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3 years ago

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A desktop computer is to be cooled by a fan whose flow rate is 0.34 m3/min. Determine the mass flow rate of air through the fan

at an elevation of 3400 m where the air density is 0.7 kg/m3. Also, if the average velocity of air is not to exceed 103 m/min, determine the diameter of the casing of the fan.

1 answer:

erik [133]3 years ago

3 0

Answer:65 mm

Explanation:

Given

Flow rate $\dot{V}=0.34\ m^3/min$

density of air $\rho =0.7\ kg/m^3$

average velocity of air $v_{avg}=103\ m/min$

Mass flow rate is given by

$\dot{m}=\rho \times \dot{V}$

$\dot{m}=0.7\times 0.34$

$\dot{m}=0.238\ kg/min$

And Flow rate can be written as

$\dot{V}=A\cdot v_{avg}$

Where A =Area of cross-section

$0.34=\frac{\pi}{4}d^2\cdot v_{avg}$

$0.34=\frac{\pi}{4}\times d^2\times 103$

$d=0.0648\ m$

$d=64.8\approx 65\ mm$

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The Reynolds number is the major parameter that relates fluid flow momentum to friction forces. How is the Reynolds number defin

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Answer:

Reynolds number determines whether a flow is laminar or turbulent flow.

Explanation:

Reynolds number is defined as ratio of inertia force to the viscous force. it is a dimension less number. Reynolds number is used to describe the type of flow in a fluid whether it is laminar flow or turbulent flow. Reynolds number is denoted by Re.

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3 years ago

Sea B = 5.00 m a 60.0°. Sea C que tiene la misma magnitud que A y un ángulo de dirección mayor que el de A en 25.0°. Sea A ⦁ B =

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Answer:

$\| \vec A \| = 6.163\,m$

Explanation:

Sean A, B y C vectores coplanares tal que:

$\vec A = (\| \vec A \|\cdot \cos \theta_{A},\| \vec A \|\cdot \sin \theta_{A})$ , $\vec B = (\| \vec B \|\cdot \cos \theta_{B},\| \vec B \|\cdot \sin \theta_{B})$ y $\vec C = (\| \vec C \|\cdot \cos \theta_{C},\| \vec C \|\cdot \sin \theta_{C})$

Donde $\| \vec A \|$ , $\| \vec B \|$ y $\| \vec C \|$ son las normas o magnitudes respectivas de los vectores A, B y C, mientras que $\theta_{A}$ , $\theta_{B}$ y $\theta_{C}$ son las direcciones respectivas de aquellos vectores, medidas en grados sexagesimales.

Por definición de producto escalar, se encuentra que:

$\vec A \,\bullet\, \vec B = \|\vec A \| \| \vec B \| \cos \theta_{B}\cdot \cos \theta_{A} + \|\vec A \| \| \vec B \| \sin \theta_{B}\cdot \sin \theta_{A}$

$\vec B \,\bullet\, \vec C = \|\vec B \| \| \vec C \| \cos \theta_{B}\cdot \cos \theta_{C} + \|\vec B \| \| \vec C \| \sin \theta_{B}\cdot \sin \theta_{C}$

Asimismo, se sabe que $\| \vec B \| = 5\,m$ , $\theta_{B} = 60^{\circ}$ , $\vec A \,\bullet \,\vec B = 30\,m^{2}$ , $\vec B\, \bullet\, \vec C = 35\,m^{2}$ , $\|\vec A \| = \| \vec C \|$ y $\theta_{C} = \theta_{A} + 25^{\circ}$ . Entonces, las ecuaciones quedan simplificadas como siguen:

$30\,m^{2} = 5\|\vec A \| \cdot (\cos 60^{\circ}\cdot \cos \theta_{A} + \sin 60^{\circ}\cdot \sin \theta_{A})$

$35\,m^{2} = 5\|\vec A \| \cdot [\cos 60^{\circ}\cdot \cos (\theta_{A}+25^{\circ}) + \sin 60^{\circ}\cdot \sin (\theta_{A}+25^{\circ})]$

Es decir,

$30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})$

$35\,m^{2} = \| \vec A \| \cdot [2.5\cdot \cos (\theta_{A}+25^{\circ})+4.330\cdot \sin (\theta_{A}+25^{\circ}})]$

Luego, se aplica las siguientes identidades trigonométricas para sumas de ángulos:

$\cos (\theta_{A}+25^{\circ}) = \cos \theta_{A}\cdot \cos 25^{\circ} - \sin \theta_{A}\cdot \sin 25^{\circ}$

$\sin (\theta_{A}+25^{\circ}) = \sin \theta_{A}\cdot \cos 25^{\circ} + \cos \theta_{A} \cdot \sin 25^{\circ}$

Es decir,

$\cos (\theta_{A}+25^{\circ}) = 0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A}$

$\sin (\theta_{A}+25^{\circ}) = 0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A}$

Las nuevas expresiones son las siguientes:

$30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})$

$35\,m^{2} = \| \vec A \| \cdot [2.5\cdot (0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A})+4.330\cdot (0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A})]$

Ahora se simplifican las expresiones, se elimina la norma de $\vec A$ y se desarrolla y simplifica la ecuación resultante:

$30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})$

$35\,m^{2} = \| \vec A \| \cdot (4.097\cdot \cos \theta_{A} +2.865\cdot \sin \theta_{A})$

$\frac{30\,m^{2}}{2.5\cdot \cos \theta_{A}+ 4.330\cdot \sin \theta_{A}} = \frac{35\,m^{2}}{4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}}$

$30\cdot (4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}) = 35\cdot (2.5\cdot \cos \theta_{A}+4.330\cdot \sin \theta_{A})$

$122.91\cdot \cos \theta_{A} + 85.95\cdot \sin \theta_{A} = 87.5\cdot \cos \theta_{A} + 151.55\cdot \sin \theta_{A}$

$35.41\cdot \cos \theta_{A} = 65.6\cdot \sin \theta_{A}$

$\tan \theta_{A} = \frac{35.41}{65.6}$

$\tan \theta_{A} = 0.540$

Ahora se determina el ángulo de $\vec A$ :

$\theta_{A} = \tan^{-1} \left(0.540\right)$

La función tangente es positiva en el primer y tercer cuadrantes y tiene un periodicidad de 180 grados, entonces existen al menos dos soluciones del ángulo citado:

$\theta_{A, 1} \approx 28.369^{\circ}$ y $\theta_{A, 2} \approx 208.369^{\circ}$

Ahora, la magnitud de $\vec A$ es:

$\| \vec A \| = \frac{35\,m^{2}}{4.097\cdot \cos 28.369^{\circ} + 2.865\cdot \sin 28.369^{\circ}}$

$\| \vec A \| = 6.163\,m$

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4 years ago

Air is to be compressed so that its volume is reduced to half of its original volume if the initial pressure is 200 kPa and the

Alex_Xolod [135]

Answer:

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Where $\gamma$ is the ratio of specific heats

The final pressure is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}}\right)^{\gamma}$

$P_{2} = (200\,kPa)\cdot (2)^{1.4}$

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3 years ago

Hey answr this sajida Yusof

Allisa [31]

What do u need? Rusbaisuwvwbs

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3 years ago

Read 2 more answers