Power law constant

Thus for any plastic where the Power Law constants are known, the clamping force can be calculated for a given radius, R, cavity depth, H, and fill time, /. 

Screw rotational speed Number of bubbles per unit volume of solution Number of bubbles per unit volume of solution initially Molar flux of volatile component at surface of gas bubble Power law constant Instantaneous molar flux of volatile component fixtm wiped film... 

It is seen that the volumetric flow ratio Qa/Qb is not equal to the layer thickness ratio y./(h — oc) as one might expect. Qa/Qb is a complicated function that depends on the pressure gradient , the Power Law constants of each phase (nA, mA, nB, and mB), and the parameters oc and 1. Therefore, for a given fluid system and flow conditions, where the pressure gradient and flow rates (QA and QB) are specified, the interface position oc can be determined from Eq. 12.3-28, when the predicted value of Qa/Qb agrees with the experimentally determined one, provided Eq. 12.3-25 is satisfied. In other words, Eqs. 12.3-25 and 12.3-28 can be used for determining values of oc and 1. 

The homogeneous non-Newtonian capillary tube-power law model has a number of limitations. The models assume a power law relationship for the emulsion, and any deviations from this rheological behavior will lead to errors. The power law constants n and K are obtained by using viscometry, and their validity in porous media is questionable. No transient permeability reduction (assumption 4) is predicted, even though experimental evidence suggests otherwise. This model is seen to have validity only for high-quality emulsions that approach steady state quickly and have small droplet-size to pore-size ratios. 

In Eq. (9), b is a constant with dimension [time ] and p a dimensionless power law constant. The resulting reactivity is then given by... 

The rheological data are given in Table 1. The second column of the table is the evaporation state of the oil in mass pereentage lost. The third column is the assessment of the stability of the emulsion based on both visual appearance and rheological properties. The power law constants, k and n, are given next. These are parameters from the Ostwald— de Waele equation which describes the Newtonian (or non-Newtonian) characteristics of the material. The viscosity of the emulsion is next and in column 7, the complex modulus which is the vector sum of the viscosity and elasticity. Column 8 lists the elasticity modulus and column 9, the viscosity modulus. In column 10, the isolated, low-shear viscosity is given. This is the viscosity of emulsion at very low shear rate. In column 9, the tan 5, the ratio of the viscosity to the elasticity component, is given. Finally, the water content of the emulsion is presented. 

A polymer solution (density = 1075 kg/m ) is being pumped at a rate of 2500 kg/h through a 25 mm inside diameter pipe. The flow is known to be laminar and the power-law constants for the solution are m = 3 Pa-s" and n = 0.5. Estimate the pressure drop over a 10 m length of straight pipe and the centre-line velocity for these conditions. How does the value of pressure drop change if a pipe of 37 mm diameter is used ... 

Before concluding this section, it is useful to link the apparent power-law index n and consistency coefficient m (equation 3.26) to the true power-law constants n and m, and to the Bingham plastic model constants rf and iib- This is accomphshed by noting that Xy, = (D/4)(—Ap/L) always gives the wall shear stress and the corresponding value of the wall shear rate Yw(= dV /dr) can be evaluated using the expressions for velocity distribution in a pipe presented in Sections 3.2.1 and 3.2.2. 

A bed consists of uniform glass spheres of size 3.57mm diameter (density = 2500 kg/m ). What will be the minimum fluidising velocity in a polymer solution of density, 1000kg/m, with power-law constants n = 0.6 and m = 0.25 Pa-s" Assume the bed voidage to be 0.4 at the point of incipient fluidisation. 

A dilute polymer solution at 25"C flows at 2m/s over a 300 nun x 300 nun square plate which is maintained at a uniform temperature of 35°C. The average values of the power-law constants (over this temperature interval) may be taken as m = 0.3 Pa-s" and n =0.5. Estimate the thickness of the boundary layer 150 mm from the leading edge and the rate of heat transfer from one side of the plate only. The density, thermal conductivity and heat capacity of the polymer solution may be approximated as those of water at the same temperature. 

A polymer solution at 25°C flows at 1.8m/s over a heated hollow copper sphere of diameter of 30 mm, maintained at a constant temperature of 55°C (by steam condensing inside the sphere). Estimate the rate of heat loss from the sphere. The thermophysical properties of the polymer solution may be approximated by those of water, the power-law constants in the temperature interval 25 < J < 55°C are given below n = 0.26 and m = 26 — 0.0566 T where J is in K. What wiU be the rate of heat loss from a cylinder 30 mm in diameter and 60 mm long, oriented normal to flow ... 

Since the same fluid is to be used in the two cases (and assuming that the power-law constants are independent of the shear rate over the ranges encountered), nii = ni2 = m, rii = U2 = n and p = p = p, therefore the equality of Reynolds number gives ... 

The thermal conductivity, heat capacity and density of the polymer solution can be taken as the same as for water. The values of the power-law constants are n = 0.36 and m = 26 — 0.0566 T Pa -s") in the range 288 < T < 323 K. Estimate the overall heat transfer coefficient and the time needed to heat one batch of hquid. 

The following shear stress-shear rate data demonstrate the effect of temperature on the power-law constants for a c ncentrated orange juice containing 5.7% fruit pulp. 

The flow characteristics of this slurry are not fully known, but the following preliminary information is available on its flow through a smaller tube, 4 mm in diameter and 1 m long. At a flow rate of 0.0018 m /h, the pressure drop across the tube is 6.9 kPa, and at a flow rate of 0.018m /h it is 10.35kPa. Evaluate the power-law constants from the data for the small diameter tube. Estimate the pressure drop in the 25 mm diameter pipe for a flow rate of 0.45 m /h. 

Flow experiments were completed on six Berea cores using polymer concentrations of 500, 1000 and 1500 ppm. The experimental data were fitted by the power-law model for most of the flow rate range as shown in Figure 4. Departure from the power-law model was observed at flow rates less than 0.009 cc/min (frontal advance rate - 0.1 ft/d) in several runs. This was expected since flow behavior should become Newtonian at low frontal advance rates. When Newtonian flow occurs, the power-law constant is 1 which means that the slope of the graph of AP versus Q on log-log paper is 45 degrees. Newtonian flow was not attained at the lowest frontal advance rate (0.022 ft/d). 

C Polymer concentration, ppm c Critical aggregation k Power-law constant, mPa s" K Huggins constant... 

It may be shown (Dodge and Metzner, 1959) that the constant K may be related to the analogous power-law constant K (Table 10.1) as follows ... 

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