Answer:
If we assumed flow was laminar, L = 1840 m
But in reality, this flow is in the turbulent region and L = 0.00000304 m = (1.197 × 10⁻⁴) inch
Step-by-step explanation:
We first check the region of flow of the fluid by computing the Reynolds number
Re = (ρvD/μ)
Listing all the parameters and converting to SI units
ρ = density of the fluid = 0.000238 slug/ft³ = 0.123 kg/m³
v = velocity of flow = (Q/A)
Q = volumetric flowrate of the air = 2.6 ft³/min = 0.0012271 m³/s
A = Cross sectional Area of the pipe = (πD²/4)
D = Pipe diameter = (1/4) inch = 0.00635 m
A = (π×0.00635²/4) = 0.00003167 m²
v = (0.0012271/0.00003167) = 38.746 m/s
μ = dynamic viscosity of the fluid = (Kinematic viscosity) × (density)
Kinematic viscosity = (1.6 × 10⁻⁴) ft²/s = (1.486 × 10⁻⁵) m²/s
μ = (1.486 × 10⁻⁵) × (0.123) = 0.0000018278 = (1.83 × 10⁻⁶) Pa.s
Re = (0.123×38.746×0.00635)/(1.83 × 10⁻⁶) = 16,556.824214903
This Reynolds number is in the turbulent flow region.
From Hagen-Poiseulle equation, the volumetric flowrate for laminar and turbulent flow is given as
Q = (πD⁴ΔP)/(128μL) for laminar flow
ΔP = (0.241 × ρ⁰•⁷⁵ × μ⁰•²⁵ × L × Q¹•⁷⁵)/D⁴•⁷⁵ for turbulent flow
Since our flow is turbulent
ΔP = (0.241 × ρ⁰•⁷⁵ × μ⁰•²⁵ × L × Q¹•⁷⁵)/D⁴•⁷⁵
L = (ΔP.D⁴•⁷⁵)/(0.241 ρ⁰•⁷⁵ μ⁰•²⁵ Q¹•⁷⁵)
Listing all the parameters and converting to SI units
L = Length of the pipe = ?
ΔP = Pressure drop across the flow channel = 90 - 75 = 15 psi = 103421.4 Pa
D = Pipe diameter = (1/4) inch = 0.00635 m
ρ = density of the fluid = 0.000238 slug/ft³ = 0.123 kg/m³
μ = dynamic viscosity of the fluid = (1.83 × 10⁻⁶) Pa.s
Q = volumetric flowrate of the air = 2.6 ft³/min = 0.0012271 m³/s
L = (0.123 × 0.00635⁴•⁷⁵) ÷ (0.241 × 0.123⁰•⁷⁵ × 0.0000018278⁰•²⁵ × 0.0012271¹•⁷⁵)
L = (4.4986 × 10⁻¹²) ÷ (1.481 × 10⁻⁶)
L = 0.0000030378 m = 0.00000304 m = (1.197 × 10⁻⁴) inch
If we assumed that flow was laminar
Q = (πD⁴ΔP)/(128μL)
L = (πD⁴ΔP)/(128μQ)
L = (π × 0.00635⁴ × 103421.4) ÷ (128 × 0.0000018278 × 0.0012271)
L = (0.0005282691) ÷ (0.0000002871)
L = 1840 m
Hope this Helps!!!