Question

At a ski resort, water at 40 °F is pumped through a 3-in.-diameter, 2001-ft-long steel pipe from a pond at an elevation of 4286 ft to a snow-making machine at an elevation of 4620 ft at a rate of 0.26 ft3/s. If it is necessary to maintain a pressure of 182 psi at the snow-making machine, determine the horsepower added to the water by the pump. Neglect minor losses.

Answer 1

Answer:

P = 24.38 hp

Explanation:

Given data:

diameter of steel pipe = 3 inc

length of pipe = 2001 ft

elevation of pond = 4286 ft

elevation of snow making machine = 4620 ft

Rate of water = 0.26 ft^3/s

Applying bernouli equation on bioth side

$\frac{P}{\rho} + \frac{v_1^2}{2g} + z_1 + h_p =\frac{P_2}{\rho} + \frac{v_2^2}{2g} + z_2 + \frac{flv^2}{2gD}$.....1

where P_2 = 182 psi

P_1 = 0, v_1 = 0

$V =V_2 = \frac{Q}{\frac{\pi}{4} D^2}= \frac{0.26}{\frac{\pi}{4} \times (3/12)^2} = 5.30 ft/s$

calculation fro friction factor

from standard table we have

$\frac{\epsilon}{D} = \frac{0.00015}{(3/12)} = 6\times 10^{-4}$

$Re =\frac{VD}[\nu} = \frac{5.30 \times (3/12)}{1.66 \times 10^{-5}} = 7.96 \times 10^4$

so for calculated R and$\frac{\epsilon}{D}$, F value = 0.0212

from equation 1

$h_p = \frac{P_2}{\rho} + Z_2 -Z_1 +(1 + f\frac{l}{D}) \frac{V^2}{2g}$

$h_p = \frac{182 \times 144/ lb /ft^2}{62.4} + 4620 - 4286 + (1 + 0.0212 \frac{2001}{(3/12)}) \frac{5.30}{2\times 32.2}$

h_p = 826.76ft so that

$P = \rho Q h_p = 62.4 \times 0.26 \times 826.76 = 13413.38 ft lb/s = 24.38 hp$