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: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the flowrate 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.

User Tvashtar
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1 Answer

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Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation


(P_1)/(\gamma)+(v_1^2)/(2g)+Z_1 = \frac{P_2{\gamma} + (v_2^2)/(2g) + Z_2 +(fLV^2)/(2\rho D)


V = (Q)/(A) = (10)/((\pi)/(4)((18)/(12))^2) = 5.66\ ft/s


\Delta P = \gamma (Z_2-Z_1) +(fLV^2)/(2\rho D)

taking f = 0.0185

at Z₁ = Z₂


\Delta P = (0.0185 * 100* 1.94* 5.66^2)/(2(18)/(12) (2))

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2


\Delta P =62.4* 2 * (1)/(144) +0.266


\Delta P= 1.13\ psi

c) When flow is downhill z₂-z₁ = -2


\Delta P =62.4* 2 * (1)/(144) +0.266


\Delta P=-0.601\ psi

User Nayburz
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