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atroni
10 days ago
14

An inventor claims to have devised a cyclical engine for use in space vehicles that operates with a nuclear fuel generated energ

y source whose temperature is 510 K and sink at 270 K that radiates waste heat to deep space. He also claims that this engine produces 4.1 kW while rejecting heat at a rate of 15,000 kJ/h. Is this claim valid?
Engineering
1 answer:
Viktor [391]10 days ago
6 0

Response:

The claim made by the inventor is not valid.

Reasoning:

We have the following parameters:

Source temperature = 510 K

Sink temperature = 270 K

Power generated = 4.1 KW

Heat disposal = 15,000 KJ/h

Heat disposal = 4.16 KW

It is known that

Heat input = Heat output + Power generated

Heat input = 4.16 + 4.1 KW

Heat input = 8.16 KW

Thus, the efficiency of the engine

\eta =\dfrac{Power\ produces}{heat\ added}

\eta =\dfrac{4.1}{8.16}

\eta =0.502

Now, let’s determine the maximum efficiency achievable with a Carnot engine

The efficiency for a Carnot engine is expressed as

\eta =1-\dfrac{T_L}{T_H}

By substituting the values

\eta =1-\dfrac{T_L}{T_H}

\eta =1-\dfrac{270}{510}

\eta =0.47

Therefore, the efficiency of the Carnot cycle falls short of the efficiency calculated for the aforementioned engine. Hence, this engine is not feasible. This indicates that the inventor's assertion lacks validity.

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Answer:

The Python code is provided below with suitable comments

Explanation:

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1 month ago
A thin-film gold conductor for an integrated circuit in a satellite application is deposited from a vapor, and the deposited gol
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Answer:

Explanation:

The equilibrium vacancy concentration can be described by:

nv/N = exp(-ΔHv/KT),

where T is the temperature at which vacancies form,

K = Boltzmann's constant,

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1 month ago
Write cout statements with stream manipulators that perform the following:
grin007 [323]

Answer:

A)cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B)cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C)cout<<fixed<<5.789E12;

D)cout<<left<<setw(7)<<67;

Explanation:

Stream Manipulators are special functions for use with the insertion (<<) and extraction (>>) operators on C++ stream objects, while the 'cout' statement outputs content to the standard output device in C++ programming.

setw: specifies the minimum width of the output field

setprecision: defines the number of decimal places for floating-point value formatting.

fixed: sets the format flag for floating-point streams.

left: left-aligns the output.

A) This statement shows the number 34.789 in a field that provides eight character spaces with two decimal precision. cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B) Here, the number 7.0 is displayed within six spaces with three decimal precision. cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

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1 month 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
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

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