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

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Discussion Overview

The discussion revolves around calculating the pressure drop in a complex fuel system with various fittings and bends. Participants explore the application of fluid dynamics equations, particularly focusing on the pressure drop calculation and the associated parameters such as flow rate, fluid velocity, and head loss. The context includes technical reasoning and mathematical calculations relevant to engineering applications.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents an equation for pressure drop and describes their method for calculating mean velocity, expressing confusion over the resulting pressure drop value.
  • Another participant inquires about the pipe's inside diameter and the calculated fluid velocity, suggesting that the fluid velocity calculation may be problematic.
  • A participant details their calculations step-by-step, including viscosity conversion, fluid velocity, Reynolds number, friction factor, and head loss, seeking feedback on potential mistakes.
  • Concerns are raised about the units used for head loss and pressure loss calculations, with suggestions to verify unit conversions and the implications for the final pressure drop result.
  • Discussions include the conversion of head loss from inches to psi, with participants sharing methods for unit conversion and the importance of using consistent units throughout the calculations.
  • Clarifications are made regarding the distinction between mass and force units, emphasizing the need for conversion factors in the calculations.

Areas of Agreement / Disagreement

Participants express uncertainty about specific calculations and unit conversions, indicating that multiple competing views exist regarding the correct approach to calculating pressure drop. There is no consensus on the final pressure drop value, as participants are still working through the calculations and conversions.

Contextual Notes

Limitations include potential errors in unit conversions, assumptions about fluid properties, and the need for clarity on the definitions of variables used in calculations. The discussion highlights the complexity of applying fluid dynamics principles to real-world systems with various geometrical features.

Who May Find This Useful

Engineers, students, and professionals involved in fluid dynamics, mechanical engineering, or related fields may find this discussion relevant, particularly those dealing with pressure drop calculations in complex piping systems.

kosig
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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?
 
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What is your pipe inside diameter? What fluid velocity did you calculate?
 
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?
 
Last edited:
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
<br /> mass flow= \rho Q(volumetric flow<br />
Ignore equation directly above, for some reason if I try and remove it, it shifts all other equations down...
<br /> V=\frac{Q}{\frac{\pi d^2}{4}}<br />

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

Does anyone see any mistakes?
 
Hi kosig,
kosig said:
5. Calculate hl:
<br /> h_l=(\frac{f l}{d} + K_l) \frac{V^2}{2 g}<br />
h_l=66.319
6. Calculate pressure loss:
dp=rho*g*hl
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
 
Yes, that is in inches. I agree, that number seems reasonable. How should I convert that from inches to psi?
 
You have the right equation. Just check units.
 
Q_Goest said:
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?
 
Last edited:
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/Gc and ft2/144in2
where Gc = (32.2 lbm ft/lbf s2)

dp = rho*g*h*Gc*(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
 
  • #10
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.
 
  • #11
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. Gc 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.
 
  • #12
Of course...I really hate having to use the US system...Thanks again!
 

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