Newton's third law of motion ( Action and Reaction ! )

When a bullet is fired from a gun equal and opposite forces are exerted on the bullet and the gun during the time the bullet is passing down the barrel

Newton's second law of motion ( momentum , force , weight , free falling )

A motortruck requires a larger force to set in motion when it is heavily loaded than when it is empty

The nature of friction: Static and dynamic friction

Friction is the name given to the force which opposes the relative sliding motion of two surfaces in contact with one another. It plays a notable part in our daily lives. For example

Newton's Laws of motion:1- Newton's first law of motion

In 1687 Newton published a book written , as was the routine in those days , in Latin and given the title , Philosophiae naturalis principia mathematica Translated , this means

Derivation of Kinetic energy formula and worked examples

Where there are no opposing forces, a moving body needs noforce to keep it moving with a steady velocity ( Newton's first law of motion ). If, however, a resultant force does act on a moving body

Wednesday, October 29, 2014

Power and Refrigeration Gas Cycles: The Brayton Cycle , The Regenerative Brayton Cycle and The Gas Refrigeration Cycle

7.5 The Brayton Cycle

The gas turbine is another mechanical system that produces power. It may operate on an open cycle when used as a truck engine, or on a closed cycle when used in a power plant. In open cycle operation, air enters the compressor, passes through a constant-pressure

combustion chamber, then through a turbine, and finally exits as products of combustion to the atmosphere, as shown in Fig. 7.4a. In closed cycle operation the combustion chamber is replaced with a heat exchanger in which energy enters the cycle from some exterior source; an additional heat exchanger transfers heat from the cycle so that the air is returned to its initial state, as shown in Fig. 7.4b.

The ideal cycle used to model the gas turbine is the Brayton cycle. It utilizes isentropic compression and expansion, as indicated in Fig. 7.5. The efficiency of such a cycle is given by

image

 

image

Figure 7.4 The Brayton cycle components. (a) Open cycle. (b) Closed cycle.

image

Figure 7.5 The Brayton cycle.

Using the isentropic relations

image

This expression for thermal efficiency was obtained using constant specific heats. For more accurate calculations the gas tables from App. E should be used.

In an actual gas turbine the compressor and the turbine are not isentropic; some losses do occur. These losses, usually in the neighborhood of 15 percent, significantly reduce the efficiency of the gas-turbine engine.

Another important feature of the gas turbine that seriously limits thermal efficiency is the high work requirement of the compressor, measured by the back work ratio,

BWR = W&comp / W&turb. The compressor may require up to 80 percent of the turbine’s

output (a back work ratio of 0.8), leaving only 20 percent for net work output. This

relatively high limit is experienced when the efficiencies of the compressor and tur- bine are too low. Solved problems illustrate this point.

EXAMPLE 7.5

Air enters the compressor of a gas turbine at 100 kPa and 25°C. For a pressure ratio of 5 and a maximum temperature of 850°C, determine the back work ratio (BWR) and the thermal efficiency for this Brayton cycle. Assume constant specific heat.

Solution

To find the back work ratio we observe that

image

EXAMPLE 7.6

Assume the compressor and the gas turbine in Example 7.5 each have an efficiency of 75 percent. Determine the back work ratio (BWR) and the thermal efficiency for the Brayton cycle, assuming constant specific heats.

Solution

We can calculate the desired quantities if we determine wcomp, wturb, and qin. The compressor work is

image

where ws,comp is the isentropic work and T2′ is the temperature of state 2′ assuming n isentropic process where state 2 is the actual state. We then have, using T= T2 rom Example 7.5,

image

Likewise, there results

image

It is obvious why the Brayton cycle cannot operate efficiently with relatively low efficiencies of the compressor and turbine. Efficiencies had to be raised into the 90’s before the Brayton cycle became a cycle that could effectively be used to produce power.

7.6 The Regenerative Brayton Cycle

The heat transfer from the simple gas-turbine cycle of the previous section is simply lost to the surroundings—either directly, with the products of combustion—or from a heat exchanger. Some of this exit energy can be utilized since the temperature of the flow exiting the turbine is greater than the temperature of the flow entering the compressor. A counterflow heat exchanger, a regenerator, is used to transfer some of this energy to the air leaving the compressor, as shown in Fig. 7.6. For an ideal regenerator the exit temperature Twould equal the entering temperature T ; and, similarly, Twould equal T . Since less energy is rejected from the cycle, the thermal efficiency is expected to increase. It is given byimage

Using the first law, expressions for q and wturb  are found to be

image

 

image

Figure 7.6 The regenerative Brayton cycle. Hence, for the ideal regenerator in which T

= T5  and q  = wturb  , the thermal efficiency can be written as

image

Using the appropriate isentropic relation, after some algebra, this can be written in the form

image

Note that this expression for thermal efficiency is quite different from that for the Brayton cycle. For a given pressure ratio, the efficiency increases as the ratio of minimum to maximum temperature decreases. But, perhaps more surprisingly, as the pressure ratio increases the efficiency decreases, an effect opposite to that of the Brayton cycle. Hence it is not surprising that for a given regenerative cycle temperature ratio, there is a particular pressure ratio for which the efficiency of the Brayton cycle will equal the efficiency of the regenerative cycle. This is shown for a temperature ratio of 0.25 in Fig. 7.7.

image

Figure 7.7 Efficiencies of the Brayton and regenerative cycles.

 

In practice the temperature of the air leaving the regenerator at state 3 must be less than the temperature of the air entering at state 5. Also, T > T . The effective- ness, or efficiency, of a regenerator is measured by

image

if we assume an ideal gas with constant specific heats. For the ideal regenerator T3 = T5

and hreg  = 1. Regenerator efficiencies exceeding 80 percent are common.

EXAMPLE 7.7

Add an ideal regenerator to the gas-turbine cycle of Example 7.5 and calculate the thermal efficiency and the back work ratio, assuming constant specific heats.

Solution

The thermal efficiency is found using Eq. (7.24):

image_thumb

This represents a 57 percent increase in efficiency, a rather large effect. Note that, for the information given, the back work ratio does not change; hence,

BWR = wcomp/wturb = 0.420.

7.7 The Combined Cycle

The Brayton cycle efficiency is quite low primarily because a substantial amount of the energy input is exhausted to the surroundings. This exhausted energy is usually at a relatively high temperature and thus it can be used effectively to produce power. One possible application is the combined Brayton Rankine cycle in which the high- temperature exhaust gases exiting the gas turbine are used to supply energy to the boiler of the Rankine cycle, as illustrated in Fig. 7.8. Note that the temperature T9 of the Brayton cycle gases exiting the boiler is less than the temperature T of the

image_thumb[1]

Figure 7.8 The combined Brayton-Rankine cycle.

Rankine cycle steam exiting the boiler; this is possible in the counterflow heat exchanger, the boiler.

To relate the air mass flow rate m& a of the Brayton cycle to the steam mass flow rate m& s of the Rankine cycle, we use an energy balance in the boiler; it gives (see Fig. 7.8),

image_thumb[2]

assuming no additional energy addition in the boiler, which would be possible with an oil burner, for example. The cycle efficiency would be found by considering the

purchased energy as Q&in , the energy input in the combustor. The output is the sum

of the net output W&GT from the gas turbine and the output W&ST from the steam turbine. The combined cycle efficiency is thus given by

image_thumb[3]

An example will illustrate the increase in efficiency of such a combined cycle.

EXAMPLE 7.8

A simple steam power plant operates between pressures of 10 kPa and 4 MPa with a maximum temperature of 400°C. The power output from the steam tur- bine is 100 MW. A gas turbine provides the energy to the boiler; it accepts air at 100 kPa and 25°C, and has a pressure ratio of 5 and a maximum temperature of 850°C. The exhaust gases exit the boiler at 350 K. Determine the thermal efficiency of the combined Brayton-Rankine cycle. Refer to Fig. 7.8.

Solution

If we neglect the work of the pump, the enthalpy remains unchanged across the pump. Hence, h2 = h1 = 192 kJ/kg. At 400°C and 4 MPa we have h3 = 3214 kJ/kg and s3 = 6.769 kJ/kg · K. State 4 is located by noting that s4 = s3 so that the quality is

image_thumb[4]

image_thumb[5]

Note that this efficiency is 59 percent higher than the Rankine cycle (see Example 6.2) and 53 percent higher than the Brayton cycle (see Example 7.5). Using steam reheaters, steam regenerators, gas intercoolers, and gas reheaters could increase cycle efficiency even more.

7.8 The Gas Refrigeration Cycle

If the flow of the gas is reversed in the Brayton cycle of Fig. 7.5, the gas undergoes an isentropic expansion process as it flows through the turbine, resulting in a substantial reduction in temperature, as shown in Fig. 7.9. The gas with low turbine exit temperature can be used to refrigerate a space to temperature T by extracting heat at rate Q&in from the refrigerated space.

Figure 7.9 illustrates a closed refrigeration cycle. (An open cycle system is

used in aircraft; air is extracted from the atmosphere at state 2 and inserted into the passenger compartment at state 1. This provides both fresh air and cooling.) An additional heat exchanger may be used, like the regenerator of the Brayton power cycle, to provide an even lower turbine exit temperature, as illustrated in Example 7.10. The gas does not enter the expansion process (the turbine) at state 5; rather, it passes through an internal heat exchanger (it does not exchange heat with the surroundings). This allows the temperature of the gas entering the

image_thumb[6]

Figure 7.9 The gas refrigeration cycle.

turbine to be much lower than that of Fig. 7.9. The temperature T after the expansion is so low that gas liquefaction is possible. It should be noted, how- ever, that the coefficient of performance is actually reduced by the inclusion of an internal heat exchanger.

A reminder: when the purpose of a thermodynamic cycle is to cool or heat a space, we do not define a cycle’s efficiency; rather, we define its coefficient of performance:

image_thumb[8]

EXAMPLE 7.9

Air enters the compressor of a simple gas refrigeration cycle at −10°C and

100 kPa. For a compression ratio of 10 and a turbine inlet temperature of 30°C,

calculate the minimum cycle temperature and the coefficient of performance using the ideal-gas equations of this section.

Solution

Assuming isentropic compression and expansion processes we find

image_thumb[9]

The COP is now calculated as follows:

image_thumb[10]

This coefficient of performance is quite low when compared with that of a vapor refrigeration cycle. Thus, gas refrigeration cycles are used only for special applications.

EXAMPLE 7.10

Use the given information for the compressor of the refrigeration cycle of Example 7.9 but add an ideal internal heat exchanger, a regenerator, as illustrated

below, so that the air temperature entering the turbine is −40°C. Calculate the

minimum cycle temperature and the coefficient of performance.

image_thumb[11]

Solution

Assuming isentropic compression we again have

image_thumb[12]

Quiz No. 1

(Assume constant specific heats)

1. Which of the following statements is not true of the diesel cycle?

(A) The expansion process is an isentropic process.

(B) The combustion process is a constant-volume process.

(C) The exhaust process is a constant-volume process.

(D) The compression process is an adiabatic process.

2. An engine with a bore and a stroke of 0.2 m × 0.2 m has a clearance of

5 percent. Determine the compression ratio.

(A) 23

(B) 21

(C) 19

(D) 17

3. A Carnot piston engine operates with air between 20 and 600°C with a low pressure of 100 kPa. If it is to deliver 800 kJ/kg of work, calculate MEP. (See Figs. 5.5 and 5.6.)

(A) 484 kPa

(B) 374 kPa

(C) 299 kPa

(D) 243 kPa

4. The heat rejected in the Otto cycle of Fig. 7.2 is

(A) Cp (T4 − T1 )

(B) Cp (T3 − T4 )

(C) Cv (T4 − T1 )

(D) Cv (T3 − T4 )

5. The maximum allowable pressure in an Otto cycle is 8 MPa. Conditions at the beginning of the air compression are 85 kPa and 22°C. Calculate the required heat addition if the compression ratio is 8.

(A) 2000 kJ/kg

(B) 2400 kJ/kg

(C) 2800 kJ/kg

(D) 3200 kJ/kg

6. The MEP for the Otto cycle of Prob. 5 is nearest

(A) 900 kPa

(B) 1100 kPa

(C) 1300 kPa

(D) 1500 kPa

7. A diesel cycle operates on air which enters the compression process at 85 kPa and 30°C. If the compression ratio is 16 and the maximum temperature is 2000°C, the cutoff ratio is nearest

(A) 2.47

(B) 2.29

(C) 2.04

(D) 1.98

8. If the power output of the diesel cycle of Prob. 7 is 500 hp, the mass flow rate of air is nearest

(A) 0.532 kg/s

(B) 0.467 kg/s

(C) 0.431 kg/s

(D) 0.386 kg/s

9. Air enters the compressor of a gas turbine at 85 kPa and 0°C. If the pressure ratio is 6 and the maximum temperature is 1000°C, using the ideal-gas equations, the thermal efficiency is nearest

(A) 52%

(B) 48%

(C) 44%

(D) 40%

10. Using the ideal-gas law, the back work ratio for the Brayton cycle of Prob.

9 is nearest

(A)

0.305

(B)

0.329

(C)

0.358

(D)

0.394

11. For the ideal-gas turbine with regenerator shown below, W& out is nearest

image_thumb[13]

(A) 950 kW

(B) 900 kW

(C) 850 kW

(D) 800 kW

12. The back work ratio for the cycle of Prob. 11 is nearest

(A)

0.432

(B)

0.418

(C)

0.393

(D)

0.341

13. Air enters the compressor of an ideal-gas refrigeration cycle at 10°C and 80 kPa. If the maximum and minimum temperatures are 250 and −50°C,

the compressor work is nearest

(A) 170 kJ/kg

(B) 190 kJ/kg

(C) 220 kJ/kg

(D) 240 kJ/kg

14. The pressure ratio across the compressor in the cycle of Prob. 13 is nearest

(A) 7.6

(B) 8.2

(C) 8.6

(D) 8.9

15. The COP for the refrigeration cycle of Prob. 13 is nearest (A) 1.04

(B) 1.18

(C) 1.22

(D) 1.49

Quiz No. 2

(Assume constant specific heats)

1. The exhaust process in the Otto and diesel cycles is replaced with a constant-volume process for what primary reason?

(A) To simulate the zero work of the actual exhaust process.

(B) To simulate the zero heat transfer of the actual process.

(C) To restore the air to its original state.

(D) To ensure that the first law is satisfied.

2. An air-standard cycle operates in a piston-cylinder arrangement with the following four processes: 1 → 2: isentropic compression from 100 kPa and 15°C to 2 MPa; 2 → 3: constant-pressure heat addition to 1200°C; 3 → 4: isentropic expansion; and 4 → 1: constant-volume heat rejection. Calculate

the heat addition.

(A) 433 kJ/kg

(B) 487 kJ/kg

(C) 506 kJ/kg

(D) 522 kJ/kg

3. A Carnot piston engine operates on air between high and low pressures of 3 MPa and 100 kPa with a low temperature of 20°C. For a compression ratio of 15, calculate the thermal efficiency. (See Figs. 5.5 and 5.6.)

(A) 40%

(B) 50%

(C) 60%

(D) 70%

4. A spark-ignition engine operates on an Otto cycle with a compression ratio of 9 and temperature limits of 30 and 1000°C. If the power output is 500 kW, the thermal efficiency is nearest

(A) 50%

(B) 54%

(C) 58%

(D) 64%

5. The mass flow rate of air required for the Otto cycle of Prob. 4 is nearest

(A) 1.6 kg/s

(B) 1.8 kg/s

(C) 2.2 kg/s

(D) 2.0 kg/s

6. A diesel engine is designed to operate with a compression ratio of 16 and air entering the compression stroke at 110 kPa and 20°C. If the energy added during combustion is 1800 kJ/kg, calculate the cutoff ratio.

(A) 3.2

(B) 3.0

(C) 2.8

(D) 2.6

7. The MEP of the diesel cycle of Prob. 6 is nearest

(A) 1430 kPa

(B) 1290 kPa

(C) 1120 kPa

(D) 1080 kPa

8. Air enters the compressor of a Brayton cycle at 80 kPa and 30°C and compresses it to 500 kPa. If 1800 kJ/kg of energy is added in the combustor, calculate the compressor work requirement.

(A) 304 kJ/kg

(B) 286 kJ/kg

(C) 232 kJ/kg

(D) 208 kJ/kg

9. The new output of the turbine of Prob. 8 is nearest

(A) 874 kJ/kg

(B) 826 kJ/kg

(C) 776 kJ/kg

(D) 734 kJ/kg

10. The BWR of the Brayton cycle of Prob. 8 is nearest (A) 0.22

(B) 0.24

(C) 0.26

(D) 0.28

11. A regenerator is installed in the gas turbine of Prob. 8. Determine the cycle efficiency if its effectiveness is 100 percent.

(A) 88%

(B) 85%

(C) 82%

(D) 79%

12. The engines on a commercial jet aircraft operate on which of the basic cycles?

(A) Otto

(B) Diesel

(C) Carnot

(D) Brayton

13. Air flows at the rate of 2.0 kg/s through the compressor of an ideal-gas refrigeration cycle where the pressure increases to 500 kPa from 100 kPa.

The maximum and minimum cycle temperatures are 300 and −20°C,

respectively. Calculate the power needed to drive the compressor fluid

using the ideal-gas equations.

(A) 198 kW

(B) 145 kW

(C) 126 kW

(D) 108 kW

14. The COP of the refrigeration cycle of Prob. 13 is nearest (A) 2.73

(B) 2.43

(C) 1.96

(D) 1.73

Power and Refrigeration Gas Cycles: The Air-Standard Cycle , The Carnot Cycle , The Otto Cycle AND The Diesel Cycle

Power and Refrigeration Gas Cycles

Several cycles utilize a gas as the working substance, the most common being the Otto cycle and the diesel cycle, used in internal combustion engines. The word “cycle” used in reference to an internal combustion engine is technically incorrect since the working fluid does not undergo a thermodynamic cycle; air enters the engine, mixes with a fuel, undergoes combustion, and exits the engine as exhaust gases. This is often referred to as an open cycle, but we should keep in mind that a thermodynamic cycle does not really occur; the engine itself operates in what we could call a mechanical cycle. We do, however, analyze an internal combustion engine as though the working fluid operated on a cycle; it is an approximation that

allows us to predict influences of engine design on such quantities as efficiency and fuel consumption. It also allows us to compare the features of the various cycles.

7.1 The Air-Standard Cycle

In this section we introduce engines that utilize a gas as the working fluid. Spark- ignition engines that burn gasoline and compression-ignition (diesel) engines that burn fuel oil are the two most common engines of this type.

The operation of a gas engine can be analyzed by assuming that the working fluid does indeed go through a complete thermodynamic cycle. The cycle is often called an air-standard cycle. All the air-standard cycles we will consider have cer- tain features in common:

• Air is the working fluid throughout the entire cycle. The mass of the small quantity of injected fuel is negligible.

• There is no inlet process or exhaust process.

• A heat transfer process replaces the combustion process with energy transferred from an external source.

• The exhaust process, used to restore the air to its original state, is replaced with a constant-volume process transferring heat to the surroundings so that no work is accomplished, as in an actual cycle.

• All processes are assumed to be in quasiequilibrium.

• The air is assumed to be an ideal gas with constant specific heats.

A number of the engines we will consider make use of a closed system with a piston-cylinder arrangement, as shown in Fig. 7.1. The cycle shown on the P-v and T-s diagrams in the figure is representative. The diameter of the piston is called the bore, and the distance the piston travels in one direction is the stroke. When the piston is at top dead center (TDC), the volume occupied by the air in the cylinder is at a minimum; this volume is the clearance volume. When the piston moves to bot- tom dead center (BDC), the air occupies the maximum volume. The difference between the maximum volume and the clearance volume is the displacement volume. The clearance volume is often implicitly presented as the percent clearance c, the ratio of the clearance volume to the displacement volume. The compression ratio r is defined to be the ratio of the volume in the cylinder with the piston at BDC to the clearance volume, that is, referring to Fig. 7.1,

image

image

Figure 7.1 The cycle of a piston-cylinder gasoline engine.

The mean effective pressure (MEP) is another quantity that is often used when rating piston-cylinder engines; it is the pressure that, if acting on the piston during the power stroke, would produce an amount of work equal to that actually done during the entire cycle. Thus,

image

In Fig. 7.1 this means that the enclosed area of the actual cycle is equal to the area under the MEP dotted line.

EXAMPLE 7.1

An engine operates with air on the cycle shown in Fig.7.1 with isentropic processes 1 → 2 and 3 → 4. If the compression ratio is 12, the minimum pressure is 200 kPa,

and the maximum pressure is 10 MPa, determine the percent clearance and the MEP, assuming constant specific heats.

Solution

The percent clearance is given by

image

But the compression ratio is r = V1 /V2 = 12. Thus,

image

To determine the MEP we must calculate the area under the P-v diagram (the work), a rather difficult task. The work from 3 → 4 is, using Pvk = C,

 

image

 

image

Equating the two expressions yields

image

7.2 The Carnot Cycle

This ideal cycle was treated in detail in Chap. 5. Recall that the thermal efficiency of a Carnot engine,

imageexceeds that of any real engine operating between the given temperatures. We will use this efficiency as an upper limit for all engines operating between T and T .

7.3 The Otto Cycle

The four processes that form the Otto cycle are displayed in the T-s and P-v diagrams of Fig. 7.2. The piston starts at state 1 at BDC and compresses the air until it reaches TDC at state 2. Instantaneous combustion then occurs, due to spark ignition, resulting in a sudden jump in pressure to state 3 while the volume remains constant (a quasiequilibrium heat addition process). The process that follows is the power stroke as the air (the combustion products) expands isentropically to state 4. In the final process, heat transfer to the surroundings occurs and the cycle is completed.

The thermal efficiency of the Otto cycle is found from

image

Noting that the two heat transfer processes occur during constant-volume processes, for which the work is zero, there results

image

image

Figure 7.2 The Otto cycle.

where we have assumed each quantity to be positive. Then

imageThus, Eq. (7.7) gives the thermal efficiency as

image

We see that the thermal efficiency in this Otto cycle is dependent only on the compression ratio r: higher the compression ratio, higher the thermal efficiency.

EXAMPLE 7.2

An Otto cycle is proposed to have a compression ratio of 10 while operating with a low air temperature of 227°C and a low pressure of 200 kPa. If the work output is to be 1000 kJ/kg, calculate the maximum possible thermal efficiency and com- pare with that of a Carnot cycle. Also, calculate the MEP. Assume constant specific heats.

Solution

The Otto cycle provides the model for this engine. The maximum possible ther- mal efficiency for the engine would be

image

image

The Otto cycle efficiency is less than that of a Carnot cycle because the heat transfer processes in the Otto cycle are highly irreversible.

The MEP is found by using the equation

image

7.4 The Diesel Cycle

If the compression ratio is large enough, the temperature of the air in the cylinder when the piston approaches TDC will exceed the ignition temperature of diesel fuel. This will occur if the compression ratio is about 14 or greater. No external spark is needed; the diesel fuel is simply injected into the cylinder and combustion occurs because of the high temperature of the compressed air. This type of engine is referred to as a compression-ignition engine. The ideal cycle used to model the engine is the diesel cycle, shown in Fig. 7.3. The difference between this cycle and the Otto cycle is that, in the diesel cycle, the heat is added during a constant-pressure process.

The cycle begins with the piston at BDC, state 1; compression of the air occurs isentropically to state 2 at TDC; heat addition takes place (this models the injection and combustion of fuel) at constant pressure until state 3 is reached; expansion occurs isentropically to state 4 at BDC; constant-volume heat rejection completes the cycle and returns the air to the original state. Note that the power stroke includes the heat addition process and the expansion process.

image

image

 

This expression for the thermal efficiency is often written in terms of the compression ratio r and the cutoff ratio r which is the ratio of volumes from TDC to the

end of combustion. We then have

image_thumbFrom this expression we see that, for a given compression ratio r, the efficiency of the diesel cycle is less than that of an Otto cycle. For example, if r = 8 and rc = 2, the

Otto cycle efficiency is 56.3 percent and the diesel cycle efficiency is 49.0 percent.As r ncreases, the diesel cycle efficiency decreases. In practice, however, a compression ratio of 20 or so can be achieved in a diesel engine; using r = 20 and rc = 2, we would find η = 64.7 percent. Thus, because of the higher compression ratios, a diesel engine typically operates at a higher efficiency than a gasoline engine.

EXAMPLE 7.3

A diesel cycle, with a compression ratio of 18, operates on air with a low pressure of 200 kPa and a low temperature 200°C. If the high temperature is limited to 2000 K, determine the thermal efficiency and the MEP. Assume constant specific heats.

Solution

The cutoff ratio rc is found first. We have

image_thumb[1]

The thermal efficiency is now calculated as

image_thumb[2]

EXAMPLE 7.4

Rework Example 7.3 using Table E.1; do not assume constant specific heats. Calculate the thermal efficiency, the MEP, and the errors in both quantities as calculated in Example 7.3.

Solution

Equation (7.10) that gives the thermal efficiency of the Otto cycle was obtained assuming constant specific heats; so it cannot be used. We must calculate the work output and the heat added. Refering to Table E.1:

image_thumb[3]

image_thumb[4]

Hence, at state 4 we find from Table E.1 that T4 = 915 K and u4 = 687.5 kJ/kg. The first law results in

image_thumb[5]

The errors are substantial primarily due to large temperature differences experienced in the diesel cycle. However, the constant specific heat analyses are of interest particularly in parametric studies where various designs are considered.