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LAB 5: SEP of an Avro RJ85 Eline Heerze (elihe376) , Pablo Herrero (pabhe590) TMAL02, Linköping University, 2018 1
Task 1: Introduction
3
In this last lab, we have computed the SEP curve for our aircraft. In order to do it, Matlab scripts of previous lab have been used like the model of the ISA atmosphere and the drag computation. The specific excess power expresses the rate of climb of the aircraft, so finding out how it changes with altitude and velocity is the aim of this lab. This plot gives an idea of the performance of the aircraft over its flight envelope as well as the several limits that are present in the flight like the absolute ceiling or the structural limit. 2
Task 2: SEP Plot
Compute the SEP of your aircraft and plot the results in [m/s] on a contour diagram (altitude [m] against Mach number) In Figure 1 the SEP can be seen. Here only thrust requirements are fulfilled, in the next graphic aerodynamic, structural, and operational limitations will be added. Specific Excess Power
15000
0
Task 3: Limitations
Add the following limitations on your diagram: Maximum dynamic pressure, Maximum Mach number, maximum standard airspace speed limitation of 250 [knots] from ground level to 3000 [m] altitude, Service ceiling with a climb rate margin of 500 [ft/min], CL max for landing and cruise and CL for cruise phase. In Figure 3 the SEP diagram is shown with the limitations stated before. Maximum dynamic pressure has been set to a value of 11,772 [N/m2] found for similar jet aircraft [1]. For the maximum Mach number the divergence Mach number has been selected (Ma = 0.73). Regarding the stall limitations we have used the cruise lift coefficient with a value of 1.41, and not the takeoff coefficient, since it was higher than the landing configuration one with a value of 3.38. These coefficients have been obtained using the stall speed (showed in Figure 2 of the aircraft provided in the operational file from the database BADA of Eurocontrol. The following equation has been used to obtain the values for the lift coefficients:
2.53
10000
0
0
Altitude [m]
5 53 2. 5 10
10
0.3
0.4
0
15
20
10
5 2.53
25
5
20
2.53
15
5000
0 0
0.1
0.2
0.5
0.6
0.7
0.8
0.9
Fig. 2 . Operational file from BADA Eurocontrol database
1
Mach [-]
Fig. 1 . Specific excess power in [m/s]
CL = It is important to mention, that the absolute obtained absolute ceiling value obtained is a little bit higher than the actual value (12,727 ft versus 10,644 ft), this could be due to the thrust model used or a low drag estimation from the actual drag.
4
2 ∗ MT OW ∗ g 2 Sre f ∗ ρ ∗Vstall
(1)
Task 4
What would the time-optimal climb path be within the valid region of your diagram? Plot it also on your SEP diagram.
1
elihe376, pabhe590
stackexchange.com/questions/12988/ which-aircraft-endure-the-highest-max-qs-aerodynamic-pre
SEP with R/C in m/s 15000 SEP Mach limit High lift config. cruise congif. Standard airspeed limit Max. dynamic pressure Profile max SEP
0
2.53 5
3 2.5
Altitude [m]
0
10000
5
2.5
3
10
10
5000
15
10
5
15
20
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Mach Number [-]
Fig. 3 . Specific excess power in [m/s] with limitations
The time optimal path is the path were the SEP is the largest for each altitude. See the black line in Figure 3 for the time-optimal path. 5
Task 5
Calculate the total time that would take to climb from the ground to the service ceiling specified above. The total time to climb from the ground to the ceiling service of 500 [ft/min] (2.53 [m/s]) is found to be 27.49 minutes. 6
Task 6
Calculate now the total time to climb to a service ceiling of 100 [ft/min]. How much does it differ from the previous time? Why? To climb to a service ceiling of 100 [ft/min] or equivalently 0.51 [m/s] a time of 41.50 minutes is needed. Is it much higher than the previous time. This is because we need to climb to a lower service ceiling, and the higher we get the lower the rate of climb is. As a merely educational purpose if we compute the needed time for climbing to the absolute ceiling where the aircraft is not able to have any vertical velocity we get a value superior of two hours. Authenticity and Plagiarism By submitting this report, the author(s) listed above declare that this document is exclusively product of their own genuine work, and that they have not plagiarized or taken advantage from any other student’s work. If any irregularity is detected, the case will be forwarded to the University Disciplinary Board.
REFERENCES
[1] Aviation S. Maximum dynamic values for several aircraft, 2015. URL
https://aviation.
2
Task 1: Introduction
3
In this last lab, we have computed the SEP curve for our aircraft. In order to do it, Matlab scripts of previous lab have been used like the model of the ISA atmosphere and the drag computation. The specific excess power expresses the rate of climb of the aircraft, so finding out how it changes with altitude and velocity is the aim of this lab. This plot gives an idea of the performance of the aircraft over its flight envelope as well as the several limits that are present in the flight like the absolute ceiling or the structural limit. 2
Task 2: SEP Plot
Compute the SEP of your aircraft and plot the results in [m/s] on a contour diagram (altitude [m] against Mach number) In Figure 1 the SEP can be seen. Here only thrust requirements are fulfilled, in the next graphic aerodynamic, structural, and operational limitations will be added. Specific Excess Power
15000
0
Task 3: Limitations
Add the following limitations on your diagram: Maximum dynamic pressure, Maximum Mach number, maximum standard airspace speed limitation of 250 [knots] from ground level to 3000 [m] altitude, Service ceiling with a climb rate margin of 500 [ft/min], CL max for landing and cruise and CL for cruise phase. In Figure 3 the SEP diagram is shown with the limitations stated before. Maximum dynamic pressure has been set to a value of 11,772 [N/m2] found for similar jet aircraft [1]. For the maximum Mach number the divergence Mach number has been selected (Ma = 0.73). Regarding the stall limitations we have used the cruise lift coefficient with a value of 1.41, and not the takeoff coefficient, since it was higher than the landing configuration one with a value of 3.38. These coefficients have been obtained using the stall speed (showed in Figure 2 of the aircraft provided in the operational file from the database BADA of Eurocontrol. The following equation has been used to obtain the values for the lift coefficients:
2.53
10000
0
0
Altitude [m]
5 53 2. 5 10
10
0.3
0.4
0
15
20
10
5 2.53
25
5
20
2.53
15
5000
0 0
0.1
0.2
0.5
0.6
0.7
0.8
0.9
Fig. 2 . Operational file from BADA Eurocontrol database
1
Mach [-]
Fig. 1 . Specific excess power in [m/s]
CL = It is important to mention, that the absolute obtained absolute ceiling value obtained is a little bit higher than the actual value (12,727 ft versus 10,644 ft), this could be due to the thrust model used or a low drag estimation from the actual drag.
4
2 ∗ MT OW ∗ g 2 Sre f ∗ ρ ∗Vstall
(1)
Task 4
What would the time-optimal climb path be within the valid region of your diagram? Plot it also on your SEP diagram.
1
elihe376, pabhe590
stackexchange.com/questions/12988/ which-aircraft-endure-the-highest-max-qs-aerodynamic-pre
SEP with R/C in m/s 15000 SEP Mach limit High lift config. cruise congif. Standard airspeed limit Max. dynamic pressure Profile max SEP
0
2.53 5
3 2.5
Altitude [m]
0
10000
5
2.5
3
10
10
5000
15
10
5
15
20
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Mach Number [-]
Fig. 3 . Specific excess power in [m/s] with limitations
The time optimal path is the path were the SEP is the largest for each altitude. See the black line in Figure 3 for the time-optimal path. 5
Task 5
Calculate the total time that would take to climb from the ground to the service ceiling specified above. The total time to climb from the ground to the ceiling service of 500 [ft/min] (2.53 [m/s]) is found to be 27.49 minutes. 6
Task 6
Calculate now the total time to climb to a service ceiling of 100 [ft/min]. How much does it differ from the previous time? Why? To climb to a service ceiling of 100 [ft/min] or equivalently 0.51 [m/s] a time of 41.50 minutes is needed. Is it much higher than the previous time. This is because we need to climb to a lower service ceiling, and the higher we get the lower the rate of climb is. As a merely educational purpose if we compute the needed time for climbing to the absolute ceiling where the aircraft is not able to have any vertical velocity we get a value superior of two hours. Authenticity and Plagiarism By submitting this report, the author(s) listed above declare that this document is exclusively product of their own genuine work, and that they have not plagiarized or taken advantage from any other student’s work. If any irregularity is detected, the case will be forwarded to the University Disciplinary Board.
REFERENCES
[1] Aviation S. Maximum dynamic values for several aircraft, 2015. URL
https://aviation.
2
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