Question

h1 h2

Consider a water pipe with a restricted section as illustrated beside. The consumption meter from the utility company indicates a constant water flow of 480 L/min. You also know that the pipe has a diameter of 40 cm at the inlet and outlet but only a diameter of 20 cm in the restricted region.

• B) What is the water column height above the restricted section of the pipe (that’s h2)?

C) What is the highest value (in kPa) of the absolute pressure in this water pipe?

D) Assume that the highest flow rate which can be reached when using this water pipe occurs when the fluid velocity in the restricted section (that’s v2) is 1 m/s at which point the pipe starts to vibrate and might, over time, fissure. What is the maximum water flow rate (in m3/min) associated with this extreme mode of operation?

Answer 1

Given,

Flow rate (Q)= 480L/min

Dia of inlet and outlet (D) = 40cm =0.4m

Dia of restricted section (d)= 20cm = 0.2m

we know

Density of water = 1000kg/m³ = 1 kg/l

So, 1m³ of water = 1000L

so, flow rate in m³/sec = (480 * 10^-3)/ 60

Q=0.008 m³/sec

Now,

B)

velocity in point 1

$v_{1}=\frac{Q}{A_{1}}$

$\Rightarrow v_{1}=\frac{0.008}{\frac{\pi D^{2}}{4}}m/sec$

$\Rightarrow v_{1}=\frac{0.008}{\frac{\pi* 0.4^{2}}{4}}m/sec$

$\Rightarrow v_{1}=0.064m/sec$

velocity at point 2

$v_{2}=\frac{Q}{A_{2}}$

$\Rightarrow v_{2}=\frac{0.008}{\frac{\pi d^{2}}{4}}m/sec$

$\Rightarrow v_{2}=\frac{0.008}{\frac{\pi* 0.2^{2}}{4}}m/sec$

$\Rightarrow v_{2}=0.255m/sec$

By Bernoulli's Theorem, we get

$P_{1}+\frac{1}{2}\rho v_{1}^{2}+P.E_{1}=P_{2}+\frac{1}{2}\rho v_{2}^{2}+P.E_{2}$

as potential energy changes with the change of height. Here, height does not change so change in potential energy is also 0.

Therefore P.E₁=P.E₂

Equation becomes

$P_{1}+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2}$

$\Rightarrow P_{1}-P_{2}=\frac{1}{2}\rho v_{2}^{2}-\frac{1}{2}\rho v_{1}^{2}$

$\Rightarrow \frac{P_{1}-P_{2}}{\rho }=\frac{1}{2}\left ( v_{2}^{2}- v_{1}^{2} \right )$

dividing acceleration due to gravity (g) from both sides we get

$\Rightarrow \frac{P_{1}-P_{2}}{\rho g }=\frac{1}{2g}\left ( v_{2}^{2}- v_{1}^{2} \right )$

Since, $P=\rho gh$

$\Rightarrow h_{1}-h_{2}=\frac{1}{2g}\left ( v_{2}^{2}- v_{1}^{2} \right )$

$\Rightarrow h_{1}-h_{2}=\frac{1}{2g}\left ( 0.255^{2}- 0.064^{2} \right )$

$\Rightarrow h_{1}-h_{2}=0.03m$

Note: we cannot find the height h₂ unless the pressure at the given point is given or pressure at point 1 is given. So, we can calculate only the difference in pressure which is equal to 0.03m

C)

$P_{absolute}=P+P_{atm}$

in order to get absolute pressure, pressure in the pipe or height of water column must be given

D)

Given,

v₂ = 1m/s

to get flow rate

we know,

$Q= AV$

$\Rightarrow Q= \frac{\pi *d^{2}}{4}* 1 m^{3}/sec$

$\Rightarrow Q= \frac{\pi *0.2^{2}}{4}m^{3}/sec$

$\Rightarrow Q= 0.0314m^{3}/sec$

$\Rightarrow Q= 0.0314*60m^{3}/min$

$\therefore Q=1.884m^{3}/min$

If any doubt feel free to comment