Question

Water flows at a rate of 30 L/min through a horizontal 7.0-cm-diameter pipe under a pressure of 80 Pa. At one point, calcium deposits reduce the cross-sectional area of the pipe to 20 cm^2. What is the pressure at this point? Consider the water to be an ideal fluid.

Answer 1

given

volume flow rate, dV/dt = 30 L/min

= 30*10^-3 m^3/(60 s)

= 5*10^-4 m^3/s

P1 = 80 Pa

A1 = pi*d1^2/4

= pi*0.07^2/4

= 3.85*10^-3 m^2

A2 = 20 cm^2

= 2*10^-3 m^2

we know, volume flowrate dV/dt = A1*v1

==> v1 = (dV/dt)/A1

= (5*10^-4)/(3.85*10^-3)

= 0.130 m/s

v2 = (dV/dt)/A1

= (5*10^-4)/(2*10^-3)

= 0.25 m/s

now use Bernoulli's theorem,

P1 + (1/2)*rho*v1^2 = P2 + (1/2)*rho*v2^2

==> P2 = P1 + (1/2)*rho*(v1^2 - v2^2)

= 80 + (1/2)*1000*(0.130^2 - 0.25^2)

= 57 pa <<<<<<<<<<<------------------------Answer