Physics, 30.07.2021 01:30 ellycleland16

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

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Answer from: nickp1233

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.

You can find another example of mass flow rate here:

I hope it helps you!

Answer from: cristalcastro901

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.

Answer from: Quest

34535543; (; ; $

Answer from: Quest

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .