Solución
Formulas
\(y\cdot=\sqrt{\frac{2KD}{h}}\)
\(t\cdot_0=\frac{y\cdot}{D}\)
\(n=\ \frac{L}{t\cdot_0}\)
\(TCU\left(y\right)=\frac{K}{\left(\frac{y}{D}\right)}+h\left(\frac{y}{2}\right)\)
A) Sustitución
\(y\cdot=\sqrt{\frac{2\left(100\right)\left(30\right)}{0.05}}=346.41\).
\(t\cdot_0=\frac{346.41}{30}=11.54=12\).
\(n=\frac{30}{12}=2.5=3\).
\(TCU\left(y\right)=\frac{100}{\left(\frac{346.41}{30}\right)}+0.05\left(\frac{346.41}{2}\right)\)
\(TCU\left(y\right)=\frac{100}{11.547}+0.05\left(173.205\right)=\)
\(TCU\left(y\right)=17.32\).
B) Sustitución
\(y\cdot=\sqrt{\frac{2\left(50\right)\left(30\right)}{0.05}}=244.94\).
\(t\cdot_0=\frac{244.94}{30}=8.16=8\).
\(n=\frac{30}{8}=3.75=4\).
\(TCU\left(y\right)=\frac{50}{\left(\frac{244.94}{30}\right)}+0.05\left(\frac{244.94}{2}\right)\)
\(TCU\left(y\right)=\frac{50}{8.164}+0.05\left(122.47\right)=\)
\(TCU\left(y\right)=12.24.\)
C) Sustitución
\(y\cdot\sqrt{\frac{2\left(100\right)\left(40\right)}{0.01}\ }=894.42\).
\(t\cdot_0=\frac{894.42}{40}=22.36=22\).
\(n=\frac{30}{22}=1.36=1\).
\(TCU\left(y\right)=\frac{100}{\left(\frac{894.42}{40}\right)}\ +\ 0.01\left(\frac{894.42}{2}\right)\)
\(TCU\left(y\right)=\frac{100}{22.3605}\ +\ 0.01\left(447.21\right)=\)
\(TCU\left(y\right)=8.94\).
D) Sustitución
\(y\cdot=\sqrt{\frac{2\left(100\right)\left(20\right)}{0.04}}=316.22\).
\(t\cdot_0=\frac{316.22}{20}=15.81=16\).
\(n=\frac{30}{16}=1.8=2\).
\(TCU\left(y\right)=\frac{100}{\left(\frac{316.22}{20}\right)}\ +0.04\ \left(\frac{316.22}{2}\right)\)
\(TCU\left(y\right)=\frac{100}{15.81}\ +\ 0.04\left(158.11\right)=\)
\(TCU\left(y\right)=12.65\).
Resultados y Conclusión