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sergiy2304 [10]

3 years ago

11

The truck travels in a circular path having a radius of 50 m at a speed of v = 4 m>s. For a short distance from s = 0, its sp

eed is increased by v # = (0.05s) m>s 2, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s = 10 m.

1 answer:

miss Akunina [59]3 years ago

6 0

Answer:

Magnitude of velocity when s is 10 m= 4.58m/s

Magnitude of acceleration when s is 10 m = 0.653 m/s²

Explanation:

Radius of circular path = 50m

Speed = 4m/s

increase in acceleration = (0.05s)m/s²

speed and acceleration when s is 10 m = ?

Magnitude of Velocity:

From third equation of motion

$v_{f}^{2}-v_{i}^{2}=2aS\\\\v_{f}=\sqrt{2aS+v_{i}^{2}} \\\\v_{f}=\sqrt{2(0.05)(10)+(4)^{2}} \\\\v_{f}=4.58\,m/s$

Magnitude of Acceleration:

$a=\sqrt{a_{t}^{2}+a_{n}^{2}}--(1)$

The tangential component of acceleration is

$a_{t}=(0.05)(10)\\a_{t}=0.5\,m/s^{2}$

The normal component of acceleration is

$a_{n}=\frac{v^{2}}{R}\\\\a_{n}=\frac{16}{50}\\\\a_{n}=0.42\,m/s^{2}$

Substituting tangential and normal component in (1)

$a=\sqrt{(0.5)^{2} +(0.42)^{2} }$

a=0.653 m/s²

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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

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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$

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nikklg [1K]

Answer:

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

Explanation:

a) By volume.

The shrinkage factor is:

$\frac{5400cu-yd}{1-0.25} =7200cu-yd$

The volume at loose is:

$V_{loose} =V_{bank} (1+swell-factor)=7200(1+0.2)=8640cu-yd$

If the Herrywampus has a capacity of 30 cubic yard:

$\frac{8640cu-yd}{30cu-yd/trip} =288trip$

b) By weight

The swell factor in terms of percent swell is equal to:

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$pounds-per-cubic-yard-loose=\frac{3000}{\frac{20}{100} +1} =2500lb/cu-yd$

The weight of backfill is:

$8640cu-yd*2500\frac{lb}{cu-yd} *\frac{1ton}{2000lb} =10800ton$

The Herrywampus has a capacity of 40 ton:

$\frac{10800}{40ton/trip} =270trip$

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

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Answer:

Line of action of axial force for a uniform stress distribution should pass through the centroid of the cross sectional area.

Explanation:

If the line of action of the force is along the centroidal axis of the cross sectional area there is no eccentricity in the line of application of force hence no moment is generated in the cross sectional area hence we get a uniform stress distribution as theorized by hookes law.

In case of eccentric force there is an additional moment in addition of the force. This induced moment induces bending in the section thus giving a non uniform stress distribution in the section.

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