All categories

2 years ago

Write a brief design of bridge

1 answer:

4 0

Answer:Brief introduction to your bridge project (Design brief): 100 words. ... The purpose of this bridge is to fit a 500mm long by 100mm gap in an area where stormy

Explanation:Brief introduction to your bridge project (Design brief): 100 words. ... The purpose of this bridge is to fit a 500mm long by 100mm gap in an area where stormy

You might be interested in

Q7. A cylindrical rod of 1040 steel originally 15.2 mm (0.60 in.) in diameter is to be cold worked by drawing; the circular cros

umka2103 [35]

Answer:

11.2mm or 0.45in

Explanation:

The percent cold work, attendant tensile strength and ductility if drawing is carried out without interruption is given by the equation you will find in the attached file.

Please go through the attached file for a step by step solution to this question.

5 0

3 years ago

Asolid rectangular rodhas a length of 90mm, made of steel material (E =207,000 MPa, Syield= 300 MPa), the cross section of the r

erastova [34]

Answer:

13.6mm

Explanation:

We consider diameter to be a chord that runs through the center point of the circle. It is considered as the longest possible chord of any circle. The center of a circle is the midpoint of its diameter. That is, it divides it into two equal parts, each of which is a radius of the circle. The radius is half the diameter.

See attachment for the step by step solution of the problem

3 0

3 years ago

Thanks for the help!

Dafna11 [192]

Answer:

the 1st one i think

Explanation:

6 0

3 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$