are indicated for all major flow streams in the power cycle. This diagram is the product of detailed calculations, which apply the laws of conservation of mass and energy to each component in the power cycle and includes the manufacturer’s expected performance. These calculations, in addition to determining flow rates and thermodynamic properties of the fluids (pressure, temperature, enthalpy, and entropy) for all components in the cycle, also predict the overall performance (efficiency and output) of the cycle. Heat balance calculations are first used to design the thermodynamic performance of power cycles.

Heat balance diagrams are prepared, along with studies for sizing, selecting and optimizing the equipment in the cycle. Heat balance diagrams are generally prepared in several stages, which represent several levels of refinement as the cycle design is developed. The steam turbine manufacturer provides a heat balance – known as a turbine vendor balance - for the components they provide, such as the turbine and feedwater system. However, the turbine vendor balance can only approximate this heat balance – by using nominal performance values – if they do not supply all the related process equipment under the scope of their contract.

While these estimates (for such items as feedwater heater performance, pump performance, condenser back pressure, auxiliary steam and water flows and variations in controllable parameters at off-design loads) are reasonable when considering the turbine portion of the conceptual project, they are prepared in greater detail later in the project.

Example 6

Calculate the heat balance for a steam plant operating on the Rankine cycle, given the following test data. Determine performance and output data (based on 1 kg of steam). Compare your results with vendor specifications provided below.

• Boiler feedwater: 50°C

• Boiler outlet conditions: 2000 kPa, 300°C

• Fuel consumption: 13.0 kg of steam/ kg of fuel oil burned (heating value of 44 200 kJ/kg)

• Boiler vendor rated efficiency: 84%

• Turbine exit conditions: 200 kPa, dryness fraction 0.85

• Plant rated efficiency: 35%

Solution 6

Using the steam tables, the heat balance results for the boiler are:

Boiler efficiency is calculated to be:

This compares favorably with the vendor rating of 84%.

Using the steam tables provided in the Academic Supplement, the heat balance results for the turbine are:

Efficiency for the cycle is.

This is substantially lower than the rated efficiency and there are likely steam and turbine efficiency losses. A more detailed heat balance is required to discover the specific problem areas.

1. Explain the three main components of every heat engine.

2. With the aid of a simple sketch, describe the operation of an engine based on the Carnot cycle.

3. With the aid of a PV diagram, describe the various stages of a steam turbine operating on the Rankine cycle.

4. An electricity generating station operates with boiler stop valve steam conditions of 3000 kPa and 370°C and a condenser backpressure of 5 kPa. Steam exhausts from turbine with 15% wetness and boiler feedwater returns at 140°C. Calculate the following:

a) Carnot efficiency

b) Rankine efficiency

5. Explain the main differences between the Otto and Diesel cycles. What practical limits are there with respect to their efficiencies?

6. Calculate the air standard efficiency of a Diesel engine with a compression ratio of 15:1. Temperature at start of compression is 20°C and, after combustion, it has reached 1200°C. Use specific heat ratio = 1.4.

7. Air enters the inlet of a gas turbine, operating on the ideal constant pressure cycle, at a temperature of 22°C and 103.25 kPa. Pressure at the discharge of the compressor is 600 kPa. If the temperature at the turbine inlet is 745°C, calculate the following, using a specific heat ratio of 1.4:

a) Temperature at the end of compression

b) Temperature at the turbine exit

c) Ideal thermal efficiency

Self-Assessment

Question 1

Rankine cycle efficiency is:

Net work output of the turbine divided by heat supplied by the boiler

Net work input to the turbine divided by heat supplied by the boiler

Net work output of the turbine divided by heat supplied to the boiler

Net heat supplied to the boiler divided by heat output of the boiler

Net heat input to the turbine divided by heat supplied to the boiler

Question 2

h₂ - h₃ over h₂ - h_f is the expression for the thermal efficiency for the:

Diesel cycle

Brayton cycle

Otto cycle

Rankine cycle

Carnot cycle

Question 3

T₁ - T₂ over T₁ is the expression used for the thermal efficiency of:

the Otto Cycle

the Rankine Cycle

the Brayton Cycle

the Carnot Cycle

the reversed Carnot cycle

Question 4

The Brayton Cycle is defined as:

a non-flow cycle

an open cycle

the air standard cycle

a pressure ratio cycle

constant volume cycle

Question 5

An engine operating on the constant volume cycle has a ratio of compression of 9.6:1. If the inlet temperature of the air is 25°C and the temperature at the end of combustion is 1250°C, the theoretical thermal efficiency would be:

53.8%

59.5%

55.4%

62.7%

47.9%