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3 years ago

5

What do weather and climate have in common? (3 points) a .Both affect average ocean temperatures.b. Both are averages measured o

ver a long period of time.
c.Both apply to a certain region.

1 answer:

aleksandrvk [35]3 years ago

4 0

Answer:

C

Explanation:

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Reynolds number E. What is the mean velocity u. (ft/s) and the Reynolds number Re = pu., D/ for 35 gpm (gallons per minute) of w

Novay_Z [31]

Answer:

The mean velocity is 13 ft/s.

The Reynolds number is 88,583 and it is dimensionless.

Explanation:

We have water flowing in a pipe of 1.05 in diameter.

The density is ρ=62.3 lb/ft and the viscosity is 1.2 cP.

The mean velocity can be calculated as

$u=\frac{Q}{A}=\frac{Q}{\pi*D^2/4}=\frac{35gpm }{3.14*(1.05in)^2/4}\\\\ u=\frac{35}{0.865}*\frac{gal}{min}\frac{1}{in^2}*\frac{231in^3}{1gal}*\frac{1}{60s} \\\\ u=156\,in/s=13\,ft/s$

The Reynolds number now can be calculated for this flow as

$Re=\frac{\rho*u*D}{\mu}$

being ρ: density, u: mean velocity of the fluid, D: internal diameter of the pipe and μ the dynamic viscosity.

To simplify the calculation, we can first make all the variables have coherent units.

<em>Viscosity</em>

$\mu=1.2cP=\frac{1.2}{100}\frac{g}{cm*s}*\frac{1lb}{453.6g}*\frac{30.48cm}{1ft}= 0.0008\frac{lb}{ft*s}$

<em>Diameter</em>

$D=1.05in*(\frac{1ft}{12in} )=0.0875ft$

Then the Reynolds number is

$Re=\frac{\rho*u*D}{\mu}\\\\Re=62.3\frac{lb}{ft^3}*13\frac{ft}{s} *0.0875ft*\frac{1}{0.0008}*\frac{ft*s}{lb}\\\\Re=88,583$

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