(1) Water Level Oscillations In A Surge Tank
The following figure depicts a reservoir R supplying a turbine T via a cylindrical pipe of length L and diameter D. A surge tank S is located just upstream of the turbine as shown.
HR (consant) and HS denote the water levels in R and S respectively.
QT (constant) and Q denote the volumetric flow rates in the turbine and pipe respectively.
The water level difference H(t) = HS HR and flow rate Q(t) are governed by the coupled first-order nonlinear ordinary differential equations (describing conservation of mass and Newton’s Second Law)
H˙ = 1 (Q − QT )
Q˙ = −g A (H + c |Q| Q) (g = 9.81 m/s2)
where AS and A denote the cross-sectional areas of the surge tank and pipe respec- tively, and c is an empirical parameter used to model the local and friction head losses between R and S.
With the set of parameter values and initial (t = 0) conditions allocated from the table* (all in SI units) use Euler’s Method (with time-step h = 0.5s) to estimate the time taken for the water level in the surge tank to reach its maximum value.
a | b | c | d | e | |
L | 100.0000 | 100.0000 | 100.0000 | 100.0000 | 100.0000 |
D | 2.2500 | 1.9500 | 2.7500 | 3.0000 | 2.5000 |
c | 0.0100 | 0.0100 | 0.0200 | 0.0200 | 0.0200 |
AS | 20.0000 | 20.0000 | 15.0000 | 25.0000 | 20.0000 |
QT | 10.0000 | 10.0000 | 10.0000 | 10.0000 | 10.0000 |
H(0) | 15.0000 | 10.0000 | 10.0000 | 12.0000 | 15.0000 |
Q(0) | 20.0000 | 15.0000 | 25.0000 | 30.0000 | 25.0000 |
Given that the system reaches an equilibrium state (i.e. stops changing as t ), write down general formulæ giving the equilibrium values
H∞ = lim H , Q∞ = lim Q .
t→∞
t→∞
[30 marks]
* The last digit of your student number determines which column of data values to use:
last digit | column |
0 , 1 | a |
2 , 3 | b |
4 , 5 | c |
6 , 7 | d |
8 , 9 | e |
(2) Solving Ax = b Using LU Decomposition
Given the matrix A and vector b allocated from the table below (use your student number as in part (1))
- Find the LU decomposition ofA.
- Usethis decomposition to solve the corresponding system of equations Ax = b.
[20 marks]
A | b | |
a | 2 0 2 −2 1 −1 4 1 4 | 0 −1 0 |
b | 1 −1 0 1 1 2 −1 5 5 | 1 1 −2 |
c | −1 1 1 −1 2 −1 −2 1 6 | 2 3 3 |
d | 1 2 −1 −2 −3 0 1 2 0 | 3 −2 1 |
e | −1 0 1 −2 1 4 0 1 1 | 1 5 1 |
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