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1MAE 5540 – Propulsion Systems II
Homework 1.2 SolutionAssignment 1.2 Date Assigned: Wednesday September 4, 2019Date Due: Friday September 20, 2019Title: 2-D Method of Characteristics (M.O.C.) Grid Solver Development Num. of Points: 10 Ø Code and Verify subroutines or scripts for
• Initial Data Line along Expansion Section Wall • Internal Flow Unit Process• Centerline Intercept Unit Process (C- characteristic line)• Wall Intercept Unit Process (C+ chracteristic line)• Minimum Length Nozzle Maximum Turning Angle
Ø Link and Sequence Unit Process Modules to Calculate M.O.C Grid
2MAE 5540 – Propulsion Systems II
Homework 1.2 Solution (2)
Solve Problem 11.1 in Anderson, page 429. (See Section 1.1 Notes for Example)... Minimum Length Nozzle with Maximum Turning Angle -- infinitesimal expansion section... Mexit = 2.0... D* = 2.0 cm
... Assume Gamma = 1.4
... Repeat with Gamma = 1.2Solve Problem 11.2 in Anderson, Page 430 but with ...
... Finite expansion section radius of Curvature equal to 1.5 x throat radius
... Mexit = 2.0
... D* = 2.0 cm ... Assume Gamma = 1.4... Repeat with Gamma = 1.2
for all parts Plot nozzle half-contoursPlot nozzle Mach number profile along upper wall and along centerlineCompare to Mach number profile calculated using A/A* equation
3MAE 5540 – Propulsion Systems II
Homework 1.2 Solution (3)
For each of the 4 Nozzle Contours above à• Use the approximate Mapping technique to solve for P,Q,S, and T of the Parabolic
Contour• Plot Contours against the derived M.O.C contours, compare shapes• Use M.O.C. values for qmax, Nozzle length LN, and Radius of Curvature Rc of the
expansion section for these calculations• Be sure to show Calculations for Xn, Yn, P, Q, S, T, etc. • Assume Nozzle exit angle is zero for each case
4MAE 5540 – Propulsion Systems II
Solution1.2, part 1• Minimum Length Nozzle … g = 1.4• Let N=6
180π
1.4 1+1.4 1−# $% &
0.5 1.4 1−1.4 1+# $% & 2.02 1−( )# $% &
0.5
# $' (% &
atan 2.02 1−( )( )0.5
( )atan−# $' (% &
Mexit = 2.0→ν(Mexit ) =γ +1γ −1
tan−1 γ −1γ +1
2.02 −1( )%&'
('
)*'
+'− tan−1 2.02 −1 =
= 26.3798o
θw Max=νexit2
== 13.1899o
5MAE 5540 – Propulsion Systems II
Solution 1.2, part 1 (2)
Minimum Length à Rc expansion = 0
θw Max=νexit2
== 13.1899o
6MAE 5540 – Propulsion Systems II
Solution 1.2, part 1 (3)
7MAE 5540 – Propulsion Systems II
Solution 1.2, part 1 (4)
• Minimum Length Nozzle … g = 1.2• Let N=6
Mexit = 2.0→ν(Mexit ) =γ +1γ −1
tan−1 γ −1γ +1
2.02 −1( )%&'
('
)*'
+'− tan−1 2.02 −1 =
= 31.46o
θw Max=νexit2
= 15.731o
Maximium Turning Angle changes!
8MAE 5540 – Propulsion Systems II
Solution 1.2, part 1 (5)
Minimum Length à Rc expansion = 0
θw Max=νexit2
== 15.731o
9MAE 5540 – Propulsion Systems II
Solution 1.2, part 1 (6)
10MAE 5540 – Propulsion Systems II
Solution 1.2, Part 1 comparisonsθw Max
=νexit2
== 13.1899o
θw Max=νexit2
=15.731o
11MAE 5540 – Propulsion Systems II
Solution 1.2, part 2 (1)
Rc expansion = 1.5 x Rthroat = 1.5 cm
θw Max=νexit2
== 13.189o
12MAE 5540 – Propulsion Systems II
Solution 1.2, part 2 (2)
13MAE 5540 – Propulsion Systems II
Solution 1.2, part 2 (3)
Rc expansion = 1.5 x Rthroat = 1.5 cm
θw Max=νexit2
== 15.732o
14MAE 5540 – Propulsion Systems II
Solution 1.2, part 2 (4)
15MAE 5540 – Propulsion Systems II
Solution 1.2, Part 2 comparisons
θw Max=νexit2
=
θw Max=νexit2
== 15.731o
= 13.189o
16MAE 5540 – Propulsion Systems II
Part 3, Approximate Contour Mappinga) Minimum Length à Rc expansion = 0, g=1.4, A/A* = 1.6875
M.O.C Nozzle Spline Fit Nozzle
θw Max=νexit2
== 13.1899o
17MAE 5540 – Propulsion Systems II
Part 3, Approximate Contour Mapping (2)
b) Non-Minimum Length à Rc expansion = 1.5 cm, g=1.4, A/A* = 1.6875
M.O.C Nozzle Spline Fit Nozzle
θw Max=νexit2
== 13.1899o
18MAE 5540 – Propulsion Systems II
Part 2, Approximate Contour Mapping (3)
c) Minimum Length à Rc expansion = 0, g=1.2, A/A* = 1.884
M.O.C Nozzle Spline Fit Nozzle
θw Max=νexit2
== 15.732o
19MAE 5540 – Propulsion Systems II
Part 2, Approximate Contour Mapping (3)
b) Non-Minimum Length à Rc expansion = 1.5 cm, g=1.2, A/A* = 1.884
M.O.C Nozzle Spline Fit Nozzle
θw Max=νexit2
== 15.732o