Respuesta :
Answer:
Engine's Efficiency = 99.9998%
Explanation:
Given :
The temperature of the gas is , TH = 9.5 *10⁸ K
The operation temperature of the gas is , TL = 864 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
TL = (864 + 273.15) K = 1137.15 K
The engine's efficiency of a Carnot engine is:
[tex]Carnot's\ Efficiency=\frac {T_H-T_L}{T_H}\times 100 \%[/tex]
So,
[tex]Carnot's\ Efficiency=\frac {(9.5\times 10^8)-1137.15}{9.5\times 10^8}\times 100 \%[/tex]
Engine's Efficiency = 99.9998%
Answer:99.99 %
Explanation:
Given
Temperature of deuterium gas[tex]\left ( T_H\right ) =9.5\times 10^8 k[/tex]
Lower temperature[tex]\left ( T_L\right )=864^{\circ}C\approx 1137 k[/tex]
We know that Engine efficiency [tex]\left ( \eta \right )[/tex] is given by
[tex]\eta =\frac{Work ouput}{heat supplied} [/tex]
and carnot efficiency[tex]\left ( \eta \right )=1-\frac{T_L}{T_H}-----------\left ( Maximum\ efficiency\right ) [/tex]
[tex]\eta =1-\frac{1173}{9.5\times 10^8}[/tex]
[tex]\eta =0.9999[/tex]
99.99%
