A growing trend in urban design is the concept of a rooftop garden. If every building in a city were to install a rooftop garden
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Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.
<u>Given the following data:</u>
- Emf = 520 Volts
- Speed = 660 r.p.m
- Number of armature conductors = 144 slots
- Number of poles = 4 poles
- Number of parallel paths = 2
To find the magnetic flux per pole:
Mathematically, the emf generated by a DC generator is given by the formula;
×
<u>Where:</u>
- E is the electromotive force in the DC generator.
- Z is the total number of armature conductors.
- N is the speed or armature rotation in r.p.m.
- P is the number of poles.
- A is the number of parallel paths in armature.
- Ф is the magnetic flux.
First of all, we would determine the total number of armature conductors:
×
×
Z = 864
Substituting the given parameters into the formula, we have;
×
×
<em>Magnetic flux </em><em>=</em><em> 0.0274 Weber.</em>
Therefore, the magnetic flux per pole is 0.0274 Weber.
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Answer:
ideal fluid follow Newtonian law
that is, shear stress is directly proportional to rate change of shear strain.
watch handwritten explanation
Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.