A nail gun operates using pressurized air, which is supplied through a 1/4-in diameter hose. The gun requires 75 psi to operate with a 2.6 ft3 /min airflow. If the air compressor develops 90 psi, determine the maximum allowable length of hose that can be used for its operation. Assume incompressible flow and a smooth pipe. Take =0.000238 slug/ft3 and = 1.6×10-4 ft2 /s.

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

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!!!

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