# Calculating the Pressure Loss of an Ejector in a Pneumatic Conveying System

As another example of calculation and dimensioning of pneumatic conveying systems we consider an ejector shown in Fig. 14.20. In fluidized bed combus­tion systems a part of the ash is circulated with the hot flue gas. The task of the ejector, is to increase the pressure of the circulating gas to compensate the pressure losses of the circulation flow. The motivation for using an ejector, rather than a compressor, is the high temperature of the flue gas. The energy

TABLE 14.2 Computed Pressure Losses in the Pneumatic Conveying System Shown In Fig. (4.19

 Position (f) Pressure P(i)(kPa) Velocity of air V(i) (m/s) Velocity diff. v(/)-c(i)(m/s) Falling veJ. w,(f)(m/s) Density of air PgPX1’^”13) Equation used I 100.0 50.0 —— —— I .200 —— ■y 102.0 49.0 7.9 17.8 1.224 Cyclone 3 105.4 47.4 7.8 17.5 1.265 Eq. (14.126) 4 108.8 46.0 22.1 17.3 1.306 Eqs. {14.128)—< 14.129)“’ 5 122.5 40.8 20.8 16.3 1.470 Eq. (14.126) 6 123.1 40.6 7.2 16.2 1.477 Eqs. (14.128H14.129Vf 7 146,8 34.1 6.6 14.9 1.762 Eq. (14.126) 8 151.8 32.9 —— —— 1.822 Eq. (14.130!}: 9 100.0 50.0 —— 1.200 —— D = 0.44 m, Th, = 9.1 kg/s, Rhs = ; 27.3 kg/s, = 3.0. Wood chips, VA — cA = 23.0 m/s (vertical)..

I’a ~ca = 8.0 m/s (horizontal), WsA = 18.0 m/s and PA = 1.0 bar.

S‘We have estimated — c, = 15 m/s, which is same as the difference of vertical and horizontal conveying velocities c(4)-c(3). T, = 0.5.

F’ c2- C| ~ 0, because c(6) = Ch»c[, = e(5). C, = 0.5 R i: = (34.1 — 6.6) m/s = 27.5 m/s

Required to increase the pressure is taken from pressurized air, which of course, also cools the flue gas. Therefore the energy balance equation is also needed in this connection.

The mass, momentum and energy balances for the mixing zone are

Mx + hi, = m,(gas flows)

(14. I 36

MJS + w3s = ms(ash)

mxvx + (m2V2 + Fhf2)][mivi + w3c3] = (P3 — pz)^3 (14.137)

WtC^Tj + Rii2CplT2 + MscpsT2 = m3cp3T3 + mscpsT3s. (.14.138)

The pressure changes in the nozzle and in the diffusor, respectively, can be esti­mated by

Ps ~ Pi = (1 + Ci^Pi^I (14.139)

 Pressurized air The velocities are

V‘ = 7TA’ 1 = 1,,5 (14.141;

F ’ini

And due to the geometry of Fig. 14.20 there are two important additional rela­tions

Px=p2 (14.142)

A1+A1 = A3. (14.143)

If we assume that T3s ~ T3, then on the basis of Eqs. (14.136) — (14.143) the ejector can be designed and the pressures calculated. An example of this is shown in Table 14.3.

The essential difference in dimensioning an ejector for pneumatic con­veying systems comes from the acceleration term of the solids, Eq. (14.137).

TABLE 14.3. Calculations for an Ejector of a Pneumatic Conveying System, According to Fig. 14.20

 Stage Calculated/chosen Equation used Remarks 1 V} =20 m/s —— Chosen 2 CI = vi ~ \$ —— Estimated 3 P2 = 3.67 kg/m3 P = MpURT) Ideal gas 4 A2 — 1.771 ■ 10-3M2 Eq. (14.141) D2 — 47.5 mm 5 Pt = 13.7 kg/m3 P = Mpl(RT) Ideal gas 6 Tii: = 0.13 kg/s Thjtfh = 7 M, == 0.38 kg/s Eq. (14.136) 8 T3 = 566 K Eq. (14.138) Assumption T-5, = T, 9 P, = 12.0 bar Eq. (14.140) Vi = 39.7 m/s Ј3 = 0.11 10 = 34.7 m/s C, = — 5 m/s Estimated 11 V 1 = 492 m/s, Eqs. (14.137), Aj = 3.70 ■ lO“5 m2 (14.142)—(14.143) 12 A3 = 1.808 • 10-3 m2 Eq. (14.143) 13 Ј’3 = 39.7 m/s Eq. (14.141) 14 PJ = 29.7 bar Eq. (14.139) B = 0.1

Rh2 = 0.13 kg/s, Rii, = 0.5 kg/s Pi — P2 = 11-5 bar, PA = 12.1 bar T, = 293 K, T2 = 1093 K

‘Rough approximation for the optimum mass ratio .

A guess at this stage. The iteration cycle consists of stages 9-13.

1. Vlaugin, G. A., The method of virtual power in continuum mechanics: Application to coupled fields. Acta Mecbanica, 35 (1980), pp. 1-70.

2. Meyer, R. E., Introduction to Mathematical Fluid Dynamics. Dover, New York, 1982. rs, Weber, M., Stromungs-Fordertechnik, Krausskopf-Verlag GmbH, Mainz, 1974.