Water at 20^◦ C flows in a 3.5-m-diameter inclined straight pipe section at a flowrate of 2.6 m^3 /sec, as illustrated in Figure EP 5.1. The pipe length between points 1 and 2 is 900 m, and the Chezy coefficient, C for the pipe is 60 m^1/2 /s^−1 . The elevations of points 1 and 2 are 1.3 m and 1.8

Question

Water at [latex] 20^{\circ} C[/latex] flows in a 3.5-m-diameter inclined straight pipe section at a flowrate of [latex] 2.6 m^{3} /sec[/latex], as illustrated in Figure EP 5.1. The pipe length between points 1 and 2 is 900 m, and the Chezy coefficient, C for the pipe is [latex] 60 m^{1/2} /s^{−1} [/latex] . The elevations of points 1 and 2 are 1.3 m and 1.8 m, respectively, and the pressure at point 1 is [latex] 3 × 10^{4} N/m^{2} [/latex] . (a) Determine the friction slope at any point in the straight pipe section. (b) Determine the head loss due to pipe friction between points 1 and 2. (c) Determine the pressure drop at point 2 due to the pipe friction.

Accepted Answer

(a) In order to determine the friction slope at any point in the straight pipe section, the Chezy equation is applied. Furthermore, in order to determine the velocity of flow in the pipe, the continuity equation is applied as follows:

[latex] D: = 3.5 m [/latex] [latex] A: = \frac{\pi .D^{2}}{4} = 9.621 m^{2}[/latex] [latex] P_{w}: = \pi .D = 10.996 m[/latex]

[latex] R_{h}:= \frac{A}{P_{w}} = 0.875 m [/latex] [latex] Q: = 2.6 \frac{m^{3}}{sec} [/latex] [latex] C: = 60 m^{\frac{1}{2} } sec^{-1}[/latex]

Guess value: [latex] S_{f}: = 0.5 \frac{m}{m} [/latex] [latex] V: = 1 \frac{m}{sec} [/latex]

Given

[latex] S_{f} = \frac{V^{2}}{C^{2}.R_{h}}[/latex]
[latex] V = \frac{Q}{h} [/latex]
[latex] \left ( \begin{matrix} V \ S_{f} \end{matrix} ight ) : = Find (V, S_{f})[/latex]

[latex] V = 0.27 \frac{m}{s} [/latex] [latex] S_{f}= 2.318 imes 10^{-5} \frac{m}{m} [/latex]

(b) In order to determine the head loss due to pipe friction between points 1 and 2, the head loss due to pipe friction equation is applied as follows:

[latex] L: = 900 m[/latex] [latex] h_{f}: = S_{f}. L = 0.021m[/latex]

(c) In order to determine the pressure drop at point 2 due to the pipe friction, the energy equation is applied between points 1 and 2 as follows:

[latex] V_{1}: = V = 0.27 \frac{m}{s} [/latex] [latex] V_{2}: = V = 0.27 \frac{m}{s} [/latex] [latex] P_{1}: = 3 imes 10^{4} \frac{N}{m^{2}} [/latex]

[latex] Z_{1}: = 1.3 m [/latex] [latex] Z_{2}: = 1.8 m [/latex]

[latex] ho : = 998 \frac{kg}{m^{3}} [/latex] [latex] g: = 9.81 \frac{m}{sec^{2}} [/latex] [latex] \gamma : = ho . g = 9.79 10^{3} \frac{kg}{m^{2}.s^{2}} [/latex]

Guess value: [latex] P_{2}: = 2 10^{4} \frac{N}{m^{2}} [/latex]

Given

[latex] \frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1} }{2. g} - h_{f} = \frac{P_{2}}{\gamma } + Z_{2}\frac{V^{2}_{2} }{2. g}[/latex]
[latex] P_{2}: = Find (P_{2})= 2.49 imes 10^{4} \frac{N}{m^{2}} [/latex]
[latex] \Delta P: = P_{1} - P_{2} = 5.099 imes 10^{3} \frac{N}{m^{2}} [/latex]

Question 5.1: Water at 20^◦ C flows in a 3.5-m-diameter inclined straight ...

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