Calculating Pressure Drop in Fuel Systems with Complex Geometry: A Case Study

[kosig]

Hello. I have been asked to calculate the pressure drop across our fuel system. It is a large system with many fittings and bends. My numbers don't seem to make sense to me Please help...
Before I begin, I use equation dp=(K*density*MeanVelocity^2)/(2*g) to calculate pressure drop.
To calculate mean Velocity I am using V=(2.951e-5*W)/(d^2*rho)
where W is flow in lb/hr d=inner diameter (in) and rho=density (lb/in^3)

As far as I can tell this gives mean velocity in ft/s which I convert to in/hr. And I use g in terms of in/hr^2 which gives me final units of lb/in^2.

First: I was told to compare the drop as mass flow (lb/hr) changes from 0-5000 lb/hr. Also the applicable temperatures are -60-300 F. For ease here I will just pick a temp and a flow. These are what the system contains and the K values I assigned each feature-
4 gradual diameter contractions .23
2 gradual expansions .17
5 sharp 90 deg bends 1.1
4 gradual 90 bends .12
1 gradual 180 bend 2.17
10 "T" which I make the 90 bend 1.1
Total K=20.41
Total length=204.88 inches
I then try to use dp=(K*density*MeanVelocity^2)/(2*g)
where K= fL/d+loss from fittings

I got that the pressure drops 6e12 psi...What am I doing wrong?

 

[edgepflow]

What is your pipe inside diameter? What fluid velocity did you calculate?

 

[kosig]

The pipe inside diameter is .715. For a mass flow of 2000 lb/hr I got a fluid velocity of 190909.94 in/hr. I think the problem is in how I am calculating my fluid velocity. How can I calculate the average velocity knowing only the flow, diameter, length, viscosity, and density?

 

[kosig]

I will go through my steps exactly so that I can hopefully get some help.
GIVEN:
temp=100 F
viscosity=1.202 cts
mass flow=2000 lb/hr
density(rho)=.0288 lb/in^3
diameter=.715 in
length=204.88 in
g=5003700480 in/hr^2
total K=20.41

STEPS:
1. convert viscosity to US: 1.202*5.58=6.70716 in^2/hr
2. Calculate fluid velocity:

mass flow= rho* Q(volumetric flow)
and
$$mass flow= \rho Q(volumetric flow$$
Ignore equation directly above, for some reason if I try and remove it, it shifts all other equations down...
$$V=\frac{Q}{\frac{\pi d^2}{4}}$$

so 2000=.0288*Q---Q=69444.44 in^3/hr V=172955.96 in/hr
3. Calculate Reynolds number:
$$R_e=\frac{V d}{v}$$
R_e=18437.977
4. Calculate friction factor:
$$\frac{1}{\sqrt(f)}=-2 \log(\frac{\frac{roughness}{d}}{3.7}+\frac{2.51}{R_e \sqrt(f)}$$
roughness=.00008 (from table)
f=.0062
5. Calculate h_l:
$$h_l=(\frac{f l}{d} + K_l) \frac{V^2}{2 g}$$
h_l=66.319
6. Calculate pressure loss:
dp=rho*g*h_l
dp=9.557e9

Does anyone see any mistakes?

 

[Q_Goest]

Hi kosig,

5. Calculate h_l:
$$h_l=(\frac{f l}{d} + K_l) \frac{V^2}{2 g}$$
h_l=66.319
6. Calculate pressure loss:
dp=rho*g*h_l
dp=9.557e9

You calculate a head loss of 66.319, but what units is that in? Generally we use feet but that looks like it should be in inches. If so, then the conversion to psi is where I'd suggest looking for a problem. I don't know what fluid you're using or I'd punch that into my program so I used water instead. It comes back with 39" of head loss or roughly 1.4 psi. Does that sound about right?

PS: I posted a paper on pipe flow that you might find useful here: https://www.physicsforums.com/showthread.php?t=179830

 

[kosig]

Yes, that is in inches. I agree, that number seems reasonable. How should I convert that from inches to psi?

 

[Q_Goest]

You have the right equation. Just check units.

 

[kosig]

You have the right equation. Just check units.

Yeah, sigh...that's what I've been doing for two days straight.

After I multiply the head loss by density and gravity I get pressure loss in lb/(in*hr^2). Which seems eerily similar to Pascals. So I convert lb to kg, into m, and hr^2 to s^2 by multiplying by 1.49569X10^-6. Then I convert Pa to psi by multiplying by the same number? Why is that?

 

[Q_Goest]

Change density to lbm/ft³, so 0.0288 lbm/in³ = 0.0288 lbm/in³ (12³ in³/ft³) = 49.76 lbm/ft³

Now change head to feet so 66.316 in = 5.51 ft

So dp = rho*g*h but you also have to convert using 1/G_c and ft²/144in²
where G_c = (32.2 lbm ft/lbf s²)

dp = rho*g*h*G_c*(ft²/144in²)

dp = 49.76 lbm/ft³ * 32.2 ft/s² * 5.51 ft * 1/(32.2 lbm ft/lbf s²)*ft²/144in²

You should get 1.90 psi

 

[kosig]

That seems reasonable. Not entirely sure what you did...but it does end in psi! Why do you cancel out gravity? Also, my thanks to you are endless. I have been canceling units for two days now.

 

[Q_Goest]

Densiity is in units of pound mass (lbm) per cubic foot, not pound force (lbf) per cubic foot. So to convert the mass units to force units, you need a conversion factor. We're not cancelling out gravity, we're just converting mass to force. G_c is a conversion factor that equals 1 just as 12 inches/ 1 foot = 1.

Note that I did also check your input conditions using water so I'm pretty sure the output is going to be close to that (1.4 psi for water) so the 1.9 psi for your fuel sounds like you got the rest of the equation right.

 

[kosig]

Of course...I really hate having to use the US system...Thanks again!