Solving Bernoulli's Equation for Fluid Mechanics in Ethanol Pipe

[escobar147]

Ethanol of density 791 kgm-3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A1= 1.20 x 10-3 m2 to A2=A1/2. The pressure difference between the wide and narrow sections of the pipe is 4120 Pa. What is the volume flow rate of the ethanol?


(Ans.: 2.24 x 10-3 m3s-1)

how is this worked out? any help would be massively appreciated.

 

[SteamKing]

Write your Bernoulli equations and start compiling the data to plug into them. Let's see some work!

 

[escobar147]

ok

bernoullis equation states that:

P1 + ρgh1 + 1/2ρ(v1)^2 = P2 + ρgh2 + 1/2ρ(v2)^2

now i believe from some other calculations that
v2 - v1 = 3.2 m/s

and we know that:

p2 - p1 = 4120 pa
ρ = 791 kgm-3
g = 9.81
the volume flow rate is given by: cross sectional area x flow velocity

but i just can't see how to find the volume flow rate with the given information

 

[escobar147]

i'm starting to think I'm using the wrong equation altogether...

 

[SteamKing]

What other calculation did you do to get v2 - v1 = 3.2 m/s? It's not shown.

 

[escobar147]

sorry i forgot i used the venturi equation:

p1 - p2 = ρ/2(v2^2 - v1^2)

 

[SteamKing]

The venturi equation is a special form of the Bernoulli equation. What you should use in addition to the Bernoulli equation is the fact that the mass flow of alcohol into the pipe is the same as the mass flow leaving the pipe.

 

[SteamKing]

Hint: the mass flow past a point = Rho * Area * velocity