Answer:
(a) the subtransient current through the breaker in per-unit and in kA rms = 71316.39kA
(b) the rms asymmetrical fault current the breaker interrupts, assuming maximum dc offset. = 152KA
Explanation:
check the attached files for explanation
Answer:
First you have to separate real and imaginary parts of Tan(x+iy)=Tan(z)=sin(z)/cos(z)
sinz=sin(x+iy)=sinxcos(iy)+cosxsin(iy)=sinxcoshy-icosx sinhy
cosz=cos(x+iy)=cosxcos(iy)-sinxsin(iy)=cosxcoshy−isinxsinhy
Now if you plug in Tan(z) and simplify (it is easy!) you get
Tan(z)=(sin(2x)+isinh(2y))/(cos(2x)+cosh(2y))= A+iB.
This means that
A=sin(2x)/(cos(2x)+cosh(2y)) and B= sinh(2y)/(cos(2x)+cosh(2y))
Now,
A/B=sin(2x)/sinh(2y)
If any questions, let me know.
Answer:
The stress in the rod is 39.11 psi.
Explanation:
The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

Replacing the diameter the area results:

Therefore the the stress results:

The equations are based on the following assumptions
1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.
Nomenclature
T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)