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:99.99 %

Step-by-step explanation:

Given

Temperature of deuterium gas
`\left ( T_H\right ) =9.5* 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 =(Work ouput)/(heat supplied)`

and carnot efficiency
`\left ( \eta \right )=1-(T_L)/(T_H)-----------\left ( Maximum\ efficiency\right )`

`\eta =1-(1173)/(9.5* 10^8)`

`\eta =0.9999`

99.99%

Answer 2

Engine's Efficiency = 99.9998%

Step-by-step 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}* 100 \%`

So,

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

Engine's Efficiency = 99.9998%