Question

The wind is blowing horizontally at 30 m / s in a storm at P o , 20°C toward a wall, where it comes to a stop (stagnation) and leaves with negligible velocity similar to a diffuser with a very large exit area. Find the stagnation temperature from the energy equation.

Answer 1

Answer:

 To=20.44 °C    

Explanation:

Given that

Velocity , v= 30 m/s

Temperature , T= 20°C

We know that specific heat capacity for air ,Cp=1.005 kJ/kg.K

By using energy conservation ,the stagnation temperature is given as

$T_0=T+\dfrac{v^2}{2C_p}$

Now by putting the values in the above equation we get

$T_0=(20+273)+\dfrac{30^2}{2\times 1.005\times 1000}\ K$

To= 293.44 K

To= 293.44 - 273 °C

To=20.44 °C

Therefore the stagnation temperature will be 20.44 °C.