Answer:
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
