Fluids: Energy equation involving head loss

Click For Summary

Discussion Overview

The discussion revolves around the application of the energy equation in fluid dynamics, specifically focusing on head loss in a pipe system. Participants explore how to incorporate head loss into the energy equation while determining the velocity of air in different sections of the pipe, using principles from the ideal gas law and Bernoulli's equation.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks guidance on determining the velocity in each section of the pipe while incorporating head loss into the energy equation.
  • Another participant suggests using volumetric throughput rate (Q) to find the velocity of air in the sections and mentions the importance of the ideal gas law for calculating air density.
  • It is noted that velocity (V) can be expressed as the flowrate (Q) divided by the cross-sectional area of each pipe.
  • Participants discuss the calculation of air density (ρ) using the ideal gas law, providing a specific value and emphasizing the need for careful unit management.
  • There is a question regarding whether the pressure (P) is absolute or gauge reading, with a participant asserting that it is absolute pressure.
  • Another participant raises a concern that if P is absolute pressure, the tank cannot exhaust to the atmosphere, as the atmospheric pressure would be greater than the pressure inside the tank.

Areas of Agreement / Disagreement

Participants generally agree on the units used for calculations and the interpretation of pressure as absolute. However, there is a disagreement regarding the implications of absolute pressure on the tank's ability to exhaust to the atmosphere, indicating unresolved aspects of the discussion.

Contextual Notes

There are limitations regarding the assumptions made about pressure readings and the specific conditions under which the calculations apply. The discussion also highlights the need for clarity in unit conversions and the definitions of terms used in fluid dynamics.

reddawg
Messages
46
Reaction score
0

Homework Statement


See attached image:

Homework Equations


p/ρg + V^2/2g + z = constant

head loss (major) = f * l/D * V^2/2g

The Attempt at a Solution


To use the energy equation while incorporating head loss, I need to determine the velocity in each section of pipe. The problem is I don't know how! I do know that the pressure drop from the tank to the atmosphere is equal to the pressure in the tank.
 

Attachments

  • problem1.JPG
    problem1.JPG
    38 KB · Views: 408
Last edited:
Physics news on Phys.org
The frictional term has to be properly combined with the bernoulli equation. To get you started, let Q represent volumetric throughput rate. This is what you will be solving for. In terms of Q, what is the velocity of the air in each of the sections. From the ideal gas law, what is the density of the air?
 
Chestermiller said:
The frictional term has to be properly combined with the bernoulli equation. To get you started, let Q represent volumetric throughput rate. This is what you will be solving for. In terms of Q, what is the velocity of the air in each of the sections. From the ideal gas law, what is the density of the air?
V is equal to the flowrate Q divided by the cross-sectional area of each pipe.

From the ideal gas law, density ρ = p/RT = 6.88*10^-5 slugs/ft^3
 
reddawg said:
V is equal to the flowrate Q divided by the cross-sectional area of each pipe.

From the ideal gas law, density ρ = p/RT = 6.88*10^-5 slugs/ft^3

You have to be careful with units here, especially R for air.

Can you show your calculation of ρ in detail?
 
SteamKing said:
You have to be careful with units here, especially R for air.

Can you show your calculation of ρ in detail?
ρ = (.5 psi)(144 in2/ft2) / (1716)(609.7)
1716 is R in British units
609.7 is 150 degrees f converted to rankine
 
reddawg said:
ρ = (.5 psi)(144 in2/ft2) / (1716)(609.7)
1716 is R in British units
609.7 is 150 degrees f converted to rankine
Is P = 0.5 psi absolute or gage reading?

The units of R here are ft-lbf / slug-°R to be precise.
 
SteamKing said:
Is P = 0.5 psi absolute or gage reading?

The units of R here are ft-lbf / slug-°R to be precise.
Yes, I agree with those units. And P is absolute pressure I think.
 
reddawg said:
Yes, I agree with those units. And P is absolute pressure I think.
If P is absolute pressure, then the tank cannot exhaust to atmosphere, since the pressure there > the supposed pressure inside the tank.
 

Similar threads

Understanding pipe ''head loss''
  • · Replies 2 ·
Replies
2
Views
3K
Engineering Fluid Dynamics: Proof of the Static Pressure Head equation
  • · Replies 2 ·
Replies
2
Views
2K
Is My Method for Calculating Pump Head Required Correct?
  • · Replies 2 ·
Replies
2
Views
2K
I have a hydrodynamic loss head question
  • · Replies 6 ·
Replies
6
Views
2K
Head Loss and Pressure Rise During Gradual Expansion
  • · Replies 8 ·
Replies
8
Views
5K
Fluid Mechanics - Head Loss Due to Friction
  • · Replies 2 ·
Replies
2
Views
5K
How Do You Calculate Fluid Velocities in Diverging Pipes?
  • · Replies 1 ·
Replies
1
Views
2K
Hydraulics help please -- Energy loss in a long pipeline
  • · Replies 12 ·
Replies
12
Views
2K
Advanced Hydrodynamic Problem - University level
  • · Replies 49 ·
2
Replies
49
Views
5K
Engineering Pump work required to reach a maximum reservoir height
  • · Replies 2 ·
Replies
2
Views
2K