Answer to Question #246910 in Molecular Physics | Thermodynamics for Anuj

Question

A pipeline 250 mm diameter and 1580 m long has upward slopes of 1 in 200 for the first 1 km length and 1 in 100 for the remaining length. The pressures at the upper end and lower end of the pipeline are 107.91 kPa and 53.95 kPa, respectively. Determine the discharge through the pipe. Given, friction factor = 0.032.

Answer 1

From continuity equation:

$A_1V_1 = A_2V_2 = Q$

For same or uniform cross-section:

$V_1=V_2$

Heat loss due to friction $= \frac{fLV^2}{2gD} = \frac{fLQ^2}{2g(\frac{\pi}{4})^2D^5}$

From Bernoulli’s equation at sections:

$\frac{p_1}{ρg} + \frac{V_1^2}{2g} +H_1 = \frac{p_2}{ρg} + \frac{V^2_2}{2g} + H_2 + Frictional \;heat\;loss$

By considering level at second section as batom

$H_2=0 \\ H_1 = (\frac{1000}{200} + \frac{580}{100}) = 5+5.8 = 10.8 \;m$

Because of uniform cross-section

$V_1=V_2=V_3 \\ \frac{p_1}{ρg} +H_1 = \frac{p_2}{ρg} + H_2 + \frac{fLQ^2}{2g(\frac{\pi}{4})^2D^5} \\ \frac{107.91}{1000 \times 9.81} + 10.8 = \frac{53.95}{1000 \times 9.81} + 0 + \frac{0.032 \times 1580 \times Q^2}{2 \times 9.81 \times (\frac{\pi}{4})^2 \times (0.250)^5} \\ 0.011 +10.8 = 0.0055 + \frac{50.56Q^2}{0.02361} \\ 10.8055 = 2141.46Q^2 \\ Q= \sqrt{0.0050458} \\ Q=0.07103 \;m^3/sec$