# Bernoulli's equation to find the flow rate

1. Apr 4, 2016

### Sall1230

1. The problem statement, all variables and given/known data
" A horizontal water pipe has a radius of 10 cm and a pressure of 8*10^4 Pa at one end. At the other end the radius is 5 cm and the pressure is 6*10^4 Pa. What is the water flow rate through this pipe?

2. Relevant equations
P1 + 0.5 * ρ * v1^2 + h1ρg = P2 + 0.5 * ρ * v2^2 + h2ρg

A1 v1 = A2 v2
3. The attempt at a solution
I tried to calculate each velocity individually,
V1= 2P/ρ = 2(8*10^4)/1000= 12.65
V2= 2(6*10^4)/1000 = 10.95
Which means that this is not true since v2 must be bigger considering the pressure dropped. So I tried this law:
P1-P2= 0.5 ρ ( v2^2 -v1^2)
The answer : (v2^2-v1^2) = 40
So I tried to plug it in with the equation Q= Av
(15*10^-2)^2 * 3.14 * 40
But it still didn't seem right

2. Apr 4, 2016

### SteamKing

Staff Emeritus
You can't do this because there isn't enough information.
This is the result of actually applying the Bernoulli equation to this pipe.
This is also correct.
This is where you went off the rails.

You use the continuity equation, A1 v1 = A2 v2, to find a relationship between v1 and v2. Once that relationship is established, then you can solve

(v22-v12) = 40

for v1 or v2, depending on your choice. Once the flow velocity is found, then the flow rate can be calculated.

3. Apr 4, 2016

### Sall1230

Ok so:
v2^2 - v1^2 = 40
A1 v1 = A2 v2
(10*10^-2)^2 v1 = ( 5*10^-2)^2 v2
0.01 v1 = 0.025 v2
v2 = 4 v1
Putting it in the equation v2^2 - v1^2 = 40
(4v1)^2 - v1^2 = 40
16v1^2 - v1^2 = 40
15v1^2 = 40
v1^2 = 40/15
v1= 1.632 m/s
Q= Av
Q= (10*10^-2)^2 * 3.14 * 1.632
Q= 0.0512 kg/m^3

Is my way correct?

4. Apr 4, 2016

### SteamKing

Staff Emeritus
Yes, except 0.0512 kg/m3 is not a flow rate, at least it doesn't have the correct units.

Remember, v1 has units of m/s.