Question

Water flows through a horiztonal pipe at a rate of 94 ft3/min. A pressure gauge placed on a 3.3 inch diameter section of the pipe reads 15 psi.
What is the gauge pressure in a section of pipe where the diameter is 5.2 inches?

Answer 1

Answer:

The gauge pressure is 1511.11 psi.

Explanation:

Given that,

Flow rate = 94 ft³/min

Diameter d₁=3.3 inch

Diameter d₂ = 5.2 inch

Pressure P₁= 15 psi

We need to calculate the pressure on other side

Using Bernoulli equation

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

We know that,

$V=Av$

$v=\dfrac{V}{A}$

Where, V = volume

v = velocity

A = area

Put the value of v into the formula

$P_{1}+\dfrac{1}{2}\rho (\dfrac{V}{A_{1}})^2=P_{2}+\dfrac{1}{2}\rho (\dfrac{V}{A_{2}})^2$

Put the value into the formula

$15+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2=P_{2}+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2$

$P_{2}=15+\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2-\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2$

$P_{2}=1525.8\ psi$

We need to calculate the gauge pressure

Using formula of gauge pressure

$P_{g}=P_{ab}-P_{atm}$

Put the value into the formula

$P_{g}=1525.8-14.69$

$P_{g}=1511.11\ psi$

Hence, The gauge pressure is 1511.11 psi.