Question

A compressible fluid flows through a compressor that increases the density from 1 kg/m3 to 5 kg/m3. The cross-sectional area of the inlet pipe is 3 m2 and that of the discharge pipe is 1 m2. The relation between the discharge volume flow rate and the inlet volume flow rate is

Answer 1

Answer:

The relation between the discharge volume flow rate and the inlet volume flow rate is $\frac{1}{5}$.

Explanation:

No matter if fluid is compressible or not, mass throughout compressor, a device that works at steady state, must be conserved according to Principle of Mass Conservation:

$\dot m_{in}-\dot m_{out} = 0$ (Eq. 1)

Where $\dot m_{in}$ and $\dot m_{out}$ are mass flows at inlet and outlet, measured in kilograms per second.

After applying Dimensional analysis, we expand the equation above as follows:

$\rho_{in}\cdot \dot V_{in} - \rho_{out}\cdot \dot V_{out} = 0$ (Eq. 2)

Where:

$\rho_{in}$, $\rho_{out}$ - Fluid densities at inlet and outlet, measured in kilograms per cubic meter.

$\dot V_{in}$, $\dot V_{out}$ - Volume flow rates at inlet and outlet, measured in cubic meters per second.

After some algebraic handling, we find the following relationship:

$\rho_{out}\cdot \dot V_{out} = \rho_{in}\cdot \dot V_{in}$

$\frac{\dot V_{out}}{V_{in}} = \frac{\rho_{in}}{\rho_{out}}$ (Eq. 3)

If we know that $\rho_{in} = 1\,\frac{kg}{m^{3}}$ and $\rho_{out} = 5\,\frac{kg}{m^{3}}$, then the relation between the discharge volume flow rate and the inlet volume flow rate is:

$\frac{\dot V_{out}}{\dot V_{in}} = \frac{1\,\frac{kg}{m^{2}} }{5\,\frac{kg}{m^{3}} }$

$\frac{\dot V_{out}}{\dot V_{in}} = \frac{1}{5}$

The relation between the discharge volume flow rate and the inlet volume flow rate is $\frac{1}{5}$.