# flow speed through pipe problem

#### ssiding

10+ Year Member
A fluid of density p flows through a horizontal pipe with negligile viscosity. The flow is streamline with constant flow rate. The diamer f the pipe at point 1 is d and the flow speed is v. If the diamer of teh pipe at point 2 is d/3, then the pressure at point 2 is

A. less that the pressure at point 1 by 4pv^s
B. less than the pressure at point 1 by 40pv^2
C. more than the pressure at point 1 by 4pv^2
D. more than the pressure at point 1 by 40pv^2

I know the pressure will less at point 2 due to increase in flow speed but I can't figure out by how much

#### w0lf137

10+ Year Member
A fluid of density p flows through a horizontal pipe with negligile viscosity. The flow is streamline with constant flow rate. The diamer f the pipe at point 1 is d and the flow speed is v. If the diamer of teh pipe at point 2 is d/3, then the pressure at point 2 is

A. less that the pressure at point 1 by 4pv^s
B. less than the pressure at point 1 by 40pv^2
C. more than the pressure at point 1 by 4pv^2
D. more than the pressure at point 1 by 40pv^2

I know the pressure will less at point 2 due to increase in flow speed but I can't figure out by how much
I hate problems like this that take forever.

flow rates on both ends have to equal: AV=AV
because A = (pi)r^2 we can also say the radius is half of the diameter (pi)(D/2)^2. This becomes A= (pi)(D^2/4). Now plug in the A in the AV=AV equation: (pi)(D^2/4)V=(pi)(D^2/4)V. The pi cancel out to become(D^2/4)V=(D^2/4)V

Obviously the A and V are different for each points but i didnt include subscripts to reduce confusion. from point A to point B the diameter is reduced to (1/3)D. in the above equation the (1/3)D becomes (1/9)D because the D is squared. Because of the decrease of 1/9 in the diameter, we have to have an increase in 9 in the velocity. So the new equation: (D^2/4)V=(1/9)(D^2/4)(9V)

Now to find out what the increase in 9V would do to the pressure we have to use that equation i forget who made it: P+.5pv^2+pgh=P+.5pv^2+pgh

Because the pipe is horizontal (no change in h) we can cancel out the pgh terms to have: P+.5pv^2=P+.5pv^2. Right away we can tell that the 9V on the right side of the equation will turn to 81V^2 cause the V is squared. so the equation becomes: P+.5pv^2=P+.5p(81)v^2. In the .5p(81)v^2 we get 40.5pv^2. Now we want to solve for P at point A in order to see how P at point B is changed.

Bring the +.5pv^2 to the other side =P+40.5pv^2 - .5pv^2
This becomes: P =P+40pv^2

As you can see at Point A the pressure is the Pressure at point B +40pv^2. So the pressure decreased by 40pv^2 from point A to Point B.

To do this faster: Automatically from your 23hr/day study for the MCAT you should have known pressure will decrease with increasing velocity so cross out C and D. Now remeber AV=AV. You need A which has r but you have D. Now you know because A=(pi)r^2, the effect on D will have a square effect on r. So 1/3 in the D becomes 1/9 in the r. The 1/9 in the r has to becomes a 9 in the v in the AV=AV. Now the 9 in the V has to be squared in that 2nd equation so it becomes 81. I first look for an answer with 81 but saw none, then i see the answers are in a pv^2 form. So divide 80/2 cause of the .5 in the .5pv^2. becoming 40.5pv^2. The left side of the equation ha a .5pv^2 canceling out my .5pv^2 in the 40.5pv^2. Either way, the answer has to be near 40pv^2.