Question 23.5: A pipe carries 2500 m³ of water per hour at a pressure of 1.......

Given Q = quantity of water

=2500 \mathrm{~m}^3 / \mathrm{h}

p=1.5 ~\mathrm{MPa}

V=30 \mathrm{~m} / \mathrm{s}

\sigma_{\mathrm{all}}=\sigma_t=\frac{400}{3}=133.3 \mathrm{~N} / \mathrm{mm}^2

we know

Q = quantity of water flow per minute

= area x velocity

=\left(\frac{\pi}{4} d^2\right) \times V

\therefore \quad \frac{2500}{60}=\frac{\pi}{4} d^2 \times 30 \times 60

Solving, we get

d=0.172 \mathrm{~m}=172 \mathrm{~mm}

Take d = diameter of pipe = 175 mm

Assuming pipe as thin one, the wall thickness is obtained from Hoop’s equation

\sigma_{\theta \theta}=\sigma_t=\frac{p d}{2 t}

It is not a seamless pipe, hence, joint efficiency should be used in the stress equation. Considering joint efficiency \eta, and corrosion allowance of 3 mm, the modified design equation becomes,

\sigma_t=\frac{p d}{2 \eta \sigma_{Q Q}}+3

taking \sigma_{Q Q}=\sigma_t=\text { allowable stress }

t=\frac{p d}{2 \sigma_t \cdot \eta}+3

=\frac{1.5 \times 175}{2 \times 133.3 \times 0.75}+3=4.3 \mathrm{~mm}

t=5 \mathrm{~mm} \text { (say) }