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3 years ago

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Air compressed in car engine from 25 C and 100 kPa in reversible and adiabatic manner. If the compression ratio = 9.058, determi

ne final temperature of air.

1 answer:

prisoha [69]3 years ago

3 0

Explanation:

It is given that,

Initial temperature, $T_1=25^{\circ}C=298\ K$

Pressure, $P_1=100\ kPa=10^5\ Pa$

Compression ratio, $r=\dfrac{V_1}{V_2}=9.058$

Let T₂ is the final temperature of air. Using the relation for reversible adiabatic process as :

$\dfrac{T_2}{T_1}=(\dfrac{V_1}{V_2})^{\gamma-1}$ ............(1)

Where,

$\gamma=\dfrac{C_p}{C_v}$

For air, $C_p=1.004$ and $C_v=0.717$

$\gamma=1.4$

So, equation (1) becomes :

$T_2=T_1\times (\dfrac{V_1}{V_2})^{\gamma-1}$

$T_2=298\times (9.058)^{1.4-1}$

$T_2=719.49\ K$

So, the final temperature of air is 719.49 K. Hence, this is the required solution.

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