Question

water flowing through a horizontal pipe of variable cross-section. The rate of volume flow is 0.4lit/sec. The pressure is 60,000pa at a point where the area of cross-section is 5x10^-5 m^2. The pressure at a point where the area of a cross-section is 10^-4 is

Answer 1

Given

Q = Flow rate = $0.4\ \text{L/s}=0.4\times 10^{-3}\ \text{m}^3/\text{s}$

$A_1$ = Area of first cross section = $5\times 10^{-5}\ \text{m}^2$

$A_2$ = Area of second cross section = $10^{-4}\ \text{m}^2$

$P_1$ = Pressure at first cross section = 60000 Pa

To find

Pressure at a point where the area of a cross-section is $10^-4\ \text{m}^2$

Solution

$\rho$ = Density of water = $1000\ \text{kg/m}^3$

$v_1$ = Velocity of water at first point

$v_2$ = Velocity of water at second point

$P_2$ = Pressure of water at second point

Flow rate is given by

$Q=A_1v_1\\\Rightarrow v_1=\dfrac{Q}{A_1}\\\Rightarrow v_1=\dfrac{0.4\times 10^{-3}}{5\times 10^{-5}}\\\Rightarrow v_1=8\ \text{m/s}$

$Q=A_2v_2\\\Rightarrow v_2=\dfrac{Q}{A_2}\\\Rightarrow v_2=\dfrac{0.4\times 10^{-3}}{10^{-4}}\\\Rightarrow v_2=4\ \text{m/s}$

From Bernoulli's relation we have

$P_2-P_1=\dfrac{\rho(v_1^2-v_2^2)}{2}\\\Rightarrow P_2=\dfrac{\rho(v_1^2-v_2^2)}{2}+P_1\\\Rightarrow P_2=\dfrac{1000(8^2-4^2)}{2}+60000\\\Rightarrow P_2=84000\ \text{Pa}$

The pressure at the required point is $84000\ \text{Pa}$.