PRESSURE LOSSES

[5umbzyyr]

Merry chrismass and a prosperous new year to you all.
please i would like to have an explanation why
1 presure drop in bend decreases when we have an increase in the operatiing Tempresure
2 pressure losses increase if we change the operating pressure from eg 10 bar to 20 bar
3 the losses decreases with change in diameter, eg from 30 to 60.
When all other condition is constant
regards

 

[Latexman]

What's the fluid?

Good luck,
Latexman

 

[katmar]

5umbzyyr,

With the changes that you are considering, it is impossible to keep all other conditions constant. For example, if you change temperature and/or pressure it is not possible to keep all of mass flowrate, volumetric flowrate and velocity constant. What are you keeping constant?

Are you concerned with gases or liquids?

Give some specific conditions and you might get some specific answers. With the questions as you have worded them I would be reluctant to venture an answer, except for number 3. If you look at the applicable equations for pressure drop you will see that for constant flowrate the pressure drop varies with the inverse of the fifth power of the diameter (for turbulent flow). If you double the diameter you get 1/32 of the pressure drop.

regards
Harvey

 

[5umbzyyr]

Thanks Katmar,
If we have a Mach number less than 0.25, compressible fluid can be treated as incompressible (graebel 2001 pg 16) that does not mean its the right way.
now , if we have a condition where the temperature is 300 celsius, diameter is 40mm and we change the pressure from 60bar to 90 bar, the losses increase why.
for the case of temperature, if we keep the daimeterand operating pressure constant and change the temperature, we have reduce pressure drop why.

regards

 

[sailoday28]

5umbzyyr (Mechanical) You must clarify your problem. It seems that your are dealing with a low mach number gas. You question use of approximation at the low mach numbers.
Are you treating your analysis at low mach numbers?
katmar (Chemical)asks what are you keeping constant?

Please provide more information as to what is constant and what your source conditions are.

Regards

 

[5umbzyyr]

Sailoday,
i am assuming the diameter and pressure does not change while i change the pressure and secondly, the diameter and tempressure does not change while i change the pressure.
i hope these help.
regards

 

[zdas04]

5umbzyyr,
You've confused me as well.

For a constant mass flow rate (or volume flow rate at standard conditions which is a good surrogate for mass flow rate) increasing pressure in a flowing fluid that is compressible at rest will lower the velocity of the fluid. Reducing the velocity will reduce the friction pressure drop (not increase it as you assert).

For a fluid that is incompressible at rest, changing the nominal pressure of the line has no effect on pressure drop (at constant mass flow rate in a constant diameter conduit) because the change in density with pressure is a second-order affect.

Grabbel limiting universe where you can treat a nominally compressible fluid as incompressible to less than 0.25 M is ultra conservative, but not wrong. If you go back to Euler's work in the 19th century, he justified treating compressible flows as incompressible by saying that in a reasonable length of pipe at "normal" flow rates the difference in density from one end to the other is insignificant (assuming that the pressure at the outlet is more than 85% of the pressure at the inlet). Without this assumption we would not have the Bernoulli Equation and all airplanes would fall out of the sky (really, my Fluid Mechanics teacher taught me that so it must be right).

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The harder I work, the luckier I seem

 

[sailoday28]

i am assuming
1)the diameter and pressure does not change while i change the pressure.......?????

2)the diameter and tempressure does not change while i change the pressure. Do you mean the process is isothermal and the source pressure is increased? If this is the second case, I would expect mass flux to increase, with an accompanying increase in losses.

 

[davefitz]

To answer the original questions, assuming the fluid is a gas :

frcit. press drop = K * {0.5 rho V^2}, where rho = density , and rho is proportional to P and inversely proportional to T using ideal gas relationship P/rho=ZRT

a)as T increases, rho decreases, so bracketed value decreases

b)as P increases, ditto

c)as diameter increases, flow area increases so V decreases , so bracketed value decreases.

 

[sailoday28]

davefitz (Mechanical)
K * {0.5 rho V^2},
=K*.5*G^2*specif vol=K*.5*G^2*RT/p where G is mass flux.
Your response to part a is T increases, losses increase. What happens to pressure, assuming mass flux remains the same. Or does mass flux increase?
Original question relates to pressure drop and losses.
I believe originator has to be much clearer on the input parameters.
Regards

 

[davefitz]

right.

DP (psi)= Nvh * sv/12 { G/1E5} ^2
where sv= spec vol(ft3/lbm) = ZRT/P
G= mass flow = W/A (lb/hr ft2)
Nvh= number of velocity heads = 0.4 for 90 degree bend

for a constant G, DP drops as P increases and DP increases as T increases

For a constant W(lb/hr), DP drops as diameter increases.

 

[katmar]

5umbzyyr,

The analysis by davefitz and sailoday28 answers your queries very well. However, I would like to caution strongly against the assumption that gas flow at Mach 0.25 can be regarded as incompressible. I do not know the book by Graebel, so I do not know the context within which this assumption was recommended. It may be acceptable with very short lines (roughly 20% error on pressure drop for a line less than 5 metre long) but with normal plant piping this assumption is going to get you into trouble.

There are plenty of tools available to do the job properly, so there is little justification for such broad assumptions.

regards
Harvey

 

[Latexman]

To support what katmar said I'll pull some information from page 171 of Volume 1 of my "The Dynamics and Thermodynamics of Compressible Fluid Flow" by Ascher H. Shapiro. If the initial Mach Number = 0.25 for a "perfect gas" with k = 1.4 and f = 0.0025, the maximum length to diameter ratio = 850 for adiabatic, constant area flow with friction.

So for a 40 mm diameter pipe, the flow becomes sonic in 34 meters of length.



Good luck,
Latexman

 

[sailoday28]

Latexman (Chemical)If we are doing an analyses of perfect gas, const specific heats, why not use the Fanno line or Shapiro or Gas Tables?---Rather than Crane. Our answers at least would be as accurate as we can read tables, etc.

However, I believe the originator might be implying that the Mach no. through out the piping is less than 0.25.
I don't have time to check Shapiro or tables right now, but believe that for M<o.25 (and adiabatic),--- incompressible flow will give same approx results.

One dimensional momentum eq.
d(ln v) +1/G^2*integral dP/v +f/(2D)dx=0
or
1/G^2*integral dP/v + ln(v/vinlet) + fL/(2d)=0
where v=specif vol
G=mass flux
F=press
f=frict factor
D=pipe or equiv diameter
x length in flow direction
L length of Pipe.
From above the integral involving v can be approximated with small changes in specific vol and ln(v/vinlet) will also basically drop out.

So assuming a Max M downstream of 0.25 will above momentum equation approximate incomp flow? Please, those with Gas tables, or Shapiro, Vol 1, comment. I believe Shapiros f=4f(crane)

Regards

 

[Latexman]

sailoday,

I used a table in Shapiro, not Crane.

Anyway, here's more info. from Table B.4 in Shapiro for Initial Mach 0 to 1:

Initial
Mach
Number 4fLmax/D
0 infinite
0.05 280
0.10 67
0.15 28
0.20 14.5
0.25 8.5
0.30 5.3
0.40 2.3
0.50 1.1
0.60 0.49
0.70 0.21
0.80 0.072
0.90 0.015
0.95 0.0033
1.00 0

Good luck,
Latexman

 

[sailoday28]

Latexman (Chemical) Thanks for taking the time for the table. I now have my Vol 1-Shapiro.
For M=0.05 (inlet) and M=0.25, 4fL/D=280-8.5= 271.5
Rho*/Rho=0.055 Rho*/Rho=.272 (For others Rho* is a parameter corresponding to M=1 )
rho(downstream)/rho(inlet) is about4.9 Therefore there is a significant change in density when we have the above 4fL/D and inlet Mach number.

Katmar's caution/warning is advisable.

I do notice that the "isentropic tables" B.2 show that static to stagnation ratios such as pressure or density show small variation for M<0.25

Regards

 

[Latexman]

Sailoday,

Excellent!

For 40 mm (1.5") diameter pipe at fully turbulent flow, f = 0.005. This change in rho occurs in 543 meters (1780 feet) of pipe (271.5 x 0.04 / 4 / 0.005).

Good luck,
Latexman