Question

In a hypothetical nuclear fusion reactor, the fuel is deuterium gas at a temperature of about 9.5 x 108 K. If this gas could be used to operate a Carnot engine with TL = 864°C, what would be the engine's efficiency?

Answer 1

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:

$Carnot's\ Efficiency=\frac {T_H-T_L}{T_H}\times 100 \%$

So,

$Carnot's\ Efficiency=\frac {(9.5\times 10^8)-1137.15}{9.5\times 10^8}\times 100 \%$

Engine's Efficiency = 99.9998%

                                     

Answer 2

Answer:99.99 %

Explanation:

Given

Temperature of deuterium gas$\left ( T_H\right ) =9.5\times 10^8 k$

Lower temperature$\left ( T_L\right )=864^{\circ}C\approx 1137 k$

We know that Engine efficiency $\left ( \eta \right )$ is given by

$\eta =\frac{Work ouput}{heat supplied}$

and carnot efficiency$\left ( \eta \right )=1-\frac{T_L}{T_H}-----------\left ( Maximum\ efficiency\right )$

$\eta =1-\frac{1173}{9.5\times 10^8}$

$\eta =0.9999$

99.99%