# Pressure in tank pipe -- Bernoulli Equations

## Homework Statement [/B]
A pressurized cylindrical tank, 5m in diameter, contains water which emerges from the pipe at point C with a velocity of 25 m/s. Point A is 10m above point B and point C is 3m above point B. The area of the pipe at point B is 0.07 m2 and the pipe narrows to an area of 0.02m2 at Point C. Assume the water is an ideal fluid in laminar flow. The density of water 1000kg/m3

## Homework Equations

Continuity Equation:
VBAB=VCAC

Bernouli's equation:
Let D = density of water
PB + (1/2)DVB2+DghB = PC + (1/2)DVC2+DghC

## The Attempt at a Solution

Let D = density of water
VBAB=VCAC
Therefore
VB=[VCAC]/AB=[25*0.02]/0.07=7.14m/s

PB + (1/2)DVB2+DghB = PC + (1/2)DVC2+DghC
Therefore
PB = PC + (1/2)DVC2+DghC -(1/2)DVB2-DghB

Except PC and PBis unknown so this approach shouldn't work. One of my peers claims it is the same as 1 ATM but I'm doubtful since this does not result in answer which is accurate. I don't see how The pressure at Point C could possibly be 1atm... 2 unknowns .. 1 equation.. not going to work :( correct me if wrong.

Upon searching the internet I came across the solution.. which does yield what I firmly believe is the correct answer .. however I don't not understand how this formula came about..

Let D = density of water

PB = [VC2/(2G) - VB2/(2G) + y)Dg
=(252/(2*9.8) - (7.142/(2*9.8) + 3)*1000*9.8
~=~ 316410 Pa ~=~ 320 kPa

It appears to be some variation of the Bernoulli formula but I need to demonstrate how to take those formulas and make them into this format... which I don't even know where to begin. I don't need to know how to derive the formulas that this one came from .. in other words.. I can just say here's the Bernoulli equation .. rearrange and combine like so and obtain this: (without having to derive the Bernoulli equation itself).

billy_joule
What are you trying to find? Pressure at B? Or Pressure at A? You don't need to consider what's going on at point B to do that.

Except PC and PBis unknown so this approach shouldn't work. One of my peers claims it is the same as 1 ATM but I'm doubtful since this does not result in answer which is accurate. I don't see how The pressure at Point C could possibly be 1atm...
.
Presumably the pipe vents to atmosphere, so point C must be at atmospheric pressure.

Let D = density of water

PB = [VC2/(2G) - VB2/(2G) + y)Dg
=(252/(2*9.8) - (7.142/(2*9.8) + 3)*1000*9.8
~=~ 316410 Pa ~=~ 320 kPa

It appears to be some variation of the Bernoulli formula but I need to demonstrate how to take those formulas and make them into this format... which I don't even know where to begin. I don't need to know how to derive the formulas that this one came from .. in other words.. I can just say here's the Bernoulli equation .. rearrange and combine like so and obtain this

That is just bernoulli's equation, you were actually very close to reaching that arrangement :
.
PB = PC + (1/2)DVC2+DghC -(1/2)DVB2-DghB

Use gauge pressure as your datum so PC = 0, and let your height datum be at point B so hB = 0. Plug all those in and collect like terms and you'll reach the solution you quoted, note that the answer is still guage, not absolute pressure.

• What are you trying to find? Pressure at B? Or Pressure at A? You don't need to consider what's going on at point B to do that.

Presumably the pipe vents to atmosphere, so point C must be at atmospheric pressure.

That is just bernoulli's equation, you were actually very close to reaching that arrangement :
.

Use gauge pressure as your datum so PC = 0, and let your height datum be at point B so hB = 0. Plug all those in and collect like terms and you'll reach the solution you quoted, note that the answer is still guage, not absolute pressure.
Changed it to 0 and bam same kPa in for gauge pressure. Thanks a ton you have no idea how many countless hours I spent on this damn problem.

billy_joule