Question 4.11: SAE 10W oil at 20°C flows at 1.1 m^3/h through a horizontal ...

• Assumptions: Laminar, steady, Hagen-Poiseuille pipe flow.

• Approach: The formulas of Eqs. (4.138) are appropriate for this problem. Note that R = 0.01 m.

\tau_{wall}=\mu\left|\frac{\partial v_z}{\partial r}\right|_{r=R}=\frac{4\mu V_{avg}}{R}=\frac{R}{2}\left(-\frac{dp}{dz}\right)=\frac{R}{2}\frac{\Delta p}{L}                                       (4.138)

• Property values: From Table A.3 for SAE 10W oil, \rho = 870  kg/m^3 and µ = 0.104 kg/(m-s).

^*In contact with air.
^{\dagger}The viscosity–temperature variation of these liquids may be fitted to the empirical expression

with accuracy of ±6 percent in the range 0 ≤ T ≤ 100°C.
^{\ddagger}Representative values. The SAE oil classifications allow a viscosity variation of up to ±50 percent, especially at lower temperatures.

• Solution steps: The average velocity follows easily from the flow rate and the pipe area:

We had to convert Q to m^3/s. The (diameter) Reynolds number follows from the average velocity:

This is less than the “transition” value of 2100; so the flow is indeed laminar, and the formulas are valid. The pressure drop is computed from the third of Eqs. (4.138):

Q=\frac{1.1}{3600}\frac{m^3}{s}=\frac{\pi R^4 \Delta p}{8\mu L}=\frac{\pi(0.01  m)^4 \Delta p}{8(0.104  kg/(m-s))(12  m)}    solve for Δp = 97,100 Pa

When using SI units, the answer returns in pascals; no conversion factors are needed. Finally, the power required is the product of flow rate and pressure drop:

• Comments: Pipe flow problems are straightforward algebraic exercises if the data are compatible. Note again that SI units can be used in the formulas without conversion factors.