Question #220035

A drug is being delivered into a patient’s arm at a rate of 319 mL min–1. The drug is being delivered from a syringe through a needle with an internal diameter of 1 mm. This drug has a density of 1050 kg m–3. Is the flow laminar or turbulent? (ηdrug = 8.90×10–4 Pa s)

Expert's answer

For the flow in a pipe (needle) the Reynold's number is what determines the laminar/turbulent flow. The Reynold's number is given as follows (see https://en.wikipedia.org/wiki/Reynolds_number#Flow_in_a_pipe):


Re=ρQDHμARe = \dfrac{\rho Q D_H}{\mu A}

where ρ=1050kg/m3\rho = 1050kg/m^3 is the density of the drug, Q=319 mL/min=31960×10−6 m3/sQ = 319\space mL/min = \dfrac{319}{60}\times 10^{-6}\space m^3/s is the volumetric flow rate, DH=1mm=10−3mD_H = 1mm = 10^{-3}m is the internal diameter of the needle, A=πDH2/4A = \pi D_H^2/4 is the cross-sectional area of the needle, and μ=8.90×10−4Pa⋅s\mu = 8.90\times 10^{-4}Pa\cdot s is the dynamic viscosity of the drug.

Thus, obtain:


Re=ρQDHμ⋅πDH24=4ρQπμDHRe=4⋅1050⋅31960×10−6π⋅8.90×10−4⋅10−3≈7990Re = \dfrac{\rho Q D_H}{\mu\cdot \dfrac{\pi D_H^2}{4}} = \dfrac{4\rho Q}{\pi\mu D_H}\\ Re = \dfrac{4\cdot 1050\cdot \dfrac{319}{60}\times 10^{-6}}{\pi\cdot 8.90\times 10^{-4}\cdot 10^{-3}} \approx 7990

Since Re>2900Re>2900, the flow is turbulent.


Answer. Turbulent.


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