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

Answer:

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

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, $\frac{\delta p}{\delta x} = 1.6 psi/ft$

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°

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

b) Flow rate assuming the pipe is vertical:

At vertical pipe position, θ = 90°

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