Question

Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes is 2 cm, and the areas of cross section at A and B are 4 cm² and 2 cm² respectively, find the rate of flow of water across any section.

Answer 1

Given:
Difference in the heights of water columns in vertical tubes = 2 cm
Area of cross section at A, a_A = 4 cm²
Area of cross section at B, a_B = 2 cm²
Now, let v_A and v_B be the speeds of water at A and B, respectively.
From the equation of continuity, we have:
$$\begin{gathered} v_A{a}_A=v_B\times a_B \\ \Rightarrow v_{\mathrm{A}}\times 4=v_{\mathrm{B}}\times 2 \\ \Rightarrow v_{\mathrm{B}}=2v_{\mathrm{A}}...\left(\mathrm{i}\right) \end{gathered}$$
From Bernoulli's equation, we have:
$$\begin{gathered} \frac{1}{2}\rho v_A^2+\rho g{h}_A+p_{\mathrm{A}}=\frac{1}{2}\rho v_{\mathrm{B}}^2+\rho g{h}_B+p_{\mathrm{B}} \\ \Rightarrow \frac{1}{2}\mathrm{\rho }{v}_{\mathrm{A}}^2+p_{\mathrm{A}}=\frac{1}{2}\mathrm{\rho }{v}_{\mathrm{B}}^2+p_{\mathrm{B}} \\ \Rightarrow p_{\mathrm{A}}-p_{\mathrm{B}}=\frac{1}{2}\mathrm{\rho }\left(v_{\mathrm{B}}^2-v_{\mathrm{A}}^2\right) \end{gathered}$$
Here,
p_A and p_B are the pressures at A and B, respectively.
h_A and h_B are the heights of water columns at point A and B, respectively.
ρ is the density of the liquid.
Thus, we have:
$$\begin{gathered} \frac{1}{2}\times 1\times \left(4v_{\mathrm{A}}^2-v_{\mathrm{A}}^2\right) \\ \Rightarrow 2\times 1\times 1000=\frac{1}{2}\times 1\times 3v_{\mathrm{A}}^2 \\ [p_{\mathrm{A}}-p_{\mathrm{B}}=2\mathrm{cm}=2\times 1\times 1000\mathrm{dyne}/{\mathrm{cm}}^2(\mathrm{water}\mathrm{column}\left)\right] \\ \Rightarrow v_{\mathrm{A}}^2=\sqrt{\frac{4000}{3}}=36.51\mathrm{cm}/\mathrm{s} \\ \therefore \mathrm{Rate}\mathrm{of}\mathrm{flow}=v_A{\mathrm{a}}_{\mathrm{A}}=36.51\times 4 \\ =146{\mathrm{cm}}^3/\mathrm{s} \end{gathered}$$
Hence, the required rate of flow of water across any section is 146 cm³/s.