Intake open Valves close Compression stroke Power Exhaust stroke FIGURE 1-26. Four-stroke cycle.
In the following paragraphs, the timing of each event is specified in terms of degrees of crankshaft travel on the stroke during which the event occurs.
It should be remembered that a certain amount of crankshaft travel is required to open a valve fully;
therefore, the specified timing represents the start of opening rather than the full-open position of the Intake Stroke During the intake stroke, the piston is pulled downward in the cylinder by the rotation of the crankshaft. This reduces the pressure in the cylinder and causes air under atmospheric pressure to flow through the carburetor, which meters the correct amount of fuel. The fuel/air mixture passes through the intake pipes and intake valves into the cylinders. The quantity or weight of the fuel/air charge depends upon the degree of throttle opening.
The intake valve is opened considerably before the piston reaches top dead center on the exhaust stroke, in order to induce a greater quantity of the fuel/air charge into the cylinder and thus increase the horsepower. The distance the valve may be opened before top dead center, however, is limited by several factors, such as the possibility that hot gases remaining in the cylinder from the previous cycle may flash back into the intake pipe and the induction system.
In all high-power aircraft engines, both the intake and the exhaust valves are off the valve seats at top dead center at the start of the intake stroke. As mentioned above, the intake valve opens before top dead center on the exhaust stroke (valve lead), and the closing of the exhaust valve is delayed considerably after the piston has passed top dead center and has started the intake stroke (valve lag). This timing is called valve overlap and is designed to aid in cooling the cylinder internally by circulating the cool incoming fuel/air mixture, to increase the amount of the fuel/air mixture induced into the cylinder, and to aid in scavenging the byproducts of combustion.
The intake valve is timed to close about 50° to 75° past bottom dead center on the compression stroke depending upon the specific engine, to allow the momentum of the incoming gases to charge the cylinder more completely. Because of the comparatively large volume of the cylinder above the piston when the piston is near bottom dead center, the slight upward travel of the piston during this time does not have a great effect on the incoming flow of gases. This late timing can be carried too far because the gases may be forced back through the intake valve and defeat the purpose of the late closing.
Compression Stroke After the intake valve is closed, the continued upward travel of the piston compresses the fuel/air mixture to obtain the desired burning and expansion characteristics.
The charge is fired by means of an electric spark as the piston approaches top dead center. The time of ignition will vary from 20° to 35° before top dead center, depending upon the requirements of the specific engine, to ensure complete combustion of the charge by the time the piston is slightly past the top dead center position.
Many factors affect ignition timing, and the engine manufacturer has expended considerable time in research and testing to determine the best setting. All engines incorporate devices for adjusting the ignition timing, and it is most important that the ignition system be timed according to the engine manufacturer's recommendations.
Power Stroke As the piston moves through the top dead center position at the end of the compression stroke and starts down on the power stroke, it is pushed downward by the rapid expansion of the burning gases within the cylinder head with a force that can be greater than 15 tons (30,000 p.s.i.) at maximum power output of the engine. The temperature of these burnning gases may be between 3,000° and
everal standards, the most oot- •poun If 1-pound = Or ıd- As the piston travels through bottom dead center apat the completion of the power stroke and starts upward on the exhaust stroke, it will begin to push ıny ng rm the burned exhaust gases out the exhaust port. The ton speed of the exhaust gases leaving the cylinder The amount of this This low the or reduced pressure speeds the flow of the fresh ground.
fuel/air charge into the cylinder as the intake valve is beginning to open. The intake valve opening is timed to occur at 8° to 55° before top dead center
RECIPROCATING ENGINE POWER AND EFFICIEN- CIES All aircraft engines are rated according to their ability to do work and produce power. This section presents an explanation of work and power and how they are calculated. Also discussed are the various efficiencies that govern the power output of a reciprocating engine.
Work The physicist defines work as 'Work is force times distance. Work done by a force acting on a body is equal to the magnitude of the force multiplied by the distance through which the force acts."
Work (W) =Force (F) × Distance (D).
in 2 min.
33,000, or 0.9 hp.
Piston Displacement When other factors remain equal, the greater the piston displacement the greater the maximum horsepower an engine will be capable of developing.
When a piston moves from bottom dead center to top dead center, it displaces a specific volume. The volume displaced by the piston is known as piston displacement and is expressed in cubic inches for most American-made engines and cubic centimeters for others.
The piston displacement of one cylinder may be obtained by multiplying the area of the cross section of the cylinder by the total distance the piston moves in the cylinder in one stroke. For multicylinder engines this product is multiplied by the number of cylinders to get the total piston displacement of the engine.
Since the volume (V) of a geometric cylinder equals the area (A) of the base multiplied by the altitude (H), it is expressed mathematically as:
V=A X H.
For our purposes, the area of the base is the area of the cross section of the cylinder or of the piston top.
To find the area of a circle it is necessary to use a number called pi. This number represents the ratio of the circumference to the diameter of any circle. Pi cannot be found exactly because it is a never-ending decimal, but expressed to four decimal places it is 3.1416, which is accurate enough for most computations.
The area of a circle, as in a rectangle or triangle, must be expressed in square units. The distance that is one-half the diameter of a circle is known as the radius. The area of any circle is found by squaring the radius and multiplying by pi. (7). The formula is expressed thus:
Where A is the area of a circle; pi is the given constant; and r is the radius of the circle, which is equal to ½ the diameter or R= 2 D Example Compute the piston displacement of the PWA inch diameter and a 5.5 inch stroke. Formulas required are:
R= 2 _ D V=A × H Total V=V x N (number of cylinders) Substitute values into these formulas and complete the calculation.
R= 2 R=5.5/2=2.75 A = 7R: A =3.1416 (2.75 × 2.75) A =3.1416 x 7.5625 = 23.7584 sq. inches V= A x H V=23.7584 x 5.5 V= 130.6712 Total V=V x N Total V=130.6712 x 14 Total V = 1829.3968 Rounded off to the next whole number total piston displacement equals 1830 cu. in.
Another method of calculating the piston displacement uses the diameter of the piston instead of the radius in the formula for the area of the base.
A=¼ (7) (D)° Substituting A =¼ × 3.1416 × 5.5 x 5.5 A=.7854 x 30.25 A =23.758 square inches.
From this point on the calculations are identical to the preceding example.
Compression Ratio All internal-combustion engines must compress the fuel/air mixture to receive a reasonable amount of work from each power stroke. The fuel/air charge in the cylinder can be compared to a coil spring, in that the more it is compressed the more work it is potentially capable of doing.
The compression ratio of an engine (sce figure 1-27) is a comparison of the volume of space in a cylinder when the piston is at the bottom of the stroke to the volume of space when the piston is at the top of the stroke. This comparison is expressed as a ratio, hence the term "compresion ratio."
Compression ratio is a controlling factor in the maximum horsepower developed by an engine, but it is limited by present-day fuel grades and the high engine speeds and manifold pressures required for takeoff. For example, if there are 140 cu. in.
of space in the cylinder when the piston is at the bottom and there are 20 cu. in. of space when the piston is at the top of the stroke, the compression ratio would be 140 to 20. If this ratio is expressed in fraction form, it would be 140/20, or 7 to 1, usually represented as 7:1.
To grasp more thoroughly the limitation placed on compression ratios, manifold pressure and its effect on compression pressures should be understood. Manifold pressure is the average absolute pressure of the air or fuel/air charge in the intake manifold and is measured in units of inches of mercury (Hg). Manifold pressure is dependent on engine speed (throttle setting) and supercharg.
ing. The engine-driven internal supercharger (blower) and the external exhaust-driven supercharger (turbo) are actually centrifugal-type air compressors. The operation of these superchargers increases the weight of the charge entering the cylinder. When either one or both are used with the aircraft engine, the manifold pressure may be considerably higher than the pressure of the outside atmosphere. The advantage of this condition is that a greater amount of charge is forced into a given cylinder volume, and a greater power results.
Clearance volume T.D. C.
4 B.D.C. Total 6 7 FIGURE 1-27. Compression ratio.
Compression ratio and manifold pressure determine the pressure in the cylinder in that portion of the operating cycle when both valves are closed.
The pressure of the charge before compression is determined by manifold pressure, while the pressure at the height of compression (just prior to ignition) is determined by manifold pressure times the compression ratio. For example, if an engine were operating at a manifold pressure of 30" Hg with a compression ratio of 7:1, the pressure at the instant before ignition would be approximately 210" Hg. However, at a manifold pressure of 60" Hg the pressure would be 420" Hg.
Without going into great detail, it has been shown that the compression event magnifies the effect of varying the manifold pressure, and the magnitude of both affects the pressure of the fuel charge just before the instant of ignition. If the pressure at this time becomes too high, premature ignition or knock will occur and produce overheating.
One of the reasons for using engines with high compression ratios is to obtain long-range fuel economy, that is, to convert more heat energy into useful work than is done in engines of low compression ratio. Since more heat of the charge is converted into useful work, less heat is absorbed by the cylinder walls. This factor promotes cooler engine operation, which in turn increases the there meation Here, again, a compromise is needed between the demand for fuel economy and the demand for maximum horsepower without knocking. Some manufacturers of high-compression engines suppress knock at high manifold pressures by injecting an antiknock fluid into the fuel/air mixture. The fluid acts primarily as a coolant so that more power can be delivered by the engine for short periods, such as at takeoff and during emergencies, when power is critical. This high power should be used for short periods only.
Indicated Horsepower The indicated horsepower produced by an engine is the horsepower calculated from the indicated mean effective pressure and the other factors which 32
affect the power output of an engine. Indicated horsepower is the power developed in the combustion chambers without reference to friction losses L = Length of the stroke in ft. or in fractions of a foot.
A = Area of the piston head or cross-sectional within the engine. area of the cylinder, in sq. in.
This horsepower is calculated as a function of the r.p.m.
actual cylinder pressure recorded during engine N = Number of power strokes per minute; 2 operation. To facilitate the indicated horsepower K = Number of cylinders.
calculations, a mechanical indicating device, at- In the formula above, the area of the piston times tached to the engine cylinder, scribes the actual the indicated mean effective pressure gives the force pressure existing in the cylinder during the complete acting on the piston in pounds. This force multioperating cycle. This pressure variation can be plied by the length of the stroke in feet gives the represented by the kind of graph shown in figure work performed in one power stroke, which, multi-
1-28. Notice that the cylinder pressure rises on plied by the number of power strokes per minute, the compression stroke, reaches a peak after top gives the number of ft.-lb. per minute of work center, then decreases as the piston moves down on produced by one cylinder. Multiplying this result the power stroke. Since the cylinder pressure varies by the number of cylinders in the engine gives the during the operating cycle, an average pressure, amount of work performed, in ft.-lb., by the engine.
line AB, is computed. This average pressure, if Since hp. is defined as work done at the rate of applied steadily during the time of the power stroke, 33,000 ft.-lb. per min., the total number of ft.-lb.
would do the same amount of work as the varying of work performed by the engine is divided by pressure during the same period. This average 33,000 to find the indicated horsepower.
pressure is known as indicated mean effective pres- Example sure and is included in the indicated horsepower calculation with other engine specifications. If the Given:
characteristics and the indicated mean effective pres- Indicated mean late the indicated horsepower rating. sure of an engine are known, it is possible to calcu- Stroke (L) effective pressure (P) = 165 Ibs./ sq. in. = 6 in. or .5 ft.
Bore = 5.5 in.
The indicated horsepower for a four-stroke-cycle engine can be calculated from the following formula, No. of cylinders (K) r.p.m. = 3,000 = 12 in which the letter symbols in the numerator are memorizing the formula: arranged to spell the word "plank" to assist in Indicated hp. = 33,000 ft. Ibs./ min. PLANK Indicated horsepower = 33.000 _ PLANK Find indicated hp.: A is found by using the equation Where: A = 1/4"D2 P = Indicated mean effective pressure in p.s.i. A = 1/4 × 3.1416 × 5.5 × 5.5 Compression Power Exhaust Intake
• Peak pressure Indicated mean effective pressure (IMEP) Spark D¡ IC E Bottom center Top center BC TC BC Piston travel FIGURE 1-28. Cylinder pressure during power cycle.
= 23.76 sq. in.
N is found by multiplying the r.p.m. by 1/2:
N = 1/2 × 3,000 = 1,500 r.p.m.
Now, substituting in the formula:
165 x .5 × 23.76 × 1,500 × 12 Indicated hp. = 33,000 ft.-lbs. / min.
= 1069.20.
Brake Horsepower The indicated horsepower calculation discussed in the preceding paragraph is the theoretical power of a frictionless engine. The total horsepower lost in overcoming friction must be subtracted from the indicated horsepower to arrive at the actual horsepower delivered to the propeller. The power delivered to the propeller for useful work is known as b.hp. (brake horsepower). The difference between indicated and brake horsepower is known as friction horsepower, which is the horsepower required to overcome mechanical losses such as the pumping action of the pistons and the friction of the pistons and the friction of all moving parts.
In practice, the measurement of an engine's
b.hp. involves the measurement of a quantity known as torque, or twisting moment. Torque is the product of a force and the distance of the force from the axis about which it acts, or Torque = Force x Distance (at right angles to the force).
Torque is a measure of load and is properly expressed in pound-inches (lb.-in.) or pound-feet (lb.-ft.) and should not be confused with work, which is expressed in inch-pounds (in.-lbs.) or footpounds (ft.-lbs.).
There are a number of devices for measuring torque, of which the.Prony brake, dynamometer, and torquemeter are examples. Typical of these devices is the Prony brake (figure 1-29), which measures the usable power output of an engine on a test stand. It consists essentially of a hinged collar, or brake, which can be clamped to a drum splined to the propeller shaft. The collar and drum form a friction brake which can be adjusted by a wheel. An arm of a known length is rigidly attached to or is a part of the hinged collar and terminates at a point which bears on a set of scales.
As the propeller shaft rotates, it tends to carry the hinged collar of the brake with it and is prevented from doing so only by the arm that bears on the scale. The scale reads the force necessary to arrest the motion of the arm. If the resulting force registered on the scale is multiplied by the length of the arm, the resulting product is the torque exerted by the rotating shaft. Example: If the scale registers 200 Ibs. and the length of the arm is Friction adjusting wheel Collar Scale 300 100- 200 pounds
• Arm Hinge Propeller shatt Drum Length of arm
3.18 feet FIGURE 1-29. Typical Prony brake.
3.18 ft., the torque exerted by the shaft is:
200 lb. × 3.18 ft. = 636 lb.-ft.
Once the torque is known, the work done per revolution of the propeller shaft can be computed without difficulty by the equation:
Work per revolution = 2T x torque.
If work per revolution is multiplied by the r.p.m., the result is work per minute, or power. If the work is expressed in ft.-lbs. per min., this quantity is divided by 33,000; the result is the brake horsepower of the shaft. In other words:
Power = Work per revolution x r.p.m.
and b.hp. = Work per revolution x r.p.m., 33,000 2n x force on the scales (lbs.) x length of arm (ft.) x r.p.m..
or b.hp. = 33,000 Example Given :
Force on scales = 200 Ibs.
Length of arm = 3.18 ft.
r.p.m. = 3,000 = 3.1416.
Find b.hp.:
Substituting in equation-
= 363.2 = 363.
As long as the friction between the brake collar and propeller shaft drum is great enough to impose an appreciable load on the engine, but is not great enough to stop the engine, it is not necessary to know the amount of friction between the collar and drum to compute the b.hp. If there were no load imposed, there would be no torque to measure, and the engine would "run away." If the imposed load is so great that the engine stalls, there may be considerable torque to measure, but there will be no
r.p.m. In either case it is impossible to measure the
b.hp. of the engine. However, if a reasonable amount of friction exists between the brake drum and the collar and the load is then increased, the tendency of the propeller shaft to carry the collar and arm about with it becomes greater, thus imposing a greater force upon the scales. As long as the torque increase is proportional to the r.p.m.
decrease, the horsepower delivered at the shaft remains unchanged. This can be seen from the equation in which 27 and 33,000 are constants and torque and r.p.m. are variables. If the change in
r.p.m. is inversely proportional to the change in torque, their product will remain unchanged.
Therefore, b.hp. remains unchanged. This is important. It shows that horsepower is the function of both torque and r.p.m., and can be changed by changing either torque or r.p.m., or both.
Friction Horsepower Friction horsepower is the indicated horsepower minus brake horsepower. It is the horsepower used by an engine in overcoming the friction of moving parts, drawing in fuel, expelling exhaust, driving oil and fuel pumps, and the like. On modern aircraft engines, this power loss through friction may be as high as 10 to 15% of the indicated horsepower.
Friction and Brake Mean Effective Pressures The IMEP (indicated mean effective pressure), discussed previously, is the average pressure produced in the combustion chamber during the operating cycle and is an expression of the theoretical, frictionless power known as indicated horsepower. In addition to completely disregarding power lost to friction, indicated horsepower gives no indication as to how much actual power is delivered to the propeller shaft for doing useful work. However, it is related to actual pressures which occur in the cylinder and can be used as a measure of these pressures.
To compute the friction loss and net power output, the indicated horsepower of a cylinder may be thought of as two separate powers, each producing a different effect. The first power overcomes internal friction, and the horsepower thus consumed is known as friction horsepower. The second power, known as brake horsepower, produces useful work at the propeller. Logically, therefore, that portion of IMEP that produces brake horsepower is called BMEP (brake mean effective pressure). The remaining pressure used to overcome internal friction is called FMEP (friction mean effective pressure). This is illustrated in figure 1-30. IMEP is a useful expression of total cylinder power output, but is not a real physical quantity; likewise, FMEP and BMEP are theoretical but useful expressions of friction losses and net power output.
Although BMEP and FMEP have no real existence in the cylinder, they provide a convenient means of representing pressure limits, or rating engine performance throughout its entire operating range.
This is true since there is a relationship between IMEP, BMEP, and FMEP.
One of the basic limitations placed on engine operation is the pressure developed in the cylinder during combustion. In the discussion of compression ratios and indicated mean effective pressure, it was found that, within limits, the increased pressure resulted in increased power. It was also noted that if the cylinder pressure was not controlled Power stroke Actual pressure FMEP
- IMEP BMEP Friction Accessories RE Impeller B FIGURE 1-30. Powers and pressures.
within close limits, it would impose dangerous internal loads that might result in engine failure.
It is therefore important to have a means of determining these cylinder pressures as a protective measure and for efficient application of power.
If the b.hp. is known, the BMEP can be computed by means of the following equation:
BMEP = b.hp. × 33,000 LANK Example Given:
b.hp. = 1,000 Stroke = 6 in.
Bore - 5.5 in.
r.p.m. = 3,000 No. of cyls. = 12.
Find BMEP :
Find length of stroke (in ft.) :
I = 0.5.
Find area of cylinder bore:
A = 1/47D2 = 0.7854 × 5.5 × 5.5 = 23.76 sq. in.
Find No. of power strokes per min.:
N = 1/2 x r.p.m.
= 1/2 × 3,000 = 1,500.
Then substituting in the equation:
BMEP = 5 x 23.76 × 1,500 × 12 1,000 × 33,000 = 154.32 Ibs. per sq. in.
Thrust Horsepower Thrust horsepower can be considered as the result of the engine and the propeller working together.
If a propeller could be designed to be 100% efficient, the thrust- and the brake-horsepower would be the same. However, the efficiency of the propeller varies with the engine speed, attitude, altitude, temperature, and airspeed, thus the ratio of the thrust horsepower and the brake horsepower delivered to the propeller shaft will never be equal. For example, if an engine develops 1,000 b.hp., and it is used with a propeller having 85 percent efficiency, the thrust horsepower of that engine-propeller combination is 85 percent of 1,000 or 850 thrust hp.
Of the four types of horsepower discussed, it is the thrust horsepower that determines the performance of the engine-propeller combination.
EFFICIENCIES Thermal Efficiency Any study of engines and power involves consideration of heat as the source of power. The heat produced by the burning of gasoline in the cylinders causes a rapid expansion of the gases in the cylinder, and this, in turn, moves the pistons and creates mechanical energy.
It has long been known that mechanical work can be converted into heat and that a given amount of heat contains the energy equivalent of a certain amount of mechanical work. Heat and work are 36
theoretically interchangeable and bear a fixed relation to each other. Heat can therefore be measured in work units (for example, ft.-lbs.) as well as in heat units. The B.t.u. (British thermal unit) of heat is the quantity of heat required to raise the temperature of 1 lb. of water 1° F. It is equivalent to 778 ft.-lbs. of mechanical work. A pound of petroleum fuel, when burned with enough air to consume it completely, gives up about 20,000 B.t.u., the equivalent of 15,560,000 ft.-lbs. of mechanical work. These quantities express the heat energy of the fuel in heat and work units, respectively.
The ratio of useful work done by an engine to the heat energy of the fuel it uses, expressed in work or heat units, is called the thermal efficiency of the engine. If two similar engines use equal amounts of fuel, obviously the engine which converts into work the greater part of the energy in the fuel (higher thermal efficiency) will deliver the greater amount of power. Furthermore, the engine which has the higher thermal efficiency will have less waste heat to dispose of to the valves, cylinders, pistons, and cooling system of the engine.
A high thermal efficiency also means a low specific fuel consumption and, therefore, less fuel for a flight of a given distance at a given power. Thus, the practical importance of a high thermal efficiency is threefold, and it constitutes one of the most desirable features in the performance of an aircraft engine.
Of the total heat produced, 25 to 30% is utilized for power output; 15 to 20% is lost in cooling (heat radiated from cylinder head fins) ; 5 to 10% is lost in overcoming friction of moving parts; and 40 to 45% is lost through the exhaust. Anything which increases the heat content that goes into mechanical work on the piston, which reduces the friction and pumping losses, or which reduces the quantity of unburned fuel or the heat lost to the engine parts, increases the thermal efficiency.
The portion of the total heat of combustion which is turned into mechanical work depends to a great extent upon the compression ratio. Compression ratio is the ratio of the piston displacement plus combustion chamber space to the combustion chamber space. Other things being equal, the higher the compression ratio, the larger is the proportion of the heat energy of combustion turned into useful work at the crankshaft. On the other hand, increasing the compression ratio increases the cylinder head temperature. This is a limiting factor, for the extremely high temperature created by high compression ratios causes the material in the cylinder to deterioriate rapidly and the fuel to detonate.
The thermal efficiency of an engine may be based on either b.hp. or i.hp. and is represented by the formula:
Indicated thermal efficiency =
1.hp. × 33,000 weight of fuel burned/ min. x heat value x 778 The formula for brake thermal efficiency is the same as shown above, except the value for b.hp. is inserted instead of the value for i.hp.
Example An engine delivers 85 b.hp. for a period of 1 hr.
and during that time consumes 50 Ibs. of fuel.
Assuming the fuel has a heat content of 18,800 B.t.u.
per Ib., find the thermal efficiency of the engine:
85 × 33,000 2,805,000 833 × 18,800 × 778 - 12,184,569 Brake thermal efficiency = 0.23 or 23%.
Reciprocating engines are only about 34% thermally efficient; that is, they transform only about 34% of the total heat produced by the burning fuel into mechanical energy. The remainder of the heat is lost through the exhaust gases, the cooling system, and the friction within the engine. Thermal distribution in a reciprocating engine is illustrated in figure 1-31.
Mechanical Efficiency Mechanical efficiency is the ratio that shows how much of the power developed by the expanding gases in the cylinder is actually delivered to the output shaft. It is a comparison between the b.hp.
and the i.hp. It can be expressed by the formula:
Mechanical uhciency = hp Brake horsepower is the useful power delivered to the propeller shaft. Indicated horsepower is the total hp. developed in the cylinders. The difference between the two is f.hp. (friction horsepower), the power lost in overcoming friction.
The factor that has the greatest effect on mechanical efficiency is the friction within the engine The friction between moving parts in an engine remains practically constant throughout an engine's speed range. Therefore, the mechanical efficiency of an engine will be highest when the engine is running at the r.p.m. at which maximum
b.hp. is developed. Mechanical efficiency of the average aircratt reciprocating engine approaches 90%.
Heat released by combustion.
40-45% is carried 15-20% is removed out with exhaust. by fins.
25-30% is converted 5-10% is removed into usetul power. by the oil.
FIGURE 1-31. Thermal distribution in an engine.
Volumetric Efficiency Volumetric efficiency, another engine efficiency, is a ratio expressed in terms of percentages. It is a comparison of the volume of fuel/air charge (corrected for temperature and pressure) inducted into the cylinders to the total piston displacement of the engine. Various factors cause departure from a 100% volumetric efficiency.
The pistons of an unsupercharged engine displace the same volume each time they sweep the cylinders from top center to bottom center. The amount of charge that fills this volume on the intake stroke depends on the existing pressure and temperature of the surrounding atmosphere. Therefore, to find the volumetric efficiency of an engine, standards for atmospheric pressure and temperature had to be established. The U.S. standard atmosphere was established in 1958 and provides the necessary pressure and temperature values to calculate volumetric efficiency.
The standard sea-level temperature is 59° F. or At this temperature the pressure of one atmosphere is 14.69 1bs./ sq. in., and this pressure will support a column of mercury 29.92 in. high.
These standard sea-level conditions determine a standard density, and if the engine draws in a volume of charge of this density exactly equal to its piston displacement, it is said to be operating at 100% volumetric efficiency. An engine drawing in less volume than this has a volumetric efficiency lower than 100%. An engine equipped with a high-speed internal or external blower may have a volumetric efficiency greater than 100%. The equation for volumetric efficiency is:
Volumetric = - Volume of charge (corrected for temperature and pressure) efficiency Piston displacement Many factors decrease volumetric efficiency; some of these are:
(1) Part-throttle operation.
(2) Long intake pipes of small diameter.
(3) Sharp bends in the induction system.
(4) Carburetor air temperature too high.
(5) Cylinder-head temperature too high.
(6) Incomplete scavenging.
(7) Improper valve timing.
Propulsive Efficiency A propeller is used with an engine to provide The engine supplies b.hp. through a rotating shaft, and the propeller absorbs the b.hp. and converts it into thrust hp. In this conversion, some power is wasted. Since the efficiency of any machine is the ratio of useful power output to the power input, propulsive efficiency (in this case, propeller efficiency) is the ratio of thrust hp. to
b.hp. On the average, thrust hp. constitutes approximately 80% of the b.hp. The other 20% is lost in friction and slippage. Controlling the blade angle of the propeller is the best method of obtaining maximum propulsive efficiency for all conditions encountered in flight.
During takeoff, when the aircraft is moving at low speeds and when maximum power and thrust are required, a low propeller blade angle will give maximum thrust. For high-speed flying or diving, the blade angle is increased to obtain maximum thrust and efficiency. The constant-speed propeller is used to give required thrust at maximum efficiency for all flight conditions.
TURBINE ENGINE CONSTRUCTION In a reciprocating engine the functions of intake, compression, combustion, and exhaust all take place in the same combustion chamber; consequently, each must have exclusive occupancy of the chamber during its respective part of the combustion cycle.
A significant feature of the gas turbine engine, however, is that a separate section is devoted to each 38
function, and all functions are performed simultane- the component parts of various engines currently ously without interruption. in use will vary slightly due to the difference in each A typical gas turbine engine consists of: manufacturer's terminology. These differences are
(1) An air inlet. reflected in the applicable maintenance manuals.
(2) Compressor section. The greatest single factor influencing the con-
(3) Combustion section. struction features of any gas turbine engine is the
(4) Turbine section. type compressor (axial flow or centrifugal flow)
(5) Exhaust section. for which the engine is designed. Later in the chap.
(6) Accessory section.
(7) The systems necessary for starting, lubrication, fuel supply, and auxiliary purposes, such as anti-icing, cooling, and pressurization.
ter a detailed description of compressors is given, but for the time being examine figures 1-32 and
1-33. Notice the physical effect the two types of compressors have on engine construction features.
It is obvious that there is a difference in their length The major components of all turbine engines are and diameter.
basically the same; however, the nomenclature of Note that in the axial-flow engine the air inlet Engine oil supply Engine mount fitting Turbo compressor
- Stainless steel fire seal Accessory case Air inlet FIGURE 1-32. Axial-flow engine.
Accessory Combustion case Compressor chamber Exhaust duct Air inlet Diffuser Turbine FIGURE 1-33. Centrifugal-How engine.
duct is one of the major engine components; on the other hand, in the centrifugal-flow engine, air enters the air inlet and is directed to the compressor inducer vanes through circumferential inlets located in front and back of the impeller. The inlets are screened to prevent entry of foreign objects that could cause serious damage to the metal components if allowed to enter the compressor.
The accessories of the two types of engines are located at different points on the engines. This is necessary because of engine construction. The front of the axial-flow engine is utilized for air entrance; consequently, the accessories must be located elsewhere.
Other than the features previously mentioned, there is little visual dissimilarity between the remaining major components of the two engines.
AIR ENTRANCE The air entrance is designed to conduct incoming air to the compressor with a minimum energy loss resulting from drag or ram pressure loss; that is, the flow of air into the compressor should be free of turbulence to achieve maximum operating efficiency. Proper design contributes materially to aircraft performance by increasing the ratio of compressor discharge pressure to duct inlet pressure.
The amount of air passing through the engine is dependent upon three factors:
(1) The compressor speed (r.p.m.).
(2) The forward speed of the aircraft.
(3) The density of the ambient (surrounding) air.
Inlets may be classified as:
(1) Nose inlets, located in the nose of the fuselage, or powerplant pod or nacelle.
(2) Wing inlets, located along the leading edge of the wing, usually at the root for singleengine installations.
(3) Annular inlets, encircling, in whole or in part, the fuselage or powerplant pod or nacelle.
(4) Scoop inlets, which project beyond the immediate surface of the fuselage or nacelle.
(5) Flush inlets, which are recessed in the side of the fuselage, powerplant pod, or nacelle.
There are two basic types of air entrances in use:
the single entrance and the divided entrance. Generally, it is advantageous to use a single entrance with an axial-flow engine to obtain maximum ram pressure through straight flow. It is used almost exclusively on wing or external installations where the unobstructed entrance lends itself readily to a single, short, straight duct.
A divided entrance offers greater opportunity to diffuse the incoming air and enter the plenum chamber with the low velocity required to utilize efficiently a double-entry compressor. (The plenum chamber is a storage place for ram air, usually asso- Oil cooler Accessory gearbox drive air outlet Oil pressure Igniter plug transmitter socket Starter motor terminals cowling de-icing socket Thermocouple-harness junction box fuel drain Overboard Hot air gate Cooling valve outlet air outlet FIGURE 1-34. Accessory location on a centrifugal-flow engine.