Question

Cooling water for a chemical plant must be pumped from a river 2500 ft from the plant site. Preliminary design cans for a flow of 600 gal/min and 6-in. Steel pipe. Calculate the pressure drop and the annual pumping cost if power cost 3 cents per kilowatt-hour. Would the use of an 8-in. Pipe reduce the power cost enough to offset the increased pipe cost? Use $15/ft of length for the installed cost of 6-in. Pipe and $20/ft for 8-in. Pipe. Annual charges are 20 percent of the installed cost.

Answer 1

Answer:

The power cost savings for the 8 inches pipe offsets the increased cost for the pipe therefore to save costs the 8-inch pipe should be used

Explanation:

The given parameters in the question are;

The distance of the river from the the site, d = 2,500 ft.

The planned flow rate = 600 gal/min

The diameter of the pipe, d = 6-in.

The pipe material = Steel

The cost of pumping = 3 cents per kilowatt-hour

The Bernoulli's equation is presented as follows;

$\dfrac{P_a}{\rho} + g\cdot Z_a + \dfrac{V^2_a}{2} + \eta\cdot W_p = \dfrac{P_b}{\rho} + g\cdot Z_b + \dfrac{V^2_b}{2} +h_f +W_m$

${P_a} = {P_b} = Atmospheric \ pressure$

$Z_a = Z_b$

Vₐ - 0 m/s (The river is taken as an infinite source)

$W_m$ = 0

The head loss in 6 inches steel pipe of at flow rate of 600 gal/min = 1.19 psi/100 ft

Therefore; $h_f$ = 1.19 × 2500/100 = 29.75 psi

$\eta\cdot W_p = \dfrac{V^2_b}{2} +h_f$

$V_b$ = Q/$A_b$ = 600 gal/min/(π·(6 in.)²/4) = 6.80829479 ft./s

$V_b$ ≈ 6.81 ft./s

The pressure of the pump = P =  62.4 lb/ft³× (6.81 ft./s)²/2 + 29.75 psi ≈ 30.06 psi

The power of the pump = Q·P ≈ 30.06 psi × 600 gal/min = 7,845.50835 W

The power consumed per hour = 7,845.50835  × 60  × 60 W

The cost = 28,243.8301 kW × 3  = $847.31  per hour

Annual cost =  $847.31 × 8766 = $7,427,519.46

Pipe  cost = $15/ft × 2,500 ft = $37,500

Annual charges = 20% × Installed cost = 0.2 × $37,500 = $7,500

Total cost = $37,500 + $7,500 + $7,427,519.46 = $7,475,519.46

For the 8-in pipe, we have;

$V_b$ = Q/$A_b$ = 600 gal/min/(π·(8 in.)²/4) = 3.83 ft./s

$h_f$ = 1.17 ft/100 feet

Total head loss = (2,500 ft/100 ft) × 1.17 ft. = 29.25 ft.

$h_p = \dfrac{V^2_b}{2 \cdot g} +h_f$

∴ $h_p$ = (3.83 ft./s)²/(2 × 32.1740 ft/s²) + 29.25 ft. ≈ 29.5 ft.

The power of the pump = ρ·g·h × Q

Power of pump = 62.4 lb/ft³ × 32.1740 ft/s² × 29.5 ft.× 600 gal/min = 3,363.8047 W

The cost power consumed per annum = 3,363.8047 W × 60 × 60 × 3 × 8766 = $3,184,608.1

The Cost of the pipe = $20/ft × 2,500 ft. = $50,000

The total cost plus charges = $3,184,608.1 + $50,000 + $10,000 = $3,244,608.1

Therefore it is more affordable to use the 8-in pipe which provides substantial savings in power costs