Question

Compute the loss of head and pressure drop in 200 ft of horizontal 6-in diameter asphalted cast iron pipe carrying water with a mean velocity of 6 ft/s. For water: rho = 1.94 slug/ft3 and μ = 2.09x10-5 slug/ft-s (ν = μ/rho). Assume ε = 0.0004 ft.

Answer 1

Answer:

The answer is below

Step-by-step explanation:

12 inches = 1 ft.

6 inches = 6 inches * (1 ft./12 inches) = 0.5 ft.

Therefore the diameter of the cast iron (D) = 6 inches = 0.5 ft.

The area of cast iron (A) = πD²/4 = π(0.5)²/4 = 0.196 ft²

The velocity (V) = 6 ft./s, the acceleration due to gravity (g) = 32.2 ft./s²

ε/D = 0.0004 ft./ 0.5 ft. = 0.0008

Using the moody chart, find the line ε/D = 0.0008 and determine the point of intersection with the vertical line R = 2.7 * 10⁵. Hence we get f = 0.02.

The head loss (h) is:

$h_f=f*\frac{L}{d}* \frac{V^2}{2g}=0.02*\frac{200\ ft}{0.5\ ft} *\frac{(6\ ft/s)^2}{2*32.2\ ft/s^2}=4.5\ ft$

The pressure drop (Δp) is:

Δp = ρg$h_f$ = $(62.4\ lbf/ft^3)(4.5\ ft)= 280\ lbf/ft^2$