Answer
diameter of parking lot = 18 in
flowrate = 10 ft³/s
pressure drop = 100 ft
using general equation
[tex]\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}[/tex]
[tex]V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s[/tex]
[tex]\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}[/tex]
taking f = 0.0185
at Z₁ = Z₂
[tex]\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}[/tex]
ΔP = 0.266 psi
b) when flow is uphill z₂-z₁ = 2
[tex]\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266[/tex]
[tex]\Delta P= 1.13\ psi[/tex]
c) When flow is downhill z₂-z₁ = -2
[tex]\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266[/tex]
[tex]\Delta P=-0.601\ psi[/tex]
