Estimate the endurance strength of a 1.5-in-diameter rod of AISI 1040 steel having a machined finish and heat-treated to a tensi

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

Mrrafil [318]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Inertia 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) Calculate the rotor's stored energy at synchronous speed.

The stored energy is represented as

$E = G \times H$

Where G stands for complex rated power and H signifies the inertia constant of the turbo-generator.

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

(b) If we suddenly increase the mechanical input to 80 MW against an electrical load of 50 MW, we shall find the rotor's acceleration while ignoring mechanical and electrical losses.

The formula for rotor acceleration is given by

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

Where M is defined as

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

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

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

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

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

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

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

(c) If the acceleration derived in part (b) persists over 10 cycles, we will calculate both the change in torque angle and the rotor speed in revolutions per minute at the end of this duration.

The change in torque angle is expressed as

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

Where t is determined from

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

Consequently,

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

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

The change in torque in rpm/s is provided by

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

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

The rotor speed in rpm at the culmination of this 10-cycle period is calculated as

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

Where P indicates the number of poles on 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

2 months ago

Given a 5x5 matrix for Playfair cipher a. How many possible keys does the Playfair cipher have? Ignore the fact that some keys m

alex41 [359]

Answer:

a. 25! = $2^{84}$ (Approximately)

b. 24!

Explanation:

a. In a Playfair cipher, there are 25 keys available because it is structured in a 5 * 4 grid. By using permutations to enumerate all potential configurations, we derive: 25! = 1.551121004×10²⁵ = $2^{84}$

Although there are 26 letters available, in the Playfair cipher, the letters 'i' and 'j' are treated as a single letter.

b. Considering any configuration of 5x5, each of the four row shifts yields equivalent configurations, amounting to five total equivalencies. Similarly, for each of these five setups, any of the four column shifts also results in equivalent arrangements. Therefore, each configuration corresponds to 25 equivalent arrangements. Consequently, the total count of distinct keys can be expressed as:

25!/25 = 24! = 6.204484017×10²³

6 0

1 month ago

In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

Daniel [329]

The value obtained is 0.60. From the given conditions, we take k = 1.4 for air and r = 16. We know that for calculations involving diesel engine efficiency, we arrive at 0.60.

7 0

1 month ago

Water flows steadily through a horizontal nozzle, discharging to the atmosphere. At the nozzle inlet the diameter is D1; at the

mote1985 [299]

The inlet gauge pressure must be 61.627 Psi.

4 0

2 months ago

A jar made of 3/16-inch-thick glass has an inside radius of 3.00 inches and a total height of 6.00 inches (including the bottom

mote1985 [299]

Response:

1. To find the volume of the glass shell (Vg), simply subtract the volume of the empty part of the jar (Ve) from the total volume of the jar (Vj):

Vg = Vj - Ve

Volume can be calculated by multiplying the base (B) with the height (h). The base of the jar is a circle, thus its area is πr^2 (where r indicates the radius).

The radius differs based on the jar's section: the inner radius for the empty part is d = 3 in, while for the total jar it includes the glass thickness a = 3 + 3/16 = 3.1875 in.

The height of the entire jar is given as h = 6 in, whereas for the empty portion, it's the total height minus the thickness of the glass h' = 6 - 0.1875 = 5.8125 in.

Now we can perform the calculations:

Vj = πa^2 • h = 191.42 in^3

Ve = πd^2 • h' = 164.26 in^3

Thus, the volume of the glass shell equals Vj - Ve, resulting in 27.16 in^3.

2. The mass of the glass jar can be determined by multiplying the density of the glass with the volume:

m = ρ • Vg

The glass density is provided in cubic feet, so we first convert it to cubic inches by dividing by 1728:

ρ = 165 lb/ft^3 / 1728 = 0.095 lb/in^3

m = 0.095 lb/in^3 • 27.16 in^3 = 2.59 lb

5. To calculate the weight and volume of the displaced water, we first need to ascertain how deep the jar sinks (H), as the volume of displaced water equals the submerged volume of the jar. The jar will descend until the gravitational force downwards equals the buoyancy force upwards. The displaced water volume is πa^2 • H, and the buoyancy is calculated as ρw • g • Vd (where ρw is the density of water, defined as 62.5 lb/ft^3 / 1728 = 0.036 lb/in^3, and Vd is the displaced water volume).

Thus, the buoyancy can be represented as:

B = ρw • g • πa^2 • H

Setting buoyancy equal to gravity:

B = m • g (where m is the mass of the jar). Therefore, we have:

ρw • g • πa^2 • H = m • g

From this, simplifying gives:

ρw • πa^2 • H = m

We can derive H:

H = m / (ρw • πa^2)

H = 2.25 inches

This indicates the jar will sink 2.25 inches into the water.

3. Calculating the volume of displaced water is straightforward. It matches the volume of the submerged jar:

Vd = πa^2 • H

Vd = 71.94 in^3

4. Lastly, to determine the weight of the displaced water:

m = ρw • Vd

m = 0.036 lb/in^3 • 71.94 in^3

m = 2.59 lb

As evident, the mass of the jar aligns with the mass of the displaced water. Following this logic could have simplified our calculations, but I chose to elaborate for clarity.

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6 0

2 months ago