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3 years ago

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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 need

s 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.

1 answer:

MArishka [77]3 years ago

8 0

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 = $\frac{V}{\frac{\pi }{4} D^2 }$

here D is diameter and V is volume

so

velocity = $\frac{1.2}{\frac{\pi }{4} 0.3^2 }$

velocity = 17 m/s

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

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According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this

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Answers:

a) $\rho_{cylinder}= 0.55 g/cm^{3}$

b) $\rho_{liq}= 1.48 g/cm^{3}$

c) When we divided both volumes (sumerged and displaced) the factor $\pi r^{2}$ is removed during calculations.

Explanation:

a) According to <u>Archimedes’ Principle:</u>

<em>A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.</em>

In this case we have a wooden cylinder floating (partially immersed) in water. <u>This object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium.</u> This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero.

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$B$ is the Buoyant force, which is the force the fluid (water in this situation) exert in the submerged cylinder, and is directed upwards.

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$m_{cylinder}g=m_{water}g$ (2)

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$\rho=\frac{m}{V}$ (3)

isolating the mass:

$m=\rho V$ (4)

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$\rho_{cylinder} V_{cylinder} g=\rho_{water} V_{water} g$ (5)

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$V_{cylinder}=\pi r^{2} h$ (6)

Where $r$ is the radius and $h=30 cm$ the total height of the cylinder.

$V_{water}=\pi r^{2} (h-h_{top})$ (7)

Where $h_{top}=13.5 cm$ is the height of the top of the cylinder above the surface of water and $(h-h_{top})$ is the height of the sumerged part of the cylinder.

Substituting (6) and (7) in (5):

$\rho_{cylinder} \pi r^{2} h g=\rho_{water} \pi r^{2} (h-h_{top}) g$ (8)

Clearing $\rho_{cylinder}$ :

$\rho_{cylinder}=\frac{\rho_{water}(h-h_{top})}{h}$ (9)

Simplifying;

$\rho_{cylinder}=\rho_{water}(1-\frac{h_{top}}{h}$ (10)

Knowing $\rho_{water}=1g/cm^{3}$ :

$\rho_{cylinder}=1g/cm^{3}(1-\frac{13.5 cm}{30cm})$ (11)

$\rho_{cylinder}= 0.55 g/cm^{3}$ (12) This is the density of the wooden cylinder

b) Now we have a different situation, we have the same wooden cylinder, which density was already calculated ( $\rho_{cylinder}= 0.55 g/cm^{3}$ ), but the density of the liquid $\rho_{liq}$ is unknown.

Applying again the Archimedes principle:

$\rho_{cylinder} V_{cylinder} g = \rho_{liq} V_{liq} g$ (13)

Isolating $\rho_{liq}$ :

$\rho_{liq}= \frac{\rho_{cylinder} V_{cylinder}}{V_{liq}}$ (14)

Where:

$V_{cylinder}=\pi r^{2} h$

$V_{liq}=\pi r^{2} (h-h_{top})$

Then:

$\rho_{liq}= \frac{\rho_{cylinder} \pi r^{2} h}{\pi r^{2} (h-h_{top})}$ (15)

$\rho_{liq}= \frac{\rho_{cylinder} h}{h-h_{top}}$ (16)

$\rho_{liq}= \frac{0.55 g/cm^{3} (30 cm)} {30 cm - 18.9 cm}$ (17)

$\rho_{liq}= 1.48 g/cm^{3}$ (18) This is the density of the liquid

c) As we can see, it was not necessary to know the radius of the cylinder (we did not need to knoe its length and width), we only needed to know the part that was sumerged and the part that was above the surface of the liquid.

This is because in this case, when we divided both volumes (sumerged and displaced) the factor $\pi r^{2}$ is removed during calculations.

6 0

4 years ago