Question

Water flows through a tube shown in figure. The area of cross section at A and B are 1 cm² and 0.5 cm² respectively. The height difference between A and B is 5 cm. If the speed of water at A is 10 cm s⁻¹ find (a) the speed at B and (b) the difference in pressures at A and B.

Answer 1

Given:
Difference in the heights of A and B = 5 cm
Area of cross section at A, a_a = 1 cm²
Area of cross section at B, a_b = 0.5 cm²
Speed of water at A, $v_A$ = 10 cms⁻¹

$$\begin{gathered} \left(a\right)\mathrm{From}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{continuity},\mathrm{we}\mathrm{have}: \\ {\overrightarrow{V}}_{\mathrm{A}}\times a_{\mathrm{A}}={\overrightarrow{V}}_{\mathrm{B}}\times a_{\mathrm{B}} \\ \Rightarrow 10\times 1={\overrightarrow{V}}_{\mathrm{B}}\times 0.5 \\ \Rightarrow {\overrightarrow{V}}_{\mathrm{B}}=20\mathrm{cm}/\mathrm{s} \end{gathered}$$
The required speed of water at cross section B is 20 cms⁻¹

$$\begin{gathered} \left(b\right)\mathrm{From}\mathrm{Bernoulli}\text{'}\mathrm{s}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \frac{1}{2}\rho v_{\mathrm{A}}^2+\rho g{h}_{\mathrm{A}}+P_{\mathrm{A}}=\frac{1}{2}\rho v_{\mathrm{B}}^2+\rho g{h}_{\mathrm{A}}+P_{\mathrm{B}} \\ \Rightarrow P_{\mathrm{B}}-P_{\mathrm{A}}=\frac{1}{2}\rho \left(v_{\mathrm{A}}^2-v_{\mathrm{B}}^2\right)+\rho g\left(h_{\mathrm{A}}-h_{\mathrm{B}}\right) \end{gathered}$$
Here,
P_A and P_B are the pressures at A and B, respectively.
h_A and h_B are the heights of points A and B, respectively.
ρ is the density of the liquid.
On substituting the values, we have:
$$\begin{gathered} P_B-P_A=\frac{1}{2}\times 1(100-400)+1\times 1000(5.0) \\ =-150+5000=4850\mathrm{Dyne}/{\mathrm{cm}}^2 \\ =485\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$
Therefore, the required pressure difference at A and B is 485 N/m².