# Pipe and Bernoulli's Equation: Solving for Mass and Gauge Pressure

• ~christina~
In summary, the conversation discusses the flow of fresh water through a horizontal drainage pipe and its exit into the atmosphere. The mass of water flowing out in one hour is found using the continuity equation, and the gauge pressure in the left section of the pipe is calculated using Bernoulli's equation. The final answer for the gauge pressure is 4.8 Pa, with the initial confusion about estimating the area difference between the two sections of the pipe.
~christina~
Gold Member

## Homework Statement

In the figure below, fresh water flows through a horizontal drainage pipe and then out into the atmosphere at a speed v1 = 15 m / s. The diameters of the left and right sections are 5.0 cm and 3.0 cm.

(a) Find the mass of water that flows into the atmosphere in one hour.

(b) What is the gauge pressure in the left section of the pipe?

http://img174.imageshack.us/img174/5987/picture1lj9.th.jpg

## Homework Equations

$$P_1 + 1/2 \rho v^2 + \rho gy_1 = P_2 + 1/2\rho v^2 + \rho g y_2$$

A1v1= A2v2

P= F/A

$$A_{circle}= \pi r^2$$

$$m= \rho \Delta V = v_1A_1 \Delta t$$

## The Attempt at a Solution

First of all I'm sort of confused with how you can estimate since A1>> A2 but I don't think I can do that here so would that change anything??

I think that i know this:

$$A_1= \pi r^2 = \pi (3cm/2)^2= 7.0685 cm^2$$

$$A_2= \pi r^2 = \pi (5cm/2)^2= 19.6349cm^2$$

$$P_1= 1.013x10^5 Pa$$ ====> I'm not completely sure it's atmospheric pressure

$$P_1= ?$$
$$V_2= ?$$

1hr=> 3600s

a) find mass of water that flow into the atmosphere in one hour

I'm not sure how to find the mass..but I think I'd use
$$m= \rho \Delta V= v_1 A_1 \Delta t$$

so would the mass be (since v1 is given...

$$m= \rho \Delta V= v_1 A_1 \Delta t = (15m/s)(0.00070685 m^2)(3600s)= 38.1699$$ I'm not sure about the units though...

b) the gauge pressure in the left section of the pipe

I think I'd use bernoulli's equation here

$$P_2 + 1/2\rho v^2 + \rho g y_2= P_1 + 1/2 \rho v^2 + \rho gy_1$$

in this case it's P1 as the atmospheric pressure so I switched the sides up.

$$P_2-P_1 = 1/2 \rho (v_1^2- v_2^2)$$

gauge pressure= $$P_2-P_1$$

BUT I'm confused since I don't have P2 but I also think I don't need it since P2-P1 = gauge pressure right?

I don't have the v2 however...but I think I can find it from the continuity equation Or do I use the mass of water found then use that to find the volume (from density) then based on the time interval that the water was flowing and the mass of water to find volume..then use that as the flow rate?

well not sure which is right so I'll use the continuity equation...

$$A_2v_2= A_1v_1$$
thus

$$v_2= (A_1v_1) / A_2$$

$$v_2= (7.0685 cm^2 / 19.6349cm^2)*(15m/s)$$

$$v_2= 5.40m/s$$

now plugging in...

$$P_2-P_1 = 1/2 \rho (v_1^2- v_2^2)$$

$$P_2- P_1= 1/2(1000kg*m ^ {-3} )(15m/s-5.40m/s ) = 4.8 Pa$$

edit: above I went and typed 1000kg m ^-3 but for some reason it keeps showing 8722-- well that's not supposed to be there.

Wait...did I find the gauge pressure?

Can someone see if I did this correctly?
and if not can you help me?

THANKS very much.

Last edited by a moderator:
~christina~ said:
First of all I'm sort of confused with how you can estimate since A1>> A2 but I don't think I can do that here so would that change anything??
That trick doesn't apply here, since it's not true that A1 >> A2 (or vice versa) like it was in that needle problem.

a) find mass of water that flow into the atmosphere in one hour

I'm not sure how to find the mass..but I think I'd use
$$m= \rho \Delta V= v_1 A_1 \Delta t$$
Right idea, but you dropped off the density in the last term.

so would the mass be (since v1 is given...

$$m= \rho \Delta V= v_1 A_1 \Delta t = (15m/s)(0.00070685 m^2)(3600s)= 38.1699$$ I'm not sure about the units though...
Fix this.
b) the gauge pressure in the left section of the pipe

I think I'd use bernoulli's equation here

$$P_2 + 1/2\rho v^2 + \rho g y_2= P_1 + 1/2 \rho v^2 + \rho gy_1$$

in this case it's P1 as the atmospheric pressure so I switched the sides up.

$$P_2-P_1 = 1/2 \rho (v_1^2- v_2^2)$$

gauge pressure= $$P_2-P_1$$

BUT I'm confused since I don't have P2 but I also think I don't need it since P2-P1 = gauge pressure right?
Yes, gauge pressure is P - Patm, which in this case is P2 - P1. So that's all you need. (But you could certainly figure out P2 if you wanted to.)

I don't have the v2 however...but I think I can find it from the continuity equation Or do I use the mass of water found then use that to find the volume (from density) then based on the time interval that the water was flowing and the mass of water to find volume..then use that as the flow rate?

well not sure which is right so I'll use the continuity equation...

$$A_2v_2= A_1v_1$$
thus

$$v_2= (A_1v_1) / A_2$$

$$v_2= (7.0685 cm^2 / 19.6349cm^2)*(15m/s)$$

$$v_2= 5.40m/s$$
Good. The continuity equation is the way to go.

now plugging in...

$$P_2-P_1 = 1/2 \rho (v_1^2- v_2^2)$$

$$P_2- P_1= 1/2(1000kg*m ^ {-3} )(15m/s-5.40m/s ) = 4.8 Pa$$
Two problems: (1) you forgot to do the squaring; (2) redo the arithmetic (you have to anyway).

Doc Al said:
That trick doesn't apply here, since it's not true that A1 >> A2 (or vice versa) like it was in that needle problem.
okay..I was thinking that it would mean that v1 aprox = v2 right? thus the bernoulli's equation would simplify.

Right idea, but you dropped off the density in the last term. Fix this.
oh..I did.so that's why I couldn't figure out the mass unit..ha, well fixing it.

$$m= \rho \Delta V= v_1 A_1 \Delta t$$

so would the mass be (since v1 is given...

$$m= \rho \Delta V= \rho v_1 A_1 \Delta t = (1000kg/m^3)(15m/s)(0.00070685 m^2)(3600s)$$= 38,169.9kg

that's quite large...

Yes, gauge pressure is P - Patm, which in this case is P2 - P1. So that's all you need. (But you could certainly figure out P2 if you wanted to.)
I guess I could by first finding the gauge pressure then solving for P2 here [P2-P1= gauge pressure]

Good. The continuity equation is the way to go.

Two problems: (1) you forgot to do the squaring; (2) redo the arithmetic (you have to anyway).

fixed I think.

$$v_2= 5.40m/s$$

$$P_2-P_1 = 1/2 \rho (v_1^2- v_2^2)$$

$$P_2- P_1= 1/2(1000kg/m^3)((15m/s)^2-(5.40m/s)^2 ) = 97,920 Pa$$

that number is quite large as well..hm...

Thanks Doc Al

Looks good.

(When considering how large a pressure is, compare it to atmospheric pressure.)

Doc Al said:
Looks good.

(When considering how large a pressure is, compare it to atmospheric pressure.)

Thanks for your help Doc Al

## 1. What is the Pipe and Bernoulli's equation?

The Pipe and Bernoulli's equation is a fundamental equation in fluid mechanics that describes the relationship between pressure, velocity, and height of a fluid flowing through a pipe.

## 2. How is the Pipe and Bernoulli's equation derived?

The Pipe and Bernoulli's equation is derived from the principles of conservation of energy and conservation of mass. It assumes that the fluid is incompressible, inviscid, and steady-state, and that there is no external work or heat transfer involved.

## 3. What is the significance of the Pipe and Bernoulli's equation?

The Pipe and Bernoulli's equation is significant because it provides a mathematical relationship between the pressure, velocity, and height of a fluid in a pipe. This equation is widely used in many applications, such as in the design of pipes, pumps, and turbines.

## 4. What are the limitations of the Pipe and Bernoulli's equation?

The Pipe and Bernoulli's equation has several limitations, including the assumptions of steady-state, incompressibility, and inviscid flow. These assumptions may not hold true in real-world scenarios, leading to errors in the calculated values. Additionally, the equation does not account for frictional losses in the pipe, which can significantly affect the flow behavior.

## 5. How is the Pipe and Bernoulli's equation applied in real-life situations?

The Pipe and Bernoulli's equation is applied in many real-life situations, such as in the design and analysis of pipelines, pumps, and turbines. It is also used in the study of aerodynamics, hydraulics, and other fields of engineering. Additionally, the equation is used in practical applications, such as measuring the flow rate of fluids in pipes or determining the pressure drop in a system.

• Introductory Physics Homework Help
Replies
12
Views
232
• Introductory Physics Homework Help
Replies
17
Views
1K
• Introductory Physics Homework Help
Replies
30
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
21
Views
1K
• Introductory Physics Homework Help
Replies
19
Views
271
• Introductory Physics Homework Help
Replies
47
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
2K