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?