Reply
Mon 19 Dec, 2005 11:03 pm
A liquid is flowing through a horizontal pipe whose radius is 0.10 m. the pipe bends straight up for 10 meters and joins another horizontal pipe whose radius is 0.20 m. It was found that the pressure in the lower pipe and the upper pipe were the same. What is the speed in the lower pipe, v1?
u are given the height (10 meters), the two radii (.10 and .20m)...i have no clue how to solve this problem..someone plz help me out..
thanks
Assume (1) inviscid, steady flow of an incompressible liquid => Area X Velocity = const.
Thus V1 = 4xV2
Apply conservation of energy with constant pressure;
==> V1 squared - V2 squared =10g
combine and solve for V1
Use conservation of mass.
That is if there is no change in density (or leaks). In this case DeltaQ=0 or Qin=Qout.
Since Q is the product of the velocity and cross sectional flow area, the velocity is a ratio.
Rap
Bernoulli saves the day.
The equation you want is
Bernoulli's Equation. The link shows everything you need.
Stumble It!
Tweet This
Bookmark on Delicious
Share on Facebook
Share on MySpace