Question

Water enters a horizontal pipe of non-uniform cross section with a velocity of 0.5 m/s and leaves other end with a velocity of 0.7 m/s the pressure of water at the first end is 1000 n/m^2 calculate the pressure at other end

Answer 1

Here we make use of Bernoulli's Theorem that,

$\dfrac {P_1}{\rho}+gh_1+\dfrac {1}{2}(v_1)^2=\dfrac {P_2}{\rho}+gh_2+\dfrac {1}{2}(v_2)^2$

Here the pipe is kept horizontal, so we have,

$h_1=h_2$

So the equation is modified as,

$\dfrac {P_1}{\rho}+\dfrac {1}{2}(v_1)^2=\dfrac {P_2}{\rho}+\dfrac {1}{2}(v_2)^2\\\\\\\dfrac {P_2}{\rho}=\dfrac {P_1}{\rho}+\dfrac {1}{2}(v_1)^2-\dfrac {1}{2}(v_2)^2\\\\\\P_2=\rho\left [\dfrac {P_1}{\rho}+\dfrac {1}{2}(v_1)^2-\dfrac {1}{2}(v_2)^2\right]\\\\\\P_2=P_1+\dfrac {\rho}{2}\left [(v_1)^2-(v_2)^2\right]$

Given,

$v_1=0.5\ \mathrm{ms^{-1}}\\\\v_2=\mathrm {0.7\ ms^{-1}}\\\\P_1=\mathrm {1000\ Nm^{-2}}$

So,

$P_2=1000+\dfrac {\rho}{2}\left [(0.5)^2-(0.7)^2\right]\\\\\\P_2=1000-0.12\rho$