SOCIETY OF HISPANIC PROFESSIONAL ENGINEERS 2001 REGIONAL ACADEMIC OLYMPIAD Written Examination
Team Name:
Dynamics
1.
Neglecting friction, estimate the angular velocity of the cylinder after the mass falls 2 m from rest.
ω
60 kg 40 cm v
40 kg (A) (B) (C) (D)
11.8 rad/s 10.5 rad/s 8.1 rad/s 5.4 rad/s
Solution: Using the work-energy equation:
1
W = =
(40 x 9.81) x 2 =
2
2
+
1 2
2
I ω ω
1 1 1 × 40 × (0.4ω ) 2 + ( × 60 × 0.4 2 ) ω 2 2 2 2
∴ ω = Answer is A.
mv
11.83 11.83 rad/s rad/s
Team Name:
Fluid Mechanics
What force is exerted on the joint if the flow rate of water is 0.01 m3/s?
2.
4 cm dia 2 cm dia V1
V2
(A) (B) (C) (D)
238 N 474 N 357 N 981 N
Team Name:
Solution:
2.
The velocities are found to be Q
V1 =
A1
V2 =
0.01
=
Q A2
π x
=
0.02
2
0.01 0.01 2
π x
= 7.96 7.96 m/s m/s
= 31.8 31.8 m/s m/s
Bernoulli's equation is used to to find the pressure at section 1. Using p2 = 0 (atmospheric pressure is zero gage), V1 2 2g
+
p1
=
γ
7.962 2 × 9.81
+
V2 2 2g
p1 9800
+
p 2 γ
=
31.82 2 × 9.81
Now, using the control volume shown, we can apply the momentum equation in the x-direction: F j 474,000 Pa ∴ p1 = 474,000
p1A1 CV
x
F j = the force of the contraction of the water
− F j = ρ Q(V 2 − V 1 ) ∴ F j = 357 N
p1 A1
2
474,000 × π × × 0.02
− F j = 1000 × 0.01(31.8 − 7.96)
Answer is C.
NOTE: Remember, all forces on the control volume must be b e included, never forget the pressure force.
Team Name:
Thermodynamics
3.
An inventor claims to have invented an engine, using a 160 degree C geothermal heat source, which operates with an efficiency of 30%. If it exhausts to the 20 degree C atmosphere, is the invention possible? (A) (B) (C)
Possible Not Possible Can not be determined
Solution:
The maximum possible efficiency, as limited by the second law, is given by ηmax
= 1 − T L = 1 − 293 = 0.323 T H
433
or 32%, therefore it meets the 30% efficiency possible..
Answer is A.
Team Name:
Thermodynamics
A nozzle accelerates air from 100 m/s, c p=1000 J/kg•K, k=1.4, 400oC and 400 kPa to a receiver where P=20 kPa. Assuming an isentropic process, find V2.
4.
(A)
560 m/s
(B)
874 m/s
(C)
950 m/s
(D)
Solution:
The First Law of Thermodynamics for steady state flow situations is
! Q! − W s ! m
2
=
V 2
− V 12 2
+ h2 − h1 + g ( z 2 − z1 )
For a nozzle, there is no work or heat transfer, however kinetic energy is included 2
0 =
V 2
− V 12
+ h2 − h1
2
Assuming ideal gas with constant c p: Q = m(h2 − h1 ) = mc p (T 2 − T 1 )
Find T2 from isentropic, ideal gas equation:
T 2
= T 1 (
P2 P1
( k −1)
)
= (400 + 273)(
k
To find the exit velocity, c p (T 2 − T 1 ) =
2
V 2
− V 1 2 2 2
1000(673 − 286) =
∴ V 2 = 874 m / s Answer is B.
V 2
− 100 2
20 400
0.4
)
1 .4
= 286 K
355 m/s
Team Name:
Thermodynamics
5.
A heat pump delivers 20,000 Kj/hr of heat with a 1.39 kW input. Calculate the Coefficient of Performance (COP). (A)
1.00
(B)
Solution:
Using the definition of the COP,
COP =
Q! H
= ! W in
Answer is D.
20,000 / 3600 1.39
= 4.00
2.00
(C)
3.00
(D)
4.00
Team Name:
Dynamics
6.
Find the velocity after the mass moves 10 m if it starts from rest. 10 m 2m
60 N
50 kg K=100 N/m
(A)
3 m/s
(B)
Solution:
Using the work-energy equation, W net
= ∆ E = 1 mv2
F • d −
2
1 2
60 × 10 −
Kx
1 2
2
= 1 mv 2 2
× 100 × 2 2 = 1 × 50 × v 2
∴ v 2 = 4m / s Answer is B.
2
4 m/s
(C)
5 m/s
(D)
6 m/s
Team Name:
Fluid Statics
7.
What is the pressure in pipe A? (Density of water = 1000 kg/m3)
oil (SG=0.86)
a
A
b 50 cm
B
30 cm
pB=8 kPa
water water (A)
4.75 kPa
(B)
5.75 kPa
(C)
Solution:
Locate points "a" and "b" so that pa=p b. Then, using p A − ρ gh = p B − ρ gh p A − 9800 × 0.5 = 8000 − 9800 × 0.3 − (9800 × 0.86) × 0.5
∴ p A = 5.75kPa Answer is B.
4.25 kPa
(D)
5.25 kPa
Team Name:
Circuits
8.
Which of the following is a correct definition for power factor in an alternating current circuit?(A) The ratio of apparent power to actual power (B) The ratio of resistance to impedance (C) The sine of the phase angle between the voltage and current of the load (D) The ratio of impedance to resistance
Solution Answer is B.
Team Name:
Civil Engineering
For problems 9 and 10 use the figure below. Each column below has the same cross-sectional and material properties. Assume: height of Col. A > height of Col. C > height of Col. B.
100 plf
B
C A
9.
For the loading shown, which column has the greatest horizontal reaction at its base? (A) (B) (C) (D)
Col. A Col. B Col. C All columns have equal horizontal reactions at their base.
Solution:
The shorter column is stiffer and therefore attracts a lager portion of the load. Answer is B.
10.
Which column has the greatest compressive capacity? (A) (B) (C) (D)
Col. A Col. B Col. C All columns have equal compressive capacities.
Solution:
The shorter column has a smaller KL/r ratio and therefore has a higher allowable compressive stress, Fa. Answer is B.
Team Name:
Statics
11.
For the truss and loading below, how many zero force members are there?
A
L
C
L
E
D L
B L
P (A) (B) (C) (D)
2 3 4 5
Solution:
Members DE, CE, AC, CD, and AB are zero force members. Answer is D.
Team Name:
Materials Science
12.
What is the hardest form of steel? (A) Pearlite (B) Ferrite (C) Bainite (D) Martensite
Solution
Hard steel is obtained by rapid quenching. Martensite has a high hardness since it is rapidly quenched. Though martensite is hard, it has low ductility. Answer is D (martensite).
Team Name:
Materials Science
13.
Plastic deformation of the crystalline structure of metals resulting in misalignment of atoms, dislocations, and large stresses and strains in small regions are characteristic of which process?
(A) (B) (C) (D)
Solution: Answer is B.
Tempering Cold forming Twinning Isostatic pressing
Team Name:
Chemistry
14. The reaction shown proceeds in a gaseous state. At equilibrium, the concentration of the components X, Y, and Z are measured to be 5.73 x 10-2 mol/L, 2.67 x 10 -2 mol/L, and 4.59 x 10 -2 mol/L, respectively. What is the equilibrium constant for this reaction? 2X ⇔ Y + 2Z
(A) (B) (C) (D)
9.8 x 10-4m ol/L -2 1.7 x 10 m ol/L -2 2.1 x 10 m ol/L -1 3.7 x 10 m ol/L
Solution:
The equilibrium constant for this reaction is
Answer is B.
[Y][Z]2
K eq
=
K eq
= 1.71x10 − 2 mol/L
[X]2
Team Name:
Chemistry
What is the enthalpy of reaction at 25°C for the combustion of ethane (C2H6)?
15.
2C 2 H 6
+ 7O 2 →
4CO
2
+ 6H 2 O
given
∆ H f (C 2 H 6 ) = − 20 . 24 kcal/mol ∆ H f (O 2 ) = 0 . 00 kcal/mol ∆ H f (CO 2 ) = − 94 . 05 kcal/mol ∆ H f (H 2 O ) = − 57 . 80 kcal/mol (A) (B) (C) (D)
-680 kcal/mol (exothermic) -340 kcal/mol (exothermic) 130 kcal/mol (endothermic) 340 kcal/mol (endothermic)
Solution:
The enthalpy of reaction is
= ∑ H f , products − ∑ H f ,reac tan ts H r = (4)(−94.05 kcal/mol) + (6)(−57.80 kcal/mol) - (2)(−20.24 kcal/mol) - (7)(0.00 kcal/mol) H r = −680.5kcal (680 kcal exothermic) H r
Answer is B.
Team Name:
Chemistry
16.
A solution is adjusted from pH 8 to pH 9. The relative concentration of the hydrogen [H+] ion has changed by a factor of what? (A) 1/100 (B) 1/10 (C) 5 (D) 10
Solution:
Explanation: The definition of pH is pH = -log10[H+] = log10(1/[H+]) Answer is B.
Team Name:
Mathematics
17.
What is the standard form of the equation of the line tangent to a circle centered at the origin with a radius of 5 at the point (3, 4)? y 5 (3, 4) (A)
x
=
(B)
y
=
(C)
y
=
(D)
y
=
−4
y−
25
3 4 3 25 x+ 4 4 −3 9 x+ 4 4 − 3 25 x+ 4 4
5
-5
-10
-5 -10
Solution:
Using the equation of a circle, x 2
+ y 2 = r 2
y 2
= 25 − x 2 ⇒ y = (25 − x 2 )2
1
−1
1 f ′( x) = (25 − x 2 ) 2 × (−2 x) = 2
f ′( x) ( 3, 4 )
=
−3 1
(25 − 9 )2
Using the equation of a line formula, y
=
−3 4
∴ y =
x + b ⇒ 4 =
−3 4
x+
Answer is D.
25 4
9 4
+b ⇒ b =
25 4
=
−3 4
− x 1 2 2
(25 − x )
10
x
Team Name:
Mathematics
18.
What are the minimum and maximum values, respectively, of the equation: 3
2
f(x) = 5x – 2x + 1
on the interval (-2,2)? (A) (B) (C) (D)
–47,33 –4,4 0.95,1 0,0.27
Solution:
The critical points are located where the first derivative is zero. f(x) f’(x) 2 15x – 4x x(15x – 4)
= 5x3 – 2x 2 + 1 = 15x2 – 4x =0 =0
x = 0 or x = 4/15 Test for the maximum, minimum, or inflection point. f’’(x) = 30x – 4 f’’(0) = -4 f’’(a) < 0 [maximum] f’’(4/15) = 4 f’’(a) > 0 [minimum] These could be local maximum and minimum. Check the endpoints of the interval and compare with the function values at the critical points. f(-2) = -47 f(2) = 33 f(0) = 1 f(4/15) = 0.95 The minimum and maximum values of the equation are at the endpoints, -47 and 33, respectively. Answer is A.
Team Name: Mathematics
Problems 19 through 21 are based on the following statement. A two-dimensional function, f(x,y), is defined as: 2
2
f(x,y) = 2x – y + 3x - y
19.
What is the gradient vector for this function?
(A) ∇ f ( x, y ) = (2 x 2
+ 3 x)i + (− y 2 − y ) jj
(B) ∇ f ( x, y ) = ( x 3
3
+
1
x )i + ( − y 2
3
2 3 (C) ∇ f ( x, y ) = (4 x + 3)i + (−2 y − 1) jj
−
1 2
2
y ) jj
(D) ∇ f ( x, y ) = (3 x + 4)i + (−2 y + 1) jj
20. What is the direction of the line passing through the point (1,-2) that has the maximum slope? (A) 4i + 2 jj (B) 7i + 3 jj (C) 7i + 4 jj (D) 9i − 7 jj
21.
What is the maximum slope at the point (1,-2)?
(A) 2.1 (B) 3.5 (C) 7.6 (D) 8.7
Team Name: Solution:
19.
It is necessary to calculate two partial derivatives.
∂ f ( x, y ) = 4 x + 3 ∂ x ∂ f ( x, y ) 2 1 = − y − ∂ y ∇ f ( x, y ) = (4 x + 3)i + (−2 y − 1) jj Answer is C.
20.
The direction of the line passing through (1,-2) with maximum slope is found by inserting x=1 and y=-2 into the gradient vector function. V = [(4)(1) + 3] i + [(-2)(-2) – 1] j = 7 i + 3 j
Answer is B.
21.
The magnitude of the slope is: V
Answer is C.
=
( 7) 2
+ (3) 2 =
58
= 7.62
Team Name: Computer Science
22.
Which data transfer path(s) is (are) used the most in typical computer operation?
Input Device
1
2
CPU
3
Output Device
4
Buffer Memory
(A) (B) (C) (D) (E)
1 2 3 1 and 2 3 and 4
Solution:
Input and output devices (e.g., keyboards and printers) are used far less frequently than memorytransfer operations. Data enters and is read (i.e., or is replaced) in buffer memory at approximately the same rates. Answer is E.
Team Name: Computer Science
23.
In the pseudo-code segment shown, how many times will the line labeled “START” execute?
START
FINISH
(A) (B) (C) (D)
I =1 J =1 J =J+I I = J^2 IF J<100 THEN GO TO START ELSE GO TO FINISH PRINT J
3 4 5 6
Solution: Loop (initially) 0 1 2 3 4
I 1 4 36 1764 3,261,636
J 1 2 6 42 1806
After four loops, J is greater than 100, so the program moves on. Answer is B.
Team Name: Computer Science
The Taylor Series approximation for the exponential function is e x = Σ(xn/n!). A spreadsheet set up like the one shown is being used to calculate e x. The cells in row 2 will be totaled to arrive at the approximation. Which formula, typed into C2 and then copied to the rest of row 2, will provide the correct total?
24.
A 0 1
1 2 3 4 5
(A) (B) (C) (D)
B 1 x
C 2
D 3
E 4
B2^B1/FACTORIAL(B1) B2*$B$2/B1 B2*$B$2/C1 B2^C1/(C1*B1*A1)
Solution:
The elements in this series, Tn, could be represented using exponents and factorial functions. In this case, absolute cell references must be used for x, but a relative reference works well for n. T n
=
x
n
n!
=
$B$2^ C1 FACTORIAL(C1)
Since this is not one of the available answers, develop a recursive definition for the series. T n+1
=
x
n +1
(n + 1)!
=
( x n )( x ) (n!)(n + 1)
x = T n n + 1 = B2($B$2/C1)
Answer is C.
Team Name: Engineering Economics
25.
How many months at an interest rate of 1 percent per month does money have to be invested before it will double in value? (A) (B) (C) (D) (E)
59 months 62 months 70 months 76 months 83 months
Solution:
Let P = $1 n F = P(1 + i)
F = $2 n $2 = $1(1.01)
i= 0.01 per month n 1.01 = 2
n= number of months
The solution may be obtained from a 1% compound interest table or by hand calculator, n = 70 months. Answer is C.
Team Name: Engineering Economics
26.
If $1,000 is deposited in a savings account that pays 6% annual interest, and all the interest is left in the account, what is the account balance after three years? (A) (B) (C) (D) (E)
$840 $1,000 $1,180 $1,191 $3,000
Solution: 3
F = P(1 + i)n = (1000)(1.06) = $1,191 Answer is D. Engineering Economics 27.
What interest rate, compounded quarterly, is equivalent to a 9.31% effective interest rate? (A) (B) (C) (D) (E)
2.25% 2.33% 4.66% 9.00% 9.31%
Solution: 4
0.0931 = (1 + i) – 1 0.25 (1 + 0.0931) = 1 + i 1.0225 = 1 + i i=2.25% per quarter i=9% annual interest Answer is (D)
Team Name: Circuits
Problems 28 through 30 refer to the following circuit. All circuit elements are ideal. 3 ohms
4 F
A
B
2H
I +
9 ohms 24 volts
6H 6 ohms
C
28.
What current is flowing through the 3 ohm resistor? (A) (B) (C) (D)
29.
2.00 A 2.67 A 3.64 A 8.00 A
What energy is stored in the capacitor? (A) (B) (C) (D)
30.
14.3 µJ 128 µJ 512 µJ 1150 µJ
What energy is stored in the 2H inductor? (A) (B) (C) (D)
7.1 J 13 J 14 J 29 J
D
Team Name: Solution:
28.
A capacitor in a DC circuit has an infinite resistance. Therefore, there is no current flow through the capacitor between nodes A and B. An inductor in a DC circuit has no resistance.
From Ohm’s law: I=
V R
=
24V 3Ω + 6 Ω
= 2.67 A
Answer is B.
29.
The energy stored in the capacitor depends on the capacitance and the voltage across the capacitor. Since there is no voltage drop across an ideal inductor, node B is grounded. The voltage at node A can be found from the current through leg AC. Since none of the current found in problem 28 flows through the capacitor, IAC = 2.67 A.
= IR = (2.67 A)(6 Ω) = 16 V VAB = VA − VB = 16 V - 0 V = 16 V VA
The energy stored in the capacitor is: 1 2 energy = CVAB 2 1 = (4x10 -6 F)(16 V) 2 2 = 512x10 -6 J Answer is C.
30.
All of the current flowing passes through the 2 H inductor. From problem 28, I = 2.67 A. The energy stored in an inductor is:
energy =
1 2
LI 2
Answer is A.
=
1 2
(2H)(2.67A) 2
= 7.13 J
Team Name: Tie Breaker Question
TB1. (A) (B) (C) (D) (E)
Which of the following Hispanic(s) is(are) part of the U.S. Space Program? Franklin Chang-Diaz Ellen Ochoa Raul Ruiz A and B B and C
Solution: Answer is D.
Team Name: Tie Breaker Question
TB2. (A) (B) (C) (D) (E)
Which of the following states is not part of SHPE Region 5? Texas Alabama Mississippi Louisiana Florida
Solution: Answer is E.
