Layers of 23-cm-thick meat slabs ( k = 0.47 W/m .K , α = 0 .13 × 10 -6 m 2 /s) and initially at a uniform temperature of 7°C are to be frozen by refrigerated air at -30°C flowing at a velocity of 1.4 m/s. The average heat transfer coefficient between the meat and the air is 20 W/m 2 K. Assuming the size of the meat slabs to be large relative to their thickness, determine how long it will take for the center temperature of the slabs to drop to -18°C. Also, determine the surface temperature of the meat slab at that time.
Layers of 23-cm-thick meat slabs ( k = 0.47 W/m .K , α = 0 .13 × 10 -6 m 2 /s) and initially at a uniform temperature of 7°C are to be frozen by refrigerated air at -30°C flowing at a velocity of 1.4 m/s. The average heat transfer coefficient between the meat and the air is 20 W/m 2 K. Assuming the size of the meat slabs to be large relative to their thickness, determine how long it will take for the center temperature of the slabs to drop to -18°C. Also, determine the surface temperature of the meat slab at that time.
Solution Summary: The author explains the time taken by the centre temperature of the slabs to drop to 18 o C and to find the surface temperature.
Layers of 23-cm-thick meat slabs
(
k
=
0.47
W/m
.K
,
α
= 0
.13
×
10
-6
m
2
/s)
and initially at a uniform temperature of 7°C are to be frozen by refrigerated air at -30°C flowing at a velocity of 1.4 m/s. The average heat transfer coefficient between the meat and the air is 20 W/m2 K. Assuming the size of the meat slabs to be large relative to their thickness, determine how long it will take for the center temperature of the slabs to drop to -18°C. Also, determine the surface temperature of the meat slab at that time.
=
The forces F₁ = 590 lb, F₂ = 380 lb, F3 = 240 lb and F
330 lb. Determine the forces in each member of the truss.
Use positive values to indicate tension and negative values to
indicate compression.
a
a
a
D
b
F₁
A
000
B.
779977
F₂V
H
G
E
F4
b
BY NC SA
2013 Michael Swanbom
Values for dimensions on the figure are given in the following
table. Note the figure may not be to scale.
Variable Value
a
6 ft
b
10.1 ft
The force in member AB is
lb.
The force in member AH is
lb.
The force in member GH is
lb.
The force in member BH is
lb.
The force in member BC is
lb.
The force in member BG is
lb.
The force in member EG is
lb.
The force in member CD is
lb.
The force in member DE is
lb.
The force in member CE is
lb.
The force in member CG is
lb.
Multiple Choice
Circle the best answer to each statement.
1. Which type of surface deviation is controlled by a cy-
lindricity tolerance but not by a circularity tolerance?
A.
B.
C.
Ovality
Taper
Lobing
D. None of the above
2. When verifying a cylindricity tolerance, the inspec-
tion method must be able to collect a set of points and
determine the:
A. Distance between two coaxial cylinders that con-
tain the set of points
B.
Cylinder that circumscribes the set of points
C. Cylinder that inscribes the set of points
D.
Distance between two coaxial circles that contain
the set of points
3. Where Rule #1 applies to a cylindrical regular feature
of size, the tolerance value of a cylindricity tolerance
applied to the feature of size must be
tolerance.
A. Less than
B. Equal to
C. Greater than
D. None of the above
the size
4. Which of the following modifiers may be applied with
a cylindricity tolerance?
A. M
B.
C. ℗
D. Ø
5. Which geometric tolerance can provide an indirect
cylindricity…
The beam AB is attached to the wall in the xz plane by a
fixed support at A. A force of
F = (−129î + 69.0ĵ + 3591) N is applied to the end of
the beam at B. The weight of the beam can be modeled with
a uniform distributed load of intensity w = 85.0 N/m acting in
the negative z direction along its entire length. Find the
support reactions at A.
Z
с
A
b
a
B
F
y
Cc 10
BY NC SA
2016 Eric Davishahl
X
Values for dimensions on the figure are given in the following.
table. Note the figure may not be to scale.
Variable
Value
a
5.60 m
b
5.00 m
C
3.70 m
A
II
=
MA = (
m
2.>
~.>
+
+
k) N
k) N-
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