Question

A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7200 cm^3/s. At one point in the pipe, where the radius is 4.00cm, the water has a pressure of 2.40x10^5 Pa. At a second point in the pipe, the water passes through a constriction where the radius is 2.00cm. What is the water's pressure as it flows through the constriction?

Answer 1

given

   water discharging from a horizontal pipe of

point A

   volume flow rate QA = A1*V1

point B

   volume flow rate is QB = A2*V2

from continuity equation A1V1 = A2 V2

given
   QA = 7200 cm^3/s = 7.2*10^-3 m^3/s

   radius rA = 4 cm = 0.04 m

area of cross section is A1 = pi*rA^2 = pi*0.04^2 m2 = 5.0265*10^-3 m2

   V1 = (7.2*10^-3)/(5.0265*10^-3) m/s = 1.4324 m/s

at point B

   area of cross section is A2 = pi*0.02^2 m2 = 1/256*10^-3 m2

V2 = QB/A2 = 7.2*10^-3/1.256*10^-3 = 5.7325 m/s

using Bernouli's equation

PA + 0.5*rho*v1^2 = PB+ 0.5*rho*v2^1

given PA = 2.40x10^5 Pa

PB = PA + 0.5*rho*v1^2 - 0.5*rho*v2^1

= PA+ (0.5*rho)(V1^2_v2^2)

= 2.4*10^5+(0.5*1000)(1.4324^2-5.7325^2) Pa

= 2.24595*10^5 Pa