Question

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle

Answer 1

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$: is the initial velocity = 20 m/s

$A_{i}$: is the inlet area of the nozzle = 60 cm²  

$\rho_{i}$: is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

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Answer 2

a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.

We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:

$\dot m_{in} = \dot m_{out}$ (1)

Where:

$\dot m_{in}$ - Inlet mass flow, in kilograms per second.

$\dot m_{out}$ - Outlet mass flow, in kilograms per second.

Given that air flows at constant rate, we expand (1) by dimensional analysis:

$\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}$ (2)

Where:

$\rho_{in}, \rho_{out}$ - Air density at inlet and outlet, in kilograms per cubic meter.

$A_{in}, A_{out}$ - Inlet and outlet area, in square meters.

$v_{in}, v_{out}$ - Inlet and outlet velocity, in meters per second.

a) If we know that $\rho_{in} = 2.21\,\frac{kg}{m^{3}}$, $A_{in} = 60\times 10^{-4}\,m^{2}$ and $v_{in} = 20\,\frac{m}{s}$, then the mass flow rate through the nozzle is:

$\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}$

$\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)$

$\dot m = 0.265\,\frac{kg}{s}$

The mass flow rate through the nozzle is 0.265 kilograms per second.

b) If we know that $\rho_{in} = 2.21\,\frac{kg}{m^{3}}$, $A_{in} = 60\times 10^{-4}\,m^{2}$, $v_{in} = 20\,\frac{m}{s}$, $\rho_{out} = 0.762\,\frac{kg}{m^{3}}$ and $v_{out} = 150\,\frac{m}{s}$, then the exit area of the nozzle is:

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

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

$A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}$

$A_{out} = 2.320\times 10^{-3}\,m^{2}$

$A_{out} = 23.202\,cm^{2}$

The exit area of the nozzle is 23.202 square centimeters.