The 60-W fan of a central heating system is to circulate air through the ducts. The analysis of the flow shows that the fan needs to raise the pressure of air by 50 Pa to maintain flow. The fan is located in a horizontal flow section whose diameter is 30 cm t both the inlet and the outlet. Determine the highest possible average flow velocity in the duct.

Respuesta :

Answer:

the highest possible average flow velocity is 17 m/s

Explanation:

given data

power  = 60 W

pressure = 50 Pa

diameter = 30 cm = 0.3 m

to find out

Determine the highest possible average flow velocity in the duct

solution

we will apply here energy balance equation that is

energy in = energy out

power  + m(Pv)1 = m (Pv)2

power = mv ( P2 -P1)

power = V ΔP

here P is pressure and V is maximum volume flow rate

put here value and find  V

60 = V ( 50)

maximum volume v = 1.2 m³/s

and

maximum average velocity is

Velocity = [tex]\frac{V}{\frac{\pi }{4} D^2 }[/tex]

here D is diameter and V is volume

so

velocity = [tex]\frac{1.2}{\frac{\pi }{4} 0.3^2 }[/tex]

velocity = 17 m/s

so the highest possible average flow velocity is 17 m/s

The highest possible average flow velocity in the duct is 17 m/s

Energy balance equation:

Here we applied the above equation i.e.

energy in = energy out

power  + m(Pv)1 = m (Pv)2

power = mv ( P2 -P1)

power = V ΔP

where,

P is pressure and V is maximum volume flow rate

Now  V should be

60 = V ( 50)

So, maximum volume v = 1.2 m³/s

Now the maximum velocity should be

[tex]= 1.2 \div (\pi \div 4 (0.3)^2)[/tex]

= 17 m/s

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