Fluids: Energy equation involving head loss

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SUMMARY

The discussion focuses on applying the energy equation to fluid dynamics, specifically incorporating head loss in pipe systems. Key equations include the Bernoulli equation, expressed as p/ρg + V²/2g + z = constant, and the head loss formula, head loss (major) = f * l/D * V²/2g. Participants emphasize the importance of calculating velocity (V) using the volumetric throughput rate (Q) and the cross-sectional area of the pipes. Additionally, the ideal gas law is utilized to determine air density (ρ), with specific attention to unit conversions and the absolute pressure of the system.

PREREQUISITES
  • Understanding of Bernoulli's equation and its applications in fluid dynamics
  • Familiarity with head loss calculations in pipe flow
  • Knowledge of the ideal gas law and its implications for density calculations
  • Proficiency in unit conversions, particularly in British units
NEXT STEPS
  • Study the derivation and applications of Bernoulli's equation in various fluid systems
  • Learn advanced head loss calculation techniques, including minor losses and friction factors
  • Explore the ideal gas law in greater detail, focusing on its application in real-world scenarios
  • Review unit conversion methods and their importance in engineering calculations
USEFUL FOR

Engineering students, fluid mechanics professionals, and anyone involved in the design and analysis of piping systems will benefit from this discussion.

reddawg
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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.
 

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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.
 

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