# Fluids: Energy equation involving head loss

1. Apr 14, 2016

### reddawg

1. The problem statement, all variables and given/known data
See attached image:

2. Relevant equations
p/ρg + V^2/2g + z = constant

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

3. 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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Last edited: Apr 14, 2016
2. Apr 14, 2016

### Staff: Mentor

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?

3. Apr 14, 2016

### reddawg

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

4. Apr 14, 2016

### SteamKing

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

Can you show your calculation of ρ in detail?

5. Apr 14, 2016

### reddawg

ρ = (.5 psi)(144 in2/ft2) / (1716)(609.7)
1716 is R in British units
609.7 is 150 degrees f converted to rankine

6. Apr 14, 2016

### SteamKing

Staff Emeritus
Is P = 0.5 psi absolute or gage reading?

The units of R here are ft-lbf / slug-°R to be precise.

7. Apr 14, 2016

### reddawg

Yes, I agree with those units. And P is absolute pressure I think.

8. Apr 14, 2016

### SteamKing

Staff Emeritus
If P is absolute pressure, then the tank cannot exhaust to atmosphere, since the pressure there > the supposed pressure inside the tank.