Water flows at a rate of 0035 m3s in a horizontal pipe whose

Water flows at a rate of 0.035 m^3/s in a horizontal pipe whose diameter is reduced from 15 cm to 8 cm by a reducer. If the pressure at the centerline is measured to be 480 kPa and 445 kPa before and after the reducer, respectively, determine the irreversible head loss in the reducer. Take the kinetic energy correction factors to be 1.05. Answer: 1.18 m The water level in a tank is 20 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end of the hose is pointed straight up. The tank is at sea level, and the water surface is open to the atmosphere. In the line leading from the tank to the nozzle is a pump, which increases the pressure of water. If the water jet rises to a height of 27 m from the ground, determine the minimum pressure rise supplied by the pump to the water line.

Solution

16-2

Q=0.035 m³/s

Flow area of 15 cm pipe=A₁=0.017671 m²

A₂=0.005027 m²

Velocities

V₁=Q/A₁=1.924 ft/s

V₂=Q/A₂=6.963 ft/s

From frictionless Bernoulli equation (with water density 1000 kg/m³)

P₁/1000g+V₁²/2g=P₂/1000g+V₂\'²/2g where V₂\'= velocity considering frictionless flow

But P₁=480 kPa

P₂=445 kPa

Hence

V₂\'²=2x9.81x[(4.81-4.45)x10⁵/1000x9.81+1.924²/(2x9.81)]

=73.7 m²/s²

Hence irreversible head loss

(V₂\'²-V₂²)/2g/(kinetic energy correction factor

=(73.7-6.963²)/(2x9.81x1.05)

=1.22 mWC

16-3

Total head at pump suction (in gage pressure)

=P_s= Total head at bottom of tank =20 m, since velocity is negligible at bottom of tank

Total head at exit of nozzle

P_d= 27m , since presssure is atmospheric at exit of nozzle

Hence the pressure rise supplied by pump

=P_d-P_s=27-20=7 m