Evaluation of the Pressure Loss Parameters on the Basis of Measured Data
We consider now a concrete example of pneumatic conveying in order to show how the parameters included in Eq, (14.126) are determined on the basis of pressure loss measurements. The two parameters, which are specific for
/P~t
(14.131) 
Vc = 
Vc = (vAcA) 1^ P
(14.132)
I. e., the parameters to be determined are the velocity difference VA — CA and the falling velocity WsA at some specified pressure PA and, of course, at a known inclination angle of the pipeline S. Then using Eqs. (14.131)(14.132), the velocities C and Ws can be calculated for each V and P. The effect of the inclination angle 8 on the velocities can be estimated with Eq. (14.83).
In principle, the velocities C and Ws can be determined by taking a series of pictures at a very high frequency of the flow through a transparent plastic tube. Because of the particle size distribution, each particle moves at a different velocity, and this makes this method difficult to apply in practice. We have therefore used an indirect method, where we have measured the pressure losses of pneumatic conveying for two mixture ratios and then fit the parameters so that Eq. (14.126) coincides as accurately as possible with measured pressure losses.
We measured the pressure losses for two different mixture ratios in a horizontal pipeline with a length of 89.6 m:
Jji = 2.45 Ap = 5.3 kPa P0= 117.5 kPa U0 = 52.6 m/s
J± = 1.577 A/? = 4.6 kPa P0 = 115.2 kPa U0 = 52.6 m/s
The temperature was 293 K in both cases. The material conveyed was partially moist wood chips with about 25% sawdust and the cubic bulk weight was 370 — 380 kg/m3. The inner diameter of the pipe was D = 0.440 m. The static pressure PQ and the corresponding velocity of air V0 are the values at the initial point (x = 0), and hence the pressures at the end of the tube (x = 89.6 m) in these two experiments were P = 117.55.3 =112.2 kPa (/a = 2.45) and P = 110.6 kPa (fj, = 1.577).
Equation (14.126) is solved numerically by Euler’s method. Equation (14.126) is written in the form dp/dx = F(v(x), c(x), ws(x), pG(jc)) = F(x) And the derivative is calculated at each point using the previous known values for V, c, ws and P. Figure 14.16 shows the principle of the solution method.
We have data from two independent measurements and two parameters to be fitted with these data. The more data we have, the more reliable will be the parameter fitting. Changing the values CA and WsA and repeating the numerical calculation of the pressure loss by Eq. (14.126), we found that the best coincidence with the empirical data presented was obtained by
(14.133) (14.134) 
VA ~CA~ 8.0 m/s (horizontal line) WsA = 18.0 m/s
For air with PA = 1.0 bar and TA = 293 K. The values (14.133)—(14.134) are valid for pneumatic conveying of slightly moist wood chips by air.
X  FIGURE 14.16 Numerical solution principle of Eq. (14.126). 
Correspondingly, for the vertical pipeline we have made a pressureloss measurement with the following results:
Ju, = 2.62 Ap = 4.3 kPa Pn = 116.5 kPa V0 = 40.7m/s
L = 14.1 m D = 0.319 m
The temperature here was also 293 K. The parameter to be fitted with this data is the velocity difference VA — cA in the vertical line. The material conveyed was wood chips as above. Making the same analysis as above with the horizontal line, the open parameter VA — cA for the vertical line was fitted in Eq. (14.126) together with (14.134) so that the pressure loss measured was achieved. The best coincidence was obtained by
VA — CA = 23.0 m/s (vertical line) (14.135)
For air with PA — 1.0 bar and TA = 293 K.
Now, when these three parameters are known, Eq. (14.126) can be used for calculating the pressure losses in pneumatic conveying of wood chips in any other conditions. Also the pressureloss curves for planning pneumatic conveying systems for wood chips can now be drawn. Figures 14.17 and 14.18 present examples of these. Similar curves can now be drawn for any pipe diameter D and for any air velocity VA by solving Eq. (14.126) numerically according to Fig. 14.16.
Using the parameters (14.133)(14.134) in Eq. (14.126), we can compute the pressure losses corresponding to the measured case, and we get for horizontal transport:
Ju, = 2.45 Ap(computed) = 5.6 kPa Ap(measured) = 5.3 kPa
/j, = 1.577 Ap(computed) = 4.8 kPa Ap(measured) = 4.6 kPa
Pressure (kPa) ■ Pressure (kPa) 














FIGURE 14.18 Pressure loss in a vertical pipeline with parameters: VA — CA = 23.0 m/s, = 18.0 m/s. PA = 1.0 bar and TA = 293 K.
The corresponding result for vertical transport was Jj. = 2.62 Ap(computed) = 4.1 kPa Ap(measured) = 4.3 kPa
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