Đáp án:
${{P}_{hp2}}=\frac{{{P}_{hp1}}}{18}$
Giải thích các bước giải:
Công suất hao phí:
${{P}_{hp}}=\dfrac{R.{{P}^{2}}}{{{U}^{2}}}=\rho .\dfrac{l}{S}.\dfrac{{{P}^{2}}}{{{U}^{2}}}$
ta có:
$\begin{align}
& \dfrac{{{P}_{hp1}}}{{{P}_{hp2}}}=\frac{\rho .\dfrac{l}{{{S}_{1}}}.\frac{{{P}^{2}}}{{{U}_{1}}^{2}}}{\rho \frac{l}{{{S}_{2}}}.\dfrac{{{P}^{2}}}{{{U}_{2}}^{2}}}=\dfrac{{{S}_{2}}}{{{S}_{1}}}.\dfrac{U_{2}^{2}}{U_{1}^{2}}=\dfrac{2.{{S}_{1}}}{{{S}_{1}}}.\dfrac{{{(3{{U}_{1}})}^{2}}}{U_{1}^{2}}=18 \\
& \Rightarrow {{P}_{hp2}}=\dfrac{{{P}_{hp1}}}{18} \\
\end{align}$
giảm 18 lần