What is the Hagen-Poiseuille equation for pressure drop in a laminar pipe?

• A. ΔP = 8 μ L Q /(π r^4)
• B. P = RT/(V - b) - a/V^2
• C. k = A e^(-Ea/RT)
• D. ln((P2^sat)/(P1^sat)) = (-ΔHvap)/R(1/T2 -1/T1)

Answer: (A)

In laminar, fully developed flow through a circular pipe, the pressure drop is caused by viscous shear and is linked to the volumetric flow rate by a simple relation. Solving the Navier–Stokes equations for a Newtonian fluid with no-slip at the wall gives a parabolic velocity profile and, after integrating across the cross-section, the volumetric flow rate Q relates to the pressure drop ΔP by Q = (π r^4 ΔP) / (8 μ L). Rearranging yields the Hagen–Poiseuille equation: ΔP = (8 μ L Q) / (π r^4). This shows why higher viscosity or a longer tube increases the pressure drop, and why a smaller radius dramatically raises ΔP due to the r^4 in the denominator. If you prefer in terms of diameter d, use r = d/2 to obtain ΔP = (128 μ L Q) / (π d^4). This formula assumes Newtonian fluid, straight, rigid tube, steady, incompressible, fully developed, and no-slip at the wall.