Question

Water is transported from the reservoir at station 1 down to another reservoir at station 2, with height difference z

1
​
−z
2
​
=15 m, as schematised below. The reservoir at station 1 is pressurised at absolute pressure of 500kPa, while the one at station 2 is at atmospheric conditions. The pipes are commercial cast iron pipes with diameter D=30 cm and length L=50 m. All localised losses are negligible and consider that the water level at station 1 and station 2 does not change over timel.

Answer 1

The pressure energy at point 1 is given by;P1=P +ρgz1Where;P is the absolute pressure in the pipeρ is the density of water, which is 1000 kg/m³g is acceleration due to gravity, which is 9.81 m/s²z1 is the height above an arbitrary datumWe are told that P1 = 500 kPa, and z1 = 0 m.Substituting these values in the above equation gives;500,000 = P + (1000 × 9.81 × 0)P = 500,000 PaThe pressure energy at point 2 is given by;P2=ρgz2Where z2 = -15 m (since z1 - z2 = 15 m)Substituting the given values into the above equation gives;P2 = 1000 × 9.81 × (-15)P2 = -1,471,500 PaThe kinetic energy and potential energy changes are negligible as they are small compared to the pressure energy change. Therefore, the Bernoulli's equation for an incompressible fluid flowing in a pipe can be used to calculate the velocity of water flowing through the pipe at station 1, as follows;P1+ 1/2 ρv12 = P2 + 1/2 ρv22Where v1 is the velocity of water at station 1.Substituting the known values into the above equation gives;500,000 + 1/2 × 1000 × v1² = -1,471,500 + 0This simplifies to;v1 = 13.32 m/sThe volume flow rate, Q is given by the following equation;Q = Av1Where A is the cross-sectional area of the pipe.Substituting the known values into the above equation gives;Q = π(0.3/2)² × 13.32Q = 0.296 m³/sTherefore, the volume flow rate of water transported from the reservoir at station 1 down to another reservoir at station 2 is 0.296 m³/s. The answer has a latex-free expression.