Water flows steadly through a horizontal pipe of non-uniform cross section. If the pressure of water is $4 * 10^6\;N/m^2$ at a point where the velocity of flow is $2\;m/s$ and cross section is $0.02\;m^2$, what is the pressure at a point where cross-section reduces to $0.01\;m^2$?

Given,
For one end of the pipe,                                                                              For another end of the pipe,
Pressure $(P_1) = 4 * 10^4 \;Ns/m^2$                                                        Pressure $(P_2) = ?$
Velocity $(v_1) = 2\;m/s$                                                                               Velocity $(v_2) = ?$
Area $(A_1) = 0.02\;m/s$                                                                             Area $(A_2) = 0.01\;m/s$

We know from the equation of the continuity equation,

$A_1 \; v_1 = A_2 \; v_2$

$v_2 = \frac{A_1 \; * \; v_1}{A_2} = \frac{0.02 \; * \; 2}{0.01}$ $= 4 \; m/s$

Again to calculate the pressure $(P_2)$,
From the Bernoulli's principle,

$P_1 + \rho\; g\;h + \frac{1}{2} \; \rho \; v_1^2 = P_2 + \rho\; g\;h + \frac{1}{2} \; \rho \; v_2^2$ 

[$\therefore \;$ For the same horizontal, $h_1 = h_2$]

$P_2 = P_1 + \frac{1}{2}(v_1^2 - v_2^2)$

$= 4 * 10^4 + \frac{1}{2} * 1000 * (2^2 - 4^2) = 4 * 10^4 - 6000 = 3.4 * 10^4\;N/m^2$

$\therefore \; P_2 = 3.4 * 10^4\;N/m^2$

Hence, the pressure of flow of water at second end is $3.4 * 10^4\; N/m^2$

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