A liquid drug, with the viscosity and density of water, is to be administered through a hypodermic needle. The inside diameter of the needle is 0.25 mm and its length is 50 mm. Determine (a) the maximum volume flow rate for which the flow will be laminar, (b) the pressure drop required to deliver the maximum flow rate, and (c) the corresponding wall shear stress.

a) Q = 4 * 10^(-7) m³/s

b) ΔP = 187.4 kPa

c) τ = 234.2 Pa

Explanation:

Let's begin by listing out the given parameters:

η = 0.00089 kg/ms, ρ = 1000 kg/m³,

d = 0.25 mm = 0.00025 m or r = 0.000125 m,

l = 50 mm = 0.05 m

a) For flow in a pipe, the Reynolds Number is less than 2300 for laminar flow; Re < 2300

Q = A * <v>

where: Q = volumetric flow rate, A = cross-sectional area, <v> = velocity

A = πr² = A = π x 0.000125² = 4.9 * 10^(-8)

Using the formula:

Re = <v> * d ÷ v

v = η ÷ ρ = 0.00089 ÷ 1000 = 8.9 * 10^(-7) m²/s

Making <v> subject of formula, we have:

<v> = Re * v ÷ d = 2299 x 8.9 * 10^(-7) ÷ 0.00025 = 8.18444

<v> = 8.18 m/s

Q = A * <v> = 4.9 * 10^(-8) * 8.18 = 4 * 10^(-7)

Q = 4 * 10^(-7) m³/s

b) flow coefficient = 64 ÷ Reynolds number

fD = 64 ÷ Re = 64 ÷ 2299 = 0.028

Pressure drop = (flow coefficient * length * density * velocity²) ÷ (2 * diameter)

ΔP = (fD * l * ρ * <v>²) ÷ (2 * d)

ΔP = (0.028 × 0.05 × 1000 × 8.18²) ÷ (2 × 0.00025)

ΔP = 187354.72 Pa

ΔP = 187.4 kPa

c) From Darcy–Weisbach equation,

Wall shear stress = coefficient of friction * density * Velocity² ÷ 8

τ = fD * ρ * <v>² ÷ 8

τ = 0.028 * 1000 * 8.18² ÷ 8

τ = 234.2 Pa

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation:

d. air pressure decreases as altitude increases.