Question

Water flows through a horizontal tube of variable cross section. The area of cross section at A and B are 4 mm² and 2 mm² respectively. If 1 cc of water enters per second through A, find (a) the speed of water at A, (b) the speed of water at B and (c) the pressure difference P_A − P_B.

Answer 1

(a) Given:
The areas of cross-sections of the tube at A and B are 4 mm² and 2 mm², respectively.
1 cc of water enters per second through A.
Let:
$\rho$ = Density of the liquid
P_A = Pressure of liquid at A
P_B = Pressure of liquid at B

(a)
$$\begin{gathered} \frac{Q_{\mathrm{A}}}{t}=\frac{1a}{\mathrm{s}}=\mathrm{Discharge} \\ \Rightarrow a_{\mathrm{A}}\times V_{\mathrm{A}}=Q_{\mathrm{A}} \\ \mathrm{Given}: \\ {a}_A=4{\mathrm{mm}}^2=4\times {10}^{-2}{\mathrm{cm}}^2 \\ \therefore 4\times {10}^{-2}\times V_{\mathrm{A}}=1\mathrm{cc}/\mathrm{s} \\ \Rightarrow V_{\mathrm{A}}=25\mathrm{m}/\mathrm{s} \\ \left(b\right)a_A\times V_{\mathrm{A}}=a_{\mathrm{B}}\times V_{\mathrm{B}} \\ \Rightarrow 4\times {10}^{-2}\times 25=2\times {10}^{-2}\times V_{\mathrm{B}} \\ \Rightarrow V_{\mathrm{B}}=50\mathrm{cm}/\mathrm{s} \end{gathered}$$

(c) By Bernoulli's equation, we have:
$$\begin{gathered} \frac{1}{2}\rho v_{\mathrm{A}}^2+P_{\mathrm{A}}=\frac{1}{2}\rho v_{\mathrm{B}}^2+P_{\mathrm{B}} \\ \Rightarrow (P_{\mathrm{A}}-P_{\mathrm{B}})=\frac{1}{2}\rho \left(v_B^2-v_A^2\right) \\ =\frac{1}{2}\times 1\times (2500-625) \\ =\frac{1875}{2}=937.5\mathrm{dyn}/{\mathrm{cm}}^2 \\ =93.75\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$