PEDAGOGÍA PARA LA FORMACIÓN EN VALORES EN LA MAESTRÍA EN DOCENCIA DE LA UNIVERSIDAD DE LA SALLE
4. MARCO TEÓRICO
4.3. RELACIÓN ENTRE ÉTICA Y MORAL
Thrust force is the force responsible for propelling the aircraft in its different flight regimes.
It is in addition to the lift, drag, and weight represent the four forces that govern the aircraft motion. During the cruise phase of flight, where the aircraft is flying steadily at a constant speed and altitude, each parallel pair of the four forces are in equilibrium (lift and weight as well as thrust and drag). During landing, thrust force is either fully or partially used in braking of the aircraft through a thrust reversing mechanism. The basic conservation laws of mass and momentum are used in their integral forms to derive an expression for thrust force.
As described in example (2.1) and Fig.2.4, the thrust generated by a turbojet is given by the relation:
T¼ _ma½ð1þ fÞue u þ Pð e PaÞAe ð3:1Þ where
Net thrust¼ T
The other types of thrusts are
Gross thrust¼ _ma½ð1þ fÞue þ Pð e PaÞAe
Momentum thrust¼ _ma½ð1þ fÞue Pressure thrust¼ (Pe Pa)Ae
Momentum drag¼ _mau
Thus: Net thrust¼ Gross thrust – Momentum drag
Or in other words, Net thrust¼ Momentum thrust + Pressure thrust – Momentum drag
If the nozzle is unchoked, thenPe¼ Pa, the pressure thrust cancels in Eq. (3.1).
The thrust is then expressed as
T¼ _ma½ð1þ fÞue u ð3:2Þ
In many cases the fuel to air ratio is negligible, thus the thrust force equation is reduced to the simple form:
T ¼ _maðue uÞ ð3:3Þ
The thrust force in turbojet engine attains high values as the exhaust speed is high and much greater than the flight speed, or:ue/u 1.
In a similar way, the thrust force for two stream engines liketurbofan (Fig.3.1) andprop fan engines can be derived. It will be expressed as
T¼ _mh½ð1þ fÞueh u þ _mcðuec uÞ þ AehðPeh PaÞ þ AecðPec PaÞ ð3:4Þ where
f ¼ _m_mf
h: fuel to air ratio
_mh: Air mass flow passing through the hot section of engine; turbine(s) _meh¼ _mhð1þ fÞ : Mass of hot gases leaving the engine
_mc: Air mass flow passing through the fan
ueh: Velocity of hot gases leaving the turbine nozzle uec: Velocity of cold air leaving the fan nozzle Peh: Exhaust pressure of the hot stream Pec: Exhaust pressure of the cold stream Aeh: Exit area for the hot stream Aec: Exit area for the cold stream
Thespecific thrust is defined as the thrust per unit air mass flow rate (T= _ma), which can be obtained from Eq. (3.4). It has the dimensions of a velocity (say m/s).
For turboprop engines (Fig.3.2), the high value of thrust is achieved by the very large quantity of the airflow rate, though the exhaust and flight speeds are very close. An analogous formula to Eq. (3.4) may be employed as follows:
T¼ _mc½ð1þ fÞue u1 þ _m0ðu1 u0Þ ð3:5Þ Fig. 3.1 An unmixed two-spool turbofan engine
where _m0is the air mass flow sucked by the propeller, while _mcis a part of the air flow crossed the propeller and then entered the engine through its intake. Here,u0, u1, andueare air speed upstream and downstream the propeller and gases speed at the engine exhaust. The exhaust nozzle is normally unchoked.
Example 3.1 Air flows through a turbojet engine at the rate of 50.0 kg/s and the fuel flow rate is 1.0 kg/s. The exhaust gases leave the jet nozzle with a relative velocity of 600 m/s. Compute the velocity of the airplane, if the thrust power is 1.5 MW in the following two cases:
1. Pressure equilibrium exists over the exit plane 2. If the pressure thrust is 8 kN
Solution
1. When the nozzle is unchoked, pressure equilibrium exists over the exit plane.
Then, thrust force is expressed as T¼ _maþ _mf
ue _mau Thrust power¼ T u Thrust power¼ _maþ _mf
ueu _mau2 1:5 106 ¼ 51ð Þ 600ð Þ u 50 u2 50u2 30, 600 u þ 1:5 106¼ 0 Fig. 3.2 Turboprop engine
or u¼30, 600 103 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 936:36 300 p
100 Thus, eitheru¼ 558.26 m/s or u ¼ 53.74 m/s
2. When the exit pressure is greater than the ambient pressure, a pressure thrust (Tp) is generated. The thrust equation with pressure thrust is then
T¼ _maþ _mf
ue _mauþ Tp
Thus, the thrust power is T u ¼ _maþ _mf
Example 3.2 A fighter airplane is powered by two turbojet engines. It has the following characteristics during cruise flight conditions:
Wing area (S)¼ 49.24 m2 Engine inlet areaAi¼ 0.06 m2 Cruise speedVf¼ 243 m/s Flight altitude¼ 35,000 ft
Drag and lift coefficients areCD¼ 0.045, CL¼ 15 CD
Exhaust total temperatureT0¼ 1005 K
Specific heat ratio and specific heat at exit areγ ¼ 1.3, Cp¼ 1100 J/(kgK) It is required to calculate:
1. Net thrust 2. Gross thrust 3. Weight
4. Jet speed assuming exhaust pressure is equal to ambient pressure ifPe¼ Pa
5. Static temperature of exhaust Te 6. Exhaust Mach number Me
Solution
At 35,000 m altitude, the properties of ambient conditions are
TemperatureT¼ 54.3C, pressureP¼ 23.84 kPa, and density 0.3798 kg/m3 The mass flow rate is _m ¼ ρaVfAi¼ 0:3798 243 0:6 ¼ 55:375 kg=m3
1. During cruise flight segment, the thrust and drag force (D) are equal. Thus for two engines and (T ) is the net thrust of each engine, then
2T¼ D ¼ ρV2ACD=2
T ¼ 0:3798 243ð Þ2 49:24 0:045=4 ¼ 12, 423 N ¼ 12:423 kN 2. Gross thrust¼ Net thrust + Ram drag
Tgross ¼ T þ _m Vf ¼ 12, 423 þ 55:3 243 ¼ 25, 879 N ¼ 25:879 kN 3. Since Weight¼ Lift, thus L ¼ W.
Moreover, lift and drag are correlated by the relation:
CL¼ 15 CD ¼ 0:675
L¼ W ¼ 15D ¼ 30T ¼ 37, 2690 N ¼ 372:69 kN
4. Assuming negligible fuel flow ratio, and sincePe¼ Pa, then the net thrust is expressed by the relation:
T¼ _m VJ Vf
VJ¼ VfþT
_m ¼ 243 þ12, 423
55:375¼ 467:3 m=s 5. Exhaust static temperature is expressed by the relation:
Te¼ T0e V2j 2Cp
¼ 1005 ð467Þ2
2 1100¼ 905:9 K 6. Sonic speed at exitae¼ ffiffiffiffiffiffiffiffi
pγRT
¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:3 287 905:7
p ¼ 581:3 m=s
Exhaust Mach number isMe¼Vaj
e¼ 0:804
Example 3.3 It is required to calculate and plot the momentum drag as well as momentum, pressure, gross, and net thrusts versus the flight speed for a turbojet engine powering an aircraft flying at 9 km (ambient temperature and pressure are 229.74 K and 30.8 kPa) and having the following characteristics,Ai¼ 0.24 m2, Ae¼ 0.26 m2,f¼ 0.02, Ue¼ 600 m/s, Pe¼ 87.50 kPa.
The flight speed varies from 500 to 4000 km/h. Consider the following cases:
The air mass flow rate is constant and equal to 40 kg/s irrespective of the variation of flight speed.
A. Air mass flow rate varies with the flight speed
B. Repeat the above procedure for altitudes 3, 6, and 12 km considering a variable air mass flow rate and a constant exhaust pressure ofPe¼ 87.50 kPa
C. Repeat the above procedure for altitudes 3, 6, 9, and 12 km considering a variable air mass and a variable exhaust pressure given by the relation:
Pe/Palt¼ 1.25
Solution
A. The mass flow rate is constant and equal to 40 kg/s at altitude 9 km The momentum thrust (Tmomentum) is constant and given by the relation
Tmomentum¼ _mað1þ fÞUe¼ 40 1:02 600 ¼ 24, 480 N The pressure thrust (Tpressure) is also constant and calculated as
Tpressure¼ Ae Pð e PaÞ ¼ 0:26 87:5 30:8ð Þ 103¼ 14, 742 N The gross thrust (Tgross) is constant and equal to the sum of momentum and pressure thrusts
Tgross¼ Tmomentumþ Tpressure¼ 39, 222 N
The momentum drag for flight speed varying from 500 to 4000 km/h is given by the relation:
Dmomentum¼ _maU¼ 40 kg=sð Þ U km=hrð Þ
3:6 ¼ 11:11 U Nð Þ
It is a linear relation in the flight speedU. Flight speed in momentum drag relation and continuity equation will be substituted in km/h.
The net thrust is then (Fig.3.3)
Tnet¼ Tgross Dmomentum¼ 39, 222 11:11 U Nð Þ
The net thrust varies linearly with the flight speed. The results are plotted in Fig.3.4. The net thrust must be greater than the total aircraft drag force during acceleration and equal to the drag at steady cruise flight. Zero net thrust results from the intersection of the gross thrust and ram drag. The flight speed corresponding to zero net thrust represents the maximum possible aircraft’s speed.
Fig. 3.3 Thrust components and drag with variable flight speed
Fig. 3.4 Thrust variations with constant mass flow rate
B. Variable air mass flow rate at altitude 9 km
The mass flow rate varies linearly with the flight speed according to the relation:
_ma¼ ρaUAi¼ Pa
The momentum thrust varies linearly with the flight speed as per the relation Tmomentum¼ _mað1þ fÞUe¼ 0:031141 U 1:02 600 ¼ 19:058 U Nð Þ The pressure thrust is constant and has the same value as in case (1)
Tpressure¼ Ae Pð e PaÞ ¼ 0:26 87:5 30:8ð Þ 103¼ 14, 742 N The gross thrust is varying linearly with the flight speed
Tgross¼ Tmomentumþ Tpressure¼ 19:058 U þ 14742 Nð Þ
The momentum drag for flight speed varying from 500 to 4000 km/h is given by the quadratic relation:
Dmomentum¼ _maU¼ 0:031141 U kg=sð Þ U km=hrð Þ
3:6 ¼ 8:65 103 U2 ð ÞN The net thrust is then
Tnet¼ Tgross Dmomentum¼ 19:058 U þ 14, 742 8:65 103 U2 ð ÞN The above relations are plotted in Fig.3.5.
C. Variable mass flow rate at altitudes 3, 6, and 12 km and constant exhaust pressure of Pe¼ 87.50 kPa
Air mass flow rate varies linearly with the flight speed according to the relation:
_ma¼ ρaUAi¼ Palt
The momentum thrust varies linearly with the flight speed as per the relation Tmomentum¼ _mað1þ fÞUe¼ 2:32 104Palt
The pressure thrust is varying with altitude pressure (refer to Table3.1):
Tpressure¼ Ae Pð e PaltÞ ¼ 0:26 87:5 Pð altÞ 103ð ÞN The gross thrust is varying linearly with the flight speed
Tgross¼ Tmomentumþ Tpressure
¼ 0:142 Palt
Talt
U þ 0:26 87:5 Pð altÞ 103ð ÞN
The momentum drag for flight speed varying from 500 to 4000 km/h is given by the quadratic relation:
Dmomentum¼ _maU
¼ 2:32 104
Palt
Talt
U kg=sð Þ U km=hrð Þ
3:6 ¼ 6:44 105Palt
Talt
U2ð ÞN Fig. 3.5 Variations of net thrust with variable mass flow rate
Table 3.1 Values of pressures and temperatures at different altitudes
Altitude (km) Pressure (kPa) Temperature (K)
3 70.122 268.66
6 47.2 249.16
9 30.762 229.66
12 19.344 216.66
The net thrust is then Tnet¼ Tgross Dmomentum
¼ 0:142 Palt
Talt
U þ 0:26 87:5 Pð altÞ 103 6:44 105Palt
Talt
U2 ð ÞN
The above relation is plotted in Fig.3.6.
D. Variable mass flow rate at altitudes 3, 6, 9, and 12 km and variable exhaust pressure based on the relation Pe/Palt¼ 1.25
The net thrust is expressed by the relation:
Tnet¼ Tgross Dmomentum
¼ 0:142 Paltð ÞPa
Taltð ÞK U þ 0:26 Pð e PaltÞ 6:44 105Paltð ÞPa
Taltð ÞK U2 ð ÞN Tnet¼ 0:142 Paltð ÞPa
Taltð ÞK U þ 0:26 0:25 Paltð Þ 6:44 10Pa 5Paltð ÞPa Taltð ÞK
U2 ð ÞN
Fig. 3.6 Net thrust variations with variable mass flow rate at different altitudes
For the case of altitude 3 km and flight speed of 600 km/h, then
Tnet¼ 0:142 70 103
268 600 þ 0:26 0:25 70 103ð Þ 6:44 10Pa 5
70 103
268 600ð Þ2 ð ÞN
Tnet¼ 22, 253 þ 4550 6055 ¼ 20, 748 N ¼ 2:0748 104 N
Figure3.7illustrates the positive net thrust for different flight speeds and altitudes of 3, 6, 9, and 12 km. It is clarified that the maximum possible flight speed for such an aircraft is nearly 2300 km/h.
It is interesting here to calculate the flight speed that provides a maximum thrust, which is obtained from the relation:∂T∂Unet¼ 0
Since
Tnet¼ 0:142 Paltð ÞPa
Taltð ÞK U þ 0:26 0:25 Paltð Þ 6:44 10Pa 5Paltð ÞPa Taltð ÞK
U2 ð ÞN
Then ∂Tnet
∂U ¼ 0:142 Paltð ÞPa
Taltð ÞK 2 6:44 105Paltð ÞPa
Taltð ÞK U ¼ 0 Fig. 3.7 Thrust variations with variable mass flow rate at different altitudes with pressure ratio at exit equals 1.25
Thus net thrust attains a maximum value at all altitudes when the flight speed is U¼ 0:142
2 6:44 105¼ 1102:5 km=hr The above value is also clear in Fig.3.7.
Example 3.4 A high bypass ratio turbofan engine is powering a civil transport aircraft flying at an altitude 11 km with a speed of 1100 km/h. The total air mass flow rate is 120 kg/s and the bypass ratio is 5.0. Exhaust speeds for the cold and hot streams are, respectively, 1460 and 2000 km/h. Both cold and hot nozzles are unchoked. Calculate the thrust force (fuel-to-air ratio is 0.012).
Solution
From Eq. (3.4), the thrust force for unchoked nozzles is T¼ _mh½ð1þ fÞueh u þ _mcðuec uÞ
Since the bypass ratioβ ¼ 5.0 and the total air mass flow is 120.0 kg/s, then _mc¼ β
1þ β _ma¼ 5
1þ 5 120 ¼ 100 kg=s
and _mh¼ 1
1þ β _ma¼ 1
1þ 5 120 ¼ 20 kg=s Then the thrust force is
T¼ 20 1 þ 0:012f ½ð Þ2000 1100 þ 100 1460 1100ð Þg 1000=3600ð Þ T¼ 15, 133 N ¼ 15:133 kN
Example 3.5 The thrust of a ramjet engine (single exhaust stream athodyd aero-engine) is expressed by the relation:
T¼ _mafð1þ fÞue ug where the exhaust speed is expressed by the relation:
ue¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2CpT0maxh1 Pð a=P0maxÞγ1γ i r
It is required to examine the effect of maximum temperatureT0maxon thrust force, by considering the following case: _ma¼ 100 kg=s, u ¼ 250 m/s, (Pa/P0max¼ 0.125), Cp¼ 1148 J/(kg.K), γ ¼ 4/3, f ¼ 0.015, and T0max¼ 1000, 1200, 1400, 1600 K
Solution
Since the exhaust speed is given by the relation:
ue¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2CpT0maxh1 Pð a=P0maxÞγ1γ i r
Then, from the above given data:
ue¼
Moreover, the thrust force is then expressed by the relation T¼ _mafð1þ fÞue ug ¼ 100 1:015½ð Þue 250
Substituting for the different values of maximum temperature, we get the following tabulated results.
It is clear from Table3.2that keeping a constant ratio between the maximum and ambient pressures, then increasing the maximum total temperature will increase the generated thrust.