Q.461. Fluid dynamics - Bernoulli's principle

Question 461.
A liquid of density 1190 kg/m3 flows with speed 1.44m/s into a pipe of diameter 0.23m. The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 8.04 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.3 atm. The acceleration of gravity is 9.8 m/s2 and
Patm = 1.013 × 105 Pa.
Applying Bernoulli’s principle, what is the pressure P1 at the entrance end of the pipe?

Answer 461.
Vb
= 1.44 * (0.23/0.05)^2
= 30.4704 m/s

According to Bernoulli's principle,
Pa + (1/2) ρVa^2 + ρgHa = Pb + (1/2) ρVb^2 + ρgHb
=> Pa - Pb
= ρ * [(1/2) (Vb^2 - Va^2) + g(Hb - Ha)]
= (1190) * [(1/2) ((30.4704)^2 - (1.44)^2) + 9.81 * 8.04]
= (1190) * (463.1858 + 78.8724) N/m^2
= 645049 N/m^2
= 645049 * 9.86923267 × 10-6 atm
= 6.366 atm
=> Pa = 6.366 + 1.3 = 7.666 atm.

For conversion of N/m^2 to atm, refer to the link as under.
Conversion Link:

Link to YA!