Question 325: Thermodynamics & KTG

Question

Let $$\eta_{1}$$ is the efficiency of an engine at $$T_{1}=447^{\circ} \mathrm{C}$$ and $$\mathrm{T}_{2}=147^{\circ} \mathrm{C}$$ while $$\eta_{2}$$ is the efficiency at $$\mathrm{T}_{1}=947^{\circ} \mathrm{C}$$ and $$\mathrm{T}_{2}=47^{\circ} \mathrm{C}$$ The ratio $$\frac{\eta_{1}}{\eta_{2}}$$ will be :

Accepted Answer

$${\eta _1} = 1 - {{420} \over {720}} = {{300} \over {720}}$$
And $${\eta _2} = 1 - {{320} \over {1220}} = {{900} \over {1220}}$$
$$ \Rightarrow {{{\eta _1}} \over {{\eta _2}}} = {{300} \over {720}} 	imes {{1220} \over {900}}$$
$$ \simeq 0.56$$

Answer

0.41

Answer

0.56

Answer

0.73

Answer

0.70

Thermodynamics & KTG

Let $\eta_{1} is the efficiency of an engine at T_{1}=447^{\circ} \mathrm{C} and \mathrm{T}_{2}=147^{\circ} \mathrm{C} while \eta_{2} is the efficiency at \mathrm{T}_{1}=947^{\circ} \mathrm{C} and \mathrm{T}_{2}=47^{\circ} \mathrm{C} The ratio \frac{\eta_{1}}{\eta_{2}}$ will be :

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