STEM Problem Detail

Problem Statement

Prompt

Provide me with a step by step solution to Example 3.5

Solution Rendering...

\begin{aligned}
1. & \text{Discharge through intake}
& \text{Daily discharge} = 150 \times 80000 = 12000000 \text{ l/day}
& \text{Intake load/hour} = \frac{12000000}{10} = 1200000 \text{ l/hr}
& = 1200 \text{ m}^{3}\text{/hr}
\therefore & Q = \frac{1200}{60 \times 60} = 0.3333 \text{ m}^{3}\text{/sec}
\\
2. & \text{Area of coarse screen in front of intake}
& \text{Area of screen, } A_{s} = \frac{Q}{v} = \frac{0.3333}{0.16} = 2.083 \text{ m}^{2}
\therefore & \text{Actual area of screens} = 2.083 / 0.7 = 2.98 \text{ m}^{2}
\therefore & \text{Available height of screen} = 1.8 - 0.3 - 0.2 = 1.3 \text{ m}
\therefore & \text{Required length of screen} = \frac{2.98}{1.3} = 2.29 \text{ m}
& \text{Provide screen of size} = 1.3 \text{ m} \times 2.3 \text{ m}
\\
3. & \text{Design of bell mouth entry}
& \text{Area of bell mouth, } A_{b} = \frac{0.3333}{0.32} = 1.042 \text{ m}^{2}
\therefore & \text{Dia. } d_{b} = \sqrt{\frac{1.042 \times 4}{\pi}} = 1.15 \text{ m}
& \text{Provide bell mouth of } 1.2 \text{ m diameter}
\\
4. & \text{Design of intake conduit}
\therefore & \text{Dia. of intake conduit, } D = \sqrt{\left(\frac{0.3333}{1.5}\right) \times \frac{4}{\pi}} = 0.532 \text{ m}
& \text{Provide } 0.5 \text{ m dia. conduit}
&V = \frac{0.3333 \times 4}{\pi (0.5)^{2}} \approx 1.7 \text{ m/sec.}
&\text{For head loss through the conduit, consider Eq. 16.10 :}
&V = 0.849 C R^{0.63} S^{0.54}
&\text{Take } C = 130 \text{ for cast iron pipes}
&\text{Also, } R = D/4 = 0.5/4 = 0.125 \text{ m.}
&\text{Hence slope } S \text{ of the energy line is given by}
&1.7 = 0.849 (130) (0.125)^{0.63} \cdot S^{0.54}
\text{or } &1.7 = 29.779 S^{0.54}
\text{From which } &S = \left( \frac{1.7}{29.779} \right)^{1/0.54} = 4.98 \times 10^{-3}
\text{But } &S = H_{L} / L
\therefore \text{ Hence loss } &H_{L} = S \cdot L = 4.98 \times 10^{-3} \left( \frac{1}{2} \times 1000 \right) = 2.49 \text{ m}
\end{aligned}

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Problem #100