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:
T¼m_a½ð1þfÞue%u* þðPe%PaÞAe ð3:1Þ where
Netthrust¼T
The other types of thrusts are
Grossthrust¼ m_a½ð1þfÞue* þðPe%PaÞAe
Momentumthrust¼ m_a½ð1þfÞue* Pressurethrust¼(Pe%Pa)Ae Momentum drag¼ m_au
Thus: Net thrust¼Gross thrust – Momentum drag
Or in other words, Net thrust¼Momentum thrust + Pressure thrust – Momentum drag
If the nozzle is unchoked, thenPe¼Pa, the pressure thrust cancels in Eq. (3.1).
The thrust is then expressed as
T¼m_a½ð1þfÞue%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 ¼m_aðue%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/u01.
In a similar way, the thrust force for two stream engines liketurbofan(Fig.3.1) andprop fanengines can be derived. It will be expressed as
T¼m_h½ð1þfÞueh%u* þm_cðuec%uÞ þAehðPeh%PaÞ þAecðPec%PaÞ ð3:4Þ where
f ¼mm__fh: fuel to air ratio _
mh: Air mass flow passing through the hot section of engine; turbine(s) _
meh¼m_hð1þfÞ: 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 thrustis defined as the thrust per unit air mass flow rate (T=m_a), 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:
T¼m_c½ð1þfÞue%u1* þm_0ðu1%u0Þ ð3:5Þ Fig. 3.1 An unmixed two-spool turbofan engine
wherem_0is the air mass flow sucked by the propeller, whilem_cis 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 m_aþm_f
$ %
ue%m_au Thrust power¼T,u Thrust power¼ m_aþm_f
$ %
ueu%m_au2 1:5,106 ¼ð Þ51 ð600Þu%50u2 50u2%30, 600uþ1:5,106¼0 Fig. 3.2 Turboprop engine
or u¼30, 6001103 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 936:36%300 p
100 Thus, eitheru¼558.26 m/s oru¼53.74 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 m_aþm_f
$ %
ue%m_auþTp
Thus, the thrust power is T,u¼ m_aþm_f
$ %
ueu%m_au2þTp,u¼ m_aþm_f
$ %
ueþTp
' (
,u%m_au2 1:5,106¼½51,600þ8000* ,u%50u2¼38, 600u%50u2
50u2%38, 600uþ1:5,106¼0 u¼38, 6001103 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1490%300 p
100 ¼38, 600134, 495
100 Thus eitheru¼731 m/s or 41 m/s
Example 3.2 A fighter airplane is powered by two turbojet engines. It has the following characteristics during cruise flight conditions:
Wing area (S)¼49.24 m2 Engine inlet areaAi¼0.06 m2 Cruise speedVf¼243 m/s Flight altitude¼35,000 ft
Drag and lift coefficients areCD¼0.045,CL¼15CD Exhaust total temperatureT0¼1005K
Specific heat ratio and specific heat at exit areγ¼1.3,Cp¼1100 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 ifPe¼Pa
5. Static temperature of exhaust Te
6. Exhaust Mach number Me
Solution
At 35,000 m altitude, the properties of ambient conditions are
TemperatureT¼ %54.3!C, pressureP¼23.84 kPa, and density 0.3798 kg/m3 The mass flow rate ism_ ¼ρaVfAi¼0:3798,243,0:6¼55:375 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,ð243Þ2 ,49:24,0:045=4 ¼ 12, 423N ¼12:423 kN 2. Gross thrust¼Net thrust + Ram drag
Tgross ¼Tþm V_ f ¼12, 423þ55:3,243 ¼25, 879 N¼25:879 kN 3. Since Weight¼Lift, thusL¼W.
Moreover, lift and drag are correlated by the relation:
CL¼ 15CD ¼0:675
L¼W ¼15D¼30T¼37, 2690N ¼372:69 kN
4. Assuming negligible fuel flow ratio, and sincePe¼Pa, then the net thrust is expressed by the relation:
T¼m V_ J%Vf
$ %
VJ¼VfþT _
m ¼243þ12, 423
55:375¼467:3 m=s 5. Exhaust static temperature is expressed by the relation:
Te¼T0e% V2j
2Cp¼1005% ð467Þ2
2,1100¼905:9 K 6. Sonic speed at exitae¼ ffiffiffiffiffiffiffiffi
pγRT
¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:3,287,905:7
p ¼581:3 m=s
Exhaust Mach number isMe¼Vaje¼0:804
Example 3.3 It is required to calculate andplot the momentum drag as well as momentum, pressure, gross, and net thrusts versus the flight speedfor 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,Ai¼0.24 m2, Ae¼0.26 m2,f¼0.02,Ue¼600 m/s,Pe¼87.50 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 ofPe¼87.50 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/Palt¼1.25
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
Tmomentum¼m_að1þfÞUe¼40,1:02,600¼24, 480 N The pressure thrust (Tpressure) is also constant and calculated as
Tpressure¼Ae,ðPe%PaÞ ¼0:26,ð87:5%30:8Þ ,103¼14, 742 N The gross thrust (Tgross) is constant and equal to the sum of momentum and pressure thrusts
Tgross¼TmomentumþTpressure¼39, 222 N
The momentum drag for flight speed varying from 500 to 4000 km/h is given by the relation:
Dmomentum¼m_aU¼40 kg=sð Þ ,Uðkm=hrÞ
3:6 ¼11:11,U ð ÞN
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)
Tnet¼Tgross%Dmomentum¼39, 222%11:11,Uð ÞN
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:
_
ma¼ρaUAi¼ Pa
RTa
UAi¼ 30:8,103 287,229:74, U
3:6,0:24 kg=sð Þ _
ma¼0:031141,U ðkg=sÞ
The momentum thrust varies linearly with the flight speed as per the relation Tmomentum¼m_að1þfÞUe¼0:031141,U,1:02,600¼19:058,U ð ÞN The pressure thrust is constant and has the same value as in case (1)
Tpressure¼Ae,ðPe%PaÞ ¼0:26,ð87:5%30:8Þ ,103¼14, 742 N The gross thrust is varying linearly with the flight speed
Tgross¼TmomentumþTpressure¼19:058,Uþ14742 Nð Þ
The momentum drag for flight speed varying from 500 to 4000 km/h is given by the quadratic relation:
Dmomentum¼m_aU¼0:031141,Uðkg=sÞ ,Uðkm=hrÞ
3:6 ¼8:65,10%3,U2 ð ÞN The net thrust is then
Tnet¼Tgross%Dmomentum¼19:058,Uþ14, 742%8:65,10%3,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 Pe¼87.50 kPa
Air mass flow rate varies linearly with the flight speed according to the relation:
_
ma¼ρaUAi¼ Palt
RTalt
UAi¼ Palt
287,Talt, U 3:6,0:24
¼2:32,10%4,Palt
Talt
U ðkg=sÞ
The momentum thrust varies linearly with the flight speed as per the relation Tmomentum¼m_að1þfÞUe¼2:32,10%4,Palt
Talt
U,1:02,600
¼0:142,Palt
Talt,Uð ÞN
The pressure thrust is varying with altitude pressure (refer to Table3.1):
Tpressure¼Ae,ðPe%PaltÞ ¼0:26,ð87:5%PaltÞ ,103ð ÞN The gross thrust is varying linearly with the flight speed
Tgross¼TmomentumþTpressure
¼0:142,Palt
Talt,Uþ0:26,ð87:5%PaltÞ ,103ð ÞN
The momentum drag for flight speed varying from 500 to 4000 km/h is given by the quadratic relation:
Dmomentum¼m_aU
¼2:32,10%4 ,Palt
Talt
Uðkg=sÞ ,Uðkm=hrÞ
3:6 ¼6:44,10%5,Palt
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 Tnet¼Tgross%Dmomentum
¼0:142,Palt
Talt,Uþ0:26,ð87:5%PaltÞ ,103%6:44,10%5,Palt
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/Palt¼1.25
The net thrust is expressed by the relation:
Tnet¼Tgross%Dmomentum
¼0:142,Paltð ÞPa
Taltð ÞK ,Uþ0:26,ðPe%PaltÞ %6:44,10%5,Paltð ÞPa
Taltð ÞK ,U2 ð ÞN Tnet¼0:142,Paltð ÞPa
Taltð ÞK ,Uþ0:26,0:25,Paltð Þ %Pa 6:44,10%5,Paltð 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 Tnet¼0:142,70,103
268 ,600þ0:26,0:25,70,103ð Þ %Pa 6:44,10%5 ,70,103
268 ,ð600Þ2 ð ÞN
Tnet¼22, 253þ4550%6055¼20, 748 N¼2:0748,104 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∂Unet¼0
Since
Tnet¼0:142,Paltð ÞPa
Taltð ÞK ,Uþ0:26,0:25,Paltð Þ %Pa 6:44,10%5,Paltð ÞPa Taltð ÞK ,U2 ð ÞN
Then ∂Tnet
∂U ¼0:142,Paltð ÞPa
Taltð ÞK %2,6:44,10%5,Paltð ÞPa
Taltð ÞK ,U¼0 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
2,6:44,10%5¼1102:5 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 T¼m_h½ð1þfÞueh%u* þm_cðuec%uÞ
Since the bypass ratioβ¼5.0 and the total air mass flow is 120.0 kg/s, then _
mc¼ β
1þβm_a¼ 5
1þ5,120¼100 kg=s and m_h¼ 1
1þβm_a¼ 1
1þ5,120¼20 kg=s Then the thrust force is
T¼f20 1½ð þ0:012Þ2000%1100* þ100 1460ð %1100Þg,ð1000=3600Þ T¼15, 133 N¼15:133 kN
Example 3.5 The thrust of a ramjet engine (single exhaust stream athodyd aero- engine) is expressed by the relation:
T¼m_afð1þfÞue%ug where the exhaust speed is expressed by the relation:
ue¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2CpT0maxh1%ðPa=P0maxÞγ%γ1i r
It is required to examine the effect of maximum temperatureT0maxon thrust force, by considering the following case: m_a¼100 kg=s, u¼250 m/s, (Pa/P0max¼0.125), Cp¼1148 J/(kg.K),γ¼4/3,f¼0.015, andT0max¼1000, 1200, 1400, 1600 K
Solution
Since the exhaust speed is given by the relation:
ue¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2CpT0maxh1%ðPa=P0maxÞγ%γ1i r
Then, from the above given data:
ue¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2CpT0maxh1%ðPa=P0maxÞγ%γ1i r
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2,1148,T0maxh1%ð0:125Þ0:25i r
¼30:5 ffiffiffiffiffiffiffiffiffiffiffi T0max
p
Moreover, the thrust force is then expressed by the relation T¼m_afð1þfÞue%ug¼100 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.