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Serggg [28]
3 years ago
7

An inventor claims to have devised a cyclical power engine that operates with a fuel whose temperature is 750 °C and radiates wa

ste heat to a sink at 0 °C. He also claims that this engine produces 3.3 kW while rejecting heat at a rate of 4.4 kW. Is this claim valid?
Engineering
1 answer:
Phantasy [73]3 years ago
5 0

Answer:

Yes

Explanation:

Given Data

Temprature of source=750°c=1023k

Temprature of sink =0°c=273k

Work produced=3.3KW

Heat Rejected=4.4KW

Efficiency of heat engine(η)=\frac{Work produced}{Heat supplied}

and

Heat Supplied {\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )

{Q_s}=3.3+4.4=7.7KW

η=\frac{3.3}{7.7}

η=42.85%

Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink

η=1-\frac{T_ {sink}}{T_ {source}}

η=1-\frac{273}{1023}

η=73.31%

Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.

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Calculate the equivalent capacitance of the three series capacitors in Figure 12-1 A) 0.060 uF B) 0.8 uF C) 0.58 uF D) 0.01 uF
Naddik [55]

Answer:

The correct answer is option (A) 0.060 uF

Note: Kindly find an attached image of the complete question below

Sources: The complete question was well researched from Quizlet.

Explanation:

Solution

Given that:

C₁ = 0.1 μF

C₂ =0.22 μF

C₃ = 0.47 μF

In this case, C₁, C₂ and C₃ are in series

Thus,

Their equivalent becomes:

1/Ceq = (1/C₁ + 1/C₂ +1/C₃

1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]

1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]

1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)

1/Ceq =[(0.1724)/(0.01034)]

1/Ceq = [(16.67)]

1/Ceq =(1/16.67) = 0.059μf

Ceq = 0.059μf ≈ 0.060μf

Therefore the equivalent capacitance of the three series capacitors is 0.060μf

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3 years ago
Multiple Choice
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Explanation:

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