Question

A diffuser has an inlet area of 60 cm2. Air enters the diffuser with a density of 0.748 kg/m2. The velocity at the inlet is 150 m/sec. It exits with a density of 2.0 kg/m2 at a velocity of 25 m/sec. Determine the area of the outlet.

Answer 1

Answer:

$A_{out} = 134.64\,cm^{2}$

Explanation:

The area of the outlet can be found by means of the Principle of Mass Conservation. A diffuser works usually as a steady state system:

$\dot m_{in} - \dot m_{out} = 0$

$\rho_{in}\cdot A_{in}\cdot v_{in}- \rho_{out}\cdot A_{out}\cdot v_{out} = 0$

The area of the outlet is:

$A_{out} =\frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}$

$A_{out} = \frac{(0.748\,\frac{kg}{m^{3}} )\cdot (60\,cm^{2})\cdot (150\,\frac{m}{s})}{(2\,\frac{kg}{m^{3}} )\cdot (25\,\frac{m}{s} )}$

$A_{out} = 134.64\,cm^{2}$