Describe SIX basic ways of installing wire ways
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Based on the Chvorinov's rule, the diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the prism casting.
<h3>How to apply the Chvorinov's rule for casting processes</h3>
The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a <em>reference</em> time. This rule states that cooling time (<em>t</em>) is directly proportional to the square of the volume (<em>V</em>), in cubic meters, divided to the surface area (<em>A</em>), in square meters. Now we proceed to model each casting:
<h3>Cylindrical casting</h3>t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²
t = C · [0.25 · D · L/(0.5 · D + L)]² (1)
<h3>Prism casting</h3>t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²
t' = C · [3 · T · L/(8 · L + 6 · T)]² (2)
<h3>Relationship between the cross sections of both castings</h3>3 · T² = 0.25π · D² (3)
Where:
- <em>t</em> - Cooling time of the cylindrical casting, in time unit.
- <em>t'</em> - Cooling time of the prism casting, in time unit.
- <em>C</em> - Cooling factor, in time unit per square meter.
- <em>D</em> - Diameter of the cylinder, in meters.
- <em>L</em> - Length of the casting, in meters.
- <em>T</em> - Width of the cross section of the prism casting, in meters.
If we know that <em>D =</em> <em>0.3 m</em>, then the thickness of the prism casting is:
<em>T ≈ 0.153 m</em>
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And (1) and (2) simplified into these forms:
<h3>Cylindrical casting</h3>t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²
t = 0.0329 · C (1b)
<h3>Prism casting</h3>t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²
t' = 0.00218 · C (2b)
Lastly we find the <em>percentual</em> difference in the solidification times of the two castings by using the following expression:
<em>r = (</em>1 <em>- t'/t) ×</em> 100 %
<em>r = (</em>1 <em>-</em> 0.00218<em>/</em>0.0329<em>) ×</em> 100 %
<em>r =</em> 93.374 %
The <em>cooling</em> time of the <em>prism</em> casting is 6.626 % of the <em>solidification</em> time of the <em>cylindrical</em> casting. The diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the <em>prism</em> casting.
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Answer:
V = 30 V
vector (E) = -10*a_p + 17.32*a_Q - 24*a_z
mag(E) = 31.24 V/m
dV / dN = 31.2 V /m
a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z
p_v = 276 pC / m^3
Explanation:
Given:
- The Volt Potential in cylindrical coordinate system is given as:
- The point P is at p = 3 , Q = 60 , z = 2
Find:
values at P for:
a.) V
b.) E
c.) E
d.) dV/dN
e.) aN
f.) rhov in free space
Solution:
a)
Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:
b)
To compute the Electric field from Volt potential we have the following relation:
E = - ∀.V
Where, ∀ is a del function which denoted:
∀ =
Hence,
Plug in the values for point P:
c)
The magnitude of the Electric Field is given by:
E = √((-10)^2 + (17.32)^2 + (24)^2)
E = √975.9824
E = 31.241 N/C
d)
The dV/dN is the Field Strength E in normal to the surface with vector N given by:
dV / dN = | - ∀.V |
dV / dN = | E | = 31.2 N/C
e)
a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:
f)
The charge density in free space:
p_v = E.∈_o
Where, ∈_o is the permittivity of free space = 8.85*10^-12
p_v = 31.24.8.85*10^-12
p_v = 276 pC / m^3
Answer:
Time required for iron rod surface temperature to cool to 200°C is 250 seconds.
Explanation:
