Cylindrical pillars similar to those of Problem 4.22 are positioned between the glass sheets with a pillar-to-pillar spacing of W . The inside surface of one glass sheet is treated with a low-emissivity coating characterized by ε 1 = 0.05 . The second inner surface has an emissivity of ε 2 = 0.95 . Determine the ratio of conduction to radiation heat transfer through a square unit area of dimension W × W for W = 10 , 20 , and 30 mm. The stainless steel pillar is located at the center of the unit area and has a length of L = 0.4 mm and diameter of D = 0.15 mm . The contact resistances and glass temperatures are the same as in Problem 4.22.
Cylindrical pillars similar to those of Problem 4.22 are positioned between the glass sheets with a pillar-to-pillar spacing of W . The inside surface of one glass sheet is treated with a low-emissivity coating characterized by ε 1 = 0.05 . The second inner surface has an emissivity of ε 2 = 0.95 . Determine the ratio of conduction to radiation heat transfer through a square unit area of dimension W × W for W = 10 , 20 , and 30 mm. The stainless steel pillar is located at the center of the unit area and has a length of L = 0.4 mm and diameter of D = 0.15 mm . The contact resistances and glass temperatures are the same as in Problem 4.22.
Solution Summary: The author explains the ratio of conduction to radiation heat transfer through a square unit area of dimension Wtimes W.
Cylindrical pillars similar to those of Problem 4.22 are positioned between the glass sheets with a pillar-to-pillar spacing of W. The inside surface of one glass sheet is treated with a low-emissivity coating characterized by
ε
1
=
0.05
. The second inner surface has an emissivity of
ε
2
=
0.95
. Determine the ratio of conduction to radiation heat transfer through a square unit area of dimension
W
×
W
for
W
=
10
,
20
, and 30 mm. The stainless steel pillar is located at the center of the unit area and has a length of
L
=
0.4
mm
and diameter of
D
=
0.15
mm
. The contact resistances and glass temperatures are the same as in Problem 4.22.
The 2-mass system shown below depicts a disk which rotates about its center and has rotational
moment of inertia Jo and radius r. The angular displacement of the disk is given by 0. The spring
with constant k₂ is attached to the disk at a distance from the center. The mass m has linear
displacement & and is subject to an external force u. When the system is at equilibrium, the spring
forces due to k₁ and k₂ are zero. Neglect gravity and aerodynamic drag in this problem. You may
assume the small angle approximation which implies (i) that the springs and dampers remain in
their horizontal / vertical configurations and (ii) that the linear displacement d of a point on the
edge of the disk can be approximated by d≈re.
Ө
K2
www
m
4
Cz
777777
Jo
Make the following assumptions when analyzing the forces and torques:
тв
2
0>0, 0>0, x> > 0, >0
Derive the differential equations of motion for this dynamic system. Start by sketching
LARGE and carefully drawn free-body-diagrams for the disk and the…
A linear system is one that satisfies the principle of superposition. In other words, if an input u₁
yields the output y₁, and an input u2 yields the output y2, the system is said to be linear if a com-
bination of the inputs u = u₁ + u2 yield the sum of the outputs y = y1 + y2.
Using this fact, determine the output y(t) of the following linear system:
given the input:
P(s) =
=
Y(s)
U(s)
=
s+1
s+10
u(t) = e−2+ sin(t)
=e
The manometer fluid in the figure given below is mercury where D = 3 in and h = 1 in. Estimate the volume flow in the tube (ft3/s) if the flowing fluid is gasoline at 20°C and 1 atm. The density of mercury and gasoline are 26.34 slug/ft3 and 1.32 slug/ft3 respectively. The gravitational force is 32.2 ft/s2.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.