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3 years ago

What energy does a curcuit board run on

Engineering

2 answers:

Citrus2011 [14]3 years ago

8 0

Answer:

energy from computer

Explanation:

son4ous [18]3 years ago

6 0

a curcuit board is powered by energy from the computers power soarce

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(20 points) A 1 mm diameter tube is connected to the bottom of a container filled with water to a height of 2 cm from the bottom

zzz [600]

Solution :

Given :

h = 2 cm

Diameter of the tube , d = 1 mm

Diameter of the hose, D = 6 mm

Between 1 and 2, by applying Bernoulli's principle, we get

As point 1 is just below the free surface of liquid, so

$P_1=P_{atm} \text{ and} \ V_1=0$

$\frac{P_{atm}}{\rho g}+\frac{v_1^2}{2g} +h = \frac{P_2}{\rho g}$

$\frac{101.325}{1000 \times 9.81}+0.02 =\frac{P_2}{\rho g}$

$P_2 = 111.35 \ kPa$

Therefore, 111.325 kPa is the gas supply pressure required to keep the water from leaking back into the tube.

Velocity at point 2,

$V_2=\sqrt{\left(\frac{111.135}{\rho g}+0.02}\right)\times 2g$

= 1.617 m/s

Flow of water, $Q_2 = A_{tube} \times V_2$

$=\frac{\pi}{4} \times (10^{-3})^2 \times 1.617$

$1.2695 \times 10^{-6} \ m^3/s$

Minimum air flow rate,

$Q_2 = Q_3 = A_{hose} \times V_3$

$V_3 = \frac{Q_2}{\frac{\pi}{4}D^2}$

$V_3 = \frac{1.2695 \times10^{-6}}{\pi\times 0.25 \times 36 \times 10^{-6}}$

= 0.0449 m/s

b). Reynolds number in hose,

$Re = \frac{\rho V_3 D}{\mu} = \frac{V_3 D}{\nu}$

υ for water at 25 degree Celsius is $8.9 \times 10^{-7} \ m^2/s$

υ for air at 25 degree Celsius is $1.562 \times 10^{-5} \ m^2/s$

$Re_{hose}=\frac{0.0449 \times 6 \times 10^{-3}}{1.562 \times 10^{-5}}$

= 17.25

Therefore the flow is laminar.

Reynolds number in the pipe

$Re = \frac{V_2 d}{\nu} = \frac{1.617 \times 10^{-3}}{8.9 \times 10^{-7}}$

= 1816.85, which is less than 2000.

So the flow is laminar inside the tube.

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