Ideal fluid flow velocities & pressure

[cheddar87]

1. Air at 0∘C flows through the tube shown in the figure. Assume that air is an ideal fluid. What is the air speed v1 at point 1?2. Formulas: A1V1=A2V2, ΔP + 1/2ρΔV2, ρgΔh3. My first through process was to calculate the change in pressure in the mercury as the change in pressure in the mercury would correspond to the change in pressure from points 1 to 2. So i did this using ΔP+ ρgh=0 and calculated the Δp= 13,328 Pascals or .1315 atms. Then this is where i got stuck. I am not sure what to do now that I have the change in pressure. I thought my next step would be:

ΔP+ 1/2ρΔV2------> ΔP= -1/2ρΔV2----> 13,328=-1/2ρΔV2
to calculate the change in velocity from points 1 to 2 but I got caught up in units and the calculations ended up being messy so I don't think I am on the right track. I know I have the areas and I am sure those will come into the equation when I find one of the velocities but I am kind of stuck at this point

 

[SteamKing]

1. Air at 0∘C flows through the tube shown in the figure. Assume that air is an ideal fluid. What is the air speed v1 at point 1?2. Formulas: A1V1=A2V2, ΔP + 1/2ρΔV2, ρgΔh3. My first through process was to calculate the change in pressure in the mercury as the change in pressure in the mercury would correspond to the change in pressure from points 1 to 2. So i did this using ΔP+ ρgh=0 and calculated the Δp= 13,328 Pascals or .1315 atms. Then this is where i got stuck. I am not sure what to do now that I have the change in pressure. I thought my next step would be:

ΔP+ 1/2ρΔV2------> ΔP= -1/2ρΔV2----> 13,328=-1/2ρΔV2
to calculate the change in velocity from points 1 to 2 but I got caught up in units and the calculations ended up being messy so I don't think I am on the right track. I know I have the areas and I am sure those will come into the equation when I find one of the velocities but I am kind of stuck at this point

We can't help unless you show all your calculations.

What units did you get caught up in? Did you try to convert to the same units?

 

[cheddar87]

I figured out the next calculation, here's what i have so far

Δp+ ρgh=0---> Δp= -ρgh----> Δp= - {13,600 kg/m3* 9.8m/s2 * -.1m)----> Δp= 13,328 pascals
and
ΔP= -1/2ρΔV2---> 2ΔP/ρ= -ΔV2----> 2 * 13,328kg/ms2/1.28kg/m3= -ΔV2-----> 26656kg/ms2/1.28kg/m3=-ΔV2-------> 20825m2/s2= -ΔV2

this is where I am stuck. I have the change in velocity but I am not quite sure how to get Velocity 1.

 

[cheddar87]

omg can someone pleaaaaaaaaaaaase help me. 4 hours later and I still can't finish the problem

 

[SteamKing]

omg can someone pleaaaaaaaaaaaase help me. 4 hours later and I still can't finish the problem

If you write the correct Bernoulli equation for this apparatus, you should get:

p₁ + ρV²₁ = p₂ + ρV²₂

You'll need to calculate (or look up) the density of air at 0° C to obtain ρ.

You have already determined what p₂ - p₁ should be from the mercury manometer.

The continuity equation can tell you what the ratio of V₁ to V₂ must be.

 

[cheddar87]

I already did all of that^^ and I have what the change in velocity is. And that's where I am stuck. I don't know how to turn the velocity difference of 144 m/s into V1

 

[cheddar87]

and I don't see how to manipulate A1V1=A2V2 to fit into this whole equation

 

[SteamKing]

and I don't see how to manipulate A1V1=A2V2 to fit into this whole equation

You don't have to fit the continuity equation into the Bernoulli equation.

The continuity equation can tell you what the ratio of V1 to V2 must be.

If you know the ratio of V1 to V2, you can eliminate one of the velocities from the Bernoulli equation and calculate the other. Since the problem asks what V1 is, V2 is the quantity you want to eliminate.