EGR 599 ADVANCED ENGINEERING MATHEMATICS II ______
LAST NAME, FIRST (3 pts)
Quiz #2 solution
Note: Your answers must be correct to 4 significant figures.
The maximum score is 25 pts.
I. Suppose a refinery develops a product, Z1, made from two raw materials X and Y. The production of 1 metric ton of Z1 involves 1 ton of X and 2.5 tons of Y and produces 1 ton of a liquid waste, W. The engineers have come up with three alternative ways to handle the waste:
* Produce a ton of a secondary product, Z2, by adding an additional ton of X to each ton of W.
* Produce a ton of another secondary product, Z3, by adding an additional ton of Y to each ton of W.
* Treat the waste so that it is permissible to discharge it.
The products yield profits of $2500, -$50, and $200/ton for Z1, Z2, and Z3 respectively. Note that producing Z2 actually creates a loss. The treatment process costs $300/ton. In addition, the company has access to a limit of 7500 and 10,000 tons of X and Y, respectively, during the production period. You want to determine how much of the products and waste must be created in order to maximize profit. Note: One tone of W is reduced for each ton of secondary product produced.
(1) The objective function P for this problem can be written as
(A) P = 2500Z1 + 50Z2 + 200Z3 + 300W (B) P = 2500Z1 - 50Z2 + 200Z3 + 300W
(B) P = 2200Z1 + 250Z2 + 500Z3 (D) None of the above
(2) The two constraints for this problem can be written as
(A) Z1 + Z2 £ 7500 and Z1 + Z3 £ 10000 (B) Z1 + Z2 £ 7500 and 3.5Z1 + Z3 £ 10000
(B) Z1 + Z2 £ 7500 and 2.5Z1 + Z3 £ 10000 (D) None of the above
II. Maximize y = 3x1 + 4x2
Subject to
4x1 + 2x2 £ 80
2x1 + 5x2 £ 120
(3) x1 = __10_____ (4) x2 = ___20____
III. The steady-state temperature (oC) associated with selected nodal points of a two-dimensional system having a thermal conductivity of 2.0 W/m×oK are shown on the right. The isothermal surface is at 180oC.
(5) The temperature at node 1 is ___155.7 oC___
T1 = 0.25(132.8 + 180 + 172.9 + 137) = 155.7 oC
(6) The temperature at node 2 is ____95.6 oC___
T2 = 0.25(45.8 + 2´103.5 + 129.4) = 95.6 oC
(7) The temperature at node 3 is ___52.2 oC___
kDy + k = k + hDy(T3 - 30)
103.5 + 67/2 - 1.5T3 = 0.5T3 - 45.8/2 + hT3 - h´30
T3 = = 52.2 oC
(8) Calculate the heat transfer rate per unit thickness normal to the page from the bottom half of the right surface to the fluid.
______
q’conv = 50´0.1{0.5(180 - 30) + (67- 30)} = 560 W/m
IV. Start with an initial guess of x = 1 and y = 1 and apply one applications of the steepest ascent method to the following function
f(x, y) = 2xy + y – 1.5x2
The new values of x and y are
(9) x = ___1/3___ (10) y = ____3___
= 2y - 3x = 2x + 1
At x(0) = 1, y(0) = 1 Þ = - 1, = 3
Search direction = - 2i + 5.5j
New position: x = x(0) + h = 1 - h
y = y(0) + h = 1 + 3h
f(x, y) = 2(1 - h)(1 + 3h) + (1 + 3h) - 1.5(1 - h)2
g(h) = 2(1 + 2h - 3h2) + (1 + 3h) - 1.5(1 - 2h + h2)
g(h) = 1.5 + 10h - 7.5h2
= 10 - 15h = 0 Þ h = 2/3
x(1) = 1 - 2/3 = 1/3
y(1) = 1 + 3h = 3
V. The solution to Laplace's equation + = 0 is given for the following system at iteration k
j\i 1 2 3 4 5 6 7
1 200 200 200 200 200 200 200
2 500 390 350 300 500
3 500 420 380 340 500
4 500 450 410 380 500
5 300 300 300 300 300 300 300
where i denotes x-direction and j denotes y-direction. Using central difference, the Laplace equation can be written as
+ = 0
(Note: Use the equation at the nodes required for the calculation)
If Dx = Dy, the value at the next iteration (k+1) for
(11) node T (i = 4, j = 3) by the Gauss-Seidel method is __360____
T4,3 = .25(300 + 3´380)
(12) node T(i = 3, j = 2) by the relaxation method with relaxation factor = 1.4 is __304.5____
T3,2 = 350 + 1.4[(200 + 300 + 380 + 390)/4 - 350]
(13) If Dx = .2, Dy =.3, the value at the next iteration (k+1) for the
node T (i = 4, j = 3) by the Gauss-Seidel method is ___367.69___
T4,3 =