Re: Pressure Drop Comparison
Dear Diego,
As far as I can understand your calculations, you have used 2 calculation algorithms and also used the observations of several researchers.
It is not clear, if those approaches are still valid when used together in one calculation.
You have calculated one system and come up with 3 pressure drops.
I modelled your installation and calculated the product loss factor for those 3 results:
36.3 kPa ----------- Product loss factor = 0.03157
52 kPa ------------- Product loss factor = 0.1176
73 kPa ------------- Product loss factor = 0.3671
(Capacity = 13 tons/hr and pumpvolume = 0.236 m3/sec)
As you already concluded, the determination of the solid friction coefficient is critical.
Tests or field data from already built installations are required.
Success
Teus ■
Teus
Re: Pressure Drop Comparison
Dear Teus,
Thanks for your quick answer. Yes, first I did the calculation with one algorithm and then with other from D. Mills Pneumatec Conveying Design Guide first with same inlet velocity, and last with same outlet velocity.
I did not understand completely your answer. I read that you have modelled the installation, and then you have imposed each one of the pressure drop and from that data you have calculated back the solid friction value. Excuseme, how is it defined that solid friction value ? For the first algorithm I used the approximate expresion: fs = 0.082*R^(-0.3) * Fr^(-0.86) * Frs^0.25 * (D/dp)^0.1 with the Froude numbers Fr = Vg/(g.D), Frs = Ut/(g.D), and the pressure drop calculation is: DPh = (4.f + fs.R)*L/D*1/2*rhog*Vg^2. Your solid friction value is the same? Then I probably have a big mistake in my calculation because the values used to obtain 36.2 kPa are from 0.00044 last pipe to 0.00066 first one, and your value is 0.03157. Probably I have some incorrect asumption. Do you think that these kind of correlation like the one presented before to calculate fs are valid enough to make estimation about pressure drop ?
As I have seen in this forum you have a propietary method for calculating pneumatic conveying systems. Are my calculation near you results or are them too far appart. In this case of course I would take yours as the more accurate
Thanks a lot, best regards from
Diego Peinado Martn.
As far as I can understand your calculations, you have used 2 calculation algorithms and also used the observations of several researchers.
It is not clear, if those approaches are still valid when used together in one calculation.
You have calculated one system and come up with 3 pressure drops.
I modelled your installation and calculated the product loss factor for those 3 results:
36.3 kPa ----------- Product loss factor = 0.03157
52 kPa ------------- Product loss factor = 0.1176
73 kPa ------------- Product loss factor = 0.3671
(Capacity = 13 tons/hr and pumpvolume = 0.236 m3/sec)
As you already concluded, the determination of the solid friction coefficient is critical.
Tests or field data from already built installations are required.
Success
Teus
■
Re: Pressure Drop Comparison
Dear Diego,
The product loss factors, which I calculated back with my own calculation algorithm
(Based on the described installation and the 3 output results) are only applicable in my
calculation program. They are not correlated whatsoever with other calculation methods.
What I did is combining the results of your 2 calculations into my computer program and
calculated 3 different loss factors (Being the only unknown parameter)
The calculated product loss factors range from 0.03157 to 0.3671 (a factor 11.6)
I used for my calculations the given pipeline, capacity and air volume.
We now can conclude:
-The 2 calculation methods are not consistent with each other (they give different outputs)
-By having a choice, whether to include velocity losses or not, might indicate that the calculation algorithm is incomplete.
-Assuming that the calculation method, which I use could give a more precise answer, then I have to choose a product loss factor between 0.03157 to 0.3671 or higher or lower.
-A product loss factor calculation, based on existing installation(s) performance data or test data, is necessary. (Testing f.i with a bulk tanker ??)
For a description of my calculation program, visit: https://news.bulk-online.com/?p=65
Assuming a product loss factor would be blind guessing.
Probably the rubber granulate collisions in the conveying pipe are highly in elastic, causing a “high” loss factor. However, we do not have figures.
Success
Teus ■
Teus
Pressure Drop Comparison
Hello, I'm comparing two methods for predicting the pressure drop in a pneumatic conveying system. The Air only pressure drop method from D. Mills, and other based on first principles.
Material: rubber granulate. Particle size dp=2 mm, particle density rhop = 1200 kg/m3. Mass flow rate mp = 13 t/h = 3.61 kg/s
The circuit is an horizontal pipe of 2m, 90º Bend, 8m vertical other 90º Bend, and 2m horizontal. Point 1 inlet, point 2 after 2m horizontal, point 3 after pipe bend, point 4 after 8m vertical, point 5 after pipe bend, and point 6 after 2 m horizontal.
I suppose at the exit p6 = 101.3 kPa, T6=300K, viscosity visg =18e-6 Pa.s, rhog6= 1.18 kg/m3.
I guess pipe internal diameter D=0.1 m, so S=pi/4*D^2=7.854e-3 m^2
For calculating terminal velocity I use rhog=1.29 kg/m3, so:
Ga=(rhog*(rhop-rhog)*g*dp^3)/ visg^2 = 3.75e5, then
Ut=sqrt(3.1*(rhop-rhog)*g*dp/rhog= = 7.52 m/s
For calculating saltation velocity I use the Rizk correlation, so:
Usalt = 19.65 m/s
I guess exit velocity V6=30 m/s
Then Point 6: exit.
p6= 101.3 kPa
rho6=1.18 kg/m3
T6=300K
V6=30 m/s, so Q6=S*V6=0.236m3/s, and mg6= 1000 kg/h, so the solid to gas mass ratio is R=13 (perhaps a bit high for a dilute phase system ?)
Point 5: horizontal pipe 2 m long.
Only horizontal pressure drop: DPh = (4.f + fs.R)*L/D*1/2*rho6*V6^2
I use a high rugosity value z = 0.5 mm, Re = 1.97e5, so f = 0.0078 (a big value).
For the solid friction coefficient fs I use a correlation for particles with dp>500 micron:
fs = 0.082*R^(-0.3) * Fr^(-0.86) * Frs^0.25 * (D/dp)^0.1 with the Froude numbers Fr = Vg/(g.D), Frs = Ut/(g.D),
so for this pipe fs = 0.00044. I suppose THIS IS the critical point, because is highly recommended to have experimental data for this value. Nevertheless, is this a good approximation?.
DPh5-6 = 939 Pa, so I calculate values for point 5
p5=p6+939Pa=102239Pa
rho5=102239/101300*1.18 = 1.19 kg/m3
V5=101300/102239*30=29.72 m/s
Point 4: pipe bend.
a 90 pipe bend pressure drop DPbend = B*(1+R)*1/2*rhog6*V6^2. Different values of B (Cambers and MArcus 1986) are 0.5, 0.75, 1.5 depending on radius to diameter ratio (>6, 4, 2). I use the biggest one so
DPbend = 1.5*(1+13)*1/2*1.19*29,72*2=11036 Pa
p4 = 113275 Pa
rho4= 1.32
V4=26.82 m/s
Point 3: vertical pipe 8 meters length.
DPv = (4.f+fs.R)*L/D*1/2*rho4*V4^2 + RHO0*H*g
The value fs is now fs = 0.00053
The particle velocity is obtained from Up/Vg = 1-0.044*dp^0.3*rhop^0.5 (Hinkle 1954) I've also seen Up/Vg=1-0.0638*dp^0.3*rhop^0.5, but in this case instead of superficial velocity is calculated with actual gas velocity and not the superficial gas velocity.
Up/Vg=0.764, so the voidage eps = 0.9813 and RHO0 = eps*rho4+(1-eps)*rhop = 23.74 kg/m3 and
DPv = 3802+1863 = 5665 Pa.
p3=118940 Pa
rho3=1.385 kg/m3
V3=25.55 m/s
Point 2: Ubend
DPbend = 1.5*(1+13)1/2*rho3*V3^2 = 9494 Pa
p2=128434 Pa
rho2=1.5 kg/m3
V2=23.66 m/s
Point 1: Acceleration and horizontal pipe of 2 meters.
DP=DPh + DPaccel
DPh=333 Pa with fs = 0.00066
With this pressure drop I take an intermediate point to calculate acceleration term.
p'=128767 Pa
rho'=1.5
V'=26.60m/s
DPaccel = 1/2*rho'*V'^2*(1+2*R*(Ut/V')) = 8715 Pa
p1 = 137483 Pa
rho1=1.6 kg/m3
V1 = 22.1
So the total pressure drop is DPT = 137483-101300 = 36.2 kPa, and the minimum velocity is 22.1 12% bigger than saltation velocity 19.65 (would this be enough ?)
If I calculate the pressure drop by the Mills Air only pressure drop method, I have:
at outlet pe=101.3 kPa, Te=300 so rhoe=1.18 kg/m3.
Imposing at inlet a velocity equal to the previously calculated Vi = 22.1 m
Guessing (after iteration) a pressure drop of 73 kPa, then Ve = 38 m/s
Equivalent length = 2 + 2*8 + 2 = 20 m, f = 0.0078, each bend k = 0.8
Rg = 287 J/kg.K
PSI= 4.f.L/D + K = 7.84
DPair = pe*(sqrt (1+PSI*Ve^2/(Rg*Te))-1) = 6.46 kPa
pi = 1/2*(pe+DPair + sqrt((pe+DPair)^2 + mp*Ti*DPair/(Rg.Te))) = 174 kPa
So the pressure drop is 73 kPa instead of 36.2 kPa. Well I could have taken into account acceleration terms in all the pipes in the first method, but even so, this difference is to big. If I impose in Mills method the exit velocity of 30 m/s then the pressure drop is 52 kPa but the velocity at the inlet is reduced to 19.86 m/s TOO CLOSE to saltation velocity.
So, please as your experience dictates, which method would be more precisse ?
In the first one should I take into account other terms ?
Are these methods consistent one to other ?
Thanks in advance and best regards, ■