An air blower handles $˙ m$ J kg/s of air at $T_{c}$ and consumes a power of $P_{0}$ kW. The air enters the blower at $V_{c}$ m/s and exits at double the velocity at inlet. Considering adiabatic conditions. the exit temperature of air is

T_{0} + (\frac{P_{0}}{m C_{F}} - \frac{30_{0}^{2}}{2000 C_{P}})

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T_{0} + (\frac{P_{0}}{m c_{p}} + \frac{3 v_{0}^{2}}{2000 c_{p}})

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T_{0} - (\frac{P_{0}}{m c_{p}} - \frac{3 v_{0}^{2}}{2000 c_{p}})

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T_{0} - (\frac{P_{0}}{m c_{p}} + \frac{3 c_{0}^{2}}{2000 c_{p}})

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Solution

The correct option is A $T_{0} + (\frac{P_{0}}{m C_{F}} - \frac{30_{0}^{2}}{2000 C_{P}})$
Using SFEE
$m [h_{0} + \frac{v_{0}^{2}}{2000} + \frac{g Z_{0}}{1000}] + Q = m [h_{1} + \frac{v_{1}^{2}}{2000} + \frac{g Z_{1}}{1000}] + W$

$Z_{0} = Z_{1}$

$Q = 0$

$W = - P_{v}$

$V_{1} = 2 V_{0}$

$m [c_{y} T_{0} + \frac{v_{0}^{2}}{2000}] = m [c_{P} T_{1} - \frac{4 V_{0}^{2}}{2000}] - T_{0}$

$P_{0} = m [c_{y} (T_{1} - T_{0}) + \frac{3 v_{0}^{2}}{2000}]$

$T_{1} = T_{0} + (\frac{P_{0}}{m c_{P}} - \frac{3 V_{0}^{2}}{2000 c_{p}})$

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