Oil (p=925kg/m^3) is flowing thru a pipe at a constant speed of 4m/s (point A) w

Question

Oil (p=925kg/m^3) is flowing thru a pipe at a constant speed of 4m/s (point A) when it comes across a vertical bend in the pipe raising it 5m and it continues flowing in the same direction (point B) . The cross sectional area of the pipe does not change. Take G= 10m/s

A) if the pressure before the oil encounters the vertical bend is P= 2 x 10^5 N/m, what is the pressure after the oil encounters the vertical bend?

B) let's assume that the cross sectional area of the pipe at point B is now half the cross sectional pipe at point A. What is the speedof the oil at point B?

C) what is the pressure of the oil at point B if the pressure at point A is the same as before?

Explanation / Answer

This question is easily solved using Bernoulli's conservation of mechanical energy:

pB + 1/2*rho*vB^2 + rho*g*zB = pA + 1/2*rho*vA^2 + rho*g*zA

where p is pressure, rho is fluid density, v is speed, g is gravitational constant and z is vertical location.

Well the question says constant speed so vB = vA. This can also be seen from a mass balance, Ff = F0 -> rho*aB*vB = rho*aA*vA -> vB = vA where aA = aB = a is the cross-sectional area.

B. So plugging in vB = vA into Bernoulli yields:

vB = 4m/s

dp = pB-pA = -1/2*rho*g*(zB-zA) = -23125 Pa or -0.23bar

where dp is the change in pressure. (1 bar = 1e5 Pa)
This shows that pressure is lost (as dp is negative) due to a solely hydrostatic effect.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.