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atroni
1 month 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]1 month 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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The rigid beam is supported by a pin at C and an A992 steel guy wire AB of length 6 ft. If the wire has a diameter of 0.2 in., d
Mrrafil [318]

Answer:

Change in length = 0.0913 in

Explanation:

Given data:

Length = 6 ft

Diameter = 0.2 in

Load w = 200 lb/ft

Solution:

We start by applying the equilibrium moment about point C, expressed as

∑M(c) = 0.............1

This can be used to find the force in AB.

10× 200 × ( 5) - (T cos(30)) × 10 = 0

Solving gives us

Tension in wire T(AB) = 1154.7 lb

We also know the modulus of elasticity for A992 is

E = 29000 ksi

And the area will be

Area = \frac{\pi }{4}\times 0.2^2

The change in length is expressed as

Change in length = \frac{PL}{AE}.........2

Substituting values results in

Change in length = \frac{1154.7 \times 6 \times 12}{\frac{\pi }{4}\times 0.2^2 \times 29000 \times 1000}

Change in length = 0.0913 in

8 0
3 months ago
An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side
Mrrafil [318]

Answer:

Total bandwidth: 8 kHz

Explanation:

Data provided:

Transmitter frequency: 3.9 MHz

Modulation up to: 4 kHz

Solution:

For the upper side frequencies:

Upper side frequencies = 3.9 × 10^{6} + 4 × 10³

Upper side frequencies = 3.904 MHz

For the lower side frequencies:

Lower side frequencies = 3.9 × 10^{6} - 4 × 10³

Lower side frequencies = 3.896 MHz

Consequently, the total bandwidth is computed as:

Total bandwidth = upper side frequencies - lower side frequencies

Total bandwidth = 8 kHz

6 0
2 months ago
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