Problem 9.69

จาก\(P_{1}+\frac{1}{2}ρv_{1}^{2}+ρgy_{1}=P_{2}+\frac{1}{2}ρv_{2}^{2}+ρgy_{2}\)

\(P_{1}+\frac{1}{2}ρv_{1}^{2}=P_{2}+\frac{1}{2}ρv_{2}^{2}; y_{1}= y_{2}\)

\(P_{1}-P_{2}=\frac{1}{2}ρ(v_{2}^{2}-v_{1}^{2})\cdots \cdots (1)\)

จาก \(Av_{1} = av_{2}\)

\(v_{2} =\frac{Av_{1}}{a}\cdots \cdots (2)\)

พิจารณาแนวเส้นปะในท่อรูปตัวยู

จาก \(P_{L}= P_{R}\)

\(P_{1} + ρgh_{1} = P_{2} + ρgh_{2} + ρ'gh\)

\(P_{1} - P_{2} = ρ'gh - ρg(h_{1} - h_{2})\)

\(P_{1} - P_{2} = ρ'gh - ρgh\)

\(P_{1} - P_{2} = (ρ' - ρ)gh\cdots \cdots (3)\)

\((3)\) และ \((2)\) แทนใน \((1)\) จะได้

\((ρ'-ρ)gh=\frac{1}{2}ρ((\frac{Av_{1}}{a})^{2} - v_{1}^{2})\)

\((ρ'-ρ)gh=\frac{1}{2}ρ((\frac{A}{a})^{2}-1)v_{1}^{2}\)

\(v_{1} =\sqrt{\frac{2gh(\frac{ρ'}{ρ}- 1)}{(\frac{A}{a})^{2} - 1}}\;\blacksquare\)