Question

Asphalt at 120 F, considered to be a Newtonian fluid with a viscosity 80000 times that of water and a specific gravity of 1.09, flow through a pipe of diameter 2.0 in. If the pressure gradient is 1.6 psi/ft determine the flowrate assuming the pipe is (a) horizontal; (b) vertical with flow up.

Answer 1

a) Flow rate if the pipe is horizontal, Q = 4.69 * 10⁻³ ft³/s

b) Flow rate if the pipe is vertical, Q = 3.30 * 10⁻³ ft³/s

Step-by-step explanation:

From the BI table, dynamic viscosity of water at 120°F is:

`\mu_(H_2O) = 1.164 * 10^(-5) lb.s/ft^2`

Pressure gradient,
`(\delta p)/(\delta x) = 1.6 psi/ft`

Pipe Diameter, D = 2 in = 2/12 ft = 0.167 ft

Dynamic viscosity of asphalt at 120°F:

`\mu = v \mu_(H_2O)\\\mu = 80000 * 1.164 * 10^(-5)\\\mu = 0.9312 lb-s/ft^2`

Specific weight of asphalt:

`\gamma = SG \gamma_(H_2O)\\\gamma = (1.09)(62.4)\\\gamma = 68.016 lb/ft^3`

Flow rate, Q, of the asphalt when the pipe is in horizontal position assuming that the flow is laminar:

Note that if the pipe is horizontal, θ = 0°

`Q = (\pi D^4)/(128 \mu) [((\delta p )/(\delta x)) - \gamma sin \theta]\\\\Q = (\pi 0.167^4)/(128 * 0.9312) [(1.6*144) - 68.016 sin (0)]\\\\Q = 4.69 * 10^(-3) ft^3 / s`

b) Flow rate assuming the pipe is vertical:

At vertical pipe position, θ = 90°

`Q = (\pi D^4)/(128 \mu) [((\delta p )/(\delta x)) - \gamma sin \theta]\\\\Q = (\pi 0.167^4)/(128 * 0.9312) [(1.6*144) - 68.016 sin (90)]\\\\Q = 3.30 * 10^(-3) ft^3 / s`