The Aluminum Electrical Conductor Handbook lists a dc resistance of 0.01558 ohm per 1000 ft at 208C and a 60-Hz resistance of 0.
1 answer:
Answer:
a) 0.01558 Ω per 1000 feet
b) 0.0923 Ω per mile
c) 3.57%
Explanation:
<u>a) Calculate and verify the DC resistance </u>
Dc resistance = R = р
for aluminum at 20°C
р = 17 Ωcmil/ft
hence R = 17 * 1000 / ( 113000 ) = 0.01527 Ω per 1000 feet
there when there is an increase in resistance of 2% spiraling
R = 1.02 * 0.01527 = 0.01558 Ω per 1000 feet
<u>b) Calculate the DC resistance at 50°C </u>
R2 = R1 ( )
where ; R1 = 0.01558 , T = 228 , t2 = 50, t1 = 20 ( input values into equation above )
hence R2 = ( 0.01746 / 0.189 ) Ω per mile = 0.0923 Ω per mile
<u>c ) Determine the percentage increase due to skin effect </u>
AC resistance = 0.0956 ohm per mile
Hence; Increase in skin effect
= ( 0.0956 -0.0923 ) / 0.0923
= 0.0357 ≈ 3.57%
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Answer:
For [1 1 0] and [1 0 1] plane, σₓ = 6.05 MPa
For [0 1 1] plane, σ = 0; slip will not occur
Explanation:
compute the resolved shear stress in [111] direction on each of the [110], [011] and on the [101] plane.
Given;
Stress direction: [1 0 0] ⇒ A
Slip direction: [1 1 1]
Normal to slip direction: [1 1 1] ⇒ B
∅ is the angle between A & B
Step 1: cos∅ = A·B/|A| |B| = ⇒ cos∅ = 1/
σₓ = τ/cos ∅·cosλ
where τ is the critical resolved shear stress given as 2.47MPa
Step 2: Solve for the slip along each plane
(a) [1 1 0]
cosλ = [1 1 0]·[1 0 0]/(·
)
note: cosλ = slip D·stress D/|slip D||stress D|
cosλ = 1/
∵ σₓ = τ/ ·
=
* 2.47MPa = 6.05MPa
Hence, stress necessary to cause slip on [1 1 0] is 6.05MPa
(b) [0 1 1]
cosλ = [0 1 1]·[1 0 0]/(·
) = 0
∵ σₓ = 2.47MPa/0, which is not defined
Hence, for stress along [1 0 0], slip will not occur along [0 1 1]
(c) [1 0 1]
cosλ = [0 1 1]·[1 0 0]/(·
)
cosλ = 1/
∵ σₓ = τ/ ·
=
* 2.47MPa = 6.05MPa
See attachment for the space diagram
Answer:
A) Ductility = 11% EL
B) Radius after deformation = 4.27 mm
Explanation:
A) From equations in steel test,
Tensile Strength (Ts) = 3.45 x HB
Where HB is brinell hardness;
Thus, Ts = 3.45 x 250 = 862MPa
From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.
Also, from image 2,at CW of 27%,
Ductility is approximately, 11% EL
B) Now we know that formula for %CW is;
%CW = (Ao - Ad)/(Ao)
Where Ao is area with initial radius and Ad is area deformation.
Thus;
%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100
%CW = [1 - (rd)²/(ro)²]
1 - (%CW/100) = (rd)²/(ro)²
So;
(rd)²[1 - (%CW/100)] = (ro)²
So putting the values as gotten initially ;
(ro)² = 5²([1 - (27/100)]
(ro)² = 25 - 6.75
(ro) ² = 18.25
ro = √18.25
So ro = 4.27 mm
Answer:
<em>screw thrust = ML</em><em> </em>
Explanation:
thrust of a screw propeller is given by the equation = p x
Re
where,
D is diameter
V is the fluid velocity
p is the fluid density
N is the angular speed of the screw in revolution per second
Re is the Reynolds number which is equal to puD/μ
where p is the fluid density
u is the fluid velocity, and
μ is the fluid viscosity = kg/m.s = M
<em>Reynolds number is dimensionless so it cancels out</em>
The dimensions of the variables are shown below in MLT
diameter is m = L
speed is in m/s = L
fluid density is in kg/ = M
N is in rad/s = L =
If we substitute these dimensions in their respective places in the equation, we get
thrust = M
= M
<em>screw thrust = ML</em><em> </em>
This is the dimension for a force which indicates that thrust is a type of force