On February 15, 2013, Asteroid 2012 DA 14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8 kilometers per second [km/s]. This is considerably closer than the orbit of geosynchronous satellites (26,200 miles). This is the closest recorded approach of an object this large. The asteroid 2012 DA 14 was estimated to have a diameter of 30 meters [m] and a specific gravity of 3. If 2012 DA 14 had hit the Earth, what is the total amount of energy that would have been released (i.e., what was the kinetic energy of the asteroid)? Express your answer in megatons [Mton]. One megaton is the energy released by one million metric tons of TNT explosive. A metric ton equals 1000 kilograms [kg], and the explosive energy of TNT is 4184 joules per gram [J/g].
On February 15, 2013, Asteroid 2012 DA 14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8 kilometers per second [km/s]. This is considerably closer than the orbit of geosynchronous satellites (26,200 miles). This is the closest recorded approach of an object this large. The asteroid 2012 DA 14 was estimated to have a diameter of 30 meters [m] and a specific gravity of 3. If 2012 DA 14 had hit the Earth, what is the total amount of energy that would have been released (i.e., what was the kinetic energy of the asteroid)? Express your answer in megatons [Mton]. One megaton is the energy released by one million metric tons of TNT explosive. A metric ton equals 1000 kilograms [kg], and the explosive energy of TNT is 4184 joules per gram [J/g].
Solution Summary: The author calculates the total amount of energy that would have been released by the asteroid, which is 0.031 Mton.
On February 15, 2013, Asteroid 2012 DA14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8 kilometers per second [km/s]. This is considerably closer than the orbit of geosynchronous satellites (26,200 miles). This is the closest recorded approach of an object this large. The asteroid 2012 DA14 was estimated to have a diameter of 30 meters [m] and a specific gravity of 3. If 2012 DA14 had hit the Earth, what is the total amount of energy that would have been released (i.e., what was the kinetic energy of the asteroid)? Express your answer in megatons [Mton]. One megaton is the energy released by one million metric tons of TNT explosive. A metric ton equals 1000 kilograms [kg], and the explosive energy of TNT is 4184 joules per gram [J/g].
You are working as an engineer in a bearing systems design company. The flow of
lubricant inside a hydrodynamic bearing (µ = 0.001 kg m¯¹ s¯¹) can be approximated
as a parallel, steady, two-dimensional, incompressible flow between two parallel plates.
The top plate, representing the moving part of the bearing, travels at a constant speed,
U, while the bottom plate remains stationary (Figure Q1). The plates are separated by
a distance of 2h = 1 cm and are W = 20 cm wide. Their length is L = 10 cm. By
applying the above approximations to the Navier-Stokes equations and assuming that
end effects can be neglected, the horizontal velocity profile can be shown to be
U
y = +h
У
2h = 1 cm
1
x1
y=-h
u(y)
=
1 dP
2μ dx
-y² + Ay + B
moving plate
-
U
stationary plate
2
I2
L = 10 cm
Figure Q1: Flow in a hydrodynamic bearing. The plates extend a width, W = 20 cm,
into the page.
(a) By considering the appropriate boundary conditions, show that the constants take
the following forms:
A =
U
2h
U
1 dP…
Question 2
You are an engineer working in the propulsion team for a supersonic civil transport
aircraft driven by a turbojet engine, where you have oversight of the design for the
engine intake and the exhaust nozzle, indicated in Figure Q2a. The turbojet engine can
operate when provided with air flow in the Mach number range, 0.60 to 0.80. You are
asked to analyse a condition where the aircraft is flying at 472 m/s at an altitude of
14,000 m. For all parts of the question, you can assume that the flow path of air through
the engine has a circular cross section.
(a)
normal
shock
472 m/s
A B
(b)
intake
engine
altitude: 14,000 m
D
exhaust nozzle→
exit to
atmosphere
472 m/s
50 m/s
B
diameter: DE = 0.30 m
EX
diameter: DF = 0.66 m
Figure Q2: Propulsion system for a supersonic aircraft.
F
a) When the aircraft is at an altitude of 14,000 m, use the International Standard
Atmosphere in the Module Data Book to state the local air pressure and tempera-
ture. Thus show that the aircraft speed of…
given below:
A rectangular wing with wing twist yields the spanwise circulation distribution
kbV1
roy) = kbv. (2)
where k is a constant, b is the span length and V. is the free-stream velocity. The wing has an
aspect ratio of 4. For all wing sections, the lift curve slope (ag) is 2 and the zero-lift angle of
attack (a=0) is 0.
a. Derive expressions for the downwash (w) and induced angle of attack a distributions
along the span.
b. Derive an expression for the induced drag coefficient.
c. Calculate the span efficiency factor.
d. Calculate the value of k if the wing has a washout and the difference between the
geometric angles of attack of the root (y = 0) and the tip (y = tb/2) is:
a(y = 0) a(y = ±b/2) = /18
Hint: Use the coordinate transformation y = cos (0)
Chapter 8 Solutions
Thinking Like An Engineer: An Active Learning Approach, Student Value Edition (4th Edition)
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