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Artyom0805 [142]
3 years ago
11

Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe

r meter of pipe (in Pa/m)
Engineering
1 answer:
Sonja [21]3 years ago
4 0

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe\left ( D\right )=0.15m

Velocity of water in pipe\left ( V\right )=15cm/s

We know viscosity of water is\left (\mu\right )=8.90\times10^{-4}pa-s

Pressure drop is given by hagen poiseuille equation

\Delta P=\frac{128\mu \L Q}{\pi D^4}

We have asked pressure Drop per unit length i.e.

\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}

Substituting Values

\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}

\frac{\Delta P}{L}=0.1898 Pa/m

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The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6*g, respectively.
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Answer:

The speed of shaft is 1891.62 RPM.

Explanation:

given that

Amplitude A= 0.15 mm

Acceleration = 0.6 g

So

we can say that acceleration= 0.6 x 9.81

acceleration,a=5.88\ \frac{m}{s^2}

We know that

a=\omega ^2A

So now by putting the values

a=\omega ^2A

5.88=\omega ^2 \0.15\times 10^{-3}

\omega =198.09\ \frac{rad}{s}

We know that

  ω= 2πN/60

198.0=2πN/60

N=1891.62 RPM

So the speed of shaft is 1891.62 RPM.

                                               

       

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8 0
4 years ago
The 150-lb man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb>ft. Determine t
irina1246 [14]

Answer:

M = 281.25 lb*ft

Explanation:

Given

W<em>man</em> = 150 lb

Weight per linear foot of the boat: q = 3 lb/ft

L = 15.00 m

M<em>max</em> = ?

Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):

∑ Fy = 0  (+↑)     ⇒    q'*L - W - q*L = 0

⇒       q' = (W + q*L) / L

⇒       q' = (150 lb + 3 lb/ft*15 ft) / 15 ft

⇒       q' = 13 lb/ft   (+↑)

The free body diagram of the boat is shown in the pic.

Then, we apply the following equation

q(x) = (13 - 3) = 10   (+↑)

V(x) = ∫q(x) dx = ∫10 dx = 10x   (0 ≤ x ≤ 7.5)

M(x) = ∫10x dx = 5x²  (0 ≤ x ≤ 7.5)

The maximum internal bending moment occurs when x = 7.5 ft

then

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8 0
3 years ago
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