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For a National Board Exam Review:

Calculate the Reynold's Number, $Re$ for water at 20 °C flowing in an open channel. The water flowing at a volumetric rate of 200 gal/sec. The channel has a height of 4 ft and a width of 8 ft. At this temperature, water has a kinetic viscosity of 1.104*10^-5 ft^2/s

The given answer is 600,000. I have tried to solve the problem like this:

$${ D_{eq} = \frac{2*4*8}{4+8} = \frac{16}{3} }$$

$${ V = \frac{Q}{A} = \frac{200 \frac{gal}{sec} * \frac{1ft^3}{7.48gal}}{4.8 ft} = 5.570409982 \frac{ft}{s}}$$

$${ Re = \frac{VD}{\mu_k} = \frac{ 5.570409982 \frac{ft}{s} \times \frac{16}{3} ft }{ 1.104\times 10^-5 \frac{ft^2}{sec} } = 2691019.315 }$$

Am I using the wrong equivalent diameter? Why am I not getting the correct answer?

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  • $\begingroup$ To start, how did you get 4.8 as the area in line 2? $\endgroup$ – Dan Aug 22 '15 at 3:56
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    $\begingroup$ Ah... the trick is that it's an open channel. The equivalent (or hydraulic) diameter is a bit different. You don't use the free surface in computing the perimeter. $\endgroup$ – Dan Aug 22 '15 at 4:01
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As Dan pointed out you got an error in your calculation.

First you start by calculating the hydraulic diameter, which is defined as

$$d_h = \frac{4 \cdot A}{U}$$

$A$ = cross sectional area of your duct

$U$ = wetted perimeter of the duct

This applies to any shape.

$$d_h = \frac{4 \cdot 8 \cdot 4}{8 + 4 + 4 } = 8 \: \mathrm{ft}$$

Your definition of the Reynolds number is correct

$$Re = \frac{v\cdot L}{\nu}$$

$L$ = characteristic length, in the case of a duct it's the hydraulic diameter $d_h$

$v$ = velocity

$\nu$ = kinematic viscosity

$$v = \frac{\dot{V}}{A} = \frac{200 \: \mathrm{\frac{gal}{s}} }{32 \: \mathrm{ft^2}} = \frac{26.74 \: \mathrm{\frac{ft^3}{s}}}{32 \: \mathrm{ft^2}} = 0.8355 \: \mathrm{\frac{ft}{s}} $$

Inserting yields

$$Re = \frac{0.8355 \:\mathrm{\frac{ft}{s}}\cdot 8 \: \mathrm{ft}}{1.104 \cdot 10^{-5} \: \mathrm{\frac{ft^2}{s}}} \cong 600 000$$

Please note that the Reynoldsnumber is dimensionless, in your final calculation this might be true, but take a closer look at your second line, the units do not equal $\mathrm{\frac{ft}{s}}$.

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    $\begingroup$ For the discrepancy, see Dan's second comment above. $\endgroup$ – Brian Drummond Aug 22 '15 at 10:04
  • $\begingroup$ @BrianDrummond Thanks! Of course, it's an open channel. Updated! $\endgroup$ – idkfa Aug 22 '15 at 10:12

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