Answer to Question #177934 in Mechanical Engineering for MOHAMED SADIQUE

Question

Air is supplied to a furnace with a uniform velocity of 0.2 m/s through a pipe of diameter of 0.5 m at 400 K and 500 kPa. Determine the mass flow rate of air.

Answer 1

Q177934

Air is supplied to a furnace with a uniform velocity of 0.2 m/s through a pipe of a diameter of 0.5 m at 400 K and 500 kPa. Determine the mass flow rate of air.

Solution:

We are given the velocity of air passing through the pipe = 0.2 m/s

and the diameter of the pipe is 0.5 m. Using this information we can find the flow rate of volume through the pipe.

But in the question, we are asked the mass flow rate of air.

We can use the gas equation PM = dRT , for finding the density of air passing through the pipe.

Once we have the density of air and volume flow rate of air, we can find the mass flow rate of air.

Step 1: To find the rate of volume of air passing through the pipe.

diameter of pipe = 0.5 m.

$$radius\space of\space pipe = \frac{diameter }{2} = \frac{0.5m}{2} = 0.25m$$

$Area\space of\space cross\space section\space of\space pipe = π r^2 = 3.142 * (0.25m)^2 = 3.142* 0.0625m^2$

$$= 0.1964 m^2$$

The flow rate of air is 0.2 m/s , Using this we can find the volume of air passing through the pipe per second.

$volume\space of\space air\space passing \space through\space the pipe per\space second$

$= velocity\space of \space air * Area\space of\space cross\space section$

$= 0.2m/s * 0.1964 m^2$

$= 0.0393 m^3/s$

Step 2: To find the density of air passing through the pipe.

$$Pressure, P = 500kPa * \frac{1000Pa}{1kPa } * \frac{1atm}{101325 pascal } = 4.935 atm.$$

$$Temperature, T = 400 K$$

$$Mean\space molar\space mass\space of\space air = 28.97 g/mol$$

$$Gas\space constant, R = 0.08206 L.atm/mol.K$$

plug all this information in the, PM = d RT, formula and find the density, d of air passing through the pipe.

$$4.935atm * 28.97 g/mol = d * 0.08206L.atm/mol.K * 400K$$

Arranging this equation for density, d we have

$$density\space of\space air, d = \frac{4.935atm * 28.97g/mol }{0.08206L.atm/mol.K * 400K }$$

$$density\space of\space air, d = \frac{142.97g }{32.824L }$$

Hence the density of air passing through the pipe is 4.356 g/L .

Convert this to g/m³. Because we have volume in m³ units.

$$density\space of\space air\space in\space g/m^3 = \frac{4.356g }{1L }* \frac{1000L}{1m^3 } = 4356 g / m^3$$

Step 3: To convert the volume flow rate to mass flow rate by using the density of air.

$$density = \frac{mass}{volume }$$

$$mass = density * volume$$

mass flow rate = density of air * volume flow rate

= 4356 g/m³ * 0.0393 m³/s

= 171g/s

Hence the mass flow rate of air is 171 grams/second.

All the quantities given in the question are in 1 significant figure, so you may have to submit your answer in 1 significant figure.

The answer will be 200grams /second or 0.2 kg/second.

Please check in which units you are asked to submit the answer.

Thank you. :)