A smooth concrete pipe with an 18-inch diameter has water flows. The flow rate is 10 ft3/s. Determine the pressure drop in a 100-ft horizontal section of the pipe. Repeat the problem if there is a 2-ft change in elevation of the pipe per 100-ft of its length (z2>z1).

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batolisis batolisis · Answer 1 · 2021-06-12T16:14:16+02:00

Answer:

i)pressure drop at 100-ft horizontal section = 0.266 psi

ii)pressure drop at +2ft and -2ft change in elevation = 1.13 psi and -0.601 psi

Explanation:

Flow rate = 10 ft^3/s

diameter of pipe ( D ) = 18 inches

Area = π * D^2

elevation ( L ) = 100 ft

i) Calculate pressure drop at 100-ft horizontal section

applying Bernoulli's equation

Δp = pg ( Z₂ - Z₁ ) + f [tex]\frac{L}{D} \frac{pV^2}{2}[/tex] ---------- ( 1 )

where : p = 1.94 slug/ft^3 , g = 32.2 ft/s^2, Z₂ = Z₁ , f = 0.0185, L=100ft , D = 18inches, V = 5.66 ft/s ( i.e. flowrate / A )

Input given values into equation 1

Δp ( pressure drop ) = 0.266 psi

ii) Calculate pressure drop at ± 2ft change in elevation

Δp = pg ( Z₂ - Z₁ ) + f [tex]\frac{L}{D} \frac{pV^2}{2}[/tex] ---------- ( 2 )

where : Z₂ - Z₁ = ± 2 ft , p = 1.94 slug/ft^3 , g = 32.2 ft/s^2, Z₂ = Z₁ , f = 0.0185, L=100ft , D = 18inches, V = 5.66 ft/s

input given values into equation above

Δp = ( 1.13 psi , -0.601 psi )