Question

A liquid of density 1150 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.99 m/s and the pipe diameter ?1 is 12.3 cm . At Location 2, the pipe diameter ?2 is 17.9 cm . At Location 1, the pipe is Δ?=8.79 m higher than it is at Location 2. Ignoring viscosity, calculate the difference Δ? between the fluid pressure at Location 2 and the fluid pressure at Location 1.

Answer #2

According to Bernoulli's theorem -

p/rho + gz + (1/2)v^2 = constant.

Also, v*A = constant

=> v*r^2 = constant.

Therefore, at location 2, flow speed of liquid = (9.99 m/s)(12.3^2/17.9^2) = 4.72 m/s

So, difference in pressure
between the fluid pressure at location 2 and fluid pressure at
location 1 is -

= rho { g(z1 - z2) + (1/2)[(v1)^2 - (v2)^2] }

= (1150 kg/m^3) * [9.8 m/s^2 (8.79 m) + (1/2)(9.99^2 -
4.72^2)m^2/s^2].

= (1150) * [86.142 + 38.76085]. = 143638 N/m^2 (Answer)

answered by: anonymous

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