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navik [9.2K]
4 years ago
10

Whats the pros about being in the army

Engineering
1 answer:
Cerrena [4.2K]4 years ago
8 0

Answer:

Benefits that last a lifetime. The Army offers you money for education, comprehensive health care, generous vacation time, family services and support groups, special pay and cash allowances to cover the cost of living.

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What happens to the duty cycle for a GMAW Gun when 75Ar/25COzgas
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Paint can shaker mechanisms are common in paint and hardware stores. While they do a good job of mixing the paint, they are also
Ymorist [56]

Answer:

A good design for a portable device to mix paint minimizing the shaking forces and vibrations while still effectively mixing the paint. Is:

The best design is one with centripetal movement. Instead of vertical or horizontal movement. With a container and system of holding structures made of materials that could absorb the vibration effectively.

Explanation:

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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 = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

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 $

Where t is given by

1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2  \\\\t = 0.2 \: sec

So,

$ \Delta  \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2 $

$ \Delta  \delta = 6.75 \: elec \: deg

The change in torque in rpm/s is given by

$ \Delta  \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ  }   $

$ \Delta  \delta =28.12 \: \: rpm/s $

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$ Rotor \: speed = \frac{120 \cdot f}{P}  + (\Delta  \delta)\cdot t  $

Where P is the number of poles of the turbo-generator.

$ Rotor \: speed = \frac{120 \cdot 50}{4}  + (28.12)\cdot 0.2  $

$ Rotor \: speed = 1500  + 5.62  $

$ Rotor \: speed = 1505.62 \:\: rpm

4 0
3 years ago
Need help right now pls
LenaWriter [7]
Put them in 2,4,1,3 and got in 100 in the test
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3 years ago
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