Question #106558
Suppose a viscous oil, whose flow is in the laminar regime is to be pumped through a 10 cm
diameter horizontal pipe over a distance of 15km at a rate of s/ 10 m
−3 3
. Viscosity of the oil is
03.0 poise. What is the required pressure drop to maintain such a flow?
1
Expert's answer
2020-03-26T14:32:52-0400

Suppose a viscous oil, whose flow is in the laminar regime is to be pumped through a 10 cm diameter horizontal pipe over a distance of 15km at a rate of 103 m3/s10^{-3}\ m^3/s . Viscosity of the oil is 3.0 poise. What is the required pressure drop to maintain such a flow? 

Solution

For a viscous oil at 20°C take ρ=100 kg/m3.\rho=100\ kg/m^3 . Given that μ=3 poise=0.3 Nsm2,\mu=3 \ poise=0.3\ \dfrac{N\cdot s}{m^2},

D=10 cm=0.1 m,L=15 km=15000 m,Q=103 m3/s.D=10 \ cm=0.1\ m, L=15\ km=15000\ m, Q=10^{-3}\ m^3/s.

Calculate the pipe velocity and Reynolds number:


V=QA=Q(πD2)/4=103 m3/s(π(0.1 m)2)/40.1273 m/sV={Q \over A}={Q \over (\pi D^2)/4}={10^{-3}\ m^3/s \over (\pi (0.1\ m)^2)/4}\approx0.1273\ m/s

Re=ρVDμ=1000kgm30.1273ms0.1 m0.3Nsm2=42.4<2100Re={\rho VD \over \mu}={1000\dfrac{kg}{m^3}\cdot0.1273\dfrac{m}{s}\cdot0.1\ m\over 0.3\dfrac{N\cdot s}{m^2}}=42.4<2100

The flow is laminar flow. Find the required pressure drop to maintain such a flow


Δp=p2p1=128μLQπD4\Delta p=p_2-p_1={128\mu LQ \over \pi D^4}

Δp=1280.3 Nsm215000 m103 m3sπ(0.1 m)4=\Delta p={128\cdot0.3\ \dfrac{N\cdot s}{m^2}\cdot15000\ m\cdot10^{-3}\ \dfrac{m^3}{s} \over \pi (0.1\ m)^4 }=

=183346.5Pa=1.833465×105Pa=0.183 MPA=183346.5 \text{Pa}=1.833465\times10^5 \text{Pa}=0.183\ \text{MPA}


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