Respuesta :
Answer:
(a) [tex]\eta_{max}=37.895\%[/tex]
(b) [tex]\eta=79.75\%[/tex]
(c) Impossibly
Explanation:
Given:
- temperature of source reservoir, [tex]T_H=950\ K[/tex]
- temperature of sink reservoir, [tex]T_L=590\ K[/tex]
- heat absorbed by the engine, [tex]Q_H=790\ kJ[/tex]
- heat rejected by the engine, [tex]Q_L=160\ kJ[/tex]
(a)
Now the maximum theoretical efficiency of the engine:
[tex]\eta_{max}=\frac{T_H-T_L}{T_H}\times 100\% [/tex]
[tex]\eta_{max}=\frac{950-590}{950}\times 100\% [/tex]
[tex]\eta_{max}=37.895\%[/tex]
(b)
Actual efficiency of the heat engine:
[tex]\eta=\frac{Q_H-Q_L}{Q_H}\times 100\% [/tex]
[tex]\eta=\frac{790-160}{790}\times 100\% [/tex]
[tex]\eta=79.75\%[/tex]
(c)
This is impossible because the actual efficiency can never be greater than the ideal (Carnot) efficiency of a heat engine.
The answers are:
(a) the maximum theoretical efficiency is 37.9%
(b) the actual efficiency is 79.7%
(c) The cycle is operating impossibly
The following data is provided to us:
The temperature of the source, [tex]T_1=950K[/tex]
The temperature of the sink, [tex]T_2=590K[/tex]
Heat absorbed by the engine, [tex]Q_1=790kJ[/tex]
Heat rejected by the engine, [tex]Q_2=160kJ[/tex]
(a) Now, the maximum efficiency of the engine is:
[tex]\eta _{max}=1-\frac{T_2}{T1}\\ \\\eta_{max}=1-\frac{590}{950}\\ \\\eta_{max}=0.379[/tex]
maximum efficiency ( [tex]\eta_{max}[/tex] ) = 37.9%
(b) actual efficiency:
[tex]\eta=1-\frac{Q_2}{Q_1}\\ \\\eta=1-\frac{160}{790}\\ \\\eta=0.797[/tex]
actual efficiency ([tex]\eta[/tex]) = 79.7%
(c) it is not possible for a Carnot engine to have more efficiency than the maximum possible efficiency.
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