me 1020 engineering programming with matlab chapter 8...
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ME 1020 Engineering Programming with MATLAB
Chapter 8 Homework Solutions: 8.1, 8.3, 8.8, 8.10, 8.12, 8.14, 8.16
Problem 8.1:
Problem 8.3:
Problem 8.8:
Problem setup:
The unknowns are π1, π2, π3, and π. Cast the equations above in terms of these unknowns.
(π 1)π + (1)π1 + (0)π2 + (0)π3 = ππ
(π 2)π + (β1)π1 + (1)π2 + (0)π3 = 0
(π 3)π + (0)π1 + (β1)π2 + (1)π3 = 0
(π 4)π + (0)π1 + (0)π2 + (β1)π3 = βππ
π₯π = [π π1 π2 π3]
π΅π = [ππ 0 0 β ππ]
π 1 = 0.036; π 2 = 4.01; π 3 = 0.408; π 4 = 0.038
ππ = 20; ππ = β10
Problem 8.10:
Problem setup: Assign the directions of the heat fluxes arbitrarily, but be consistent when deriving the
equations.
R R R
R R
R R R
R R R
R R R
Ta T1 T2 T3
T4
T7
T6 T5
T8 T9 Tb
qa1
q12
q23
q45
q56
q78
q89
q9b
q14
q47
q25
q36
q58
q69
Heat in = Heat out: Heat balance equations.
ππ1 = π12 + π14
π12 = π23 + π25
π23 = π36
π14 = π47 + π45
π45 + π25 = π56 + π58
π36 + π56 = π69
π47 = π78
π78 + π58 = π89
π69 + π89 = π9π
Fourierβs Law of Heat Conduction: Heat transfer is proportional to the temperature difference.
ππ1 =1
π (ππ β π1)
π12 =1
π (π1 β π2)
π23 =1
π (π2 β π3)
π14 =1
π (π1 β π4)
π45 =1
π (π4 β π5)
π25 =1
π (π2 β π5)
π56 =1
π (π5 β π6)
π36 =1
π (π3 β π6)
π47 =1
π (π4 β π7)
π78 =1
π (π7 β π8)
π58 =1
π (π5 β π8)
π89 =1
π (π8 β π9)
π69 =1
π (π6 β π9)
π9π =1
π (π9 β ππ)
Substitute the π expressions into the heat balance equations. The interior temperatures are the unknowns.
(3)π1 + (β1)π2 + (0)π3 + (β1)π4 + (0)π5 + (0)π6 + (0)π7 + (0)π8 + (0)π9 = ππ
(β1)π1 + (3)π2 + (β1)π3 + (0)π4 + (β1)π5 + (0)π6 + (0)π7 + (0)π8 + (0)π9 = 0
(0)π1 + (β1)π2 + (2)π3 + (0)π4 + (0)π5 + (β1)π6 + (0)π7 + (0)π8 + (0)π9 = 0
(β1)π1 + (0)π2 + (0)π3 + (3)π4 + (β1)π5 + (0)π6 + (β1)π7 + (0)π8 + (0)π9 = 0
(0)π1 + (β1)π2 + (0)π3 + (β1)π4 + (4)π5 + (β1)π6 + (0)π7 + (β1)π8 + (0)π9 = 0
(0)π1 + (0)π2 + (β1)π3 + (0)π4 + (β1)π5 + (3)π6 + (0)π7 + (0)π8 + (β1)π9 = 0
(0)π1 + (0)π2 + (0)π3 + (β1)π4 + (0)π5 + (0)π6 + (2)π7 + (β1)π8 + (0)π9 = 0
(0)π1 + (0)π2 + (0)π3 + (0)π4 + (β1)π5 + (0)π6 + (β1)π7 + (3)π8 + (β1)π9 = 0
(0)π1 + (0)π2 + (0)π3 + (0)π4 + (0)π5 + (β1)π6 + (0)π7 + (β1)π8 + (3)π9 = ππ
π₯π = [π1 π2 π3 π4 π5 π6 π7 π8 π9]
ππ = 150; ππ = 20
Problem 8.12:
Problem setup:
Reactor A:
(6 hr
ton) (π₯
ton
wk) + (2
hr
ton) (π¦
ton
wk) + (10
hr
ton) (π§
ton
wk) = 35
hr
wk
Reactor B:
(3 hr
ton) (π₯
ton
wk) + (5
hr
ton) (π¦
ton
wk) + (2
hr
ton) (π§
ton
wk) = 40
hr
wk
6π₯ + 2π¦ + 10π§ = 35
3π₯ + 5π¦ + 2π§ = 40
Discussion of results:
The row-reduced echelon form is:
(1)π₯ + (0)π¦ + 1.9176π§ = 3.9583
(0)π₯ + (1)π¦ β 0.7500π§ = 5.6250
These equations can be reduced:
π₯ = β1.9176π§ + 3.9583
π¦ = 0.7500π§ + 5.6250
For this problem, we must have π₯ β₯ 0, π¦ β₯ 0, and π§ β₯ 0.
for π§ β₯ 0, π₯ β€ 3.9583
for π§ β₯ 0, π¦ β₯ 5.6250
for π₯ β₯ 0, β 1.9176π§ + 3.9583 β₯ 0 or π§ β€ 2.0648
for π¦ β₯ 0, 0.7500π§ + 5.6250 β₯ 0 or π§ β₯ β7.5 (not a valid solution)
Ranges of variables:
0 β€ π₯ β€ 3.9583
π¦ β₯ 5.6250
0 β€ π§ β€ 2.0648
Part c): Profit analysis:
π = 200π₯ + 300π¦ + 100π§
π = 200(β1.9176π§ + 3.9583) + 300(0.7500π§ + 5.6250) + 100π§
π = 2479 β 58.52π§
For maximum profit, set π§ = 0. This gives π = 2479.
π₯ = 3.9583
π¦ = 5.6250
Part d): Profit analysis:
π = 200π₯ + 500π¦ + 100π§
π = 200(β1.9176π§ + 3.9583) + 500(0.7500π§ + 5.6250) + 100π§
π = 3604 + 91.48π§
For maximum profit, set π§ to its maximum value: π§ = 2.0648. This gives
π = 3604 + 91.48(2.0648) = 3793
π₯ = β1.9176(2.0648) + 3.9583 = 0
π¦ = 0.7500(2.0648) + 5.6250 = 7.1736
Problem 8.14:
Problem 8.16:
Problem setup:
Part a):
π¦ = ππ₯2 + ππ₯ + π
Point 1: (π₯, π¦) = (1,4): 4 = π(1)2 + π(1) + π; (1)π + (1)π + (1)π = 4
Point 2: (π₯, π¦) = (4,73): 73 = π(4)2 + π(4) + π; (16)π + (4)π + (1)π = 73
Point 3: (π₯, π¦) = (5,120): 120 = π(5)2 + π(5) + π; (25)π + (5)π + (1)π = 120
π₯π = [π π π]
Part b):
π¦ = ππ₯3 + ππ₯2 + ππ₯ + π
Point 1: (π₯, π¦) = (1,4): 4 = π(1)3 + π(1)2 + π(1) + π; (1)π + (1)π + (1)π + (1)π = 4
Point 2: (π₯, π¦) = (4,73): 73 = π(4)3 + π(4)2 + π(4) + π; (64)π + (16)π + (4)π + (1)π = 73
Point 3: (π₯, π¦) = (5,120): 120 = π(5)3 + π(5)2 + π(5) + π; (125)π + (25)π + (5)π + (1)π = 120
π₯π = [π π π]
The exact solution (in terms of d) is:
(1)π + (0)π + (0)π + (0.05)π = 0.25
(0)π + (1)π + (0)π + (0.05)π = 3.5
(0)π + (0)π + (1)π + (1.45)π = 0.25
π + 0.05π = 0.25 or π = β0.05π + 0.25
π + 0.05π = 3.5 or π = β0.05π + 3.5
π + 1.45π = 0.25 or π = β1.45π + 0.25
π¦ = (β0.05π + 0.25)π₯3 + (β0.05π + 3.5)π₯2 + (β1.45π + 0.25)π₯ + π
The least squares solution is (inexact solution):
π¦ = (0.2414)π₯3 + (3.5862)π₯2 + (0)π₯ + 0.1724