Re: Fly Ash Conveying System
It all depends on your Fly Ash, not all of them can be transferred at SLR 30-60. Before designing the system make sure you can achieve these conveying conditions with this fly ash. Since 500m is a long distance you might struggle especially with 10 bends. For 6tph a 4m3 blow tank doesn’t look right 1.3m3 blowtank doing 6 cycles /hr should be more then sufficient. As far as stepping is concerned give some air flow rates. ■
Re: Fly Ash Conveying System
Dear JimLee,
First of all, the flyash particle size distribution is missing in your subject, as well as the particle density.
This information is needed to determine the suspension velocity.
As you have calculated the pneumatic conveying installation, you must also have calculated the conveying pressure.
What pressure did you obtain?
From the loading ratio you give (30-60) and the capacity (12.5 tons/hr)
it is possible to calculate the air mass flow and from that the free air flow.
Together with the pipeline diameter of 150 mm the end velocity ranges between
5.45 m/sec and 2.72 m/sec.
The fly ash I have experience with (unloading sea going vessels), the minimum end velocity should be at least 8 to 10 m/sec., resulting in at least 10 m3/min of convey air.
If you clarify to the forum members how you did calculate and with what figures, we can comment and learn.
best regards
teus ■
Teus
Re: Fly Ash Conveying System
TO TEUS
fly ash mean partical diameter :18um (approx)
fly ash partical density: 2100kg/m^3
the suspention velocity is about 1 m/s ,so that minimum conveying velocity is about 4-5m/s .
fist of all ,i should say sorry to all of you that i didn't make myself clear!
12.5t/h is the real product capacity!
so i took 200%*12.5t/h as the design capacity!
from fist supply fly ash hopper to third supply fly ash hopper is assigned 12t 8t 5t to convey per hour each.if the system is able to meet 25t/h requirment,so each hopper will be assigned 28.8minute 19.2minute 12minute at most!
the volumatic concentration ratio is:
i :
if u=30
VCR=u*airdensity/bulkdensity=0.048
1/4*D^2*(VCR/(1+VCR))*V*bulkdensity*3600=25*10^3
where D--pipeline bore
V-- mean velocity of particl(fly ash)
when V=8m/s D=180mm
V=12m/s D=146mm
V=15m/s D=131mm
V=20m/s D=113mm
conveying time equation
t= (Volu*4*(1+VCR)/(pi*D^2*VCR)+1000)/(V*60)
where Volu--Volume of conveyed fly ash of each hopper per hour
(for exampel :fist fly ash hopper's Volu=12000/750=16m^3)
when V=8m/s D=180mm t=30.7minute>28.8 (abandon)
D=200 t=25.26minute<28.8 (accept)
V=12m/s D=146mm t=30.38minute>28.8 (abandon)
D=150mm t=28.8minute=28.8 (accept)
V=15m/s D=131mm t=29.92minute>28.8 (abandon)
D=150mm t=23minute<28.8 (accept)
V=20m/s D=113mm t=29.9minte>28.8 (abandon)
D=125mm t=24.56minte<28.8 (accept)
ii:
u=60
the calculation method just as above
V=8m/s D=130mm t=30.8minute>28.8 (abandon)
D=150 t=23.6<28.8 (accept)
V=12m/s D=106mm t=33.7minute>28.8 (abandon)
D=120mm t=23.8minute=28.8 (accept)
V=15m/s D=95mm t=29.7minute>28.8 (abandon)
D=100mm t=26.9minute<28.8 (accept)
V=20m/s D=82mm t=29.7minte>28.8 (abandon)
D=100mm t=20minte<28.8 (accept)
according to the above calculation resulting,i take 125/150 as the system's pipeline bore! pressure drop is not took in consideration in this calculation!i just want to know whether it can meet the requirment or not if this system work well! if it can meet the requirement then i will calculate the pressure drop under this calculation resulting.
thank you for any suggestion and comment!
BR ■
Re: Fly Ash Conveying System
Dear Jim Lee,
I modeled your pneumatic conveying installation and calculated a capacity table
for the fly-ash as you described it.
In your formula for the calculation of the velocity, you use the bulk density of the fly-ash, but I think that you should use the particle density.
Then the airflow calculation through this calaculation or direct from the assumed loading ratio correlate much better.
JimLee jlfa d 11-03-2006
Pressure discharge Fly-ash
Convey length = 500 m
Nu of bends= 10
Pump vol = .193 + .193 m^3/s
q-convey = 0.193 m^3/s
Dia begin = 125 mm Dia end = 150 mm
Kettle volume = 5.0 m^3
Fly-ash vol.= 4.0 m^3
C-Vessel = 100 tons/hr
Kettle Fly-ash content = 3.9 tons
Pipevolume = 7.59 m^3
---------Pipeline--------- Syst
Press----Cap-----mu----Cap------ v-begin----- v-end-------kWh/ton-------res.time
2.000-----29-----34------26 ---------6.4-----------12.7---------1.16------------68.48
1.750-----27-----32------24---------7.0------------12.7---------1.11------------64.71
1.500-----25-----29------23 ---------7.6-----------12.6---------1.05------------60.87
1.250-----23-----27------21 ---------8.4----------12.6----------1.00------------56.95
1.000-----20-----24------19 ---------9.4----------12.6----------0.96------------52.94
0.750-----17-----20------16---------10.7---------12.5----------0.93------------48.83
0.500-----13-----15------12 ---------12.4---------12.5----------0.97------------44.49
Whether your calculated airflows, capacities and pressures correlate with this calculation,
I cannot determine. I am interested in your calculation outcome.
success
teus ■
Teus
Fly ash conveying system
hello all guys
Tody i calculated a pneumaitc conveying system of fly ash!
fist of all,i supposed the solids loading ratio is approx between 30-60kg/kg (I don't exactly know whether it is right or not.)
the basic references of the system just as the following:
capacity:12.5t/h
total length:500m
vertical length :30
bulk density:750kg/m^3
bends: 10 pics
this system has 3 electrostatic precipitator hoppers,so there are 3 blow tanks.
the supply fly ash is 6t/h 4t/h 2.5t/h in turn
my design result:
the blow tanks volumatric volume is 4m^3 3m^3 2m^3 in turn3
the pipeline bore is 125/150,the stepped position is about at 1/3 of whole pipeline.
IS MY CALCULATION RESULTING RIGHT?
Can it satisfied the requrement? ■