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, ■