Frictional flow equations for highly viscous fluids 1

[mcoll]

Hi everyone,

Can anyone point me in the right direction for calculating a pressure requirement for my system.
I'm extruding a polymer (pneumatically) from a reservoir through a connection tube and out through a (circular) die, the discharge rate required is 25mm/s.

1.Being of such high viscosity I assume a laminar flow.
2.There is no heat transfer across the boundary of the system so that q = 0 and the heat energy generated by material shearing and friction at the wall is negligible.(This is not true, but I intend to deal with this aspect as and when I have grasped the dynamics of this basic system)
3.The change in height across the system is small so that z1 = z2
4.The material is considered uncompressible.
5.There is no work done by the fluid so that w=0.

I have therefore derived that:
K.½r.u² -- Inlet Losses
K.½r.u3² -- Outlet Losses
4FL/D .½r.u² -- Friction losses at wall (shear)

Where; ½r.u² = Dynamic Pressure (r - rho-density)
K = 0.5 (90 Degree)

Therefore:
4F1L1/D1 .½r.u1² + K1.½r.u1² + 4F2L2/D2 .½r.u2² + K2.½r.u2² + 4F3L3/D3 .½r.u3² + K3.½r.u3²

(can supply a diagram for anyone who is sufficiently interested).

The figures I get back are clearly wrong, even when I calulate for water (still laminar flow).

Can anyone reccommend another way (possibly other than an enengy equlibrium equasion) to calculater a force/pressure required at the piston (D1) to achieve the stated discharge rate?

Thanks in advance for any help, and well done for reading all of this post!

MC

 

[CHD01]

You don't need to re-invent the wheel. Use conventional pressure drop versus flow equations available in any text - see Crane Flow of Fluids Manual for inexpensive and excellent reference (see formula below). The equation is based on laminar flow with the Fanning friction factor = 64/Reynolds No.

Your problem will be to determine the physical properties of the polymer. In particular, you need to know the flow viscosity of the polymer. This is further complicated by the fact that APPARENT viscosity will change with the flowrate (i.e. - it is related to shear).

dP = 0.000034 * Viscosity * Length * Flowrate /

diameter^4 * density

Your "APPARENT" viscosity at flow conditions will be probably be very high - like 1.7 million centipoise or so. Again - this will hard to calulate and usually is best obtained from lab pilot data.

If you will give me die hole diameter, hole length, fluid viscosity, specific gravity and flow rate per hole; I will calculate the pressure drop for you. If you can give me the specific heat of the polymer I can also give you the temperature rise of the polymer due to shear. The more you learn, the less you are certain of.

 

[rbcoulter]

I'm having difficulty reading your equations. I would
try an orifice equation like in the Crane handbook.
I believe polymers are shear thinning so their viscosity is reduced under shear. Need to know the viscosity at the process shear rate to determine Reynold's number through die, IMO.

 

[CHD01]

Yes viscosity is reduced under shear, but even so - viscosity will be extremely high versus water. The more you learn, the less you are certain of.

 

[mcoll]

Thanks for your input guys.

Info will be with you shortly CHD01.


My excuse is:
I've done a series of design degrees, having completed them I now realise that I should have/would have preferred to have done MechEng.
Fluids is fairly new to me (learning from Bernhard Massey at the moment! - well trying)


Thanks again.

 

[mcoll]

Right, the data I have is:

D1=50mm X 250mm long
D2=6.35mm X 550mm long
D3=0.5mm X 30mm long

Volume flow rate (V*) = U.(3.142Xr²)
V*= 0.025 X (1.96X10-7) = 4.9 X 10-9 m³/s
(2.5 X 10-2 m³/s = 25mm/s = Q)
Therefore:
*.U3 = V*/A3 = 2.5 X 10-2 m³/s

*.U2 = V*/A2 = 1.55 X 10-4 m³/s

*.U1 = V*/A1 = 2.5 X 10-6 m³/s

Specific Gravity = 1.04 (density=1040kg/m3)
Specific heat capacity = 1200JKg-1 K-1
Process temp = 250oc
***(Polymer is an extrusion grade ABS)***
The dynamic/kinematic viscosity is as you say very difficult to assertain, in fact it is not a method of measurement used in the polymer industry (unless anyone can tell me any different?). As such I will not have a figure until a 'capillary rheometery test' is done (which is why I attemped to prove/disprove the original formula using data for water). Could you maybe use your engineering judgement for now???

It's not necessary for you to do this for me (though I really appreciate your help and effort), I don't want to take up too much of anyones time, and I want to be able to do and understand this myself. But if you could help me towards a formula/forulea and understanding of the system I would be really grateful.
Of course, if you want to have a go at it (and don't mind doing that) then please go ahead!!!

Regards,

MC

 

[CHD01]

You appear to be trying to vary capillary diameter and length while keeping the velocity constant. This presents alot of problems.

The usual objective is to first determine capillary size based on the diameter pellet you wish to produce, then determine the correct L/D ratio for the capillary that results in good product (this is critical since viscosity and shear are related and directly affect product quality), then determine the number of capillaries you need based on the total product flowrate required. Then you have to look at all this and determine if you have a reasonable solution for a pelletizer/die plate combination. One other thing, not discussed, you will also need to calculate the pelletizer speed and determine the number of knives per revolution in order to get the length of pellet desired.

Now for your examples:

Converting to english units:

D1 = 1.97in x 9.84 in long
D2 = 0.25in x 21.65in long
D3 = 0.02 in x 1.18 in long

U1 = 8.83xE-5 ft3/sec
U2 = 0.00547 ft3/sec
U3 = 0.883 ft3/sec

Density = 64.93 lbs/ft3

Sp Ht = 0.29 BTU/lb-Deg F

Unknown in these cases is viscosity, I will assume 1.7XE6 for all cases. This could be way off and with the ridiculous capillary diameter and length cases given it will in fact be dramatically different.

With this data then, the dp is:

dP1 = 1 psi way too low for extrusion of polymer
DP2 = 5.26E5 psi impossible case
dP3 = 1.13E11 psi impossible case

The more you learn, the less you are certain of.

 

[mcoll]

Clearly I have much to learn.
Thanks for your help all the same, much appreciated.

MC