Question

Water is pumped up from the Colorado River to supply Grand Canyon Village, located on the rim of the canyon. The river is at an elevation of 564 m, and the village is at an elevation of 2,095 m. Imagine that the water is pumped through a single long pipe 19.0 cm in diameter, driven by a single pump at the bottom end.

(a) What is the minimum pressure at which the water must be pumped if it is to arrive at the village?
____________ MPa

(b) If 4,700 m³ of water is pumped per day, what is the speed of the water in the pipe? Note: Assume the free-fall acceleration and the density of air are constant over this range of elevations. The pressures you calculate are too high for an ordinary pipe. The water is actually lifted in stages by several pumps through shorter pipes.
_____________m/s

Answer 1

Given

the elevations of the river is h1 = 564 m
and of the village is h2 = 2095 m

the elevation difference is h = h2-h1 = 2095-564 m = 1531 m

we know that the 1 m depth of the water has the pressure is

   P1 = rho*g*h

   P1 = 1000 kg/m^3*9.8 m/s^2*1 m = 9800 Pa = 9.8 k Pa

so the pressure created by the water at height of 1531 m is

   P2 = 9.8*1531 kPa

   P2 = 15003.8 kPa

(a) the minimum pressure at which the water must be pumped if it is to arrive at the village is P2 = 15003.8 kPa

   P2 = 15.0038 MPa

(b)
the amount of water pumped per day is 4700 m^3

the amount of water pumped per second or the rate of flow is Q = 4700 m^3/(24*60*60)s

   Q = 0.0544 m^3/s

we know that Q = A*v

given area of the pipe is A = pi*r^2

here r is radius of the pipe r = 9.5 cm =0.95 m

   A = pi*0.095^2 m^2

   A = 0.028353   m^2

now the speed is v = Q/A

   v = 0.0544 m^3/s / 0.028353 m^2
  
   v = 1.9187   m/s