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 )
