Question

a 6.0cm diameter horizontal pipe gradually narrows to 4.5cm. when water flows through this pipe at a certain rate , the gauge pressure in these two sections is 32.0kPa and 24.0kPa respectively. evaluate the volume flow rate

Answer 1

ANSWER

Q = 0.0077 m³/s

EXPLANATION

We have a pipe with two ends that have different diameter:

The given information is,

• d1 = 6 cm

,

• d2 = 4.5 cm

,

• P1 = 32,000 Pa

,

• P2 = 24,000 Pa

By Bernoulli's law - which states that the total presure and kinetic energy throughout the flow is constant. Since this is an horizontal pipe we'll have only two terms in the equation,

$P_1+\frac{1}{2}\cdot\rho\cdot v^2_1=P_2+\frac{1}{2}\cdot\rho\cdot v^2_2$

Also, we know that the volume flow rate is also constant,

$Q=A_1\cdot v_1=A_2\cdot v_2$

In both equations the speed of the water is missing. Let's solve the second equation for v2,

$v_2=v_1\cdot\frac{A_1}{A_2}$

The sectional area of the pipe is a circle, so the equation is,

$A=\pi\cdot\frac{d^2}{4}$

Replace into the equation for v2,

$v_2=v_1\cdot\frac{\pi\cdot\frac{d^2_1}{4}}{\pi\cdot\frac{d^2_2}{4}}$

Note that π and 4 get cancelled out because of the fraction,

$v_2=v_1\cdot\frac{d^2_1}{d^2_2}$

Now replace with the diameters. Because they are in a fraction we can use the diameters in centimeters, and the units cm² get cancelled out,

$v_2=v_1\cdot\frac{6^2}{4.5^2}$$v_2=\frac{16}{9}v_1$

The next step is to replace v2 by this equation into Bernoulli's equation,

$P_1+\frac{1}{2}\cdot\rho\cdot v^2_1=P_2+\frac{1}{2}\cdot\rho\cdot(\frac{16}{9}v_1)^2$

And then solve for v1. First put all the terms containing v1 on the same side of the equation,

$\frac{1}{2}\cdot\rho\cdot v^2_1-\frac{1}{2}\cdot\rho\cdot(\frac{16}{9}v_1)^2=P_2-P_1$

Take out 1/2ρv1² as a common factor,

$\frac{1}{2}\cdot\rho\cdot v^2_1(1-(\frac{16}{9})^2)=P_2-P_1$

Solve the parenthesis,

$-\frac{175}{81}\cdot\frac{1}{2}\cdot\rho\cdot v^2_1=P_2-P_1$

And solve for v1,

$v_1=\sqrt[]{\frac{P_2-P_1}{-\frac{175}{81}\cdot\frac{1}{2}\cdot\rho}}$

Replace P1 and P2. ρ is the density of water, which is about 1000kg/m³.

$v_1=\sqrt[]{\frac{24,000Pa-32,000Pa}{-\frac{175}{81}\cdot\frac{1}{2}\cdot1000\frac{kg}{m^3}}}$

To check the units, remember that Pa = kg/m·s².

$v_1=\sqrt[]{\frac{-8000\frac{\operatorname{kg}}{m\cdot s^2}}{-\frac{175}{81}\cdot\frac{1}{2}\cdot1000\frac{kg}{m^3}}}$$v_1=\sqrt[]{7.41\frac{\frac{\operatorname{kg}}{m\cdot s^2}}{\frac{kg}{m^3}}}$$v_1\approx2.72\sqrt[]{\frac{m^3}{m^{}\cdot s^2}}=2.72\sqrt[]{\frac{m^2}{s^2}}=2.72m/s$

Now, knowing the velocity of the water at one end of the pipe, we can find the volume flow rate,

$Q_{}=A_1\cdot v_1$

To find the are, now we do have to use the diameter in meters. d1 = 6cm = 0.06m,

$Q=\pi\cdot\frac{0.06^2}{4}\cdot2.72m/s$$Q\approx0.0077m^3/s$

Hence, the volume flow rate is 0.0077 m³/s.