A 1" pipe buried 2' deep runs 40' from house to outhouse. Frost depth = 3',...

A 1" pipe buried 2' deep runs 40' from house to outhouse. Frost depth = 3', so a trickle must flow.

a) Say it's laminar flow, at 75% of the velocity needed to make Re=2100. Find that velocity, gal/day carried. Use p=1.94slug/ft^3, m= 2.2*10^-5 lb-sec/ft^2 in this and the following Re problems.

b) Find the head loss hl=hp and the pressure drop (psi) over the l=40'

Solutions

Expert Solution

Answer a) Re = 2100 = pvD/m where p is density of water p = 1.94 slug/ft3 = 1.94 *32.174 lb/ft3 = 62.417 lb/ft3

D = dia of pipe = 1" = 1/12 ft

Dynamic visosity coeffecient = m = 2.2 *10^-5 lb-sec/ ft2

Thus velocity of flow = v1 = 0.0088 ft/sec

Thus 75% of this velocity = v2 = 0.75* 0.0088 ft/sec = 0.0066 ft/sec which is the velocity that has been asked to be calculated.

Flow = Area * velocity =( 3.142/4)*(1/144) * 0.0066 ft3/s = 3.602 * 10^-5 ft3/sec = 3.602*10^-5 * 646316.883 gal/day (US gallon) = 23.28 gal/day

Answer b) for calculating head loss calculating friction factor f = 64/Re

For laminar flow velocity = 0.0066 ft/s thus Re = 62.417*0.0066*(1/12) /(2.2*10^-5) = 1560.425 thus f= 64/1560.425 = 0.041

Thus using Darcy weisbach equation head loss for 40' length of pipe = flv^2/2gD = 0.041*40*0.0066*0.0066/(2*32.2*1/12)= 1.33*10^-5 ft

Pressure drop = (128*mlQ)/(3.142*D^4) = 128*2.2*10^-5 * 40*3.602*10^-5 /( 3.142*(1/12)^4)= 0.0267 lb/ft2 = 0.0267/144 = 0.000185 psi

Ally Wells answered 3 months ago

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