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Answer:
The fraction of heat that is converted to work is 0.7667 or 76.67% of the heat of burned gases.
Explanation:
Given:
Temperature of hot burned gases is, [tex]T_H=1200\ K[/tex]
Temperature of cold air is, [tex]T_C=280\ K[/tex]
Now, the efficiency of the airplane engine in terms of temperature is given as:
[tex]\eta=\dfrac{T_H-T_C}{T_H}[/tex]
Plug in the given values and solve for [tex]\eta[/tex]. This gives,
[tex]\eta=\frac{1200-280}{1200}\\\\\eta=\frac{920}{1200}=0.7667[/tex]
Now, efficiency in terms of work and heat input is given as:
[tex]\eta=\frac{Work}{Heat\ in}\\\\\eta=\frac{W}{Q_H}[/tex]
As input heat is from hot burned gases. So, [tex]Heat\ in=Q_H[/tex]
Now, plug in the given value of efficiency and express "Work" in terms of input heat, [tex]Q_H[/tex]. This gives,
[tex]0.7667=\frac{Work}{Q_H}\\\\Work=0.7667Q_H[/tex]
Therefore, the fraction of heat that is converted to work is 0.7667 or 76.67% of the heat of burned gases.
Fraction of the heat converted into work is 76.66%
Given that;
Efficient airplane engine provides work as heat TH = 1200 K
Burned gases heat TL = 280 K
Find:
Fraction of the heat converted into work
Computation:
Fraction of the heat converted into work = [(TH - TL) / TH]100
Fraction of the heat converted into work = [(1200 - 280) / 1200]100
Fraction of the heat converted into work = [(1200 - 280) / 12]
Fraction of the heat converted into work = [920 / 12]
Fraction of the heat converted into work = 76.66%
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