Question

A liquid of density 1230 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.39 m/s and the pipe diameter d1 is 11.5 cm. At Location 2, the pipe diameter d2 is 15.3 cm. At Location 1, the pipe is Δy=9.59 m higher than it is at Location 2. Ignoring viscosity, calculate the difference ΔP between the fluid pressure at Location 2 and the fluid pressure at Location 1.

Answer 1

To develop this problem it is necessary to use the continuity equations and Bernoullie's theorem.

It is known from Bernoullie's theorem that

$P_1 + \rho gh_1+\frac{1}{2} \rho v_2^2 =P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2$

Where

P = Pressure

g = Gravity

h= Height

v = Velocity

$\rho$ = Density

On the other hand we have that the continuity equation is given by

$A_1v_1 = A_2 v_2$

Where A is the Cross-sectional area and v the velocity.

For our values we know that

$A_1v_1 = A_2 v_2$

$(\frac{\pi d_1^2}{4})v_1 =(\frac{\pi d_2^2}{4})v_2$

$d_1^2v_1=d_2^2v_2$

$(11.5cm)^2(9.39)=(15.3)^2v_2$

$v_2 = 5.305m/s$

Using Bernoulli's expression we can now find the pressure difference,

$P_1 + \rho gh_1+\frac{1}{2} \rho v_2^2 =P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2$

$P_1-P_2=-\rho gh_1-\frac{1}{2}\rho v_2^2 +\rho gh_2+\frac{1}{2} \rho v_2^2$

$P_1-P_2 = \rho g (h_1-h_2)+\frac{1}{2}\rho(v_1^2-v_2^2)$

$P_1-P_2 = (1.3*10^3)(9.8)(9.59)+\frac{1}{2}(1.3*10^3)((9.39)^2-(5.305)^2)$

$P_1-P_2 = 1.612*10^5Pa$