A tube well screen has an outside diameter of $0.20\text{ m}$ and length of $10.0\text{ m}$. The open slot area is $15\%$ of the outer cylindrical surface area. If the well discharges $0.03\text{ m}^3\text{/s}$, the entrance velocity of water through the screen slots in $\text{m/s}$ is ________ (round off to 3 decimal places).
Answer Key:
0.032
Step-by-Step Solution & Explanation
1. Outer surface area of the screen $A_{cyl}$:
$A_{cyl} = \pi D L = \pi \times 0.20 \times 10.0 = 2.0 \pi \approx 6.2832\text{ m}^2$
2. Open slot area $A_{open}$ ($15\%$):
$A_{open} = 0.15 \times 6.2832 \approx 0.94248\text{ m}^2$
3. Entrance velocity $V_e$:
$V_e = \frac{Q}{A_{open}} = \frac{0.03\text{ m}^3\text{/s}}{0.94248\text{ m}^2} \approx 0.03183\text{ m/s} \approx 0.032\text{ m/s}$
(Well design criteria recommend $V_e \le 0.03\text{ m/s}$ to minimize encrustation and head loss).