Question

A pipe 300m long has a slope of 1 in 100 and tapers from 1.2m diameter at the high end to 0.6m diameter at the low end quantity of water flowing is 5400 litres per minute. If the pressure at the high end is 68.67kPa (0.7 kg(f)/cm²),find the pressure at the low end. Negelect losses

Answer 1

Answer:

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Answer 2

Answer:

The pressure at the low end will be 97.964 kPa

Step-by-step explanation:

As per the question we have given that

Length of the pipe, L = 300 m

The slope of the pipe = 1 in 100

Diameter at the high end, d₁ = 1.2 m

Diameter at the low end, d₂ = 0.6 m

The volume flow rate, Q = 5,400 L/min

Pressure at the high end, P = 68.67 kPa

Density of water = 997 kg/m³

Elevation of pipe is given as ($Z_1$) = $\[300m \times \frac{1}{{100}} = 3m\]$

$Z_2$ = 0

Therefore, according to the continuity equation

Q = A₁·v₁ = A₂·v₂

$\[\sqrt[n]{x}Q = 5400L/\min = 5400\frac{L}{{\min }} \times \frac{{1{m^3}}}{{1000L}} \times \frac{{1\min }}{{60{\text{secons}}}} = 0.09{m^3}/s\]$

Now,

$\[\begin{gathered} 0.09 = \frac{\pi }{4} \times {1.2^2} \times {\nu _1} \hfill \\ {\nu _1} = 0.0796 \hfill \\ 0.09 = \frac{\pi }{4} \times {0.6^2} \times {\nu _2} \hfill \\ {\nu _2} = 0.318 \hfill \\ \end{gathered} \]$

Now, applying the Bernoulli's equation

$\[\begin{gathered} \frac{{{P_1}}}{{\rho .g}} + \frac{{\nu _1^2}}{{2.g}} + {Z_1} = \frac{{{P_2}}}{{\rho .g}} + \frac{{\nu _2^2}}{{2.g}} + {Z_2} \hfill \\ {\text{Substituting the values and we get :}} \hfill \\ {{\text{P}}_2} = 97964.46Pa \hfill \\ {{\text{P}}_2} = 97.964kPa \hfill \\ \end{gathered} \]$

Therefore, the pressure at the low end is 97.964kPa