Flow and Flow Rate: Determine Rate & Pressure Change

[missnola2a]

Homework Statement

A horizontal water main with a cross-sectional area of 220 cm2 necks down to a pipe of area 55 cm2. Meters mounted in the flow on each side of the transition coupling show a change in gauge pressure of 85 kPa. Determine the flow rate through the system, taking the fluid to be ideal.

The flowmeter in Fig. P113 shows a difference in height of 6.0 cm on an oil pipeline 240 cm2 in cross section., If the throat of the device has a diameter of 14 cm, what is the pipleline flow rate? (Assume the liquid to be ideal.)

Homework Equations

I am lost, I don't know what equation to use... the book doesn't supply a flow rate eq.
A1V1=A2V2 but these are all knowns?

The Attempt at a Solution

 

[rl.bhat]

A1V1 = A2V2
So A1/A2 = V2/V1. ...(1)
According to Bernoulli dequation
P1 + 1/2ρV1^2 = P2 + 1/2ρV2^2
P1 - P2 = 1/2ρV2^2 - 1/2ρV1^2...(2)
From eq.1 and 2, find the value of V1 and V2.
Flow rate is AV.

 

[missnola2a]

hi,

this is very confusing.

the question is

A horizontal water main with a cross-sectional area of 220 cm2 necks down to a pipe of area 55 cm2. Meters mounted in the flow on each side of the transition coupling show a change in gauge pressure of 85 kPa. Determine the flow rate through the system, taking the fluid to be ideal.

the answer is to be in m3/s

I know A1 and A2 which I take 220 cm2 (1m/ (100cm)(100cm)) and the same for 55cm2.

where does the change in pressure come in?

what do I use for flow rate? DeltaA*DeltaP?? if flow rate through the system is AV, which A do I use which V do I use?

 

[djeitnstine]

missnola, you have 2 equations with 2 unknowns, V1 and V2 at the entrance and exit respectively. rl.bhat showed you the Bernoulli equation needed to find V1 and V2. The role pressure plays in here is the key to solving the problem. You have a relationship between V1 and V2

Specifically $$V_1 = \frac{A_2}{A_1}V_2$$. so you place that in the Bernoulli equation and you get a term that looks like $$\frac{1}{2} \rho \left(\frac{A_2}{A_1} \right)^2 V_2 ^2$$.

Do you see where this is going?

 

[missnola2a]

a pipe that carries a fluid ρ = 873 kg/m3 lies along the slope of a smooth hill which is 140 m high. The speed of the liquid at the bottom of the hill is 6.5 m/s. The diameter of the crude oil pipe line at the top of the hill reduces by a factor 5. Please answer the following:
Hint
(a) By what factor the area of the oil pipe line reduces at the top of the hill?

(b) What is the ratio of speeds of the oil from the bottom of the hill to the top of the hill?

(c) How much potential energy per unit volume the oil gains as it moves from the bottom to the top of the hill?
J/m3
(d) What is the change in pressure in the pipe line during this motion along the hill?




OK ... so I first of all am confused by this because I can't tell if the pipe starts at the bottom of the hill and goes up or vice versa.

This may not matter mathematically, but I am a conceputal person.

That said,
a1v1=a2v2

(pie drops out)
r1^2 v1= r2^2 v2

this is where i am a little stuck, should it be 5r1=r2
OR r2=r1/5

this changes the equation a lot. after I get v2 (or v1) I put it into origional

P1 + 1/2 rho V1^2 + rho*g*Y1 = P2 + 1/2 rho V2^2

*** assuming that PE for 2 is 0, but again, i am confused as to where in the system Y=0

thanks for your help

 

[djeitnstine]

How the question is stated the water runs uphill...

A good indicator is that they give you the velocity at the bottom of the hill first as +6.5 then ask you about the velocity uphill.

Also why are you stuck?

Ask yourself, how do you define section 1 and section 2. Is section 1 at the bottom and 2 at the top? If so then the answer to that question becomes obvious from the wording of the question.

Your first statement 5r1=r2 would say, the radius at the top is five times larger than the bottom. Which is not what the statement says..

The second statement r2=r1/5 says that the radius at the top is 1/5th the radius at the bottom, which would be correct if you used diameter.

Hint you can replace r by d in your other equations (a1v1=a2v2 in particular) if you understand that $$A = \frac{\pi}{4} d^2$$

 

[djeitnstine]

Also when you choose your reference frame for the PE anywhere you choose can be 0 as long as you are consistent with the relative distances. Ex. in $$\rho g Y_1$$, Y1 can be 0 wherever you choose at the top or at the bottom at the hill, if the hill is height H, and you choose 0 at the top, the bottom's PE has to be $$\rho g (-H)$$