SAJJAD KHUDHUR ABBASCeo , Founder & Head of SHacademyChemical Engineering , Al-Muthanna University, IraqOil & Gas Safety and Health Professional – OSHACADEMYTrainer of Trainers (TOT) - Canadian Center of Human Development

Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING

(Part 2)

Content:Conversion factorsCalculation of UsaltCalculation of PH1Calculation of PVP BendsCalculation of ΔPH2 and ΔP total Calculation of PgCalculation of ConclusionsComparison of Results:

DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING A plastics production plant wants to increase the capacity through an existing conveying system. The existing system has 6 inch ID pipes and is configured as shown in the diagram below. The High Density Polyethylene (HDPE) particles have an average size of 4 mm. The conveying gas is at 68oF. The existing blower can produce 1375 SCFM. The desired capacity increase is from 20,000 lbm/hr to 30,000 lbm/hr. Can the existing blower and pipe system meet this increase in capacity? Assume the pressure drop across the cyclone is 5 inches of water. The pressure drop across the blower inlet pipe and silencers is 0.3 psi. The pipe bends have R/D = 6. Pipe roughness is k = 0.00015 ft. The particles have density pρ = 59 lbm/ft3. Terminal velocity of the particles is = 30.6 ft/s.

ΔPH1

ΔPH2

ΔPv

Conversion factors

LH1 325 ft 99 mLH2 100 ft 30.48 m LV 50 ft 15.24 mρf 0.0753 Ibm /ft 3 1.2 kg/m3ρp 59 Ibm /ft 3 947.3 kg/m3

D 6 in 0.1524 mMp 30000 Ibm/hr 3.78 kg/sx 4 mm 0.004 m Gas flowrate

1375 ft3/min 0.649 m3/s

Calculation of Usalt:

 

)

)

 

Calculation of PH1:

 

 

Calculation of PV:

P Bends:

Fro two bends in the system:

∆p Bend = 7.5 M Of vertical pipe

∆PBend

∆PBend

Calculation of ΔPH2 and ΔP total :

 

 

Calculation of Pg:Pressure drop in cyclone = 5 inches water = 0.127 m water = 1254.87 Pa

Calculation of :

Inlet pressure to blower Pin

Conclusions:Calculation of gas velocity supplied by blower:

 The velocity in the pipeline supplied by blower is ( ), hence the velocity everywhere in the pipeline exceeds the saltation velocity (21.67 m/s).  Assuming the blower is capable of the 52.397 kPa pressure increase, the velocity provided by the flow rate of 1375 SCFM exceeds the saltation velocity everywhere in the pipe line; hence the blower and pipe system is capable of conveying 3.78 kg/s of solids.

Item Results by Martin Rhods

equations

Solved example results

BTU SI unit

U salt 71.2 ft/s 21.7 m/sΔPH1 1.729

psig11921 Pa

ΔPV 0.465 psig

3206 Pa

ΔPH2 0.473 psig

3261.2 Pa

ΔP Bend 0.559 psig

3845.2 Pa

ΔP Total 3.226 psig

22242.5 Pa

ΔP blower

4.44 pig 30612.7 Pa

Comparison of Results:

It is obvious from the above table that the pressure loss calculated by Martin Rhods equations more than the results in the solved example. That difference may because the Martin Rhods equations depend on the saltation velocity and the equations in the solved example depend on gas velocity in each point.

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