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7nadin3 [17]
3 years ago
14

It is said that Archimedes discovered his principle during a bath while thinking about how he could determine if King Hiero’s cr

own was actually made of pure gold. While in the bathtub, he conceived the idea that he could determine the average density of an irregularly shaped object by weighing it in air and also in water. If the crown weighed 34.7 N in air and 31.5 N in water, determine if the crown is made of pure gold. The density of gold is 19,300 kg/m3. The density of water is 1000 kg/m3.
Engineering
1 answer:
arlik [135]3 years ago
3 0

Answer: The crown is not made of pure gold.

Explanation:

Archimedes discovered that any solid, of any shape, when submerged in a liquid receives an upward force, equal to the weight of the volume of the liquid removed by the solid, which is equal to the solid volume.

So, if any body is weighed in air, the normal force will be equal to the gravity force (which we call weight) which can be expressed as follows:

Fg = m g = δ V g = 34.7 N

When submerged in water, the normal force is equal to the difference between the actual weight, and the upward force due to Archimedes' principle, called buoyant force, as follows:

Fn = Fg - Ep = δx. V. g - δH20 . V. g = 31.5 N

Dividing Fg between Fn, and simplifying common terms, we have:

δx / (δx - δh20) = 34.7 / 31.5 = 1.10

Solving for δx, we get the following value:

δx = 11,000 Kg/m3, less dense than pure gold, so we can conclude that the crown was not made of pure gold.

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Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

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Electrical power = Pelec = 50 MW

Number of poles = P = 4

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(c) rotor speed = 1505.62 rpm

Explanation:

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E = G \times H

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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

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$ \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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