Question

Determine the pressure difference in N/m^2, between two points 800m apart in horizontal pipe-line, 150mm diameter, discharging water at the rate of 12.5litres per second. Take the frictional coefficient,f, as being 0.008.

Answer 1

Answer:

Explanation:

Concept:

We will use the concept of fluid mechanics to solve this question.

Given:

The two points are $800m$ apart in horizontal pipe-line.

The diameter is $150mm$ .

Water is discharging at the rate of $12.5litres$ per second.

The frictional coefficient is $f=0.008$ .

To Find:

We have to calculate the pressue difference in $N/m^{2}$ .

Solution:

The diameter will be $150mm=150 \times 10^{-3} m$

The average velocity of flow is $\bar{v}$ .

The formula of average velocity of flow is $\bar{v}=$ discharge flow rate$/$cross section area.

Put the vaue in the above formula.

$\bar{v}=\frac{ 12.5^{-3} \times 10^{-3}/\pi \times (\frac{150}{2} \times 10^{-3})^{2} }$

After further calculation we will get $\bar{v}=0.71 \mathrm{~m} / \mathrm{s}$

The formula of pressure difference $(\Delta P)=\frac{2fL\rho \bar{v}^{2} }{d}$

The value of $L=800m$

the value of $\rho = 997 \mathrm{~kg} / \mathrm{m}^{3}$

and the value of $f=0.008$

$\Delta P=\frac{2 \times 0.008 \times 800 \times 997 \times(0.71)^{2}}{150 \times 10^{-3}}$

After further calculation, the value of pressure difference is $4.28 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$.

So, the pressure difference in $\mathrm{~N} / \mathrm{m}^{2}$ is $4.28 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$ .

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