Question

Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.85 × 10⁵ Pa and the pipe radius is 2.60 cm. At the higher point located at y=2.50 m, the pressure is 1.25 × 10⁵ Pa and the pipe radius is 1,70 cm.

Answer 1

I will assume the lower point to be at zero fluid potential.

Applying equation of continuity to the lower and upper end of the pipe we must have,

$A_1v_1=A_2v_2$

A denotes the area and v denotes the velocity

1 denotes the lower end and 2 denotes the upper end

For the radius of the pipes on both ends be r1 and r2 we must have

$\pi r_1^2v_1=\pi r_2^2v_2$

$or,\ v_1=\left (\frac{r_2}{r_1} \right )^2v_2$

Putting the values,

$v_1=\left (\frac{1.7}{2.6} \right )^2v_2$

Now let us apply Bernoulli's principle at the bottom and upper end such that

$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2+\rho gy$

P denotes the pressure, $\rho$ is the density of water, g is the acceleration due to gravity

Putting the values,

$1.85\times10^5+\frac{1}{2}\times1000\times v_1^2=1.25\times10^5+\frac{1}{2}\times1000\times v_2^2+1000\times9.8\times2.5$

$or,\ 1.85\times10^5+500v_1^2=1.25\times10^5+500v_2^2+24500$

$or,\ 500(v_2^2-v_1^2)=(1.85-1.25)\times10^5-24500=35500$

Using equation (i)

$(v_2^2-0.4275^2v_2^2)=71$

$or,\ v_2=\sqrt{\frac{71}{1-0.4275^2}}\ m/s=9.32\ m/s$ Ans(b)

This is the velocity in the upper end.

a) Using equation (i)

$v_1=\left (0.4275\times9.32 \right )\ m/s=3.98\ m/s$

This is the velocity in the lower end

c) Flow rate through the pipe is nothing but the product of area and velocity.

$Q=A_1v_1$

$=\pi r_1^2\times v_1$

Putting the values,

$Q=\pi \times \left ( \frac{2.6}{100} \right )^2\times 3.98\ m^3/s=8.45\times10^{-3}\ m^3/s$