Bernoulli's equation to find the flow rate

In summary: So the units of v1^2 are m2/s2.So, the units of v1^2/v2^2 is (m2/s2)/(m2/s2) = dimensionless.Therefore, the units of (v2^2-v1^2) are also dimensionless.Ask yourself, where do the units of kg/m3 come in?
  • #1
Sall1230
10
0

Homework Statement


" 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?

Homework Equations


P1 + 0.5 * ρ * v1^2 + h1ρg = P2 + 0.5 * ρ * v2^2 + h2ρg[/B]
A1 v1 = A2 v2

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
 
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  • #2
Sall1230 said:

Homework Statement


" 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?

Homework Equations


P1 + 0.5 * ρ * v12 + h1ρg = P2 + 0.5 * ρ * v22 + h2ρg[/B]
A1 v1 = A2 v2

The Attempt at a Solution


I tried to calculate each velocity individually,
V1= 2P/ρ = 2(8*104)/1000= 12.65
V2= 2(6*104)/1000 = 10.95

You can't do this because there isn't enough information.
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 ( v22 -v12)
This is the result of actually applying the Bernoulli equation to this pipe.
The answer : (v22-v12) = 40
This is also correct.
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
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
SteamKing said:
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.

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
Sall1230 said:
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?
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.
 

FAQ: Bernoulli's equation to find the flow rate

What is Bernoulli's equation and how is it used to find flow rate?

Bernoulli's equation is a fundamental principle in fluid dynamics that relates the pressure, velocity, and elevation of a fluid at a specific point in time. It is used to predict the flow rate of a fluid through a pipe or channel by balancing the energy at different points in the system.

What are the assumptions made in Bernoulli's equation?

The assumptions made in Bernoulli's equation are that the fluid is incompressible, the flow is steady, and there is no friction or viscosity in the system. Additionally, the fluid must be flowing along a streamline and the fluid properties must be constant throughout the system.

What are the units for the different variables in Bernoulli's equation?

The pressure in Bernoulli's equation is typically measured in Pascals (Pa), the velocity in meters per second (m/s), and the elevation in meters (m). The resulting flow rate is usually measured in cubic meters per second (m^3/s).

How is Bernoulli's equation used to find the flow rate for a fluid?

To use Bernoulli's equation to find the flow rate, you first need to identify the different points in the system where the pressure, velocity, and elevation are known. Then, you can plug these values into the equation and solve for the flow rate. It is important to ensure that the units are consistent and the assumptions of the equation are met.

Can Bernoulli's equation be used for any type of fluid?

Bernoulli's equation can be used for any type of fluid, as long as the assumptions are met. However, for compressible fluids, the equation needs to be modified to account for changes in density. It is also important to note that for highly viscous fluids, the assumption of no friction may not be valid and the equation may need to be adjusted accordingly.

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