Question 8.10: Heavy fuel oil flows from A to B through a 100 m horizontal ...
Heavy fuel oil flows from A to B through a 100 m horizontal steel pipe of 150 mm diameter. The pressure at A is 1.08 MPa and at B is 0.95 MPa. The kinematic viscosity is 412.5 × 10^{-6} m ^{2}/s, and the relative density of the oil is 0.918. What is the flow rate in m ^{3}/s?
The Bernoulli′s equation between A and B
\frac{1.08 \times 10^{6}}{918 \times 9.81}+\frac{V^{2}}{2 g}+0=\frac{0.95 \times 10^{6}}{918 \times 9.81}+\frac{V^{2}}{2 g}+0+f \frac{100 V^{2}}{2 g \times 0.150}
or 119.925-105.49=\frac{666.67 f V^{2}}{2 g}
Both V and f are unknown so we have to follow another approach. If laminar flow exists, then from Eq. (8.48),
-\left(\frac{ d p}{ d z}\right)=\frac{8 \mu\left(v_{z}\right)_{ av }}{R^{2}} (8.48)
\frac{p_{1}-p_{2}}{l}=\frac{8 \mu V_{ av }}{R^{2}}
or V_{\text {av }}=\frac{\left(p_{1}-p_{2}\right) D^{2}}{32 \mu l}=\frac{(1080-950) 1000(.150)^{2}}{32\left(412.5 \times 10^{-6} \times 918\right)(100)}
=\frac{130 \times 1000 \times 0.0225}{32 \times 412.5 \times 918 \times 100} \times 10^{6}=2.41 m/s
\operatorname{Re}=\frac{2.41 \times(150 / 1000)}{412.5 \times 10^{-6}}=876.36
Hence, laminar flow assumption is valid.
Q=A V_{ av }=\frac{\pi}{4}(0.15)^{2} \times 2.41=0.0425 m ^{3} / s