Question

$p_1 = p_2 + f \frac{\ell}{D} \frac{\rho V^2}{2}$ $V = Q/A = 4Q/(\pi D^2) = 4(2.0 \text{ ft}^3/\text{s})/\pi D^2$ $p_1 - p_2 = (0.5)(144) \text{ lb/ft}^2$ $= f \frac{(100 \text{ ft})}{D} (0.00238 \text{ slugs/ft}^3) - \frac{1}{2} \left( \frac{2.55 \text{ ft}}{D^2 \text{ s}} \right)^2$ $D = 0.404 f^{1/5}$

          $p_1 = p_2 + f \frac{\ell}{D} \frac{\rho V^2}{2}$

$V = Q/A = 4Q/(\pi D^2) = 4(2.0 \text{ ft}^3/\text{s})/\pi D^2$

$p_1 - p_2 = (0.5)(144) \text{ lb/ft}^2$

$= f \frac{(100 \text{ ft})}{D} (0.00238 \text{ slugs/ft}^3) - \frac{1}{2} \left( \frac{2.55 \text{ ft}}{D^2 \text{ s}} \right)^2$

$D = 0.404 f^{1/5}$

p1 = p2 + f (ℓ)/(D)(ρ V^2)/(2)

V = Q/A = 4Q/(π D^2) = 4(2.0 ft^3/s)/π D^2

p1 - p2 = (0.5)(144) lb/ft^2

= f ((100 ft))/(D) (0.00238 slugs/ft^3) - (1)/(2)( (2.55 ft)/(D^2 s))^2

D = 0.404 f^1/5

Added by Jessica C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Chai Santi

09/23/2021

Step 1

The Darcy-Weisbach equation is given by: \[ \Delta P = \frac{fL\rho V^2}{2D} \] Where: - ΔP = pressure drop (0.50 psi per 100ft) - f = friction factor - L = length of the pipe (100ft) - ρ = density of air (0.00238 slugs/ft^3) - V = velocity of air (2.0 ft/s) - D Show more…

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Air is at standard temperature and pressure flows through a horizontal, galvanized iron pipe (e = 0.0005 ft) at a rate of 2.0. Determine the minimum pipe diameter if the pressure drop is to be no more than 0.50 psi per 100ft of pipe. Diagram: Find: Use a similar method to determine: (a) the pipe diameter if the liquid is cold seawater and the pipe is 600 m long. (b) Determine the pipe’s total cost if the price is $9/kg of pipe material. epV2 V=Q/A=4Q/D=42.03/s/D P1-p=0.5144lb/t (100ft 0.00238 slugs/ft3 2.55 ft D=0.404f15