*Air-Standard Diesel Cycle

The air-standard Diesel cycle is the ideal cycle that approximates the Diesel combustion engine

Process Description 1-2 Isentropic compression 2-3 Constant pressure heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection

The P-v and T-s diagrams are

*Thermal efficiency of the Diesel cycle Now to find Qin and Qout. Apply the first law closed system to process 2-3, P = constant. Thus, for constant specific heats

*Apply the first law closed system to process 4-1, V = constant (just as we did for the Otto cycle)Thus, for constant specific heatsThe thermal efficiency becomes

*What is T3/T2 ?where rc is called the cutoff ratio, defined as V3 /V2, and is a measure of the duration of the heat addition at constant pressure. Since the fuel is injected directly into the cylinder, the cutoff ratio can be related to the number of degrees that the crank rotated during the fuel injection into the cylinder.

*What is T4/T1 ?Recall processes 1-2 and 3-4 are isentropic, soSince V4 = V1 and P3 = P2, we divide the second equation by the first equation and obtain

*Therefore,

An air-standard Diesel cycle has a compression ratio of 20 and a maximum temperature of 2200 K. At the beginning of the compression process, air is at 95 kPa and 20C. Using the cold air standard assumptions temperature, determine (a) the thermal efficiency and (b) the mean effective pressure.

*Properties The properties of air at room temperature are cp = 1.005 kJ/kgK, cv = 0.718 kJ/kgK, R = 0.287 kJ/kgK, and k = 1.4 (Table A-2).DIESEL CYCLE EXAMPLE

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Example 2An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 2. At the beginning of the compression process, air is at 100 kPa, 27 C and 0.0018 m3. Utilizing the cold-air-standard assumptions,determine the temperature and pressure of air at the end of each process,(b) the net work output, the thermal efficiency of this cycle, and(d) the mean effective pressure for the cycle.

State 1: air at p1 = 100 kPa, T1 = 300 K and V1 = 0.0018 m3(a)V3 = rc V2 = 2 (0.0001) = 0.0002 m3V4 = V1 = 0.0018 m3Process 1-2: Isentropic compression

= 1906.6 KProcess 3-4: Isentropic expansionProcess 2-3: Constant-pressure heat additionp3 = p2 = 5719.8 kPa= 791.7 K

= 0.0021(1.005)(1906.6 593.3) = 2.77 kJQin = m(h3 h2) = mcp(T3 T2)(b)Qout = m(u4 u1) = mcv(T4 T1)= 0.0021(0.718)(791.7 300) = 0.74 kJTable A-2cp = 1.005 kJ/kgKcv = 0.718 kJ/kgK

Wnet = Qnet = Qin Qout= 2.77 0.74 = 2.03 kJ(c)(d)= 1194.1 kPa

By considering heat transfer to the air undergoing the power cycle as occurring in two steps: constant volume followed by constant pressure, the air-standard Dual cycle aims to mimic the pressure-volume variation of actual internal combustion engines more closely than achievable with the Otto and Diesel cycles.Air-Standard Dual Cycle

The air-standard Dual cycle consists of five internally reversible processes in series:Process 1-2: isentropic compression.Process 2-3: constant-volume heat addition to the air from and external source.Process 3-4: constant-pressure heat addition to the air from an external source.Process 4-5: isentropic expansion.Process 5-1: constant-volume heat transfer from the air.Air-Standard Dual CycleAs for the Diesel cycle, the Dual cycle has a two-step power stroke: process 3-4 followed by process 4-5.

Using closed system energy balances for each of the processes, the following expression for thermal efficiency for the air-standard Dual Cycle can be developed:Air-Standard Dual Cycle(Eq. 9.14)Note: As for the Diesel cycle, enthalpy appears only for notational convenience and does not signal use of control volume concepts. Like the Otto and Diesel cycles, thermal efficiency increases with increasing compression ratio.

Air-Standard Dual CycleThe specific internal energies and temperatures at each principal state are determined using methods similar to those used for the Otto and Diesel Cycles.Areas on the T-s and p-v diagrams of the Dual cycle can be interpreted as heat and work, respectively, as in the cases of the Otto and Diesel cycles.

*The compression ratio of an ideal dual cycle is 14. Air is at 80 kPa and 253 K at the beginning of the compression process. The pressure ratio for the constant volume process is 1.5 and the cut-off ratio is 1.2. Assuming constant specific heats for air, determine the thermal efficiency of the cycle.The properties of air at room temperature are cp = 1.005 kJ/kgK, cv = 0.718 kJ/kgK, R = 0.287 kJ/kgK, and k = 1.4 (Table A-2).DUAL CYCLE EXAMPLE

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Indicated power (IP)Brake power (bp)Friction power (fp) and mechanical efficiency, mBrake mean effective pressure (bmep), thermal efficiency and fuel consumptionVolumetric efficiency, vCriteria of Performance*

Defined as the rate of work done by the gas on the piston as evaluated from an indicator diagram obtained from the engine using the electronic engine indicator.For four-stroke engine,And for two-stroke engine,Indicated Power*ip = work done per cycle cycle per minuten is the number of cylinders.

Indicated Power*Indicated mean effective pressure, pi given by:,For one engine cylinder Work done per cycle = pi A L Where A = area of piston L = length of stroke

Power output = (work done per cycle) x (cycle per minute)

For four-stoke engines, the number of cycles per unit time is N/2 and for two-stroke engines the number of cycles per unit time is N, where N is the engine speed.

Brake power is a way to measure the engine power output.The engine is connected to a brake (or dynamometer) which can be loaded so that the torque exerted by the engine can be measured.The torque is obtained by reading off a net load, w at known radius, r.Brake Power*

ThereforeBrake power is also can be expressed asThen the brake mean effective pressure (Pb) is*

Friction Power*The difference between the Ip and bp is the friction power (fp). It is the power that overcome the frictional resistance of the engine parts.

Power input to the shaft is usually bigger than the indicated power due to frictional losses or the mechanical efficiency.Mechanical Efficiency*

Brake Mean Effective Pressure*From the definition of Brake power Since for 4 stroke engine andSince and Pi are difficult to obtain, they may be combined and replaced by a brake mean effective pressure, PbEquating this equation to another definition of bp: So: Its observed that bmep is proportional to torque.

Brake Thermal Efficiency*The power output of the engine is obtained from the chemical energy of the fuel supplied. The overall engine efficiency is given by the brake thermal efficiency, mf = mass flow fuel , Qnet,v = net calarofic value of the fuel.

sfc is the mass flow rate of fuel consumed per unit power output and is a criterion of economical power production.Specific Fuel Consumption*

Volumetric efficiency is only used with four-stroke cycle engine, which have a distinct induction process. The parameter used to measure the effectiveness of an engines induction process is the volumetric efficiency.Volumetric Efficiency*

An engine operating at 2400 rpm consumes 12 ml of fuel (s.g. 0.85) in 60 second. The engine indicates a load of 30 N on the pony brake system and the brakes torque arm is 20 cm. Determine (a) the brake power, (b) the mass flow rate of fuel, and (c) the specific fuel consumption.Example 5.4Solution:*

A four-cylinder petrol engine has a bore of 57 mm and a stroke of 90 mm. Its rated speed is 2800 rpm and it is tested at this speed against a brake which has a torque arm of 0.356 m. The net brake load is 155 N and the fuel consumption is 6.741 l/h. The specific gravity of the petrol used is 0.735 and it has a net calorific value of 44,200 kJ/kg. The engine is tested in an atmospheric condition at 101.325 kPa and 15 oC at air-fuel ratio of 14.5/1. Calculate for this speed, the engine torque, the bmep, the brake thermal efficiency, the specific fuel consumption and the volumetric efficiency of the engine.Example 5.4*

Real Case*

*Brayton Cycle

The Brayton cycle is the air-standard ideal cycle approximation for the gas-turbine engine. This cycle differs from the Otto and Diesel cycles in that the processes making the cycle occur in open systems or control volumes. Therefore, an open system, steady-flow analysis is used to determine the heat transfer and work for the cycle.

Search the web for information on new gas turbine technologies. One site that gives a link to the GE 7FA group of gas turbines that operate on natural gas ishttp://www.webshots.com/explains/sports/gas-turbine.html

A second link gives an overview of the applications of gas turbines.http://www.energysolutionscenter.org/DistGen/Tutorial/CombTurbine.htm#Heat_Recovery

Brayton Cycle Analysis

We assume the working fluid is air and the specific heats are constant and will consider the cold-air-standard cycle.

*The closed cycle gas-turbine engine

*Shown here are the compressor and turbine stages of an industrial gas turbine under repair. The compressor has eighteen stages of compression and the turbine has three stage of expansion.CompressorTurbinePhoto courtesy of Progress Energy Carolinas, Inc.

*The T-s and P-v diagrams for the Closed Brayton Cycle

Process Description 1-2 Isentropic compression (in a compressor)2-3 Constant pressure heat addition 3-4 Isentropic expansion (in a turbine)4-1 Constant pressure heat rejection

*Thermal efficiency of the Brayton cycleNow to find Qin and Qout.Apply the conservation of energy to process 2-3 for P = constant (no work), steady-flow, and neglect changes in kinetic and potential energies.The conservation of mass givesFor constant specific heats, the heat added per unit mass flow is

*The conservation of energy for process 4-1 yields for constant specific heats (lets take a minute for you to get the following result)The thermal efficiency becomes

*Recall processes 1-2 and 3-4 are isentropic, soSince P3 = P2 and P4 = P1, we see thatThe Brayton cycle efficiency becomesIs this the same as the Carnot cycle efficiency? Since process 1-2 is isentropic,

*where the pressure ratio is rp = P2/P1 and Extra Assignment

Evaluate the Brayton cycle efficiency by determining the net work directly from the turbine work and the compressor work. Compare your result with the above expression. Note that this approach does not require the closed cycle assumption.

*Example 9-2

The ideal air-standard Brayton cycle operates with air entering the compressor at 95 kPa, 22oC. The pressure ratio rp is 6:1 and the air leaves the heat addition process at 1100 K. Determine the compressor work and the turbine work per unit mass flow, the cycle efficiency, the back work ratio, and compare the compressor exit temperature to the turbine exit temperature. Assume constant properties.

Apply the conservation of energy for steady-flow and neglect changes in kinetic and potential energies to process 1-2 for the compressor. Note that the compressor is isentropic.The conservation of mass gives

*For constant specific heats, the compressor work per unit mass flow isSince the compressor is isentropic

*The conservation of energy for the turbine, process 3-4, yields for constant specific heats (lets take a minute for you to get the following result)Since process 3-4 is isentropic

*Since P3 = P2 and P4 = P1, we see thatWe have already shown the heat supplied to the cycle per unit mass flow in process 2-3 is

*The net work done by the cycle is The cycle efficiency becomes

*The back work ratio is defined asNote that T4 = 659.1 K > T2 = 492.5 K, or the turbine outlet temperature is greater than the compressor exit temperature. Can this result be used to improve the cycle efficiency?

What happens to th, win /wout, and wnet as the pressure ratio rp is increased? Consider the T-s diagram for the cycle and note that the area enclosed by the cycle is the net heat added to the cycle. By the first law applied to the cycle, the net heat added to the cycle is equal to the net work done by the cycle. Thus, the area enclosed by the cycle on the T-s diagram also represents the net work done by the cycle.

*Let's take a closer look at the effect of the pressure ratio on the net work done.

*Note that the net work is zero whenFor fixed T3 and T1, the pressure ratio that makes the work a maximum is obtained from:This is easier to do if we let X = rp(k-1)/kSolving for X

*Then, the rp that makes the work a maximum for the constant property case and fixed T3 and T1 isFor the ideal Brayton cycle, show that the following results are true.When rp = rp, max work, T4 = T2When rp < rp, max work, T4 > T2When rp > rp, max work, T4 < T2The following is a plot of net work per unit mass and the efficiency for the above example as a function of the pressure ratio.

*Regenerative Brayton Cycle

For the Brayton cycle, the turbine exhaust temperature is greater than the compressor exit temperature. Therefore, a heat exchanger can be placed between the hot gases leaving the turbine and the cooler gases leaving the compressor. This heat exchanger is called a regenerator or recuperator. The sketch of the regenerative Brayton cycle is shown below.

*We define the regenerator effectiveness regen as the ratio of the heat transferred to the compressor gases in the regenerator to the maximum possible heat transfer to the compressor gases.

*For ideal gases using the cold-air-standard assumption with constant specific heats, the regenerator effectiveness becomes Using the closed cycle analysis and treating the heat addition and heat rejection as steady-flow processes, the regenerative cycle thermal efficiency isNotice that the heat transfer occurring within the regenerator is not included in the efficiency calculation because this energy is not heat transferred across the cycle boundary.

Assuming an ideal regenerator regen = 1 and constant specific heats, the thermal efficiency becomes (take the time to show this on your own)

*When does the efficiency of the air-standard Brayton cycle equal the efficiency of the air-standard regenerative Brayton cycle? If we set th,Brayton = th,regen thenRecall that this is the pressure ratio that maximizes the net work for the simple Brayton cycle and makes T4 = T2. What happens if the regenerative Brayton cycle operates at a pressure ratio larger than this value?

*For fixed T3 and T1, pressure ratios greater than this value cause T4 to be less than T2, and the regenerator is not effective.

What happens to the net work when a regenerator is added?

What happens to the heat supplied when a regenerator is added?

The following shows a plot of the regenerative Brayton cycle efficiency as a function of the pressure ratio and minimum to maximum temperature ratio, T1/T3.

*Example 9-3: Regenerative Brayton Cycle

Air enters the compressor of a regenerative gas-turbine engine at 100 kPa and 300 K and is compressed to 800 kPa. The regenerator has an effectiveness of 65 percent, and the air enters the turbine at 1200 K. For a compressor efficiency of 75 percent and a turbine efficiency of 86 percent, determine(a) The heat transfer in the regenerator.(b) The back work ratio.(c) The cycle thermal efficiency.

Compare the results for the above cycle with the ones listed below that have the same common data as required. The actual cycles are those for which the turbine and compressor isentropic efficiencies are less than one. (a) The actual cycle with no regeneration, = 0.(b) The actual cycle with ideal regeneration, = 1.0.(c) The ideal cycle with regeneration, = 0.65.(d) The ideal cycle with no regeneration, = 0.(e) The ideal cycle with ideal regeneration, = 1.0.

We assume air is an ideal gas with constant specific heats, that is, we use the cold-air-standard assumption.

*The cycle schematic is the same as above and the T-s diagram showing the effects of compressor and turbine efficiencies is below.

*Summary of Results

Cycle typeActualActualActualIdealIdealIdealregen0.000.651.000.000.651.00comp0.750.750.751.001.001.00turb0.860.860.861.001.001.00qin kJ/kg578.3504.4464.6659.9582.2540.2wcomp kJ/kg326.2326.2326.2244.6244.6244.6wturb kJ/kg464.6464.6464.6540.2540.2540.2wcomp/wturb0.700.700.700.4530.4530.453th24.0%27.5%29.8%44.8%50.8%54.7%

*Compressor analysis

The isentropic temperature at compressor exit isTo find the actual temperature at compressor exit, T2a, we apply the compressor efficiency

*Since the compressor is adiabatic and has steady-flowTurbine analysis

The conservation of energy for the turbine, process 3-4, yields for constant specific heats (lets take a minute for you to get the following result)

*Since P3 = P2 and P4 = P1, we can find the isentropic temperature at the turbine exit. To find the actual temperature at turbine exit, T4a, we apply the turbine efficiency.

*The turbine work becomesThe back work ratio is defined as

*Regenerator analysis

To find T5, we apply the regenerator effectiveness.

*To find the heat transferred from the turbine exhaust gas to the compressor exit gas, apply the steady-flow conservation of energy to the compressor gas side of the regenerator.

*Using qregen, we can determine the turbine exhaust gas temperature at the regenerator exit.

*Heat supplied to cycle

Apply the steady-flow conservation of energy to the heat exchanger for process 5-3. We obtain a result similar to that for the simple Brayton cycle.Cycle thermal efficiency

The net work done by the cycle is

*The cycle efficiency becomesYou are encouraged to complete the calculations for the other values found in the summary table.

*Other Ways to Improve Brayton Cycle Performance

Intercooling and reheating are two important ways to improve the performance of the Brayton cycle with regeneration.

*The T-s diagram for this cycle is shown below. Sketch the P-v diagram.

*One stage of a two-stage, intercooled refrigerant compressor in a chilled water refrigeration system

*Typical shell and tube heat exchanger used for the intercooler, evaporator, and condenser for a chilled water refrigeration system.

*Forced draft cooling tower for cooling water used in the intercooler and condenser heat exchangers of the chilled water refrigeration system

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*Intercooling

When using multistage compression, cooling the working fluid between the stages will reduce the amount of compressor work required. The compressor work is reduced because cooling the working fluid reduces the average specific volume of the fluid and thus reduces the amount of work on the fluid to achieve the given pressure rise.

To determine the intermediate pressure at which intercooling should take place to minimize the compressor work, we follow the approach shown in Chapter 7.

For the adiabatic, steady-flow compression process, the work input to the compressor per unit mass is

*For the isentropic compression process Notice that the fraction kR/(k-1) = Cp.Can you obtain this relation another way? Hint: apply the first law to processes 1-4.

*For two-stage compression, lets assume that intercooling takes place at constant pressure and the gases can be cooled to the inlet temperature for the compressor, such that P3 = P2 and T3 = T1.

The total work supplied to the compressor becomes To find the unknown pressure P2 that gives the minimum work input for fixed compressor inlet conditions T1, P1, and exit pressure P4, we set

*This yieldsor, the pressure ratios across the two compressors are equal.Intercooling is almost always used with regeneration. During intercooling the compressor final exit temperature is reduced; therefore, more energy must be supplied in the heat addition process to achieve the maximum temperature of the cycle. Regeneration can make up part of the required heat transfer.

To supply only compressed air, using intercooling requires less work input. The next time you go to a home supply store where air compressors are sold, check the larger air compressors to see if intercooling is used. For the larger air compressors, the compressors are made of two piston-cylinder chambers. The intercooling heat exchanger is often a pipe with a attached fins that connects the large piston-cylinder chamber with the smaller piston-cylinder chamber. Often the fly wheel used to drive the compressor has spokes shaped like fan blades that are used to increase air flow across the compressor and heat exchanger pipe to improve the intercooling effect.

*Extra Assignment

Obtain the expression for the compressor total work by applying conservation of energy directly to the low- and high-pressure compressors.Reheating

When using multistage expansion through two or more turbines, reheating between stages will increase the net work done (it also increases therequired heat input). The regenerative Brayton cycle with reheating was shown above.

The optimum intermediate pressure for reheating is the one that maximizes the turbine work. Following the development given above for intercooling and assuming reheating to the high-pressure turbine inlet temperature in a constant pressure steady-flow process, we can show the optimum reheat pressure to be or the pressure ratios across the two turbines are equal.

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